Grade 8 Math

IL FOUNDATION SERIES

MATHEMATICS

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ISBN 978-81-985385-1-2 Second Edition

Contents Module 1 1. Rational Numbers

01

2. Linear Equations in One Variable

34

3. Understanding Quadrilaterals

51

4. Data Handling

79

5. Squares and Square Roots

120

6. Cubes and Cube Roots

142

7. Comparing Quantities

156

8. Algebraic Expressions and Identities

175

9. Mensuration

204

1

RATIONAL NUMBERS

1.1 INTRODUCTION TO RATIONAL NUMBERS 1.1.1 Rational numbers p Any number that is written in the form of , where 'p' and 'q' are integers and q ≠ 0 is called a q rational number. The set of rational numbers is denoted by 'Q'. A rational number may be positive, zero, or negative.

Example:

1 2 −2 0 −5 , , , , 2 2 3 1 11

1.1.2 Positive rational numbers A rational number is said to be positive when both the numerator and the denominator are either positive or negative integers.

Example:

1 2 −1 −3 , , , ,……….

2 5 −2 −13

1.1.3 Negative rational numbers A rational number is said to be negative if its numerator and denominator have opposite signs. −1 −2 3 7 Examples: , , , ,……. .

2 3 −5 −11 Note: 1. The expression

−3 3 3 can be written as or - . −4 4 4

0 0 0 , , and so on.

1 2 7 3. When zero is divided by any integer (except zero), the result is always zero.

2. Zero is a rational number, and it can be written in the form of

4. The division by zero is not defined.

5. Every natural number is a rational number, but not every rational number is a natural number.

6. Similarly, every integer is a rational number, but not every rational number is an integer.

1.1.4 Standard form of a rational number p A rational number is said to be in standard form if 'p' and 'q' are integers having no common q divisor other than 1, where 'q' is positive.

3 −5 8 Example: , etc. , 4 6 11 1

RATIONAL NUMBERS

Note: p p p×m p p×m is a rational number, and for any m ≠ 0, then = . Here , are called q q q×m q q×m equivalent rational numbers.

Property 1: If

Example: −5 ( −5 ) × 2 −10 −5 ( −5 ) × 3 −15 −5 −10 −15 = = (or) = = ⇒ , , are equivalent rational numbers.

3 3×2 6 3 3×3 9 3 6 9 p p÷m p is a rational number, and 'm' is a common divisor of 'p' and 'q', then = q q÷m q 10 ÷ ( −5 ) 10 −2 −12 −12 ÷ 12 −1 = = ; = = ......... etc.

Thus, we can write −15 ( −15 ) ÷ ( −5 ) 3 24 24 ÷ 12 2 Property 2: If

Property 3: Two rational numbers in the standard form can be compared by finding their cross products. p r For any two rational numbers and q s p r i) if ps > rq, then > q s p r < q s p r iii) if ps = rq , then = q s ii) if ps < rq, then

1.2 OPERATIONS ON RATIONAL NUMBERS 1.2.1 Addition and subtraction of rational numbers Addition

The operation of addition on rational numbers is similar to that of addition of fractions. There are two different cases: Case - I: When the rational numbers have the same denominators. a b a+b + = c c c

Example: Add Solution: 5 3 5+3 8 + = = 7 7 7 7 2

5 3 + .

7 7

IL Foundation Series Class 8

Case - II: When the rational numbers have different denominators.

Follow these steps: 1. Write the rational numbers in standard form.

2. Find the LCM of the denominators.

3. Express each rational number with a common denominator which is the LCM of the denominators.

4. Now, add the numbers as in case - I. 5 3 Example: Add + 2 5 Solution: LCM of denominators = 10 5 5 × 5 25 = = 2 2 × 5 10 3 3×2 6 = = 5 5 × 2 10 Therefore,

5 3 25 6 25 + 6 31 + = + = = 2 5 10 10 10 10

Subtraction

The operation of subtraction on rational numbers is similar to that of subtraction of fractions. There are two different cases: Case - I: When the rational numbers have the same denominators. a b a + ( −b ) − = c c c Example: Subtract

5 3 − .

7 7

Solution: 5 3 5 −3 5 − 3 2 − = = = 7 7 7 7 7

3

RATIONAL NUMBERS

Case - II: When the rational numbers have different denominators.

Follow these steps: 1. Write the rational numbers in standard form.

2. Find the LCM of the denominators.

3. Express each rational number with a common denominator, which is the LCM of the denominators.

4. Now, subtract the numbers as in case - I. Example: Subtract 3 - 1 .

2 5 Solution: LCM of denominators = 10 3 3 × 5 15 = = 2 2 × 5 10 1 1× 2 2 = = 5 5 × 2 10 3 1 15 2 15-2 13 Therefore,= = = 2 5 10 10 10 10 1.2.2 Multiplication and division of rational numbers Multiplication

When we multiply two rational numbers, we multiply the numerator with the numerator and the denominator with the denominator. a c a×c a c For any two rational numbers and , we define × = . b d b×d b d Example:

−3 4 −3 × 4 −12 × = = 7 5 7 ×5 35

Multiplication of a rational number by zero

When any rational number is multiplied by zero, the result is zero. a a a For any rational number , we have × 0 = 0 × = 0 . b b b 7 7×0 0 = = 0.

Example: × 0 = 8 8 8 Division

If

a c c and are two rational numbers such that ↑≠ 00, then b d d

a c a dd c ÷ = ×  is the reciprocal of . b d b cc d 4

IL Foundation Series Class 8

Here,

a c ad is called the dividend, is called the divisor, and the result is . b d bc

Example: If the multiplicative inverse of

x +1 2 is , then find x. x 3

Solution: x 2 = ⇒ 3x = 2x + 2 ⇒ x = 2 x +1 3 SOLVED EXAMPLES Example 1: Add

16  −5   −7  ,   , and  .

9  12   18 

Solution:  −7  16  −5  ,   , and  18 .

9  12    As the LCM of 9, 12, 18 is 36.

Given rational numbers are

Thus, 16 16 4 64 = × = 9 9 4 36 −5 −5 3 −15 = × = 12 12 3 36 −7 −7 2 −14 = × = 18 18 2 36

Now, 16  −5   −7  + +  9  12   18  16 5 7 = − − 9 12 18 64 15 14 = − − 36 36 36 64 − 15 − 14 = 36 35 = 36

5

RATIONAL NUMBERS

Example 2: Subtract

−4 3 from .

7 5

Solution: According to the question,

3  −4  −  5  7 

3 4 3× 7 + 4×5 + = 5 7 35 21 + 20 41 = = 35 35 =

Example 3: Multiply

2 3 by .

7 5

Solution: Given numbers, Now, 2 3 = × 7 5 =

2 3 by .

7 5

2×3 7 ×5

=

6 35

Example 4: Multiply

 −3  5 by  .  4  9

Solution:

 −3  5 and  .  4  9 5 −3 5 × ( − 3 ) = × = 9 4 9× 4 − 15 −5 = = 36 12

Given numbers,

Example 5: The product of two numbers is Solution:

Let the other number be x.

Then, according to the question:

6

−28 −4 . If one of the numbers is , find the other.

27 9

IL Foundation Series Class 8

(−4 )

− 28 9 27 ( −28 ) ( − 4 ) ÷ ⇒x= 27 9 ( − 28 ) 9 ⇒x= × 27 ( −4 ) x×

=

×9 ) ( − 28× 27×( −4 ) ( −28×9 ) ⇒x= 27×( −4 ) ⇒x=

⇒x=

7 3

1.3 PROPERTIES OF RATIONAL NUMBERS 1.3.1 Closure property of rational numbers Closure property of addition of rational numbers 1 2 3 10 13 1 2 Let and be two rational numbers. Then, + = + = is also a rational number.

5 3 15 15 15 5 3

If we add any two rational numbers, we always get a rational number.

In general, the sum of any two rational numbers is a rational number. i.e., if x and y are two rational numbers, then ( x + y ) is also a rational number. This property is called the closure property under addition.

Closure property of multiplication of rational numbers

Let

−3 8 and are two given rational numbers.

4 5

Then,

−3 8 −6 is also a rational number. × = 4 5 5

∴ The product of any two rational numbers is a rational number, i.e., if x and y are two rational numbers, then x × y is also a rational number. This property is known as the closure property of rational numbers under multiplication.

7

RATIONAL NUMBERS

1.3.2 Commutative property of rational numbers Commutative property of addition of rational numbers

2 1 and are two rational numbers.

7 4 2 1 8 7 15 Then, + = + = 7 4 28 28 28 If

also,

1 2 7 8 15 + = + = 4 7 28 28 28

i.e., if x and y are two rational numbers, then x + y = y + x.

This property is known as the commutative property under addition.

Commutative property of multiplication of rational numbers

Let

−3 7 and be any two given rational numbers.

4 11

Then,

−3 7 ( −3 ) × 7 −21 × = = 4 11 4 × 11 44

7 × ( −3 ) −21 and 7 × −3 = = 11 4 11 × 4 44 ∴

7 −3 −3 7 × = × 4 11 11 4

i.e., if x and y are any two rational numbers, then x × y = y × x .

This property is known as the commutative property of rational numbers under multiplication.

1.3.3 Associative property of rational numbers Associative property of addition of rational numbers

Let

−3 −7 5 , and be given rational numbers.

4 4 4

Then,  −3 + −7  + 5 = −10 + 5 = −5 4  4 4 4 4  4 also, −3 +  −7 + 5  = −3 +  −2  = −5 4  4 4 4  4  4  −3 −7  5 −3  −7 5  ∴ + + +  + = 4  4 4  4 4  4

i.e., if x, y and z are any three rational numbers, then

( x + y ) + z = x + ( y + z ) . This property is known as the associative property of addition.

8

IL Foundation Series Class 8

Associative property of multiplication of rational numbers

−2 3 −8 , , and are the three given rational numbers.

3 4 5       Then,  −2 × 3 × −8 =  −2 × 3 × −8  = −6 × −8 = 48 = 4  3 × 4   5  12  3 4  5 5 60 5 and

−2  3 −8  −2  3×( −8 )  −2  ( −24 )  −2×( −24 ) 48 4 = = × × = = × × = 3  4 5  3  4×5  3  20  3×20 60 5

 −2 3  −8 −2  3 −8  hence,  × × = × ×   3 4  5 3  4 5  i.e., if x, y and z are any three rational numbers, then ( x × y ) × z = x × ( y × z ) This property is known as the associative property of multiplication.

SOLVED EXAMPLES Example 1: For x =

−7 , verify that − ( − x ) = x.

4

Solution: Given, x =

−7 4

  −7    7  −7 ∴− ( − x ) = −  −  =x  = −  = 4 4   4  Hence -(-x)=x Example 2: For x = Solution:

1 2 and y = , verify that x + y = y + x.

3 7

For x =

1 2 and y = 3 7

∴

x+y =

1 2 7 6 7 + 6 13 + = + = = 3 7 21 21 21 21

Now,

y+x=

2 1 6 7 13 + = + = 7 3 21 21 21

Heence, x + y = y + x.

9

RATIONAL NUMBERS

Example 3: From the sum of

5 −8 1 −5 and subtract the sum of and .

9 27 54 18

Solution: −8 5  −8  5 8 15 − 8 7 5 = + = = and = − 27 9  27  9 27 27 27 9 −5 1  −5  1 5 1 − 15 −7 1 = + = Sum of and = − = 18 54  18  54 18 54 27 54 Sum of

−7 7 from 27 27 7  −7  7 7 7 + 7 14 − = + = i.e. = 27  27  27 27 27 27 Now, we have to subtract

Example 4: Verify the associative property for addition for the rational numbers Solution: −7 5 −4 , b= , c= Let a = be the given rational numbers.

9 6 3 −7 5   −4   −7 × 2 + 5 × 3   −4  + + = +  18  9 6  3     3 

( a + b ) + c = 

 −14 + 15   −4  1  −4  = +  = +   18   3  18  3  1 + ( −4 ) × 6 1 + ( −24 ) −23 = = 18 18 18 7  5 ( −4 )  −7  5 + ( −4 ) 2  a + (b + c) = − +  + + =  9 6 3  9  6  =

=

7  −3  7  −1  −7  5 + ( − 8 )  +  = − +  = − +  9  6 9  6  9  2  

=

−7 × 2 + ( −1 ) × 9 −14 + ( −9 ) −23 = = 18 188 18

..........(1) …….. ( 1 )

..........(2) …….. ( 2 )

From (1) and (2), ( a + b ) + c = a + ( b + c ) is true for rational numbers.

10

−4 −7 5 , , and .

3 9 6

IL Foundation Series Class 8

Example 5: Verify the commutative property for multiplication for the rational numbers −8 .

33

−10 and 11

Solution: −10 −8 Let a = and b = be the given rational numbers.

11 33 −10  −8  80 ..........(1) × ……. ( 1 ) = 11  33  363 80 a×b = 363 −8  −10  80 ..........(2) b×a = × ………( 2 ) = 33  11  363 80 b×a = 363

a×b =

From (1) and (2), a × b = b × a Hence, multiplication is commutative for rational numbers.

1.3.4 The role of zero (additive identity and inverse) Additive identity

2 2 2 2 be the given number, then + 0 = 0 + = i.e., if x is a rational number, then x + 0 = 0 + x = x .

3 3 3 3 '0' is called the additive identity of rational numbers.

Let

Additive inverse (existence of negative of rational number)

If

3 3  −3   −3  3  −3  is a rational number, then there exists   such that +  = + = 0 4 4  4   4  4  4 

i.e., if x is a rational number, then x + ( − x ) = ( − x ) + x = 0.

( − x ) is known as the additive inverse of the rational number x and vice versa.

1.3.5 The role of 1 (multiplicative identity and inverse) Multiplicative identity

−5 −5 −5 −5 is a rational number, then ×1 = 1× = 3 3 3 3 i.e., if x is any rational number, then x × 1 = 1 × x = x.

'1' is known as the multiplicative identity of rational numbers.

11

RATIONAL NUMBERS

Multiplicative inverse

For any two rational numbers a, b if a × b = b × a = 1, then a, b are said to be reciprocals to each other.

Example: The reciprocal of 3 is If

1 1 1 and the reciprocal of is 3, since 3 × = 1 i.e., 3 3 3

p p q is a rational number, then × = 1. q q p −1

q p p We call as the multiplicative inverse of and denote it as   when p, q ≠ 0. p q q Note: b a 1. is the reciprocal of . a b 2. Zero has no reciprocal.

3. Reciprocal of '1' is '1' and reciprocal of '( −1 )' is '(-1)'. −1

1 a a a 4. As a −1 = , we denote the reciprocal of by   . Clearly   b a b b

−1

b = . a

5. '0' [zero] has no multiplicative inverse.

Example: If the multiplicative inverse of Solution:

x +1 2 is , then find x. x 3

According to the question, x 2 = ⇒ 3x = 2x + 2 ⇒ x = 2 x +1 3 1.3.6 Distributive properties of addition and subtraction Distributive property of multiplication over addition

For any three rational numbers x, y, and z, x × ( y + z ) = ( x × y ) + ( x × z ) , this property is known as the distributive property of multiplication over addition.

Distributive property of multiplication over subtraction

For any three rational numbers x, y, and z, x × ( y − z ) = ( x × y ) − ( x × z ) , this property is known as the distributive property of multiplication over subtraction.

1 −2 −5 Example: Verify the property x × ( y + z ) = ( x × y ) + ( x × z ) by taking x = , y = , z = .

2 3 6

12

IL Foundation Series Class 8

Solution: 1 −2 −5 Given, x = , y = , z = 2 3 6 1  −2 −5  1  −2 5  x×( y + z ) = × + −   = × 2  3 6  2  3 6 1  −4 − 5  1  −9  −3 = ×  = × = 2  6  2  6  4 1  2  1  −5  Now, ( x × y ) + ( x × z ) = ×  −  + ×   2  3 2  6  =

1 × ( −2 ) 1 × ( −5 ) + 2×6 2×3

=

−2 −5 + 6 12

=

−2 5 − 6 12

=

−4 − 5 12

−9 −3 = 12 4

Thus, x × ( y + z ) = ( x × y ) + ( x × z ). =

1.4 COMPARISON OF RATIONAL NUMBERS To compare any two rational numbers, we follow the below steps: 1. Express each of the two given rational numbers with a positive denominator.

2. Take the LCM of these positive denominators.

3. Express each rational number obtained in step 1 with this LCM as the common denominator.

4. Compare the numerators of rational numbers obtained in step 3. The number having the greater numerator is the greater rational number.

Example: Compare

−3 5 and .

7 −11

Solution: We first rewrite the second rational number with a positive denominator.

13

RATIONAL NUMBERS

5 × ( −1 ) 5 −5 = = −11 ( −11 ) × ( −1 ) 11 LCM of denominators 7 and 11 is 77.

Now, we rewrite the given rational numbers so that they have a common denominator 77. −3 −3 × 11 −33 = = 7 7 × 11 77 And

−5 −5 × 7 −35 = = 11 11 × 7 77

Comparing the numerators, we get: − 33 > − 35 − 33 − 35 ∴ > 77 77 − 3 −5 ⇒ > 7 11 SOLVED EXAMPLES Example 1: Find the multiplicative inverse of

1 1 + .

4 6

Solution: 1 1 3+2 5 + = = 4 6 12 12 −1 −1 −1 12 1 1  3+2   5  ∴ +  =  =    = 5 4 6  12   12  We have,

Example 2: Arrange the rational numbers

−14 10 4 in ascending order. , , 20 −16 −6

Solution: We first rewrite the given rational numbers so that their denominators are positive.

10 × ( −1 ) 4 × ( −1 ) 10 −10 4 −4 = = and = . = −16 ( −16 ) × ( −1 ) 16 −6 ( −6 ) × ( −1 ) 6 LCM of the denominators 20, 16, and 6 is 240.

Now, we rewrite them so that they have a common denominator 240.

14

IL Foundation Series Class 8

−14 −14 × 12 −168 = , = 20 20 × 12 240 −10 −10 × 15 −150 = = 16 16 × 15 240 and

−4 −4 × 40 −160 = = 6 6 × 40 240

Comparing the numerators, we get − 168 < −160 < −150 −168 −160 −150 ∴ < < 240 240 240 4 10 −14 ⇒ < < 20 −6 −16

Example 3: Verify the distributive property a × ( b + c ) = ( a × b ) + ( a × c ) for the rational numbers −1 2 −5 a = , b = , and c = .

2 3 6 Solution: −1 2 −5 Given the rational number a = , b = and c = .

6 2 3 −1  2  −5   −1  ( 2 × 2 ) + ( −5 × 1 )  −1  4 + ( −5 )  −1  −1  a×( b + c ) = × +   = ×  = ×  = ×  6 2  3  6   2  6 2    2  6  a×( b + c ) =

1 12 −1 2   −1  −5   −2 5 ( −2 × 2 ) + 5 × 1 −4 + 5 = ×  +  × + = = 12 12  2 3   2  6   6 12

( a × b ) + ( a × c ) =  ( a×b )+( a×c ) =

1 12

From (1) and (2), we have a × ( b + c ) = ( a × b ) + ( a × c ) which is true.

Hence, multiplication is distributive over addition for rational numbers.

Example 4: Verify the identity property for addition and multiplication for the rational numbers 15 −18 and .

19 25 Solution:

Given number,

15 −18 and 19 25

15

RATIONAL NUMBERS

15 15 0 15 + 0 15 +0 = + = = 19 19 19 19 19 −18 −18 0 −18 + 0 −18 +0 = + = = 25 25 25 25 25 The identity property for addition is verified.

15 15 ×1 15 ×1 = = 19 19 19 −18 −18 ×1 −18 ×1 = = 25 25 25 The identity property for multiplication is verified.

Example 5: Verify the additive and multiplicative inverse property for the rational numbers −7 17 and .

17 27 Solution:

Given numbers,

−7 17 and .

17 27

−7 7 −7 + 7 0 + = = =0 17 17 17 17 17  17  17 + ( −17 ) 0 = =0 + −  = 27  27  27 27 Thus, the additive inverse for rational numbers is verified. −7 17 × =1 17 −7 17 27 × =1 27 17 Thus, the multiplicative inverse for rational numbers is verified.

1.5 IRRATIONAL NUMBERS 1.5.1 Irrational numbers A number that cannot be expressed as a terminating decimal or a repeating decimal is called an irrational number.

These are generally denoted by 'P or Q'.

16

IL Foundation Series Class 8

Thus, non-terminating, non-repeating decimals are irrational numbers.

Example: Type 1: i) Clearly, 0.010010001... is a non-terminating and non-repeating decimal and therefore, it is irrational. ii) 1.41412413414... and 1.42442444244442... are irrational numbers.

Type 2: If 'm' is a positive integer, not a perfect square, then m is irrational. Thus, 2 , 3 , 5 , 7 , 8 , 10 , 11 etc., are all irrational.

If 'm' is a positive integer, not a perfect cube, then 3 m is an irrational number. Thus, 3 2 , 3 3 , 3 4 , 3 5 , 3 7 , 3 9 etc., are all irrational numbers.

Type 3: 'π' is an irrational number, and 'e' is an irrational number. π and e are called transcendental numbers.

1.5.2 Properties of irrational numbers Irrational numbers satisfy the commutative, associative, and distributive laws under addition and multiplication. Multiplication is distributive over addition.

Note: 1. The sum of two irrationals need not be an irrational.

(

)

(

)

Example: Both 2 + 3 and 4 − 3 is irrational.

(

) (

)

But 2 + 3 + 4 − 3 = 6 is rational.

2. The difference of two irrationals need not be an irrational.

(

)

(

)

Example: Both 5 + 2 and 3 + 2 is irrational.

(

) (

)

But 5 + 2 − 3 + 2 = 2, is rational.

3. The product of two irrationals need not be irrational.

Example: 3 is an irrational. But 3 × 3 = 3, is rational.

4. The quotient of two irrationals need not be irrational.

Example: Both 2 3 and 3 are irrational. But

2 3 = 2, which is rational.

3

(

)(

)

5. If 'a' is a rational number and b is irrational, then each of a + b , a − b , a b , and

a are b

irrational numbers.

17

RATIONAL NUMBERS

(

)(

)

Example: Each one of 4 + 3 , 8 − 5 , 5 3 and

3 is irrational.

2

Example: The statement, 'The product of two irrational numbers is always an irrational number’ can be negated by the counter-example.

Solution: Take 2 − 5 , 2 + 5 both are irrational. Their product:

(

= 2− 5

)( 2 + 5 )

= 22 − ( 5 )2 = 4 − 5 = −1

is a rational number.

1.6 DECIMAL REPRESENTATION OF REAL NUMBERS All rational numbers can be converted into decimal numbers. There are two types of decimals.

1. Terminating decimals: Terminating decimal is the decimal number in which there is an enddigit. There is a finite number of digits after the decimal point.

13 7 7 For example: = 0= .104, 0.28, = 0.875 125 25 8 These are examples of terminating decimals.

2. Non-terminating repeating decimal: In a non-terminating and repeating decimal, a single digit or a block of digits repeat themselves infinitely after the decimal point.

2 For example: = 0= .666... 0. 6 3 1 = 0= .0909..... 0.09 11 Note: 1) If the denominator of a fraction in its simplest form contains only 2 or 5 as prime factors, then only the fraction can be written as a terminating decimal.

2) A fraction can be written as a terminating decimal, if the denominator can be written in the form of 10 n, since 2 and 5 are the only prime factors of 10.

Example: Find whether the following rational number is a terminating or non-terminating decimal.

4 1 i) −5 ii) 5 3

18

IL Foundation Series Class 8

Solution: 4 −29 −58 i) −5 = = = −5.8 5 5 10 Thus, the above rational number represents a terminating decimal.

1 = 0.3333…

3 Thus, the above rational number represents a non-terminating recurring decimal. ii)

1.7 DENSITY PROPERTY AND RATIONAL NUMBERS ON NUMBER LINE 1.7.1 Density property If x and y are any two rational numbers such that x < y, then x < x+y lies between x and y.

2 1 1 Example: Let and be two rational numbers.

2 3

x+y < y (or) the rational number 2

1  1 1  1 3+2  1 5 5 Then, ×  +  =  = × = 2  2 3  2  6  2 6 12 We can write Now, clearly Similarly,

1 1× 6 6 1 1× 4 4 = = = and = 2 2 × 6 12 3 3 × 4 12

4 5 6  1 5 1 5 1 1 < < < , i.e., lies between and  ∴ < 12 3 2 12 12 12  3 12 2

1  1 5  1  6 + 5  1 11 11 × +  = × = × = 2  2 12  2  12  2 12 24

which lies between

1 1 and and so on.

2 3

Following the above process, we can find an infinite number of rational numbers between two given rational numbers. This property is called the 'density property' of rational numbers.

1.7.2 Rational numbers on a number line To represent rational numbers on the number line, we begin by drawing a line and indicating a point to denote the rational number zero. Subsequently, we mark positive rational numbers to the right of 0, while negative rational numbers are marked to the left of 0.

19

RATIONAL NUMBERS

Example: Represent the rational number Solution:

1 on a number line.

4

1 on the number line, we first draw a number line and mark a point 'O' representing 4 '0' as shown below: To represent

0 O Now, we find a point, say P, on the number line, which represents the numerator 1 of the rational 1 number . It is clear that point P will be on the right side of 0.

4 Divide the segment OP into four equal parts.

Let A, B, and C be the points of division as shown below.

1 0

4

O

A

1 B

C

P

Then OA = AB = BC = CP By construction, each part OA, AB, BC and CP represent 1 4 1 ∴ The point A represents the rational number .

4 Similarly,

th

of the segment 'OP'.

−1 can be represented on the number line on the left-hand side of O. 4

Example: Represent all the rational numbers between - 2 and 2, having five as the denominator.

Solution: The rational numbers between - 2 and 2, having five as the denominator can be represented on the number line as:

(-ve)

(+ve)

-2 -9 -8 -7 -6 -1 -4 -3 -2 -1 0 1 2 3 4 1 6 7 8 5

20

5

5 5

5

5

5

5

5

5

5

5

5

5 5

9 5

2

IL Foundation Series Class 8

1.8 CONCEPT OF INFINITY A mathematical infinity is the conceptual representation of such a numberless quantity. In many instances, it is treated as if it were a number that counts or measures an infinite number of terms, but it is not the same sort of number as a natural number.

The common symbol for infinity is '∞'. Actual infinite quantities would be things such as ordinal numbers and cardinal numbers. Infinity is best described as a concept or an idea rather than a concrete number.

For example, a list of natural numbers 1, 2, 3, 4....... no matter how long you count for, it can never reach the end of all numbers.

Example: Below, you can find some fun facts about some mathematical concepts that stretch infinitely!

1. The sequence of natural numbers is infinite: {1, 2, 3, ......}.

2. A line or a line segment consists of infinite points.

3. The number pi (π) goes on forever. (3.14159...)

4. Certain fractions are finite, but they are infinite when written as decimal numbers. ( 5. The number of prime numbers is infinite.

1 is 0.333...).

3

Properties of infinity: Here are some important properties to remember when working with infinity.

Addition property: If any number is added to infinity, the sum is also equal to infinity. ∞ + ∞ = ∞, − ∞ + −∞ = −∞ Subtraction property: Subtracting infinity from infinity will result in an indeterminate form. (The term 'indeterminate' means an unknown value or anything that can’t be defined). ∞ − ∞ = Indeterminate form Multiplication property: If a number is multiplied by infinity, then the value of the product is also equal to infinity. ∞ × ∞ = ∞, − ∞ × ∞ = −∞, − ∞ × −∞ = ∞ Special Properties: If x is any integer, then: x + ( −∞ ) = −∞, x + ∞ = ∞, x − ( −∞ ) = ∞, x − ∞ = −∞ For x > 0 : x × ( −∞ ) = −∞, x × ∞ = ∞ For x < 0 : x × ( −∞ ) = ∞, x × ∞ = −∞

21

RATIONAL NUMBERS

1.9 ABSOLUTE VALUE 1.9.1 Absolute value of an integer The absolute value of an integer is the numerical value of the integer regardless of its sign.

If 'a' is an integer, then its absolute value is denoted by a and is defined as a = a if a ≥ 0, = −a if a < 0 Example: 6 = 6 and −6 = − ( −6 ) = 6.

Here, i) a = a, if a is positive. ii) a = 0, if a is zero. iii) a = −a, if a is negative.

SOLVED EXAMPLES Example 1: Find the absolute value of the following: i) -76 Solution:

ii) +50

iii) -100

i) −76 = 76 ii) +50 = 50 iii) −100 = 100 Example 2: Write a rational number between 5 and 6 .

Solution:  5 = 2.236… and 6 = 2.449…

∴ A rational number between 2.24 and 2.44 (approximately) is 2.3 or 2.31 or 2.32 etc.

Note: Take the lower limit slightly greater than 5 and the upper limit slightly lesser than 6 . ⇒ One number between 5 and

22

6 = 2.3

IL Foundation Series Class 8

43 Example 3: The decimal expansion of the rational number 4 3 will terminate after how many 2 ×5 places of decimal?

Solution: Given rational number is

43 .

2 × 53 4

Now, 43 43 × 51 = 24 × 53 24 × 53 × 51

Thus,

=

43 × 5 ( 215 ) 215 = = 24 × 54 (2 × 5)4 104

=

215 = 0.0215 10000

43 will terminate after 4 places of decimal.

2 × 53 4

Example 4: Insert 10 rational numbers between

−3 8 and 11 11

Solution: −3 8 and .

11 11 From the given rational numbers, the numerator lies between - 3 to 8.

Given rational numbers are

Thus, −3 < −2 < −1 < 0 < 1 < 2 < 3 < 4 < 5 < 6 < 7 < 8 −3 −2 −1 0 1 2 3 4 5 6 7 8 < < < < < < < < < < < 11 11 11 11 11 11 11 11 11 11 11 11 −3 8 Hence, the 10 rational numbers between and are 11 11 Now,

−2 −1 0 1 2 3 4 5 6 7 , , , , , , , , , 11 11 11 11 11 11 11 11 11 11 .

Example 5: Find two irrational numbers between 2 and 2.5.

Solution: We know, if a and b are two distinct positive rational numbers such that ab is not a perfect square of a rational number, then ab is an irrational number lying between a and b.

23

RATIONAL NUMBERS

Thus, an irrational number between 2 and 2.5 is: = 2 × 2.5 = 2×

5 = 5 2

Again, an irrational number between 2 and 5 is: = 2× 5 = 2 5 So, the required irrational numbers between 2 and 2.5 are 5 and 2 5 .

QUICK REVIEW •

Rational numbers are closed under the operations of addition, subtraction, and multiplication.

•

The addition and multiplication operations are: 1) Commutative for rational numbers 2) Associative for rational numbers

•

The rational number 0 is the additive identity for rational numbers.

•

The rational number 1 is the multiplicative identity for rational numbers.

•

The additive inverse of the rational number

• •

a a is − and vice-versa. b b a c a c The reciprocal or multiplicative inverse of the rational number is if × = 1. b d b d Distributivity of rational numbers: For all rational numbers a, b and c, a ( b + c ) = ab + ac and a ( b − c ) = ab − ac

• •

•

24

Rational numbers can be represented on a number line. x+y If x and y are any two rational numbers such that x < y, then x < < y (or) the rational 2 x+y number lies between x and y.

2 A number which cannot be expressed as a terminating decimal or a repeating decimal is called an irrational number. These are generally denoted by P or Q.

IL Foundation Series Class 8

WORKSHEET - 1 I.

OPERATIONS ON RATIONAL NUMBERS 1. Express the following rational numbers in standard form.

142 i) −355 2. Subtract

ii)

−16 −256

iii)

42 −105

iv)

−88 −121

−5 −3 from .

8 7

−7 5 to get ?

8 9 −3 −9 4. The sum of two rational numbers is . If one of them is , find the other number.

5 10 3. What number should be added to

5. Simplify the following. −7 3 i) + 9 4

ii)

−3 5 + −11 9

iii)

−9 −4 + 20 60

6. The sum of two rational numbers is -8, if one of the numbers is

iv)

−8 −11 + 27 81

−10 , find the other number.

7

1 −5 7. What should be subtracted from , so as to get ?

3 12 8. The sum of two rational numbers is

−1 −11 . If one of the numbers is . Find the other number.

3 3

9. Multiply: 22  9  i)   by −27  −11 

 −8   −5  ii)   by    25   16 

 −2   33  iii)   by    9   54 

10. Divide the following: i)

2 −4 by 3 5

ii) -4 by

−3 5

11. The product of two rational numbers is

iii)

−6 by -15 7

iv)

−1 3 by 8 4

−8 −4 . If one of the numbers . Find the other.

9 15

3.

2 12. Find ( x + y ) ÷ ( x − y ), if= x = ,y 3 2 −18 2 1 −2 13. From the sum of and subtract the sum of and .

51 17 85 15

25

RATIONAL NUMBERS

1 1 1 14. The perimeter of a triangle is 15 m . If the length of two of its sides are 5 m and 3 m, find 2 5 2 the length of the third side of the triangle.

1 1 of his income on paying rent, of his income on food, clothing, and education 3 2 and saves the remaining income. What part of his income does he save?

15. Ravi spends

II. PROPERTIES AND COMPARISON OF RATIONAL NUMBERS 1. Verify the commutative property of multiplication x × y = y × x of rational numbers if −1 2 x= ,y= .

5 7 2. Verify the associative property of multiplication x × ( y × z ) = ( x × y ) × z of rational numbers if −11 1 7 x= ,y= ,z= .

4 3 2 3. Verify the distributive property x × ( y + z ) = ( x × y ) + ( x × z ) of rational numbers if 7 9 x = −3, y = , z = .

2 9 4. Verify that x + y = y + x if x =

−8 5 ,y= .

9 7

5. Verify that ( x + y ) + z = x + ( y + z ) if x = 6. Verify:

−2 7 −3 ,y= ,z= .

3 8 5

3  −5 7   3 −5  7 + + = + + 4  6 8 4 6  8

7. Find the multiplicative inverse of the following: i) 9 iv) -1

ii)

7 −16

iii)

−10 −3

1 2 v) + 5 5

4 3 = ,y .

5 9 5 3 9. Verify that: (x − y)−1 ≠ x −1 − y −1, if= . x = ,y 5 7 19 −8 10. Verify that (x × y)−1 = x −1y −1 , if x = , y = .

17 31 8. Verify that (x + y)−1 ≠ x −1 + y −1, where= x

11. Verify that (x + y)−1 ≠ x −1 + y −1 , if x =

26

−15 5 ,y= .

26 13

IL Foundation Series Class 8

12. Fill in the blanks with the correct symbol out of >, ==and, < . −8 7 −5 −6 −11 33 i)  ii)  iii)  9 13 6 7 13 −39

iv)

−7 6  −10 7 −5 −5 , 6 −7

13. Which of the two rational numbers is greater? −5 3 i) , 11 11

ii)

−7 −3 , 9 5

iii)

−7 3 , 11 −5

iv)

iii)

2 3

iv) 0

14. Write the additive inverse of the following: −2 i) 5 15. Find the product of

ii)

−7 9

25 5 and the multiplicative inverse of .

14 7

III. IRRATIONAL NUMBERS AND ABSOLUTE VALUE

( )( 2 2 − 7 7 ) 2. Solve: ( 5 + 7 ) ( 13 − 11 ) 3. Solve: ( 2 6 + 7 7 ) + ( 13 2 − 4 7 ) 4. Solve: ( 7 7 ) × ( −4 7 ) 1. Solve: 2 2 + 7 7

5. What is the absolute value of −6 ?

6. What is the absolute value of 0 ?

IV. DENSITY PROPERTY AND DECIMAL REPRESENTATION OF RATIONAL NUMBERS 1. State whether the following rational numbers are terminating or repeating decimals. −17 13 i) ii) 15 8 2. Express each of the following in decimal form.

7 156 i) ii) 2 20 3. Find a rational number between

−1 1 and .

2 2

−2 11 and .

19 19 5. Find three rational numbers between 2 and 3.

4. Find ten rational numbers between

6. Write the following in decimal form and state the kind of decimal expansion each has: 36 1 i) ii) 100 11 27

RATIONAL NUMBERS

2 5 and .

3 3 −3 9 8. Insert two rational numbers between and .

13 13 V. RATIONAL NUMBERS ON THE NUMBER LINE 7. Find four rational numbers between

1. Represent the following rational numbers on the same number line.

1 2 −1 i) , , and 2 3 6

0 −8 7 ii) , , and 5 3 6

1 3 5 iii) − , , and 2 4 3

4 5 iv)− and 3 3

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. Among the following, a correct statement is:

a) The additive identity under natural number is 0. b) The additive inverse of the number '1' does not exist. c) The multiplicative inverse of the number '0' does not exist. d) The multiplicative inverse of the number '1' does not exist.

2. If a)

x a x a x × = 1 where and are rational numbers, then the multiplicative inverse of is: y b y b y a b

b)

−x y

c)

−y x

d)

b a

p p 3. The property illustrated by the relation + 0 = , where p and q are integers but not equal to q q zero is: a) commutative property of rational numbers under addition. b) associative property of rational numbers under addition. c) multiplicative identity of rational numbers. d) additive identity of rational numbers.

28

IL Foundation Series Class 8

4. Among the following relations exhibits, the commutative property of rational numbers under multiplication is:  −21  6 6  −21  a)   × = ×  5  7 7  5 

 3 8  −6 3  8 −6  b)  ×  × = × ×   5 7  7 5  7 11 

1 1 c) 7 × 1 = 7 2 2

9 6 =1 d) − × 8 −9

5. The product of the additive inverse of ( −0.8 ) and the multiplicative inverse of 0.2 is: a) 4

b) -5

c)

1 4

d) −

1 5

6. Among the following relations illustrated, the associative property of rational numbers under addition is: a)

1 −7 −7 1 + = + 2 5 5 2

b)

1  2 −11   1 2  −11 + +  = + + 3 5 8  3 5 8

c)

−6 −6 + 0 = 7 7

 −8  8 d)  + = 0  9  9

6   6 3  7. The expression  ×  +  −4 ×  is an equivalent form of which of the following 11   11 2   expression?

3  6  a) ×  − 4  2  11  c)

6 3  ×  + 4  11  2 

3 6  b)  −4 ×  × 2  11  d)

6 3  × − 4  11  2 

8. Rational numbers are closed under addition because: a) any two rational numbers can be added in any order. b) the sum of any two rational numbers is again a rational number. c) the sum of any three rational numbers is independent of the way the three numbers are grouped together. d) the sum of 0 and any rational number is the same rational number.

9. The sum of the pair of positive numbers is always: a) positive

b) negative

c) neither positive nor negative

d) zero

29

RATIONAL NUMBERS

10. Among the following, a terminating decimal is: 1 1 1 b) c) 3 5 7 11. Among the following, a non-terminating decimal is:

a)

a)

1 16

b)

1 25

c)

1 32

a is: b a b a) b) c) 1 b a 13. The product of a non-zero number and its multiplicative inverse is:

d)

1 6

d)

1 26

d)

−a b

12. Multiplicative inverse of

a) -1

b) 0

c) 1

d) undefined

 −2  14. Additive inverse of −   is:  3  a)

2 3

b)

−2 3

c)

3 2

d)

−3 2

15. Multiplication of a rational number with '0' is: a) additive identity

b) least whole number

c) 0

d) all the above

2 3 and is: 3 4 21 b) 16

16. A rational number between a)

17 24

c)

24 17

d)

15 7

c)

−105 4

d)

4 105

 15 5   7 2  17.  +  ÷  −  =  16 4   12 3  a)

−4 105

b)

105 4

18. The reciprocal of a negative rational number: a) is a positive rational number b) is a negative rational number c) can be either a positive or a negative rational number d) does not exist

30

IL Foundation Series Class 8

 −9 8  × = 19.   16 15  a)

−3 10

b)

−4 15

c)

−9 25

20. Which of the following numbers is in standard form? −12 −49 −9 a) b) c) 26 71 16 21. The smallest among rational numbers a)

−4 5

b)

−8 11

−2 −4 −8 −5 , , , and is: 3 5 11 9 −2 c) 3

d)

−2 5

d)

28 −105

d)

−5 9

22. Rational numbers are not closed under: a) addition

b) subtraction

23. The equidistant rational number between a)

5 24

b)

c) multiplication

d) division.

1 1 and is: 3 2

5 6

c)

5 12

d)

5 8

II. FILL IN THE BLANKS 1. The number of rational numbers between any two distinct rational numbers is ________.

2. Between any two rational numbers, there exist infinitely many rational numbers. This property is known as __________. −64 is __________.

125 −16 −15 4. The product of two numbers is . If one of the numbers is , then the other number is 35 14 ___________.

3. The standard form of

5. The number should be added to

7 −4 to get ___________.

12 15

−4 6. The sum of two numbers is . If one of the numbers is -5, then the other number is 3 ____________.

7. The number that should be added to

−5 −2 to get is _________.

7 3

8. The number that should be subtracted from

−3 to get -2 is _____________.

5 31

RATIONAL NUMBERS

9. The rational number which is equal to its negative is ____________.

10. The value of 'a' for which two rational numbers

3 a are equivalent is ___________. , 7 42

11. The reciprocal of a positive rational number is __________.

12. The product of a rational number with its reciprocal is ____________. 5 13.   6

−1

= __________.

1 is ____________. a 15. The product of two negative rational numbers is always __________.

14. The reciprocal of

5 5 16. −4 × = × _____________.

9 9 III. SUBJECTIVE QUESTIONS 1 1 1 1 1. Simplify: − + − .

2 3 4 5 2. Subtract:

7 −4 from .

9 9

3. Find the difference between the sum of 4. If

4 6 3 2 , and , .

3 7 5 3

p 3 p 3 + = 0, then what can you say about with respect to ? q 10 q 10

−2 4 and .

3 5 6 −7 6. Multiply by the reciprocal of .

13 16 5. Find a rational number between

3  35  + 10 .

7. Find the value of ×  5  24  8. The sides of a triangle are 13 9. Represent

1 3 3 m, 11 m and 3 m. Find its perimeter.

8 2 4

−2 on a number line.

9

10. Write any two rational numbers between -1 and 0.   −25  14  5 11. Evaluate:   ×  + .   28  15  6

32

IL Foundation Series Class 8

12. Write the rational numbers

−5 −5 −5 , and in descending order.

7 2 4

8 4 1 13. Evaluate:  ÷  ÷ . 6 3 2 14. Find a rational number between:

1 1 i) and ii) 2 and 3 4 3 2 15. Find the rational numbers that should be added to to get -1.

9 −2 −5 16. Find the rational numbers that should be subtracted from to get .

3 6 1 1 17. The sum of two fractions is 5 , one of them is 4 , what is the other fraction?

7 7 1 1 18. A tin holds 16 litres of oil. How many such tins will be required to hold 313 litres of oil?

2 2 1 19. From a rope of length 40 m 50 cm, how many pieces of rope can be made of length 2 m each?

4 20. Find the product of the rational numbers 21. Divide: i)

3 −5 by 11 11

−35 37 −99 . , , 111 45 56

2 8 ii) 1 by 7 3 9

22. The cost of 12 cricket bats is `3000. What will be the cost of one bat?

25 5 23. Find the product of and the multiplicative inverse of .

14 7 3 9 24. Find the value of  ÷  × 12. 4 8  21 21   12  25. Find the value of  −  ÷  .  13 13   19   −5   −8  26. Find the value of  ×  × ( −36 ).  16   15 

33

2

2.1

LINEAR EQUATIONS IN ONE VARIABLE

INTRODUCTION TO LINEAR EQUATIONS

Expressions of the form 2 + 3; 12 + ( −5 ) ; 2 × 3 etc., are called numerical expressions.

Two numerical expressions joined by 'is equal to' or 'is greater than' or 'is less than' are called mathematical sentences.

A mathematical sentence that can be verified as either true (or) false but not both is called a mathematical statement.

A sentence that cannot be verified as either true or false is called an open sentence.

An open sentence containing the sign 'is equal to’ is called an equation.

2.1.1 Equation A statement of equality of two algebraic expressions involving one (or) more variables is called an equation.

Example: 2 x + 5 = 25 represents an equation.

In the equation 2 x + 5 = 25, 2 x + 5 is called LHS, and 25 is called RHS. 2.1.2 Linear equation An equation containing only linear polynomials is said to be a linear equation.

Or An equation in which the highest power of the variables present in the equation is one is called a linear equation.

Example: i) 2 x − 3 = 5 is a linear equation in one variable. ii) 2 x − 3y = 5 is a linear equation in two variables.

Standard Form

A linear equation in one variable means that the equation has only one variable in it. It means that this linear equation has one solution. The standard form or the general form of linear equations in one variable is written as ax + b = 0. Here, x is a variable, a is a coefficient, and b is a constant.

Where a and b are integers, and x is the single variable. Both 'a’ and 'b’ are not equal to zero.

34

IL Foundation Series Class 8

For example, 3 x + 6 = 12 is the standard form of a linear equation with a single variable x.

Example: Rewrite the linear equation in standard form, 2y = −5 x + 7.

Solution: To rewrite the given equation in standard form, we will transpose the term −5x on the left-hand side. This means it will become 2y + 5 x = 7. Now, we can arrange the terms on the lefthand side as per the order given in the standard form. This will make it 5 x + 2y = 7. This equation is in its standard form.

2.2

SOLVING LINEAR EQUATIONS IN ONE VARIABLE

2.2.1 Solving linear equations in one variable Root or solution of the equation

A number which, when replaced for the variable of an equation, makes it a true statement is called the root/solution of the equation.

In a true mathematical statement, it is LHS = RHS. The root of the equation satisfies the equation.

Example: Solve the equation 9 x = 20 + 4 x .

Solution: Given, 9 x = 4 x + 20 ⇒ 9 x − 4 x = 20 ⇒ 5 x = 20 ⇒ x = 4, which is the required solution.

SOLVED EXAMPLES Example 1: Solve the equation 3 ( x − 1 ) = 8 Solution: Given, 3 ( x − 1 ) = 8 ⇒ 3x − 3 = 8 ⇒ 3 x = 11 11 ⇒ x = , which is the required solution.

3

35

LINEAR EQUATIONS IN ONE VARIABLE

Example 2: Solve the equation 5 x − 7 = 2 x + 8 Solution: Given 5 x − 7 = 2 x + 8 ⇒ 5x − 2x = 8 + 7 ⇒ 3 x = 15 ⇒ x = 5, which is the required solution.

Example 3: Solve the equation 0.8 x + 1.25 = 2 x + 0.05 Solution: Given, 0.8 x + 1.25 = 2 x + 0.05 ⇒ 2 x − 0.8 x = 1.25 − 0.05 ⇒ 1.2 x = 1.20 ⇒ x =1, which is the required solution.

Example 4: Solve the equation 5 ( x + 43 ) = 2 ( 3 x + 4 ) Solution: Given, 5 ( x + 43 ) = 2 ( 3 x + 4 ) ⇒ 5 x + 215 = 6 x + 8 ⇒ 6 x − 5 x = 215 − 8 ⇒ x = 207 , which is the required solution.

2.2.2 Reducing equations to a simpler form Reducing equations is a method of solving a complex equation and writing the equation into a simpler form. Not all equations are linear, but they can be solved by putting them into the form of linear equations by performing some mathematical operations like cross-multiplication.

After reducing these non-linear equations to linear form, they can be solved, and the value of the variable can be calculated easily.

Steps for reducing equations to a simpler form

Step 1: If the given equation is non-linear, it cannot be directly solved. Therefore, first, we will need

to simplify the given equation by using the cross-multiplication technique.

Step 2: Cross-multiply both sides of the equation, i.e., the denominator on one side is multiplied by

36

the numerator on the other side.

IL Foundation Series Class 8

Step 3: Use the distributive law to open the brackets.

Step 4: Bring all the variables on one side (LHS) and constants on the other side of the equation (RHS).

Step 5: Solve the rest of the equation as a linear equation in one variable.

Example: Solve the equation

6x + 1 x −3 +1 = .

3 6

Solution: 6x + 1 x −3 +1 = 3 6 Multiplying both sides of the equation by 6, we get: Given,

⇒

6( x − 3 ) 6 ( 6x + 1 ) + 6( 1 ) = 6 3

⇒ 2 ( 6x + 1 ) + 6 = x − 3 ⇒ 12 x + 2 + 6 = x − 3 ⇒ 12 x − x = − 3 −6 − 2 ⇒ 11x = − 11 ⇒ x = −1 Example:

5( 1 − x ) + 3( 1 + x ) =8 1 − 2x

Solution:

5(1 − x ) + 3(1 + x ) =8 1 − 2x By cross multiplication, we get: We have,

⇒ ( 5 ( 1 − x ) ) + ( 3 ( 1 + x ) ) = 8 × ( 1 − 2x ) ⇒ 5 − 5 x + 3 + 3 x = 8 − 16 x ⇒ 8 − 2 x = 8 − 16 x

Transposing 8 to RHS, it becomes -8 and −16x to LHS it becomes 16x. ⇒ 16 x − 2 x = 8 − 8 ⇒ 14 x = 0 ⇒x=0

37

LINEAR EQUATIONS IN ONE VARIABLE

Example: Solution: We have,

y − ( 4 − 3y ) 1 = 2y − ( 3 + 4y ) 5 y − ( 4 − 3y ) 1 = 2y − ( 3 + 4y ) 5

⇒

y − 4 + 3y 1 = 2y − 3 − 4y 5

⇒

−4 + 4y 1 = −2y − 3 5

By cross multiplication, we get: ⇒ 5 × ( 4y − 4 ) = 1 × ( −2y − 3 ) ⇒ 20 y − 20 = −2y − 3 Transposing - 20 to RHS, it becomes 20, and transposing −2y to LHS, it becomes 2y. ⇒ 20 y + 2y = +20 − 3 ⇒ 22y = 17 ⇒y=

17 22

Example: Solve the equation

x x x + + + 10000 = x 2 4 5

Solution: x x x + + + 10000 = x 2 4 5 x x x + + − x = −10000 2 4 5 10 x + 5 x + 4 x − 20 x ⇒ = −10000 20 19 x − 20 x ⇒ = −10000 20 −x = −10000 ⇒ 20 ⇒ x = 200000 ⇒

38

IL Foundation Series Class 8

2.3 APPLICATIONS OF LINEAR EQUATIONS IN ONE VARIABLE 2.3.1 Applications of equations in solving verbal (or) word problems Equations are frequently used to solve practical problems. You have already learnt the method and variables involved in solving a word problem. Let us recall them.

Step 1: Read the problem carefully and note down what is given and what is required.

Step 2: Select a letter, say x, y , or z to represent the unknown quantity asked for.

Step 3: Represent the word statements of the problem in the symbolic language step by step.

Step 4: Look for quantities which are equal as per the conditions given and form an equation.

Step 5: Solve the equation obtained in step (4).

Step 6: Check the result to make sure that your answer satisfies the requirements of the problem.

Example: What should be added to twice the rational number

−7 3 to get ?

3 7

Solution: Twice the rational number

−7 ⎛ −7 ⎞ −14 is 2 × ⎜ = ⎝ 3 ⎟⎠ 3 3

3 Suppose x is added to this number, which gives .

7 − 14 3 ⎛ ⎞ = ; That is, x + ⎜ ⎝ 3 ⎟⎠ 7 14 3 = 3 7 3 14 14 ⇒x= + (transposing to RHS) 7 3 3 ( 3 × 3 ) + ( 14 × 7 ) = 9 + 98 = 107 ⇒x= 21 21 21 ⇒x−

Thus,

107 3 ⎛ −7 ⎞ should be added to 2 × ⎜ to give . ⎟ ⎝ 3 ⎠ 21 7

39

LINEAR EQUATIONS IN ONE VARIABLE

SOLVED EXAMPLES 7 Example 1: Solve the equation 5 x − 2 ( 2 x − 7 ) = 2 ( 3 x − 1 ) + .

2 Solution: Let us open the brackets.

LHS: 5 x − 4 x + 14 = x + 14 RHS: 6 x − 2 +

7 4 7 3 = 6x − + = 6x + 2 2 2 2

The equation is, x + 14 = 6 x + ⇒ 14 = 6 x − x + ⇒ 14 − ⇒x =

3 2

3 2

25 3 = 5x ⇒ = 5x 2 2

5 25 1 × ⇒x= 2 5 2

5 ∴ The required solution is x = .

2 1 1 Example 2: S olve the equation x + x = x − 7.

2 4 Solution: 1 1 We have, x + x = x − 7 2 4 Taking LCM on the LHS, we have: 2x + x x − 7 ⇒ = 4 1 On cross-multiplying, we have: ⇒ 2x + x = 4 ( x − 7 ) ⇒ 3 x = 4 x − 28 ⇒ 3 x − 4 x = −28 ⇒ − x = −28 ⇒ x = 28, which is the required answer.

40

IL Foundation Series Class 8

Example 3: The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. What can be the original number?

Solution: Let us take the two-digit number such that the digit in the units place is b.

The digit in the tens place differs from b by 3, i.e., b + 3. So, the original number is 11b + 30.

Now, the interchanging of digits will be 10 b + ( b + 3 ) = 11 b + 3.

If we add these two-digit numbers, their sum is:

( 11b + 30 ) + ( 11b + 3 ) = 11b + 11b + 30 + 3 = 22b + 33 It is given that the sum is 143. ∴ 22 b + 33 = 143 ⇒ 22 b = 110 ⇒ b =

110 ⇒b =5 22

The units digit is 5, the tens digit is 5 + 3 which is 8. ∴ The number is 85.

Example 4: Arjun is twice as old as Shriya. Five years ago, his age was three times Shriya’s age.

Find their present ages.

Solution: Let us take Shriya’s present age to be x years.

Then, Arjun’s present age would be 2x years.

Shriya’s age five years ago was ( x − 5 ) years.

Arjun’s age five years ago was ( 2 x − 5 ) years.

It is given that Arjun’s age five years ago was three times Shriya’s age. ⇒ 2 x − 5 = 3 ( x − 5 ) ⇒ 2 x − 5 = 3 x − 15 ⇒ 10 = x So, Shriya’s present age = x = 10 years. ∴ Arjun’s present age = 2 x × 10 = 20 years.

5 1 Example 5: A person travelled of a distance by train, by bus, and the remaining 15 km by boat.

8 4 Find the total distance travelled by him.

Solution: Let the total distance travelled by him = x km.

41

LINEAR EQUATIONS IN ONE VARIABLE

5 1 Distance travelled by train = x km and the distance travelled by bus = x km.

8 4 5 1 5x + 2x 7 x ∴ Total distance travelled by train and bus in km = x + x = = 8 4 8 8 Remaining distance in km = x −

7x x = 8 8

He travelled this distance by boat. ∴ We are given that the remaining distance = 15 km. x ∴ = 15 or x = 15 × 8 = 120 8 ∴ Total distance travelled = 120 km.

Example 6: The denominator of a rational number is greater than the numerator by 10. If the numerator is increased by 1, and the denominator is decreased by 1, then find the expression for the new denominator.

Solution: Let us assume numerator be x. x x + 10 As per the condition given in the question, the numerator is increased by 1 and the denominator is decreased by 1.

So, denominator = x + 10. Then, the rational number =

Now, new rational number = =

Numerator +1 Denominator −1 x +1 x + 10 − 1

x +1 x+9 Hence, the new denominator is x + 9. =

Example 7: Solve the equation

2 x − 17 ⎛ x −1⎞ −⎜ x− ⎟ = 12 ⎝ 2 3 ⎠

Solution: 2 x − 17 ⎛ x −1⎞ We have, −⎜ x− ⎟ = 12 (removing brackets) ⎝ 2 3 ⎠

42

IL Foundation Series Class 8

⇒

2 x − 17 x − 1 12 −x+ = 2 3 1

⇒

3 ( 2 x − 17 ) − 6 x + 2 ( x − 1 ) 72 = (taking LCM 6 in denominator) 6 6

⇒ 6 x − 51 − 6 x + 2 x − 2 = 72 ( ∴ Cancelling 6 from both the denominators) ⇒ 2 x − 53 = 72 ⇒ 2 x = 72 + 53 ⇒ 2 x = 125 ⇒x =

125 2

⇒ x = 62

1 2

1 Hence, x = 62 is the required answer.

2

QUICK REVIEW •

statement of equality of two algebraic expressions involving one (or) more variables is called A an equation.

•

number which, when replaced for the variable of an equation, makes it a true statement is A called the root/solution of the equation.

In a true mathematical statement, it is LHS = RHS The root of the equation satisfies the equation.

•

A linear equation may have any rational number for its solution.

•

Just as numbers, variables can also be transposed from one side of the equation to the other.

•

ccasionally, the expressions forming equations must be simplified before we can solve them by O usual methods. Some equations may not even be linear, to begin with, but they can be brought to a linear form by multiplying both sides of the equation by a suitable expression.

•

he utility of linear equations is in their diverse applications; different problems on numbers, T ages, perimeters, combinations of currency notes, etc., can be solved using linear equations.

43

LINEAR EQUATIONS IN ONE VARIABLE

WORKSHEET - 1 I. SOLVING LINEAR EQUATIONS IN ONE VARIABLE 1. Solve the following linear equations. i) 7 x = 4 − 3 x

ii) 4 p + 12 = 36 − 2 p

iii) −7 t = 3t − 30

iv) 9y + 5 = 15y − 1

v) 3 ( t − 3 ) = 5 ( 2t + 1 )

vi) 15 ( t − 4 ) − 2 ( t − 9 ) + 5 ( t + 6 ) = 0

vii) 3 ( 5 z − 7 ) − 2 ( 9 z − 11 ) = 4 ( 8 z − 13 ) − 17

viii) 0.25 ( 4 f − 3 ) = 0.05 ( 10 f − 9 )

ix) 0.8 x + 1.25 = 2 x + 0.05

x) ( 0.3 ) x − 1.8 = x − 1.16

xi) 1.3t + 2.7 t = 0.5

xii) 5 ( x + 43 ) = 2 ( 3 x + 4 )

2. Solve the following equations by reducing them to linear forms. x −5 x −3 1 − = 2 5 2 x 1 x 1 iii) − = + 2 5 3 4 8 x 17 5 x v) x + 7 − = −

3 6 2 3t − 2 2t + 3 2 vii) − = −t 4 3 3 i)

8x − 3 = 2 3x z 4 = xi) z + 15 9 ix)

x + 5 x − 5 25 + = 2 3 6 n 3n 5n + = 21 iv) − 2 4 6 x −5 x −3 vi) = 4 5 m −1 m −2 viii) m − = 1− 2 3 ii)

9x = 15 7 − 6x 3y + 4 −2 = xii) 2 − 6y 5

x)

II. APPLICATION OF LINEAR EQUATIONS IN ONE VARIABLE 1. One number is 3 less than two times the other. If their sum is increased by 7, the result is 37.

Find the number.

1 1 2. of a number of butterflies in a garden are on jasmines and of them are on roses. Three 5 3 times the difference between the butterflies on jasmines and roses is on lilies. If the remaining one is flying freely, find the total number of butterflies in the garden.

44

IL Foundation Series Class 8

4 of one of its equal sides. If the perimeter of the triangle is 3 400 cm, then find the lengths of its sides.

3. The base of an isosceles triangle is

4. There are some lotus flowers in a pond, and some bees are hovering around. If one bee lands on each flower, one bee will be left. If two bees land on each flower, one flower will be left. Find the number of bees and the number of flowers in the pond.

5. In a two-digit number, the tens digit is twice the units digit. The number formed by interchanging the digits is 36 less than the original number. Find the number.

6. In a two-digit number, the units digit is 2. If the digits are interchanged, the new number 3 formed is times the old number. Find the number.

8 7. In a two-digit number, the units digit is twice the tens digit. If the sum of the digits is added to the whole number, the result is equal to 30. Find the number.

1 1 of his property to his son, to his daughter and the remaining to his wife. If the 3 4 wife’s share was `40000, find the total value of his property.

8. Murthy left

9. Sundari gave

1 1 1 of her property to 5 social organisations, to 3 religious trusts, and to her 8 15 24

two sons. She donated the remaining property, valued at `1,15,000 for the construction of a school building in the name of her husband. Find the total value of her property.

10. Three prizes are to be distributed at a sports meet so that their total cost is `2500. The value 3 1 of the second prize is of the first prize and the value of the third prize is of the first prize.

4 2 Find the value of each of the three prizes.

5 11. A milkman has some buffaloes, cows, and goats. He has goats equal to times the number 2 3 of cows and cows equal to times the number of buffaloes. If there are 150 animals with the 2 milkman, how many of each category does he have with him?

12. A positive number is 5 times another number. If 21 is added to both numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

13. The sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. Find the two-digit number.

45

LINEAR EQUATIONS IN ONE VARIABLE

14. One of the two digits of a two-digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

15. Shobu’s mother’s present age is six times Shobu’s age. Five years from now, Shobu’s age will be one-third of his mother’s present age. What are their present ages?

16. A grandfather is ten times older than his granddaughter. He is also 54 years older than her.

Find their present ages.

17. Half of a herd of deer is grazing in the field, and three-fourths of the remaining are playing nearby. The remaining 9 are drinking water from the pond. Find the number of deer in the herd.

18. A man’s age is three times his son’s age. Ten years ago, he was five times his son’s age. Find their present ages.

19. A man is 35 years old, and his son is 7 years old. In how many years will the son be half as old as his father?

1 20. The present age of A is twice that of B. 30 years from now, A's age will be 1 times that of B. 2 Find the present ages of A and B. 21. The present age of a boy is one-sixth the age of his father. 5 years later, the sum of their ages will be 45 years. Find their present ages.

22. The present ages of A and B are 20 years and 5 years respectively. After how many years will the age of A be twice that of B? 23. Sobha’s age 13 years ago is half her age 13 years later. Find her present age.

24. Mr Rao is 50 years old, and his daughter Sowjanya is 20 years old. How many years ago was Mr Rao three times the age of his daughter?

25. The ages of Hari and Harry are in the ratio 5 : 7. Four years from now, the ratio of their ages will be 3 : 4. Find their present ages.

26. The denominator of a rational number is greater than its numerator by 8. If the numerator 3 is increased by 17 and the denominator is decreased by 1, the number obtained is . Find the 2 rational number.

27. Sobha drove her car at a speed for the first 4 hours and increased the speed by 10 km/hr for the next two hours. If the total distance travelled by her was 500 km, find the speeds at which Sobha drove her car at different times.

28. Ramaswamy walked to the town at a speed of 7 km/hr and came back by bicycle at a speed of 1 12 km/hr. The total time taken to reach the town and return back was 2 hrs. Find the time 2 taken by him to travel on foot and bicycle.

46

IL Foundation Series Class 8

29. A postman takes tappalas on foot from village P to village Q at a speed of 4 km/hr. If he walks at a speed of 5 km/hr, then he reaches the village 7 minutes earlier. Find the distance between the two villages.

30. A man rowed upstream for 4 hours and returned to his starting point in 2 hours. If he rowed at the rate of 5 km/hr in still water, what was the rate of flow of the stream?

WORKSHEET - 2 I. MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. The standard form of a linear equation in one variable x is: a) ax + b = 0

b) ax 2 + bx + c = 0

c) ax 3 + bx 2 + cx + d = 0

d) ax 4 + bx 3 + cx 2 + dx + e = 0

2. The degree of the equation x 2 − 2 x + 1 = x 2 − 3 is: a) 1

b) 2

c) 0

d) 3.

3. The equation form of the statement, 'a number added to 10, the number becomes 20' is x = 20 10 4. Seven times the number is 42. This statement in the form of an equation is: x a) x + 7 = 42 b) 7 x = 42 c) = 42 d) x − 7 = 42 7 20 − x is: 5. The root of the equation 3 x = 7 5 10 20 5 a) b) c) d) − 7 7 21 7 a) x − 10 = 20

b) x + 10 = 20

c) 10 x = 20

d)

6. The largest number of the three consecutive numbers is x + 1, then the smallest number is: a) x + 2

b) x + 1

c) x

d) x − 1

7. If x is an even number, then the consecutive even number is: a) x + 1

b) x + 2

c) 2x

d) x + 3

8. If x is an odd number, then the largest odd number preceding x is: a) x − 1

b) x − 2

c) x − 3

d) x − 4

9. The difference between two numbers is 21. If the larger number is x, then the smaller number is: a) 21 + x

b) 21 − x

c) x − 21

d) − x − 21

47

LINEAR EQUATIONS IN ONE VARIABLE

10. The root of the equation 2y = 5 ( 3 + y ) is: 1 1 c) -5 d) − 5 5 11. In a two-digit number, the units digit is x and the tens digit is y. Then the number is:

a) 5

b)

a) 10y + x

b) 10x + y

x x 1 + 4 = + 3 , then x = 3 2 2 1 −1 a) b) 3 3

c) 10y − x

d) 10x − y

c) -3

d) 3

c) 100

d)

100 8

c) 7

d)

12 7

12. If

13. If 0.5 x − 2.5 = 77.5 − 0.3 x, then x = a) 80 14. If

b) 10

( 3 x − 1 ) − ( 1 + x ) = 3 − ( x − 1 ) , then x = 5

2

a) 2

2

b) 3

15. The sum of three numbers is 60, and their ratio is 1 : 2 : 3. Then the largest number is: a) 60

b) 10

c) 20

d) 30

16. The length of a rectangle is 4 cm more than its breadth. If the perimeter of the rectangle is 4.16 m, then its breadth is: a) 102 m

b) 1.02 m

c) 1020 cm

d) 104 cm

17. The perimeter of a triangle is 18 m, and the ratio of their sides is 2 : 3 : 4. Then the smallest side is: a) 2 m

b) 4 m

c) 6 m

d) 8 m

18. If the sum of four consecutive integers is 46, then the four integers are: a) 11, 12, 13, 14

b) 10, 11, 12, 13

c) 6, 7, 8, 9

d) 20, 21, 22, 23

19. If a number is tripled and the result is increased by 5, we get 50. Then the number is: a) 10

b) 15

c) 20

d) 25

20. A number is such that it is as much greater than 84 as it is less than 108. Then the number is: a) 86

b) 96

c) 106

d) 116

21. A number is 56 greater than the average of its one-third, quarter and one-twelfth. Then the number is: a) 62

48

b) 72

c) 82

d) 92

IL Foundation Series Class 8

22. If 0.16 ( 5 x − 2 ) = 0.4 x + 7, then x = a) 16.3 23. If x + 7 −

b) 17.3

c) 18.3

d) 20.3

c) 6

d) 7

8 x 17 5 x = − , then x = 3 6 8

a) 4

b) 5

3 2 24. part of a number is 5 more than its part. This statement in the form of an equation is: 4 3 2 3 2 3 3 2 3 −2 x a) x − x = 5 b) x − 5 = x c) x = x + 5 d) x − 5 = 3 4 3 4 4 3 4 3 25. If two angles are complementary and one angle is 10 greater than the other, then the smaller angle of the two is: a) 40

b) 50

c) 90

d) 180

II. FILL IN THE BLANKS 1. Two times a number is as much greater than 30 as three times the number is less than 60.

Then the number is ______________.

2. One number is greater than the other number by 3. The sum of the two numbers is 23. Then the two numbers are ___________.

3. If 9 is added to two times a number, we get 67, then the number is ____________.

4. The root of the equation ( 2 x − 1 ) + ( x − 1 ) = x + 2 is ______________.

5. The base of an isosceles triangle is 6 cm, and its perimeter is 16 cm. The length of each of the equal sides is ______________.

6. If cx + d = 0, then x = ______________. x 7 7. If + 1 = , then x = ______________.

3 15 8. The sum of a two-digit number and the number obtained by reversing the digits is a multiple of _______________.

III. SUBJECTIVE QUESTIONS 1. Two numbers are in the ratio 5 : 8. If the sum of the numbers is 572, find the numbers.

2. Find the value of x, for which the expressions 3 x − 4 and 2 x + 1 become equal.

3. Kamalesh’s present age is thrice of Rupesh. If Rupesh’s age three years ago was 'x', then find Kamalesh’s present age in terms of x.

4. '9 is subtracted from the product of p, and 4 and the result is 11'. Express this statement in the form of a linear equation.

5. The sum of three consecutive multiples of 7 is 357. Find the smallest multiple.

49

LINEAR EQUATIONS IN ONE VARIABLE

6. Solve the equation

−4 3 y= .

3 4

7. The speed of a boat in still water is x km/hr, the speed of the stream is 4 km/hr, and the speed of downstream is 10 km/hr. Express this statement in the form of a linear equation.

8. The adjacent angles of a parallelogram are x and y. Express the sum of the adjacent angles of the parallelogram in the form of a linear equation.

9. Solve the equation 0.35 x − 0.25 = 0.3 x.

10. Find the two consecutive integers whose sum is 63.

50

3

3.1

UNDERSTANDING QUADRILATERALS

INTRODUCTION

3.1.1 Polygons A closed plane shape bounded by three or more line segments is called a polygon.

Number of sides

Name

Number of sides

Name

3

Triangle

14

Tetrakaidecagon (or) Tetradecagon

4

Quadrilateral

15

Pendedecagon (or) Quindecagon

5

Pentagon

16

Hexadecagon

6

Hexagon

17

Heptadecagon

7

Septagon/ Heptagon

18

Octadecagon

8

Octagon

19

Enneadecagon

9

Nonagon

20

Icosagon

10

Decagon

⋅

⋅

11

Hendecagon

⋅

⋅

12

Dodecagon

⋅

⋅

13

Triskaidecagon (or) Tridecagon

n

n-gon

Example:

Diagonal: A diagonal is a line segment connecting two non-adjacent vertices (corners) of a polygon.

The number of diagonals in a polygon with n sides can be determined by the formula:

n( n − 3 ) .

2

51

UNDERSTANDING QUADRILATERALS

Example:

A quadrilateral has two diagonals.

A pentagon has five diagonals.

3.1.2 Classification of polygons Based on interior angles

Convex polygon: It is a polygon whose interior angles are less than 180. All diagonals of a convex polygon lie inside the closed figure.

Example:

Concave polygon: Also known as a non-convex polygon, it is a polygon with at least one interior angle measuring more than 180. Some diagonals of a concave polygon lie outside the closed figure.

Example:

Based on the lengths of the sides

Regular polygon: It is a polygon with all sides and interior angles of equal measure. Thus, a regular polygon is both equilateral and equiangular.

Examples: An equilateral triangle, square, regular pentagon.

Irregular polygon: It is a polygon that has sides of unequal length and angles of unequal measure.

Any polygon that is not regular is thus an irregular polygon.

52

IL Foundation Series Class 8

Examples: A scalene triangle, rectangle, trapezoid.

3.1.3 Sum of angles in polygons Interior angles in a polygon

The sum of the measures of the internal angles of a polygon is given by ( n − 2 ) × 180°, where n is the number of sides of the polygon.

The interior angle of a regular polygon can be measured by using the formula: Interior angle =

( n − 2 ) × 180° , where n is the number of sides. n

Exterior angles in a polygon

The sum of the measures of the external angles of any polygon is 360°.

The exterior angle of a regular polygon can be measured by using the formula: Exterior angle =

360 , where n is the number of sides. n

SOLVED EXAMPLES Example 1: Find the number of diagonals of a polygon with the following sides: i) 7 sides

ii) 20 sides

Solution: We know that the number of diagonals in a polygon with n sides is given by the formula:

n( n − 3 ) .

2

i) The number of diagonals in a polygon with 7 sides: =

7( 7 − 3 ) 7 × 4 = = 14 2 2

ii) The number of diagonals in a polygon with 20 sides: 20 ( 20 − 3 ) 20 × 17 = = = 170 2 2 Example 2: Find the measure of x in the given figure.

90°

50° x° 110°

53

UNDERSTANDING QUADRILATERALS

Solution: We know that the sum of the measures of the external angles of any polygon is 360°. x + 90 + 50 + 110 = 360° ⇒ x + 250 = 360° ⇒ x = 110° Example 3: Find the number of sides of a regular polygon that has exterior angles of 45 each.

Solution: The total measure of all exterior angles = 360 The measure of each exterior angle = 45 Therefore, the number of exterior angles =

360° =8 45°

The polygon has 8 sides.

3.2

QUADRILATERALS

Quadrilateral: A quadrilateral is a closed figure formed by four line segments such that no two line segments cross each other except at their endpoints.

Quadrilaterals can be classified into two types.

1. Convex quadrilateral: A quadrilateral in which the measure of each interior angle is less than 180 is known as a convex quadrilateral.

C D

A

B

2. Concave quadrilateral: A quadrilateral in which the measure of one of the interior angles is more than 180 is known as a concave quadrilateral.

C B

D

A 54

IL Foundation Series Class 8

3.2.1 Elements of a quadrilateral C D

A

B

In a quadrilateral ABCD, i) Four sides: AB, BC , CD, DA. ii) Four angles: ∠A, ∠B, ∠C , ∠D . iii) Four vertices: A, B, C , D . iv) Two diagonals: AC , BD .

In a quadrilateral ABCD, i) Adjacent sides: AB and BC ; BC and CD; CD and DA; DA and AB . ii) Opposite sides: AB and CD; AD and BC . iii) Opposite angles: ∠A and ∠ C ; ∠B and ∠D .

3.2.2 Properties of a quadrilateral i) The sum of the interior angles in a quadrilateral is 360.

Example: In a quadrilateral ABCD,∠A + ∠B + ∠C + ∠D = 360. ii) Each diagonal divides the quadrilateral into two triangles.

Example: D C A B

The diagonal AC divides the quadrilateral ABCD into DABC and DADC . iii) In general, a convex quadrilateral is treated as a quadrilateral. iv) In a convex quadrilateral, both diagonals lie in the interior, whereas in a concave quadrilateral, one diagonal lies in the interior and the other lies in the exterior of the quadrilateral.

55

UNDERSTANDING QUADRILATERALS

3.3

TYPES OF A QUADRILATERAL

3.3.1 Trapezium Trapezium: A Trapezium is a quadrilateral in which one pair of the opposite sides is parallel.

(

)

Note: The parallel sides AB , CD are called the bases of the trapezium, and the other two sides are called its non-parallel sides (legs) (BD, AC ).

D

C

B

A Isosceles trapezium

A trapezium in which the non-parallel sides are equal to each other is known as an isosceles trapezium.

In the isosceles trapezium ABCD, AB  CD, AD = BC. Note: In an isosceles trapezium, the diagonals are equal.

C

D 3

4

2 A

1

B

3.3.2 Kite A quadrilateral having two pairs of equal adjacent sides but unequal opposite sides is called a kite.

Quadrilateral ABCD is a kite with AB = BC and AD = CD .

D A

O

B

56

C

IL Foundation Series Class 8

Properties

i) The diagonals of a kite are perpendicular to each other, i.e., BD ⊥ AC . ii) OA = OC . iii) ∠A = ∠C. iv) Diagonal BD bisects ∠B and ∠D. v) Diagonal BD divides the kite into two congruent triangles.

3.3.3 Parallelogram A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram.

D

C

O A

B

Properties

= ( AO OC = , BO OD) . i) The diagonals of a parallelogram bisect each other ii) In the parallelogram ABCD, both pairs of opposite sides are equal, i.e., AB = CD; BC = AD. iii) The opposite angles are equal, i.e., ∠A = ∠C ; ∠B = ∠D. iv) The adjacent angles are supplementary, i.e., the sum of the adjacent angles is 180°.

Theorem 1: The opposite sides of a parallelogram are equal.

D

A

C

B

Let us consider a parallelogram ABCD in which AB  CD and BC  AD . Join AC. Now, in DABC and DCDA

∠ CAB = ∠ ACD

[Alternate angles]

∠ ACB = ∠ CAD

[Alternate angles]

AC = CA

[Common] 57

UNDERSTANDING QUADRILATERALS

∴ ∆ABC ≅ ∆CDA

[ASA congruency]

Hence, AB = CD and BC = AD

[C.P.C.T.]

Theorem 2: The opposite angles of a parallelogram are equal.

D

C

A

B

Let us consider a parallelogram ABCD in which AB  CD and BC  AD .

Now, AB  DC and AD is a transversal. ∴ ∠ A + ∠ D = 180 ----------(i) [The sum of the measures of the interior angles on the same side of a transversal is 180. ] Again, AD  BC and AB is a transversal. ∴ ∠ A + ∠ B = 180 ----------(ii) From (i) and (ii), we have:

∠D =∠B Similarly, ∠ A = ∠ C .

Theorem 3: The diagonals of a parallelogram bisect each other.

D

C

O A

B

Let us consider a parallelogram ABCD in which AB  CD and AD  BC . Join AC and BD such that they intersect at O. In DAOB and DCOD , we have:

58

∠ DCO = ∠ OAB

[Alternate angles]

∠ CDO = ∠ OBA

[Alternate angles]

IL Foundation Series Class 8

AB = CD

[Opposite sides of a parallelogram]

∴ ∆AOB ≅ ∆COD

[ASA congruency]

Hence, AO = OC and BO = OD

[C.P.C.T.]

Theorem 4: Each diagonal of a parallelogram divides it into two congruent triangles.

D

C

A

B

Let us consider a parallelogram ABCD where AB  CD and AD  BC .

Join AC. In DABC and DCDA , we have: ∠DCA = ∠BAC

[Alternate angles]

∠DAC = ∠ACB

[Alternate angles]

AC = CA

[Common]

∴ ∆ABC ≅ ∆CDA

[ASA congruency]

Similarly, by drawing the other diagonal BD, we can prove that ∆ABD ≅ ∆CDB. 3.3.4 Rectangle Rectangle: If one of the angles of a parallelogram is a right angle, then all the angles are right angles. Such a parallelogram is called a rectangle.

(OR) A parallelogram in which one angle is a right angle is called a rectangle.

D

C O

A

B

59

UNDERSTANDING QUADRILATERALS

Properties

A rectangle satisfies all the properties of a parallelogram. i) The lengths of the diagonals of a rectangle are equal and bisect each other. ii) The opposite sides of a rectangle are equal. iii) The opposite angles of a rectangle are equal.

Theorem 5: Each angle of a rectangle is a right angle.

D

C

A

B

Let us consider a rectangle ABCD. Then, = AB DC = , BC AD ----------(i)

and ∠ A = ∠ B = ∠ C = ∠ D ----------(ii) Now, AB  DC and AD is the transversal. ∴ ∠ A + ∠ D = 180 ⇒ ∠ A + ∠ A = 180 ⇒ ∠ A = 90

[From (ii)]

Hence, ∠ A = ∠ B = ∠ C = ∠ D = 90

[From (ii)]

Theorem 6: The diagonals of a rectangle are equal.

D

C

A

B

Let us consider a rectangle ABCD with diagonals AC and BD. In DABC and DBAD , we have:

60

∠ ABC = ∠ BAD

[Each right angle]

AB = AB

[Common]

AD = BC

[Opposite sides of a rectangle]

So, ∆ABC ≅ ∆BAD

[SAS congruency]

⇒ AC = BD

[C.P.C.T.]

IL Foundation Series Class 8

3.3.5 Rhombus A parallelogram in which two adjacent sides are equal is called a rhombus.

D

A

C

O

B

Properties

i) Each diagonal of a rhombus divides it into two congruent isosceles triangles. ii) The opposite angles are equal, and the sum of any two adjacent angles is 180. iii) The diagonals bisect each other perpendicularly. iv) The diagonal AC bisects ∠A and ∠C ; the diagonal BD bisects ∠B and ∠D. Theorem 7: The diagonals of a rhombus bisect each other at right angles.

D

C

O A

B

Let us consider a rhombus ABCD with diagonals AC and BD. In DAOB and DBOC , we have: AB = BC

[Sides of a rhombus]

BO = OB

[Common]

AO = OC

[Diagonals bisect each other]

∴ ∆AOB ≅ ∆COB

[SSS congruency]

Hence, ∠ AOB = ∠ BOC

[C.P.C.T]

But, ∠ AOB + ∠ BOC = 180

[Linear pair]

2∠ AOB = 180 ∴ ∠ AOB = ∠ BOC = 90

61

UNDERSTANDING QUADRILATERALS

3.3.6 Square Square: A rectangle in which adjacent sides are equal is called a square.

(OR) A rhombus in which one of its angles is a right angle is called a square.

D

C

A

B

Properties

i) All sides are equal. ii) Each angle is equal to 90. iii) The diagonals are equal and are mutual perpendicular bisectors. iv) Each diagonal divides the square into two congruent right-angled isosceles triangles. v) The quadrilateral formed by joining successively the midpoints of the sides of a square is a square.

Theorem 8: The diagonals of a square are equal.

D

C

A

B

Let us consider a square ABCD with diagonals AC and BD. In DABD and DBAC , we have: AD = BC

[Sides of the square]

AB = BA

[Common]

∠ DAB = ∠ ABC

[Each 90 ]

∴ ∆ABD ≅ ∆BAC

[SAS congruency]

AC = BD

[C.P.C.T.]

Note: A parallelogram is said to be a square if all its sides and angles are equal.

62

IL Foundation Series Class 8

Family of quadrilaterals QUADRILATERAL

TRAPEZIUM (One pair of the opposite) sides are parallel

PARALLELOGRAM (Both pairs of opposite) sides are parallel

KITE (Two pairs of equal adjacent sides but unequal opposite sides)

RHOMBUS (Adjacent sides are equal)

SQUARE (Adjacent sides are equal, and one angle is 90O)

ISOSCELES TRAPEZIUM (Non-parallel sides are equal)

RECTANGLE (One angle is 90O)

Properties of quadrilaterals

Properties of quadrilaterals

Rectangle Square

Parallelogram

Rhombus

Trapezium Kite

All sides are equal

No

Yes

No

Yes

No

No

Opposite sides are equal

Yes

Yes

Yes

Yes

No

No

Opposite sides are parallel

Yes

Yes

Yes

Yes

Yes

No

All angles are equal

Yes

Yes

No

No

No

No

Opposite angles Yes are equal

Yes

Yes

Yes

No

No

Sum of two adjacent angles is 180°

Yes

Yes

Yes

Yes

No

No

Diagonals bisect Yes each other

Yes

Yes

Yes

No

No

Diagonals bisect No perpendicularly

Yes

No

Yes

No

No

63

UNDERSTANDING QUADRILATERALS

Properties of quadrilaterals

Rectangle Square

Parallelogram

Area (A)

Length × breadth

Length × height 1 × 2 product of

Side × side = side2

Rhombus

diagonals (or) 1 d1 × d2 2 Perimeter (P)

Sum of the lengths of its sides or 2× (length + breadth)

Sum of the lengths of its sides or 4 × side

Sum of the lengths of its sides

Trapezium Kite 1 Sum × 2 of parallel sides × distance between them

Sum of the Sum of the lengths of lengths of its sides its sides or 4 × side

1 product × 2 of diagonals (or) 1 d1 × d2 2

Sum of the lengths of its sides

Quadrilaterals formed by joining the midpoints of sides

The quadrilateral formed by joining the midpoints of the sides

Type of quadrilateral

Parallelogram

Parallelogram

Rectangle

Rhombus

Rhombus

Rectangle

Square

Square

Quadrilateral

Parallelogram

SOLVED EXAMPLES Example 1: In a quadrilateral, three angles are 76 , 54 and 108. Find the fourth angle.

Solution: Let the fourth angle of the quadrilateral be x°.

The three angles of a quadrilateral are 76 , 54 and 108.

The sum of the interior angles of a quadrilateral = 360° ⇒ 76° + 54° + 108° + x° = 360° 64

IL Foundation Series Class 8

⇒ 238° + x° = 360° ⇒ x° = 122° Hence, the fourth angle of the quadrilateral is 122°.

Example 2: Find the values of x and y if the quadrilateral is a parallelogram.

E

4y

F

6x - 12 D

2x + 36 6y - 42

G

Solution: The opposite sides of a parallelogram are equal.

EF = DG ⇒ 4y = 6y − 42 ⇒ −2y = −42 ⇒ y = 21 DE = FG ⇒ 6 x − 12 = 2 x + 36 ⇒ 4 x = 48 ⇒ x = 12 So, when x is 12 and y is 21, DEFG is a parallelogram.

Example 3: Find the measure of each interior angle.

B 2x° x° A

C 2x° x° D

Solution: In quadrilateral ABCD, the sum of the measures of the angles is 360°.

∠ A + ∠ B + ∠ C + ∠ D = 360° ⇒ x + 2 x + 2 x + x = 360° ⇒ 6 x = 360° 65

UNDERSTANDING QUADRILATERALS

⇒ x = 60° The measure of each angle is:

∠ A = 60° ∠ B = 2 × 60° = 120° ∠ C = 2 × 60° = 120° ∠ D = 60° Example 4: Prove that, in a parallelogram, the sum of any two adjacent angles is 180 (or) in a parallelogram, two adjacent angles are supplementary.

Solution: et ABCD be a parallelogram. Then, ∠A, ∠B ; ∠B , ∠C ; ∠C , ∠D and ∠D, ∠A are the four pairs of L adjacent angles.

D

A

C

B

We have to prove that: ∠ A + ∠B = 180°

∠B + ∠C = 180°

∠ C + ∠D = 180°

∠D + ∠A = 180°

I n a parallelogram ABCD, we have AD  BC , transversal AB intersects with them at A and B, respectively. ∴∠ A + ∠B = 180° [The sum of the co-interior angles is 180°.]

Similarly, we can prove that: ∠ B + ∠C = 180°, ∠C + ∠D = 180°, ∠D + ∠A = 180° Example 5: Two adjacent angles of a parallelogram are equal. What is the measure of each angle?

Solution: Let the measure of each angle be x.

Since adjacent angles of a parallelogram are supplementary. ∴ x + x = 180  ⇒ 2 x = 180 180° ⇒ x= 2 ⇒ x = 90°

66

IL Foundation Series Class 8

Hence, the measure of each angle is 90.

Example 6: In a parallelogram ABCD, ∠D = 115. Determine the measure of ∠A and ∠B. Solution: The sum of any two consecutive angles of a parallelogram is 180.

In a parallelogram ABCD, ∠ A + ∠D = 180° (The sum of the two adjacent angles of a parallelogram is 180°.) ⇒ ∠A + 115° = 180° ⇒ ∠A = 65° Now, ∠A + ∠B = 180° (The sum of the two adjacent angles of a parallelogram is 180°) ⇒ 65° + ∠B = 180° ⇒ ∠B = 115° Example 7: Two adjacent angles of a parallelogram are in the ratio 2 : 3. Find the measures of all the angles.

Solution: Let ABCD be a parallelogram and ∠A, ∠B are two adjacent angles such that ∠A : ∠B = 2 : 3 Let ∠A = 2 x, ∠B = 3 x ⇒ 2 x + 3 x = 180 (∴∠A + ∠B = 180° ) ⇒ 5 x = 180 180 ⇒x = 5 ⇒ x = 36

( ) ∠B = 3 x = 3 ( 36 ) = 108

∴ ∠A = 2 x = 2 36 = 72

∠C = 72 (∵ ∠A = ∠C ) ∠D = 108 (∵ ∠B = ∠D )

67

UNDERSTANDING QUADRILATERALS

QUICK REVIEW

68

•

A closed plane shape bounded by three or more line segments is called a polygon.

•

A diagonal is a line segment connecting two non-adjacent vertices (corners) of a polygon.

•

The number of diagonals in a polygon with n sides can be determined by the formula:

•

A convex polygon is a polygon that has all its interior angles less than 180.

•

concave polygon is a polygon that has at least one of its interior angles measuring more than A 180.

•

A regular polygon is a polygon that has all sides and interior angles of equal measure.

•

An irregular polygon is a polygon that has sides of unequal length and angles of unequal measure.

•

he sum of the measures of the internal angles of a polygon is given by ( n − 2 ) × 180°, where n T is the number of sides of the polygon.

•

The sum of the measures of the external angles of any polygon is 360°.

•

quadrilateral is a closed figure formed by four line segments such that no two line segments A cross each other except at their endpoints.

•

The sum of the interior angles of a quadrilateral is 360.

•

A trapezium is a quadrilateral in which one pair of opposite sides are parallel.

•

trapezium in which the non-parallel sides are equal to each other is known as an isosceles A trapezium.

•

quadrilateral that has two pairs of equal adjacent sides but unequal opposite sides is called a A kite.

•

A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram.

•

I f one of the angles of a parallelogram is a right angle, then all the angles are right angles. Such a parallelogram is called a rectangle.

•

A parallelogram in which two adjacent sides are equal is called a rhombus.

•

A rectangle in which adjacent sides are equal is called a square.

n( n − 3 ) .

2

IL Foundation Series Class 8

WORKSHEET - 1 I.

POLYGONS 1. In the figure, find the value of x.

85°

20°

x° 92° 89°

2. The ratio between an exterior and interior angle of a regular polygon is 1 : 5. Find the number of sides of the polygon.

3. Find the measure of each angle of a regular octagon.

4. Find the measure of an exterior angle of a regular pentagon and an exterior angle of a regular decagon. What is the ratio between these two angles?

5. Find the number of diagonals of a polygon with the following sides: i) 7 sides

ii) 11 sides

iii) 15 sides

iv) 20 sides

6. Find the sum of the interior angles of: i) Hexagon

ii) Nonagon

iii) Decagon

7. State the name of a regular polygon that has: i) 3 sides

ii) 4 sides

iii) 6 sides

8. Find x in the following figures: 125°

i)

60°

x°

ii) 70° x°

125°

9. Find the measure of each exterior angle of a regular polygon with: i) 9 sides

ii) 15 sides

iii) 36 sides 69

UNDERSTANDING QUADRILATERALS

10. Find the measure of an interior angle of a regular polygon with 9 sides.

11. How many sides does a regular polygon have if the measure of one of its exterior angles is 24°?

12. How many sides does a regular polygon have if each of its interior angles is 165?

13. i) What is the minimum measure of an interior angle possible for a regular polygon? Why? ii) What is the maximum measure of an exterior angle possible for a regular polygon?

14. If each exterior angle of a regular polygon is 30, find the number of sides it has.

II. QUADRILATERALS AND TYPES OF QUADRILATERALS 1. The angles of a quadrilateral are x, x − 10 , x + 30, and 2x. Find the angles.

2. The angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4. Find the measure of each angle.

3. In a quadrilateral, three angles are 60 , 80 , 100. Find the fourth angle.

4. In a quadrilateral, two angles are 80 , 120, and the remaining two angles are equal. Find the measure of each angle.

5. In a quadrilateral ABCD, ∠A + ∠C = 180°, find the value of ∠B + ∠D. 6. ABCD is a kite with AB = BC = 8 cm and AD = CD = 5 cm. The diagonals intersect at O such = AO 3= cm, DO 4 cm. Find the area of the kite. that D 5cm A

5cm C

O

8cm

8cm

B

7. If ABCD is an isosceles trapezium and ∠A = 40°, then find all the remaining angles of the trapezium.

8. In a rectangle ABCD, the diagonals bisect each other at O and ∠AOB = 120, find ∠OAB and ∠OAD. 9. In a kite ABCD, AB = BC and AD = DC , and ∠A = 115°, find ∠D + ∠B.

70

IL Foundation Series Class 8

10. In the given figure, PQRS is a square, and STR is an equilateral triangle. Find the value of a.

P

Q T

N

a S

R

11. In the following figure, ABCD is a parallelogram. Find the value of r.

D

C s°

r°

3p° q° 7p°

5p°

A

B

12. The angles of a quadrilateral ABCD are x  , (x +1) , (x + 2), and (x + 3), taken in the same order. Determine the type of quadrilateral ABCD. 13. The adjacent sides of a rectangle are in the ratio 5 : 12 . If the perimeter of the rectangle is 34 cm, find the length of the diagonal.

14. ABCD is a parallelogram. Find x.

D

A

C

5x-1

3x+5

B

15. 'Every square is a rectangle, but every rectangle is not a square'. Is this statement true or false?

Justify your answer.

16. The adjacent sides of a parallelogram differ by 30 cm. If its perimeter is 160 cm, find the length of its sides.

71

UNDERSTANDING QUADRILATERALS

17. In the given figure, AB  DC and AD  BC . Find ∠ DAC. A

B

40°

50° C

D

18. The diagonals PR and QS of a rectangle PQRS intersect at O. If ∠QPO = 30, then find ∠SQR. 19. In a square PQRS , PQ = ( 5a − 17 ) cmand QR = ( 2a + 4 ) cm. Find PS. 20. If ABCD is a kite, find the values of x and y.

B

10

x+12

C

A 102° D

3y+1

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. The sum of the exterior angles of a quadrilateral is:

a) 90

b) 180

c) 270

d) 360

2. The sum of the interior angles of a quadrilateral is: a) 1 right angle

b) 2 right angles

c) 4 right angles

d) 8 right angles

c) 3 triangles

d) 4 triangles

3. Each diagonal divides the quadrilateral into: a) 5 triangles

b) 2 triangles

4. In the given figure, ABCD is a quadrilateral, the adjacent sides are: C D

A

a) AB , BC 72

b) AB , CD

B

c) AC , BD

d) BC , AD

IL Foundation Series Class 8

5. The number of diagonals in a decagon is: a) 25

b) 35

c) 45

d) 55

6. Statement A: In a quadrilateral ABCD, the diagonals are AC and BD. Statement B: In a quadrilateral ABCD,∠A and ∠C are opposite angles. a) Both A and B are true.

b) Both A and B are false.

c) A is true, but B is false.

d) A is false, but B is true.

7. Statement A: The sum of the interior angles in a quadrilateral is 180.

Statement B: A quadrilateral has only one diagonal. a) Both A and B are true.

b) Both A and B are false.

c) A is true, but B is false.

d) A is false, but B is true.

8. Assertion (A): If the three angles of a quadrilateral are 55 , 65 and 150 , then the fourth angle is 90.

Reason (R): The sum of all interior angles of a quadrilateral is 360. a) Both A and R are correct, and R is the correct explanation of A. b) Both A and R are correct, but R is not the correct explanation of A. c) A is correct, but R is incorrect. d) A is incorrect, but R is correct.

9. A quadrilateral in which one pair of opposite sides are parallel is called a: a) Trapezium

b) Parallelogram

c) Rectangle

d) Square

10. The diagonals of a kite are ____________ to each other. a) Perpendicular

b) Parallel

c) Equal

d) Not equal

11. In a parallelogram, if two adjacent angles are equal, then it is called a: a) Rectangle

b) Square

c) Rhombus

d) Trapezium

12. S tatement A: A quadrilateral in which one pair of the opposite sides are parallel is called a trapezium.

Statement B: A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram. a) Both A and B are true.

b) Both A and B are false.

c) A is true, but B is false.

d) A is false, but B is true.

13. The diagonals of a parallelogram ___________ each other. a) Bisect

b) Trisect

c) Perpendicular

d) Both a and c

73

UNDERSTANDING QUADRILATERALS

14. If one angle of a rhombus is a right angle, then it is a: a) Trapezium

b) Square

c) Rhombus

d) Parallelogram

15. The number of diagonals of a regular hexagon is: a) 2

b) 0

c) 4

d) 9

16. The sum of the angles of a convex polygon with 7 sides is: a) 900

b) 1080

c) 1440

d) 720

17. The sum of the angles of a convex polygon with n number of sides is: a) ( n − 2 ) × 180

b) ( n + 2 ) × 180

c) ( 2n − 4 ) × 180

d) ( 2n + 4 ) × 180

c) Regular hexagon

d) Regular octagon

18. The name of a regular polygon with 3 sides is: a) Equilateral triangle b) Square 19. The name of a regular polygon with 4 sides is: a) Regular hexagon

b) Rectangle

c) Equilateral triangle d) Square

20. The sum of the measures of the external angles of any polygon is: a) 90

b) 180

c) 360

d) 720

21. If the measure of each of the four angles of a quadrilateral is equal, then the measure of each angle is: a) 45

b) 30

c) 60

d) 90

22. Two adjacent angles of a quadrilateral measure 130 and 40. The sum of the remaining two angles is: a) 190

b) 180

c) 360

d) 90

23. The measures of the two angles of a quadrilateral are 110 and 100, and the remaining two angles are equal. The measure of each of the remaining two angles is: a) 30

b) 60

c) 75

d) 45

24. The number of sides of a regular polygon, whose each exterior angle has a measure of 45, is: a) 4

b) 6

c) 8

d) 10

25. The measure of each exterior angle of a regular polygon with 9 sides is: a) 30

b) 40

c) 60

d) 45

26. The measure of each exterior angle of a regular polygon with 15 sides is: a) 30

b) 45

c) 60

d) 24

27. The number of sides a regular polygon has, if the measure of an exterior angle is 24, is: a) 6

74

b) 9

c) 15

d) 12

IL Foundation Series Class 8

28. The number of sides a regular polygon has, if each of its interior angles is 165, is a) 12

b) 24

c) 9

d) 6

29. In a regular polygon of n sides, the measure of each interior angle is: a)

360 n

 2n − 4   b)   90  n 

c)

( n − 2 ) 90 n

d) 2n right angles

30. The sum of the interior angles of a polygon is 10 right angles. The number of sides it has is: a) 5

b) 6

c) 7

d) 8

31. Among the following statements, the incorrect statement is: a) All the angles of a rectangle are equal. b) No angle of a rectangle can be obtuse. c) The diagonals of a rectangle bisect each other. d) The opposite sides of a rectangle are not equal.

32. Among the following statements, the incorrect statement is: a) A square is a rectangle whose adjacent sides are equal. b) A square is a rhombus whose one angle is a right angle. c) The diagonals of a square bisect each other at right angles. d) The diagonals of a square do not divide the whole square into four equal parts.

33. Among the following statements, the false statement is: a) All rectangles are parallelograms.

b) All squares are rectangles.

c) All parallelograms are rectangles.

d) All rhombuses are parallelograms.

34. Among the following statements, the true statement is: a) All rectangles are squares. b) All parallelograms are rhombuses. c) All rhombuses are squares. d) Each parallelogram is a trapezium.

35. One angle of a parallelogram is a right angle. The name of the quadrilateral is: a) Square

b) Rectangle

c) Rhombus

d) Kite

36. Two adjacent sides of a rectangle are equal. The name of the quadrilateral is: a) Square

b) Kite

c) Rhombus

d) None of these

37. If one angle of a parallelogram is 60, then the measure of the opposite angle is: a) 60

b) 120

c) 30

d) none of these

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UNDERSTANDING QUADRILATERALS

38. If all the sides of a parallelogram are equal and the adjacent angles are of 120 , 60, then the name of the quadrilateral is: a) Rectangle

b) Square

c) Rhombus

d) Kite

39. The incorrect statement in the following in the case of a kite is: a) The diagonals are perpendicular to each other. b) The diagonals bisect each other. c) Only one pair of the opposite angles is equal. d) Both pairs of adjacent sides are equal.

II. MULTIPLE CHOICE QUESTIONS WITH MULTIPLE CORRECT ANSWERS 1. In a quadrilateral ABCD , the side adjacent to BC is: a) AB

b) AD

c) DC

d) AC

2. The diagonals are equal in a/an: a) Trapezium

b) Isosceles trapezium c) Rectangle

d) Rhombus

3. If the bisectors of angles A and B of a quadrilateral ABCD meet at O, then ∠AOB is equal to: a) ∠C + ∠D

1 1 b) ∠C + ∠D 3 6

c)

1 ( ∠C + ∠D ) 3

d)

2∠C + 2∠D 4

4. The diagonals bisect each other in a: a) Parallelogram

b) Square

c) Rectangle

d) Trapezium

5. The ratio of the measure of an exterior angle of a regular nonagon to the measure of one of its interior angle is: a) 1 : 4

b) 2 : 7

c) 2 : 8

d) 4 : 14

III. FILL IN THE BLANKS 1. The sum of the measures of the four angles of a quadrilateral is __________.

2. The number of diagonals in a triangle is __________.

3. If the length of a side of a rhombus is 6 cm, then the perimeter of the rhombus is __________.

4. In a parallelogram, if ∠ A : ∠ B = 1 : 2, then ∠ A = __________.

5. The quadrilateral which is equiangular but not equilateral is __________.

6. A quadrilateral has three acute angles, each 80. The fourth angle is __________.

7. A quadrilateral with one pair of the opposite sides are parallel is a __________.

8. The sum of angles at a point is __________.

9. The polygon that has the sum of the angles as 720 is called a __________.

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IL Foundation Series Class 8

10. __________ pair of the opposite sides are equal in an isosceles trapezium.

11. The number of diagonals in a hexagon is __________.

IV. SUBJECTIVE QUESTIONS 1. The angles of a quadrilateral are 110 , 72 , 55 and x . Find the value of x.

2. A quadrilateral has all four angles of the same measure. What is the measure of each angle?

3. The four angles of a quadrilateral are of the ratio 3 : 5 : 7 : 9. Find the angles.

4. The sides of a quadrilateral are produced in order. What is the sum of the four exterior angles?

5. Find the number of sides of a regular polygon when each of its angles has a measure of: i) 160

ii) 135

iii) 175

iv) 150

6. Find the measure of each exterior angle of a regular pentagon.

7. Find the sum of the interior angles of a polygon with 15 sides.

8. Find the number of sides of a regular polygon whose: i) the sum of interior angles is 2700. ii) each of the interior angles is 108.

9. The exterior angles of a pentagon are in the ratio 6 : 3 : 4 : 3 : 2. Find all its interior angles.

10. The measure of the angles of a hexagon are x  , (x − 5) , (x − 5) , (2 x − 5) , (2 x − 5) , (2 x + 20).

Find the value of x.

11. Find the value of x in each of the following figures:

i)

x°

80°

x°

75°

120° 105°

ii) 50°

110°

115°

12. Four angles of a quadrilateral are (2 x + 10) , (4 x − 15) , 2 x  and (3 x − 20), respectively.

Find the value of x.

13. The measures of two adjacent angles of a quadrilateral are 125 and 65, and the other two angles are equal. Find the measure of each of the equal angles.

14. One angle of a quadrilateral is a right angle. If the other three angles are equal, then find the measure of the equal angle.

15. Two adjacent angles of a parallelogram are ( 3 x − 4 )° and ( 3 x + 10 )°. Find the angles.

16. The ratio of the sides of a parallelogram is 3 : 5 and its perimeter is 48 cm. Find the sides of the parallelogram.

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UNDERSTANDING QUADRILATERALS

17. In a rhombus ABCD, ∠ABC = 60, find ∠ACD. 18. A rectangle PQRS has ∠RPQ = 40, find ∠SQR. S

R O

400 Q

P

19. The sides of a rectangle are in the ratio 1 : 2, and its perimeter is 24 cm. Find the sides.

20. In the following figure, ABCD is a trapezium in which AB  DC . Find the measure of ∠ C .

D

C

1200 A

78

B

4

4.1

DATA HANDLING

INTRODUCTION

Data handling involves the collection and presentation of information, which can then be utilised to find results and conduct further studies. In our daily lives, we come across various forms of information.

Example: i)

The runs scored by a batsman in the last 10 test matches.

ii) The number of wickets taken by a bowler in the last 10 ODIs. iii) The marks scored by students in your class in the Mathematics unit test. iv) The number of storybooks read by each of your friends, and so on.

All such collected information is referred to as 'data’. Data is usually collected within the context of a situation we want to study, and it is most often collected in numerical forms.

Consider an example where 20 students in a class have preferences for different fruits: Banana - 8 students Orange - 3 students Apple - 5 students Guava - 4 students This collective information forms a data set.

The branch of mathematics that deals with the collection, organisation, analysis, and interpretation of data is called statistics.

4.2

REPRESENTATION OF DATA

Data can be represented in two main ways: graphically and in tabular form.

Graphical representations include: i)

Pictograph

ii) Bar graph iii) Double-bar graph iv) Histogram v) Polygon 79

DATA HANDLING

vi) Pie chart Tabular representations include: i)

Raw data

ii) Frequency distribution table 4.2.1 Pictograph Representing data using pictures of objects is called pictograph.

In a pictograph: i)

Pictures representing the same objects are of the same shape and size.

ii) Identical pictures are assigned identical values. iii) A half picture represents half the value of a full picture.

Example: The colours of fridges preferred by people living in a locality are Blue

-

50 people

Green

-

30 people

Red

-

55 people

White

-

20 people

The pictograph for this data is Scale: Colour Blue

Green Red White

Note: denotes

80

1 of 10.

2

= 10 people Number of people

IL Foundation Series Class 8

Advantages of a pictograph

i)

It is visually appealing to the reader.

ii) It provides a quick overview of the data.

Disadvantages of a pictograph

i)

It is time-consuming to create.

ii) It can be challenging to draw. iii) It can be difficult to represent fractional data accurately.

4.2.2 Bar graph Representing data using bars of uniform width drawn horizontally (or) vertically with equal spacing between them is called a bar graph. The lengths of the bars are proportional to the data they represent. Bar graphs can be of two types: i) horizontal bar graphs and ii) vertical bar graphs.

For example, the sale of shirts in a readymade shop from Monday to Saturday is shown in the following horizontal bar graph.

Scale: 1 unit length = 5 shirts Saturday

Days

Friday Thursday Wednesday Tuesday Monday 0

5 10 15 20 25 30 35 40 45 50 55 60 65

Number of shirts sold

Number of students

The number of students in a particular class of a school is shown in the following vertical bar graph: 80 70 60 50 40

Scale: 1 unit length = 10 students

30 20 10 0 2000 2001 2002 2003

Years

81

DATA HANDLING

4.2.3 Double-bar graph A double-bar graph is used to compare two different sets of data of the same type using bar graphs, which can be either horizontal or vertical.

Marks obtained by a student

For example, the marks obtained by a student in various subjects in two different academic years can be shown using a double-bar graph.

2005-06

80

2006-07

70 60 50 40 30 20 10 Maths

SST

Science English

Hindi

Subjects

SOLVED EXAMPLES Example 1: The following pictograph represents the number of wristwatches manufactured by a factory in a particular week.

Scale: Days

= 100 wristwatches

Number of wristwatches manufactured

Monday Tuesday Wednesday Thursday Friday Saturday

i) On which day were the least number of wristwatches manufactured? ii) On which day were the maximum number of wristwatches manufactured?

82

IL Foundation Series Class 8

iii) Find out the approximate number of wristwatches manufactured in the week.

Solution: From the pictograph, we notice that i) The least number of wristwatches, i.e., 550 watches, were manufactured on Saturday. ii) The maximum number of wristwatches, i.e., 800 watches, were manufactured on Thursday. iii) The approximate number of wristwatches manufactured in the particular week = 600 + 750 + 650 + 800 + 600 + 550 = 3950 Example 2: The following table shows Imran’s family's monthly expenditures on various items.

Items

Expenditure (in `)

House rent

3000

Food

3400

Education

800

Electricity

400

Transport

600

Miscellaneous

1200

Represent the data in the form of bar diagrams: i) Horizontal ii) Vertical Solution: Steps to draw a bar diagram: 1. Draw two perpendicular lines, one horizontal and one vertical.

2. Along the horizontal line, mark the 'items’, and along the vertical line, mark the corresponding 'expenditure'. This gives us the vertical bar graph. For the horizontal bar graph, we mark the 'items’ along the vertical line and the corresponding 'expenditure' along the horizontal line.

3. Take bars of the same width, keeping the uniform gap between them.

4. Choose the scale as 1 unit length = `200 and mark the corresponding values.

5. Calculate the lengths of bars as follows and complete the bar graph.

83

DATA HANDLING

Item

Length of bar

House Rent

3000 ÷ 200 = 15 units

Food

3400 ÷ 200 = 17 units

Education

800 ÷ 200 = 4 units

Electricity

400 ÷ 200 = 2 units

Transport

600 ÷ 200 = 3 units

Miscellaneous

1200 ÷ 200 = 6 units

i) Horizontal bar graph: 1 unit length = ₹ 200 Miscellaneous Transport Electricity

0

200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000 3200 3400 3600

Education Food House rent

Expenditure (in ₹)

ii)Vertical bar graph:

84

Scale : 1 unit length = ₹ 200

0

X

Food Education Electricity Transport Miscellaneous

3600 3400 3200 3000 2800 2600 2400 2200 2000 1800 1600 1400 1200 1000 800 600 400 200

House rent

Expenditure (in ₹)

Y

IL Foundation Series Class 8

Example 3: A mathematics teacher wants to see whether the new teaching technique she applied after the quarterly test was effective (or) not. She takes the scores of 5 weakest children in the quarterly test (out of 25) and in the half-yearly test (out of 25).

The considered scores are as follows: Students

Ashish

Arun

Kavish

Maya

Rita

Quarterly

10

15

12

20

9

Half-yearly

15

18

16

21

15

Draw the double-bar graph and conclude whether the new technique was successful.

Solution: sing the scale as 1 unit = 5 marks, let us draw the bar graph vertically for both quarterly and halfU yearly tests in the same graph as follows: Scale: 1 unit = 5 marks

Y

Quarterly Half-yearly

25

Marks

20 15 10 5 0

Arun

Maya Rita

X

Students

The double-bar graph clearly shows that the marks obtained by the 5 children in the half-yearly test were more than those in the quarterly test. Hence, the new technique was successful.

4.2.4 Organising data Data is often available in an unorganised form, known as raw data. For example, 5, 9, 1, 5, 0, 2, 5, 2, 9, 0, 3. To analyse this data, it needs to be organised and turned into organised data.

Example: A group of students was asked for their favourite subject among Art (A), Mathematics (M), Science (S), and English (E). The results were listed as A, M, S, E, M, A, E, M, E, A, S, A, S, S, M, A, E, A, S, M, S, A. This is a raw data. The organised data would look like this:

85

DATA HANDLING

Subject

Tally marks

Number of students

Art

7

Mathematics

5

Science

6

English

4

This organised data in the table form is called a frequency table. requency: The number of times a particular entry repeats in each data is called its frequency. In F the above example, the number of students preferring each subject is the frequency of the respective subject.

4.2.5 Grouping data When the size of the data is large, it is convenient to organise it into small groups called class intervals, and we find their frequencies using tally marks. The class intervals are the groups of type 0 − 10, 10 − 20, 20 − 30, etc. ote: If 20 is an entry in the data, then it belongs to 20 − 30, not the 10 − 20 interval. This is, by N convention, giving priority to the higher class.

Example: The marks obtained by 40 students of a class in a test are : 18, 8, 12, 0, 8, 16, 12, 5, 23, 2, 16, 23, 2, 10, 20, 12, 9, 7, 6, 5, 3, 5, 13, 21, 13, 15, 20, 24, 1, 7, 21, 16, 13, 18, 23, 7, 3, 18, 17, 16.

Minimum mark = 0 and maximum mark = 24 We can organise this data into class intervals 0 − 5, 5 − 10, 10 − 15, 15 − 20, 20 − 25.

Marks

Tally marks

0 −5

6

5 − 10

10

10 − 15

7

15 − 20

9

20 − 25

8 Total =

This is called grouped frequency distribution.

86

Number of students (frequency)

40

IL Foundation Series Class 8

Class limits: In a grouped frequency distribution, the class intervals have limits called class limits.

The lowest value is called the Lower-class limit, and the highest value is called the upper-class limit.

Example: For the class interval 15 - 20, the lower-class limit = 15, and the upper-class limit = 20.

Width or size: The width or size of a class interval is defined as the difference between its class limits.

Width = Upper-class limit - Lower-class limit Example: For the class interval 15 - 20, width = 20 − 15 = 5.

4.2.6 Histogram A histogram is the graphical representation of a grouped frequency distribution, with classes taken along the x-axis and frequencies taken along the y-axis. The data is presented in the form of rectangles whose widths are identical and whose heights are proportionate to the frequencies.

Example: Let us consider the marks obtained by 43 students given below: Marks

Number of students

10-20

3

20-30

1

30-40

4

40-50

9

50-60

5

60-70

10

70-80

5

80-90

3

90-100

3

The histogram for this data is:

Number of students

Y 10 9 8 7 6 5 4 3 2 1

Scale: i) On the x-axis: 1 unit = 10 marks ii) On the y-axis: 1 unit = 1 student

0 10 20 3040 50 60 70 80 90 100

X

Marks 87

DATA HANDLING

Difference between histogram and vertical bar graph

There are significant differences between a histogram and a vertical bar graph: i) Gaps between bars are used in bar graphs, whereas no gap is used in histograms between rectangles. ii) The width of a bar in a bar graph is not recognised, whereas the widths of rectangles in histograms represent the size of classes and must be constant.

SOLVED EXAMPLES Example 1: For which of these would you use a histogram to show the data? a) The number of letters for different areas in a postman’s bag. b) The height of competitors in an athletics meet. c) The number of cassettes produced by 5 companies. d) The number of passengers boarding trains between 7:00 am and 7:00 pm at a station. Give reasons for each.

Solution: The histogram can be drawn in cases (b) and (d) but cannot be drawn in cases (a) and (c).

Reasons: a) In this case, the data is the name of the area assigned to a number of letters. The areas’ names cannot be organised as groups (class intervals). b) In this case, the data is the heights of the competitors and can be organised into groups such as 110 cm − 120 cm, 120 cm − 130 cm , etc. c) In this case, the data is the names of companies assigned to a number of cassettes. The companies names cannot be organised as groups (class intervals). d) In this case, the data is the time 7 am - 7 pm, which can be organised into groups such as 7 am 8 am, 8 am - 9 am, and 6 pm - 7 pm.

Example 2: The shoppers who come to a departmental store are marked as Man (M), Woman (W), Boy (B), and Girl (G). The following list gives the shoppers who came during the first hour of the morning: W, W, W, G, B, W, W, M, G, G, M, M, W, w, W, W, G, B, M, W, B, G, G, M, W, W, M, M, W, W, W, M, W, B, W, G, M, W, W, W, W, G, W, M, M, W, W, M, W, G, W, M, G, W, M, M, B, G, G, W. Make a frequency distribution table and draw its bar graph.

88

IL Foundation Series Class 8

Solution: The frequency table is: Shopper

Number of shoppers

Tally marks

Man

15

Woman

28

Boy

5

Girl

12 Total

60

The bar graph for this data is:

Number of shoppers

Y

Scale: 1 unit = 5 shoppers 28

30 25 20 15 10

15

12 5

5 0

M

W

B

G

X

Shoppers Example 3: The weekly wages of 30 workers in rupees are : 830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806, 840. Prepare a grouped frequency distribution with class intervals 800 - 810, 810 - 820, etc. Also, draw a histogram to represent the data and answer the following: i) Which group has the maximum number of workers? ii) How many workers earn `850 and more? iii) How many workers earn less than `850?

Solution: The minimum wage = `804 89

DATA HANDLING

The maximum wage = `898 ∴ The class intervals chosen are 800 − 810, 810 − 820, ….., 890 − 900. ∴ The grouped frequency distribution is:

Weekly wage (in `)

Tally marks

Number of workers

800 − 810

3

810 − 820

2

820 − 830

1

830 − 840

9

840 − 850

5

850 − 860

1

860 − 870

3

870 − 880

1

880 − 890

1

890 − 900

4 Total =

30

Scale : i) On x-axis: 1 unit = ₹ 10

Y

ii) On y-axis: 1 unit = 2 workers

8 6 4

900

890

880

870

860

850

840

830

820

0

810

2 800

Number of workers

10

X

Wages in ₹

From the histogram, we notice that i) The group of workers with weekly wages between `830 - `840 has the maximum number of workers.

90

IL Foundation Series Class 8

ii) The number of workers with weekly wages of `850 or more = 1 + 3 + 1 + 1 + 4 = 10 iii) The number of workers with weekly wages less than `850 = 3 + 2 + 1 + 9 + 5 = 20 ote: In graphs, the N graph.

symbol is used to hide some range of values that are useless in the

4.2.7 Pie chart A circle graph (or) pie chart is a pictorial representation of the numerical data using nonintersecting adjacent sectors of a circle. The area of each sector is proportional to the magnitude of the data it represents.

Angle of the sector in a pie chart:

The angle of a sector is directly proportional to the area of the sector. However, the area of the sector is proportional to the value of the component represented by it. So, the angle of a sector ( θ ) is proportional to the value of the component represented by the sector. The central angle of a sector in a circle graph is obtained by: Value of component ⎡ ⎤ Central angle of a component (sector) = ⎢ × 3660 ⎥ ⎣ Sum of all component values ⎦ Construction of a pie chart:

The construction of a pie chart involves the following steps: 1. Obtain the data and calculate the sum of all component values. In the case of a frequency distribution, the frequencies are the component values.

2. Divide each component value by the sum of all component values and multiply the fraction by 360° to obtain the central angles of sectors representing the components.

3. Draw a circle of convenient radius using a compass and draw a radius that shows the 12 o’clock position in a clock.

4. Draw a sector with the greatest angle starting from the 12 o’clock position in a clockwise direction.

5. Continue drawing the sectors in a clockwise direction with decreasing central angles.

Label the sectors with the component names and their values.

Note: The sectors should be drawn in a decreasing order of angles or component values, either in a clockwise or an anticlockwise direction.

91

DATA HANDLING

SOLVED EXAMPLES Example 1: A survey was made to find the types of music that a certain group of young people liked in a city. The adjoining pie chart shows the findings of this survey. From this pie chart, answer the following: i) If 20 people liked classical music, how many young people were surveyed?

Semi-Classical 20%

ii) Which type of music is liked by the maximum number of people?

Classical 10%

iii) If a cassette company were to make 1000 CDs, how many of each type would they make?

Solution From the pie chart, we notice that: i) 10% of the young people liked classical music. ∴ If 20 people liked classical music, then 10% of all people = 20 people 10 ∴ × Number of young people surveyed = 20 100 Number of young people surveyed = 20 × 100 = 200 10 ii) 40% of the young people liked light music. ∴ Light music is liked by the maximum number of people. iii) Total number of CDs = 1000 ∴ Number of CDs with light music = 40% of1000 =

40 × 1000 = 400 100

Number of CDs with folk music = 30% of1000 30 × 1000 = 3000 100 Number of CDs with semi-classical music = 20% of 1000 20 = × 1000 100 =

= 200 Number of CDs with classical music = 10% of 1000 92

Light 40%

Folk 30%

IL Foundation Series Class 8

=

10 × 1000 100

= 100 Example 2: A group of 360 people were asked to vote for their favourite season from the three seasons: monsoon, winter, and summer. i) Which season got the most votes? ii) Find the central angle of each sector. iii) Draw a pie chart to show this information.

Season

No. of votes

Summer

90

Monsoon

120

Winter

150

Solution: i) The season that got the most votes is winter. ii) Sum of the number of all the votes = 90 + 120 + 150 = 360 Season

Number of votes

In fractions

Fraction of 360° (central angle)

Summer

90

90 1 = 360 4

1 × 360 = 90 4

Monsoon

120

120 1 = 360 3

1 × 360 = 120 3

Winter

150

150 5 = 360 12

5 × 360 = 150 12

iii) Pie chart

Summer

90⁰

Winter 1500

Monsoon 120⁰

93

DATA HANDLING

Example 3: Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.

Colour

Number of people

Blue

18

Green

9

Red

6

Yellow

3

Total

36

Solution: Total number of people = 36 Colour

Number of people

In fraction

Fraction of 360° (central angle)

Blue

18

18 1 = 36 2

1 × 360 = 180 2

Green

9

9 1 = 36 4

1 × 360 = 90 4

Red

6

6 1 = 36 6

1 × 360 = 60 6

Yellow

3

3 1 = 36 12

1 × 360 = 30 12

Pie chart

(3) llow

Ye

300

Red (6)

600

900 Green (9)

94

Blue (18) 1800

IL Foundation Series Class 8

4.3 PROBABILITY Probability tells the level of uncertainty. It measures the chance or likelihood of an event happening.

4.3.1 Getting a result The occurrence of an event is studied under a given situation. These situations are called experiments, and the possible results are known as outcomes and the set of all possible outcomes in the experiment is called sample space. It is usually denoted by the letter S . Sample space can be written using the set notation, {}.

Example 1: Tossing a coin is the experiment.

Possible outcomes are head and tail.

Sample space, S = {head, tail} Example 2: Tossing a die is the experiment.

Possible outcomes are the numbers 1, 2, 3, 4, 5, and 6.

Sample space, S = {1, 2, 3, 4, 5, 6} Example 3: Picking a card from a deck of cards is the experiment.

Possible outcomes would be the 52 unique cards in a standard deck.

The sample space, therefore, would include all 52 cards.

A deck of playing cards consists of 52 cards, which are divided into 4 suits of 13 cards each. They are black spades, red hearts, red diamonds, and black clubs.

The cards in each suit are 'Ace, King, Queen, Jack, 10, 9, 8, 7, 6, 5, 4, 3, and 2'.

Types of experiments

There are two types of experiments: i) Deterministic experiment

ii) Random experiment

Deterministic experiment: This is an experiment which, when repeated under similar conditions, gives the same result or outcome.

Example: In scientific endeavors, such as measuring the diameter of a wire, the process is inherently deterministic. No matter how many times the experiment is repeated, the result remains the same (within experimental error).

Random experiment: This is an experiment which, when repeated under similar conditions, may not give the same result or outcome.

Example: Tossing a coin does not give the same result when repeated several times. So, tossing a coin is a random experiment.

95

DATA HANDLING

Note: In this chapter, we deal only with random experiments.

4.3.2 Equally likely outcomes When the outcomes of an experiment have equal chances of occurring, then such outcomes are called equally likely outcomes.

Example: 1. In tossing a fair coin, the possible outcomes are head and tail. Both these outcomes have an equal chance of occurring, and hence, these outcomes are equally likely.

2. Drawing a card from a well-shuffled deck of cards: each card has an equal chance of being drawn.

However, it is important to note that not all experiments have equally likely outcomes.

For example, if a die has the number 4 on two faces and the numbers 1, 2, 3, and 5 on the other four faces, these outcomes are not equally likely because the probability of getting a 4 is higher than the other numbers.

4.3.3 Probability of an event The probability of an event is a measure of the likelihood of its occurrence. It can be defined in two ways: Experimental probability This is calculated as the number of times the event occurred divided by the total number of trials. It is used when we have data on the past occurrences of the event.

Probability of an event =

Number of times the event happened Total number of trials

Theoretical probability This is used when past history is not available, but the possible outcomes are known. It is calculated as the number of outcomes favourable to the event divided by the total number of outcomes.

Probability of an event =

Number of outcomes favourable to the event Total number of outcomes

Probability scale

The probability of any event ranges from 0 to 1.

If there is no chance of an event occurring, its probability is 0. Such an event is called an impossible event.

If an event is certain to happen, its probability is 1. Such an event is called a certain event or a sure event.

96

IL Foundation Series Class 8

4.3.4 Outcomes as events In any experiment, an event can be defined as a single outcome or a collection of outcomes. i) When you toss a coin, the outcomes 'Head’ and 'Tail’ are each considered an event. ii) Similarly, when you roll a die, getting any of the numbers 1, 2, 3, 4, 5, or 6 is an event. iii) An event can also be a collection of outcomes. For instance, getting an even number when rolling a die is an event, as the outcomes 2, 4, or 6 are all considered even.

Example: What is the probability of getting an even number when rolling a die?

Solution: here are 3 even numbers (2, 4, 6) out of a total of 6 possible outcomes. So, the probability is T 3 1 = 6 2 Example: Consider a bag containing 4 red balls and 2 yellow balls (all balls are identical except for their colour). If a ball is drawn from the bag without looking, what is the probability of drawing a red ball? Is it higher or lower than the probability of drawing a yellow ball?

Solution: here are a total of 6 balls in the bag. Drawing a red ball has 4 favourable outcomes. Therefore, the T probability of drawing a red ball is 4 2 = 6 3 Similarly, the probability of drawing a yellow ball is 2 1 = 6 3 Hence, the probability of drawing a red ball is higher than that of drawing a yellow ball.

4.3.5 Chance and probability related to real life Probability in weather forecasting

Meteorologists use probability to predict weather changes.

Example: If there’s a 40% chance of rain, it means that historically, it has rained 40 out of 100 times in similar weather patterns.

Probability in sports

Consider a cricket player who has scored a half-century in 20 out of 50 matches. The probability that this player will score a half-century in a future match is 40%.

20 = 0.4 50

97

DATA HANDLING

Probability in insurance

Let’s say an insurance company found that out of 1000 policy holders like a certain individual, 100 have filed a claim in the past year. If this individual applies for insurance, the company might set their premium based on a 10% chance of a claim being filed.

100 = 0.1 1000 Probability in games

1 because a standard die has six faces, and each 6 face ( 1, 2, 3, 4, 5, 6 ) is equally likely to land face up.

In a game of dice, the probability of rolling a 6 is

SOLVED EXAMPLES Example 1: Find the outcomes of the following events and find their probabilities when a die is thrown: i) a prime number is obtained. ii) a non-prime number is obtained. iii) a number greater than 5 is obtained. iv) a number not greater than 5 is obtained.

Solution: When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, 6. i) Favourable outcomes of the event, 'a prime number is obtained' are 2, 3, 5.

Probability =

Number of favourable outcomes 3 1 = = Total number of outcomes 6 2

ii) Favourable outcomes of the event, a non-prime number is obtained' are 1, 4, 6.

Probability =

Number of favourable outcomes 3 1 = = Total number of outcomes 6 2

iii) Favourable outcome of the event, 'a number greater than 5 is obtained' is 6.

Probability =

Number of favourable outcomes 1 = Total number of outcomes 6

iv) Favourable outcomes of the event, 'a number not greater than 5 is obtained' are 1, 2, 3, 4, 5.

Probability =

98

Number of favourable outcomes 5 = Total number of outcomes 6

IL Foundation Series Class 8

Example 2: Find the probability of getting a head when a fair coin is tossed. Also, find the probability of getting a tail.

Solution: When a fair coin is tossed, the possible outcomes are head and tail. We know that Number of favourable outcomes Total number of outcomes ∴ The probability of getting a head = 1 2 1 The probability of getting a tail = .

2 Example 3: If you have a spinning wheel with 3 green sectors, 1 blue sector, and 1 red sector, then what is the probability of getting a green sector? What is the probability of getting a non-blue sector?

Probability =

Solution: Total number of outcomes = 3 + 1 + 1 = 5 Number of favourable outcomes Total number of outcomes 3 ∴ The probability of getting a green sector = 5 Probability =

The probability of getting a non-blue sector = The probability of getting a green or red sector 3 +1 4 = 5 5 Example 4: Find the probability of getting an ace when a card is drawn from a well-shuffled deck of 52 playing cards. =

Solution: In a deck of 52 playing cards, there are 4 aces (one each of Hearts, Diamonds, Clubs, and Spades).

So, the sample space(S) for this experiment is: S = {Ace of Hearts, Ace of Diamonds, Ace of Clubs, Ace of Spades} ∴ Probability of getting an ace =

1 Number of favourable outcomes 4 == = .

52 13 Total number of outcomes

99

DATA HANDLING

QUICK REVIEW •

rouped data can be presented using histograms. A histogram is a type of bar diagram G where the class intervals are shown on the horizontal axis, and the heights of the bars show the frequency of the class interval. Also, there is no gap between the bars as there is no gap between the class intervals.

•

ata can also be presented using a circle graph or pie chart. A circle graph shows the D relationship between a whole and its part.

•

There are certain experiments whose outcomes have an equal chance of occurring.

•

A random experiment is one whose outcome cannot be predicted exactly in advance.

•

The outcomes of an experiment are equally likely if each has the same chance of occurring.

•

Probability of an event =

Number of outcomes that make an event , when the outcomes are Total number of ou utcomes of the experiment

equally likely.

•

One or more outcomes of an experiment make an event.

WORKSHEET - 1 I.

REPRESENTATION OF DATA – PICTOGRAPH, BAR GRAPH, DOUBLE-BAR GRAPH 1. The number of girl students in each class of a co-educational middle school is depicted by the following pictograph: Scale:

Classes I II III IV V VI VII VIII

100

= 4 girls

Number of girl students

IL Foundation Series Class 8

Observe the pictograph and answer the following questions: i) Which class has the minimum number of girl students? ii) Is the number of girls in class VI less than the number of girls in class V? iii) How many girls are there in class VII? 2. Two hundred students of Class VI and VII were asked to name their favourite colour to decide upon the colour of their school building. The results are: Favourite colours

Red

Green

Blue

Yellow

Orange

Number of students

43

19

55

49

34

Represent the data as a bar graph and answer the following: i) Which is the most preferred colour, and which is the least preferred? ii) How many colours are there in all? What are they?

3. Sales of English and Hindi books in the years 1995, 1996, 1997, and 1998 are given below: Years

1995

1996

1997

1998

English

350

400

450

620

Hindi

500

525

600

650

Draw a double-bar graph and answer the following questions: i) In which year was the difference in the sales of the two language books the least? ii) Can you say that the demand for English books has increased? Justify.

Number of students in Class VIII

4. The following bar graph represents the number of students in class VIII of a given school.

350 300 250 200 150 100 50 0

2003-04 2004-05 2005-06 2006-07 2007-08 Academic years

101

DATA HANDLING

i) What is the information given by the bar graph? ii) In which year is the increase in the number of students maximum? iii) In which year is the maximum number of students in the class? iv) State true or false: 'The number of students during 2005 - 06 is twice that of 2003 - 04'.

5. Use the bar graph given below and answer the following.

Students

Y

12 10 8 6 4 2 0

Pets owned by students of class VIII

Dogs

Cats

Rabbits

Hamsters

Others

X

Pet animals

i) Which is the most popular pet? ii) How many students have a dog as a pet?

6. The double-bar graph shows the result of a survey to test water-resistant watches made by different companies A, B, C, and D. Y

Scale: 1 unit = 10 watches

40

Watches

30 Number tested

20

Number that leaked 10 0

A

B

C

Companies

102

D

X

IL Foundation Series Class 8

i) Work out a fraction of the number of watches that leaked to the number tested for each company. ii) Based on these fractions, tell which company has better watches.

7. The following graph shows the results of students in a school.

Number of students

Result in a school

350 300 250

Pass Fail

200 150 100 50 0

1991-92

1992-93

1993-94 1994-95 1995-96

Year

i) Which year has the lowest difference in the number of students who passed from those who failed? ii) How many times is the number of failed students the same? iii) What is the approximate percentage of students who failed for 5 years? iv) When is the percentage increase in the total number of students maximum in comparison to the previous year? v) What is the average number of students who failed in the school in the last 5 years?

II. REPRESENTATION OF DATA – HISTOGRAM 1. Draw a histogram for the following grouped frequency distribution: Class interval Frequency

0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 Total 2

10

21

19

7

1

60

2. Draw a histogram for the following distribution of test scores for 32 students in a class.

Test scores Number of students

40 − 50

50 − 60

60 − 70

70 − 80

80 − 90

3

0

4

11

8

90 − 100 Total 6

32

Using the histogram, answer the following: i) How many students scored between 50 − 60? ii) Which group of test scores contained the maximum number of students?

103

DATA HANDLING

3. Construct a frequency histogram for the following data showing mileage (kilometre/litre) versus the number of cars.

Class Interval

15 − 19

19 − 23

23 − 27

27 − 31

31 − 35

35 − 39

39 − 43

5

11

8

5

7

3

1

Frequency

Using the histogram, answer the following questions. i) In which interval is the greatest frequency found? ii) What is the number of cars (frequency) reporting mileage between 27 and 31 kilometres per litre? iii) For what interval are the fewest cars reported? iv) How many of the cars reported mileage of 31 kilometres per litre or more? v) What percentage of cars reported mileage from 23 to 27 kilometres per litre?

4. The following histogram shows the frequency distribution of the ages of 22 teachers in a school. i) How many teachers are in the oldest and youngest age groups? ii) In which age group do most teachers fall, and in which do they fall the least? iii) What is the size of the classes?

Y 7

Number of teachers

6 5 4 3 2 1 0

20 25 30 35 40 45 50 55

X

Ages (in years)

5. The weights (in grams) of 35 mangoes are given below: 30, 40, 45, 32, 43, 50, 55, 62, 70, 70, 61, 62, 53, 52, 50,, 42, 35, 37, 53, 55, 65, 70, 73, 74, 45, 46, 58, 59, 60, 62, 74, 34, 35, 70, 68

104

IL Foundation Series Class 8

Construct a grouped frequency distribution using equal class intervals, one of which is 40 - 45 (45 is not included).

Also, draw a histogram for this distribution.

6. The following is the distribution of ages (in years) of 25 persons working in an office: Age (in years)

20 − 30

30 − 40

40 − 50

50 − 60

Total

Number of persons

7

10

6

2

25

i) What is the class size? ii) What is the lower limit of 30 − 40? iii) What is the upper limit of 40 − 50? iv) What is the frequency of the class interval 50 − 60?

7. Observe the histogram and answer the following questions:

Y

7

Number of Girls in Class VIII

7 6 5

4

4

3

3

2

2 1 0

2

1

1

125 130 135 140 145 150 155 160

X

Heights in cm i) What information is being given? ii) Which group contains the maximum number of girls? iii) How many girls have a height of 145 cm or more? iv) If we divide the girls into the following categories, then how many would there be in each?

150 cm and more - Group 'A’ 140 cm to less than 150 cm - Group 'B’ Less than 140 cm - Group 'C’ 105

DATA HANDLING

8. Draw a histogram for the daily-earnings of 46 general stores given in the following table.

Daily earnings (in `)

Number of general stores

1450 − 1500

4

1500 − 1550

10

1550 − 1600

9

1600 − 1650

18

1650 − 1700

5

III. REPRESENTATION OF DATA – PIE CHART 1. The monthly salary of a person is `15000. The central angle of the sector, representing his expenses on food and house rent on a pie chart, is 60. What is the amount he spends on food and house rent?

2. The pie chart shows the expenditure (in %) on various items and savings of a family during a month. use Ho t n Re 10%

T 5% ran s

po

Children Education15%

rt

Others 20%

Food 25%

Savings 15%

Clothes 10%

i) On which item was the expenditure maximum? ii) On which item is the expenditure equal to the total savings of the family? iii) If the monthly savings of the family is ` 3000, what is the monthly expenditure on clothes?

3. On a particular day, the sales (in ` ) of different items of a baker’s shop are given below:

106

Ordinary bread

:

320

Fruit bread

:

80

Cakes and pastries

:

160

IL Foundation Series Class 8

Biscuits

:

120

Others

:

40

Total

:

720

Draw a pie chart for this data.

4. The number of students in a hostel speaking different languages is given below. Display the data in a pie chart.

Language

Hindi

English

Marathi

Tamil

Bengali

Total

Number of students

40

12

9

7

4

72

5. The pie chart represents expenditure on different items in the construction of a flat in Delhi.

Brick 500 Cement

Labour 1000

750 450

1000 Timber

Steel

If the expenditure incurred on cement is `112500, find i) The total cost of the flat. ii) The expenditure incurred on labour.

6. The pie chart given below represents the distribution of proteins in parts of the human body.

Now, answer the following questions.

Muscles 1 3 Hormones, enzymes, and other proteins

1 Skin 10 Bones 1 6

107

DATA HANDLING

i) What is the ratio of the distribution of proteins in the muscles to that of proteins in the

bones? ii) What is the central angle of the sector (in the above pie chart) representing skin and bones together? iii) What is the central angle of the sector representing hormones, enzymes, and other

proteins?

7. The pie chart gives the marks scored by a student. If the total marks obtained were 540, then answer the following:

S.

cs ati

m the

Sci

Ma

en

ce

0

900 65

800

700

Sc

ish

gl

ien

En

ce

550

Hindi i) In which subject did the student score 105 marks? ii) How many more marks did the student get in Mathematics compared to Hindi? iii) Examine whether the total marks in Social Science and Mathematics exceed the sum of the marks obtained in Science and Hindi.

IV. PROBABILITY 1. Akash rolls a die. Find the probabilities of getting a number i) less than 7

ii) less than 1

iii) which is even.

2. The diagram shows a spinner. If this spinner is spun, then find the probability that the pointer stops on 'D'.

A

B C

A D

108

IL Foundation Series Class 8

3. What is the probability of choosing a vowel from the alphabet?

4. In a school, only 3 out of 5 students can participate in a competition. What is the probability of the students not making it to the competition?

5. A coin is tossed 200 times, and the head appears 120 times. What is the probability of getting a head in this experiment?

6. Anil and Sunil are playing with 5 cards, as shown in the figure. What is the probability of Anil picking up a card that has the number 2 on it without seeing it?

4

1

2

3

2

2 7. A bag has 4 red balls and 2 yellow balls that are identical. A ball is thrown from the bag without looking into it. What is the probability of getting a red ball? Is it more or less than the probability of getting a yellow ball?

8. The numbers 1 to 10 are written on ten slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking into it. What is the probability of getting i) a number 6?

ii) a number less than 6?

iii) a number greater than 6?

iv) a one-digit number?

9. Rashmi collected the data regarding the heights of the students in her class and prepared the following table: Heights

4′ 4′′ − 4′ 6′′

4′ 6′′ − 4′ 8′′

3

5

Number of students

4′ 8′′ − 5′

5′ − 5′ 2′′

25

8

5′ 2′′ − 5′ 4′′ 2

A student is to be selected randomly from her class for some competition. Find the probability of selection of a student with height between 4′ 8′′ − 5′.

10. Sonia picks up a card from the given cards.

R 1

Y 2

Y 3

R 4

B 5

B 6

G 7

Y 8

R 9

G 10

109

DATA HANDLING

Find the probability of getting i) an odd number

ii) a Y card

iii) a G card

iv) a B card bearing a number greater than 7

11. Classify the following statements as an experiment (Ex) or as an event (Ev): i) Tossing two coins simultaneously. ii) Drawing a pencil from a bag containing pens and pencils. iii) Landing the pointer of a spinner on an even number. iv) A cricket team winning the match.

12. Classify the following statements under appropriate headings. i) Getting the sum of angles of a triangle as 180ϒ. ii) India winning a cricket match against Pakistan. iii) Sun setting in the evening. iv) Getting 7 when a die is thrown. v) Sun rising from the West. vi) You win a racing competition.

Certain to happen

Impossible to happen

May or may not happen

13. Sowmya and Ramya are friends. They both were born in 2013. What is the probability that they have: i) the same birthdays?

ii) different birthdays?

14. Here is an extract from a mortality table.

110

Age (in years)

Number of persons surviving out of a sample of one million

60

16090

61

11490

62

8012

63

5448

64

3607

65

2320

IL Foundation Series Class 8

i) Based on this information, what is the probability of a person aged 60 dying within a year? ii) What is the probability that a person aged 61 will live for 4 years?

15. The ages (in years) of the workers of a factory are given below: Age (in years)

10 − 19

20 − 29

30 − 39

40 − 49

50 − 59

60 and more

5

42

33

15

10

5

Number of Workers

A worker is chosen at random. What is the probability that the age of the chosen worker is: i) 40 years or more? ii) Between 30 and 39 years? iii) 39 years or less? iv) What is the probability that the number of workers is greater than 10?

16. Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is i) 8?

ii) 13?

iii) less than or equal to 12?

17. If a pair of dice is rolled, find the probability of getting 5 on at least one of them.

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. The pictures used in a pictograph are

a) Similar

b) Identical

c) Different

d) Cartoons

2. The graph that shows two sets of data simultaneously for comparison is a) Bar graph

b) Histogram

c) Double-bar graph

d) Ogive graph

3. For the data 5, 9, 11, 7, 5, 3, 9, 7, 11, 3, 11, the entry with highest frequency is a) 3

b) 5

c) 7

d) 11

4. The size of a class is 8, and its upper limit (not included) is 18. Then, the lowerLower-class class limit is a) 8

b) 10

c) 18

d) 26

5. Consider a histogram with the scale on the vertical axis: 1 unit = 500. If the height of a rectangle is 3.2 units, then the frequency of the class interval represented by it is a) 1600

b) 3.2

c) 166 (approx)

d) 600

111

DATA HANDLING

6. The widths of the rectangles in a histogram represent a) Lower limit of class b) Upper limit of class c) Class size d) Frequency 7. In a pie chart, the central angle of a sector is 72°. Then, the fraction of the circle represented by the sector is 1 1 1 1 a) b) c) d) 2 3 4 5 8. The monthly medical expenditure of a family with a monthly income of `25,000 is `2,500. The angle of the sector representing it on a pie chart is a) 18°

b) 36°

c) 9°

d) 27°

9. There are 4 sectors in a pie chart. The angles for three of them are 45°, 70° , and 130°. Then the angle of the fourth sector is a) 110°

b) 115°

c) 120°

d) 125°

10. Which of the following is not a random experiment? a) Picking a book from a pile of different books without looking at it. b) Picking a ball from an urn containing 10 red balls without looking at it. c) Tossing a fair coin. d) Picking a card from a well-shuffled deck of 52 cards.

11. The probability of getting a red ace when a card is drawn from a well-shuffled deck of 52 cards is 1 1 2 4 a) b) c) d) 26 13 13 13 12. For a fair coin, the outcomes are a) 3 in number

b) Equally likely

c) Unequally likely

d) Predictable

13. A fair die shows 4 when it is rolled once. If it is rolled once again, then the probability of getting 4 again is 1 1 a) 0 b) 1 c) d) 4 6 14. The proportion of the sector for red in the pie chart is

Yel lo

w

Red 45⁰

Green

45⁰

Blue

a) 112

1 2

b)

1 4

c)

1 8

d)

1 3

IL Foundation Series Class 8

15. A die is thrown. The probability of getting an even prime number is a)

1 6

b)

1 4

c)

1 3

d)

1 2

16. A child has a block in the shape of a cube with one letter written on each face, as shown below: A

B

C

D

E

A

. The cube is thrown once. Then, the probability of getting an 'A'

is a)

1 3

b)

1 6

c)

1 2

d)

1 4

17. Observe the pie chart and find the sector that has the greatest angle.

A

C

30%

30% 40%

B a) A

b) B

c) C

d) B, C

18. Observe the histogram and find the number of students getting marks 4 to less than 6.

Y 9

Number of students

10

a) 2

8

8

7

6 4

4

2

2 0

2 4 6 8 10 Marks obtained

b) 4

X

c) 6

d) 8

c) 15

d) 20

c) 30

d) 35

19. The size of the class interval 15 − 20 is a) 5

b) 10

20. The upper limit of the class interval 25 − 30 is a) 20

b) 25

21. The frequencies of the class intervals 20 − 25 and 30 − 35 are 20 and 20, respectively. Their difference is a) 0

b) 10

c) 20

d) 5 113

DATA HANDLING

22. The lower limit of the class interval 35 − 40 is a) 20

b) 25

c) 30

d) 35

23. The probability of drawing a red card from a well-shuffled deck of 52 playing cards is a)

1 4

b)

1 2

c)

1 26

d)

1 13

24. In the pie chart representing the percentage of students having an interest in reading various kinds of books, the central angle of the sector representing students reading novels is 54°.

The percentage of such students is a) 15%

b) 18%

c) 20%

d) 22%

25. A number is selected at random from the first 20 natural numbers. Then, the probability of getting a number that is a multiple of 3 is 1 3 2 7 b) c) d) 3 10 5 10 26. The probability of drawing a number card from a well-shuffled deck of 52 playing cards is

a)

a)

3 52

b)

10 52

c)

5 52

d)

9 13

27. Two dice are tossed once. The probability of getting an even number on the first die or a total of 8 is a)

1 2

b)

7 12

c)

5 12

d)

1 3

28. In a single throw of two dice, the probability of neither a doublet nor a total of 9 is a)

12 18

b)

12 18

c)

13 18

d)

3 12

29. A number is selected from the first 50 natural numbers. The probability that it is a multiple of 3 or 5 is: a)

13 25

b)

21 50

c)

12 50

d)

23 50

d)

1 7

30. The probability that a non-leap year has 53 Sundays is a)

2 7

b)

5 7

c)

6 7

31. The value of the components with central angle 18 in a pie chart is 90 . Then, the value of the components with central angle 22 is a) 94

b) 90

c) 110

d) 96

32. If a pair of dice is thrown, then the number of possible outcomes is a) 6

114

b) 12

c) 24

d) 36

IL Foundation Series Class 8

33. The probability of getting exactly 2 heads when 3 coins are tossed simultaneously is a)

1 4

b)

3 8

c)

2 3

d)

5 8

34. Three unbiased coins are tossed together. Then, the probability of getting at least two heads is a)

3 8

b)

1 2

c)

7 8

d)

7 8

35. Three unbiased coins are tossed together. Then, the probability of getting at most one tail is a)

1 2

b)

3 8

c)

7 8

d) None

36. Two coins are tossed 1000 times, and the outcomes are recorded as below: Number of heads

2

1

0

Frequency

200

550

250

Based on this information, the probability of at most one head is a)

1 5

b)

1 4

c)

4 5

d)

3 4

37. The probability of an event not happening is 0.63. The probability of that event happening is: a) 0.36

b) 3.6

c) 3.7

d) 0.37

II. FILL IN THE BLANKS 1. Tally marks are used to find _____________.

2. The height of a rectangle in a histogram shows the ________________ of the class.

3. Size of the class 150-175 is ________________.

4. The sixth-class interval for grouped data whose first two class intervals are 10-15 and 15-20 is ________________.

5. In the class intervals 10-20 and 20-30, respectively, 20 lies in the class ________________.

6. A pie chart shows the relationship between a whole and ______________.

7. The quantitative measure of certainty is called ________________.

8. The average of the lower-class limit and the upper-class limit is called _________________.

9. The number of times a particular entry occurs in a data is given by its _______________.

10. The probability of an event lies between and ____________________ 11. The probability of getting a number less than or equal to 6 in a throw of a dice is ______________.

12. The probability of ______________ event is 0.

115

DATA HANDLING

13. The bag contains 3 blue and 2 red marbles. A marble is drawn at random. The probability of drawing a red marble is __________________.

14. An experiment whose outcomes cannot be predicted in advance is called ____________.

15. If a number 7 is repeated n times in a data, then the frequency of the number 7 is ________________.

III. SUBJECTIVE QUESTIONS 1. In a village, six fruit merchants sold the following number of fruit baskets in a particular season: Name of fruit merchants

Number of fruit baskets

= 100 fruit baskets

Rahim Lakhanpal Anwar Martin Ranjit Singh Joseph

Observe this pictograph and answer the following questions: i) Which merchant sold the maximum number of baskets? ii) How many fruit baskets were sold by Anwar? iii) The merchants who have sold 600 or more baskets are planning to buy a godown for the next season. Can you name them?

2. The bar graph shows the number of tourists who visited the different hill stations during the vacations in a particular year. Study the bar graph and answer the following: Y Darjeeling

Hill Stations

Manali Shimla Mussoorie Nainital Monday 0 1000 2000 3000 4000 5000 6000 7000 Number of Tourists

116

X

IL Foundation Series Class 8

i) Write the scale along the horizontal axis. ii) Which hill station was visited by the maximum number of tourists? iii) How many tourists visited Darjeeling?

3. The number of Mathematics books sold by a shopkeeper on six consecutive days is shown below: Days

Sunday Monday Tuesday

Number of books sold

65

40

Wednesday

Thursday

Friday

50

20

70

30

Draw a bar graph (horizontal or vertical) representing the data.

4. The histogram depicts the marks of 40 students of class VIII obtained in a test of mathematics.

Number of students

Y 16 14 12 10 8 6 4 2 0

5 10 15 20 25 30 35

X

Marks i) What is the class size? ii) How many students obtained less than 5 marks? iii) How many students obtained 20 or more marks but less than 25? iv) If 25 or more marks make the first division, how many first divisions are there? v) How many students obtained 10 or more but less than 20 marks?

117

DATA HANDLING

5. Yearly savings (in `) of 30 students of class VIII are as under: 38

42

40

35

72

27

57

40

59

80

84

73

65

40

76

44

38

60

58

38

54

39

50

71

83

45

38

80

77

62

Construct a grouped frequency table with 30 − 35 as one of the class intervals (35 not included). Draw a histogram for the grouped frequency distribution.

6. The adjoining pie chart shows the interest of 300 students of a school in various games.

Tennis

Cricket

780 900 Fo

otb

660 all

360

90

0

y

Hocke

Basketball

a) How many of them like:

i) Tennis

ii) Basketball

iii) Hockey

iv) Game other than cricket

b) Find the ratio of the students who like to play cricket to that of basketball.

7. Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.

118

Colours

Number of People

Blue

18

Green

9

Red

6

Yellow

3

Total

36

IL Foundation Series Class 8

8. A die is thrown. Find the probability of getting i) a prime number

ii) 4

iii) a multiple of 3.

9. An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.

10. A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is: i) a black king

ii) a spade

iii) a red card

iv) a face card

119

5

SQUARES AND SQUARE ROOTS

5.1 INTRODUCTION TO SQUARES AND SQUARE ROOTS 5.1.1 Square numbers We are familiar with the square of a number.

This means the square of the number 2 is defined as 2 raised to the power 2.

That is, square of 2 = 22 = 2 × 2 = 4, square of 3 = 32 = 3 × 3 = 9, etc.

So, the square of number 3 is defined as 3 raised to the power 2 and so on. These statements can also be expressed by saying that 'the square of 2 is 4' or 'the square of 3 is 9', etc.

In other words, the square of a number is obtained by multiplying the number by itself.

Thus, if a is a number, then the square of a can be written as a2 = a × a.

The given table gives us the squares of numbers from 1 to 20.

Number

Square

Number

Square

1

12 = 1

11

112 = 121

2

22 = 4

12

122 = 144

3

32 = 9

13

132 = 169

4

42 = 16

14

142 = 196

5

52 = 25

15

152 = 225

6

62 = 36

16

162 = 256

7

72 = 49

17

172 = 289

8

82 = 64

18

182 = 324

9

92 = 81

19

192 = 361

10

102 = 100

20

202 = 400

Table 5.1 Numbers and their squares Clearly, we can see that 1, 4, 9, 16, 25, etc., are the squares of numbers 1, 2, 3, 4, 5 and so on.

The numbers 1, 4, 9, 16, 25, etc., are called perfect squares or square numbers.

Perfect square: A natural number is called a perfect square or a square number if it is the square of a natural number.

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IL Foundation Series Class 8

Thus, if a natural number n is a perfect square, then there exists a natural number m such that, m2 = n.

Example: 4 is a perfect square because there is a natural number 2 such that 22 = 4.

Steps to check a perfect square: Let us now try to find if a given number is a perfect square.

To do so, we follow these two steps in the given order.

1. Write the number as the product of its prime factors.

2. If the number is a perfect square, the prime factors can be grouped into pairs of equal factors.

Otherwise, the number is not a perfect square.

Example: Is 16 a perfect square?

Solution: The given number 16 can be written as the product of its prime factors.

We have 16 = 2 × 2 × 2 × 2, as we can see that the prime factors can be grouped as pairs.

∴ 16 is a perfect square.

Method of squaring numbers: The squares of small numbers, like 3, 4, 5, 6, 7, etc., are easy to find.

But can we find the square of 26 quickly?

The answer is not so easy, and we may need to multiply 26 by 26.

There is a way to find this without having to multiply 26 × 26.

We know: 262 = (20 + 6)2 = (20 + 6)(20 + 6) = 20(20 + 6) + 6(20 + 6) = 202 + 20 × 6 + 6 × 20 + 62 = 400 + 120 + 120 + 36 = 676 Example: Find the squares of the following numbers. i) 93

ii) 211

Solution: i) 93 = 90 + 3 932 = (90 + 3)2 = 90(90 + 3) + 3(90 + 3) = 902 + 90 × 3 + 3 × 90 + 32 = 8100 + 270 + 270 + 9 = 8649

121

SQUARES AND SQUARE ROOTS

ii) 211 = 200 + 11 2112 = (200 + 11)2 = 200(200 + 11) + 11(200 + 11) = 2002 + 200 × 11 + 11 × 200 + 112 = 40000 + 2200 + 2200 + 121 = 44521

5.2 PROPERTIES OF SQUARE NUMBERS Property - 1: The squares of the first 20 numbers end with 0, 1, 4, 5, 6 and 9, as shown in the previous table. This means that a number ending with 2, 3, 7 or 8 at the units place is never a perfect square. Thus, we can say that numbers like 22, 73, 87, 68, etc., are not perfect squares. At the same time, not all numbers ending with 0, 1, 4, 5, 6, and 9 are perfect squares.

It is not necessary that if a number ends with 0, 1, 4, 5, 6 or 9, then it must be a perfect square.

Property - 2: A number which ends with an odd number of zeroes is never a perfect square. For example, 10, 1000, 100000, etc., are not perfect squares, whereas 100, 10000, etc., are perfect squares.

Property - 3: The squares of even numbers are always even numbers, whereas squares of odd numbers are always odd.

For example: 12 = 1

22 = 2

32 = 9

42 = 16

52 = 25

62 = 36

7 2 = 49, etc.

82 = 64, etc.

Property - 4: Let us observe the following patterns.

12 = 1 22 = 4 = 3 × 1 + 1;

22 = 4= 4 × 1

32 = 9 = 3 × 3 + 0; 32 = 9 = 4 × 2 + 1 42 = 16 = 3 × 5 + 1; 42 = 16= 4 × 4 52 = 25 = 3 × 8 + 1; 52 = 25 = 4 × 6 + 1 62 = 36= 3 × 12;

62 = 36= 4 × 9, etc.

Note: 1. The square of a natural number (except 1) is either a multiple of 3 or exceeds a multiple of 3 by 1.

2. The square of a natural number (except 1) is either a multiple of 4 or exceeds a multiple of 4 by 1.

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IL Foundation Series Class 8

Based on these observations, we can say that if p2 is a multiple of 3, then p is also a multiple of 3.

Property - 5: Observe the following statements: 1 is a square number, but 2 × 1 = 2 is not a square number.

4 is a square number, but 2 × 4 = 8 is not a square number.

9 is a square number, but 2 × 9 = 18 is not a square number, etc.

So, if n is a perfect square number, then 2n can never be a perfect square.

In other words, if n = q2 for some natural number q, then we cannot find a natural number p such that 2q2 = p2.

Thus, two natural numbers, p and q, such that p2 = 2q2 do not exist.

Property - 6: A perfect square leaves a remainder of 0 or 1 when divided by 3.

Property - 7: For every natural number n, we have: (n + 1)2 − n 2 = [(n + 1) + n ][(n + 1) − n ] = (n + 1) + n By using the identity a 2 − b 2 = (a + b)(a − b) Hence, (n + 1)2 − n 2 = 2n + 1 This helps us in writing the difference between two consecutive squares.

For example: 52 − 42 = 5 + 4 = 9 52 − 42 = 2 × 4 + 1 = 9 152 − 142 = 15 + 14 = 29 152 − 142 = 2 × 14 + 1 = 29 Property - 8: The square of a natural number n is equal to the sum of the first n odd natural numbers.

1 + 3 + 5 + 7+ ... + (2n - 1) = n2 For example: 12 = 1 = The first odd number 22 = 4 = 1 + 3 = sum of the first two odd numbers 32 = 9 = 1 + 3 + 5 = sum of the first three odd numbers 42 = 16 = 1 + 3 + 5 + 7 = sum of the first four odd numbers 52 = 25 = 1 + 3 + 5 + 7 + 9 = sum of the first five odd numbers Thus, n2 = 1 + 3 + 5 + 7 + ... + (2n - 1)= sum of the first n odd numbers

123

SQUARES AND SQUARE ROOTS

Property - 9: Let us observe the following pattern for the squares of numbers with 1 as all the digits.

Increasing up to 4

112 = 121

11112 = 1 2 3 4 3 2 1 Decreasing up to 1

1112 = 12321

111112 = 1 2 3 4 5 4 3 2 1

Increasing up to 5 Decreasing up to 1

Thus, we can say that the squares of numbers having all digits as 1 follow a set of patterns that is given above. Also, the sum of digits of the number on the right-hand side is a perfect square.

For example: 1 + 2 + 1 = 4 = 22 1 + 2 + 3 + 2 + 1 = 9 = 32 Property - 10: Yet another interesting pattern is given below: 121 × (1+2+1) = 484 = 222 12321 × (1+2+3+2+1) = 110889 = 3332 The above pattern can also be written as follows: 112 × ( sum of digits in 112 ) = 222 1112 × ( sum of digits in 1112 ) = 3332 Hence, 1111111112 × (sum of digits in 1111111112) = 9999999992 SOLVED EXAMPLES Example 1: Why are the following numbers not perfect squares? i) 1057

ii) 23453

iii) 7928

iv) 222222

Solution: 1057 ends with 7, 23453 ends with 3, 7928 ends with 8 and 222222 ends with 2. As no perfect square ends with 2, 3, 7, or 8. So, the numbers given in the question are not perfect squares.

Example 2: What will be the units of digits of the squares of the following numbers? i) 81

124

ii) 272

iii) 3853

iv) 52698

v) 99680

IL Foundation Series Class 8

Solution: i) The units digit of 81 is 1. So, the units digit of its square will also be 1, as the units digit of squares of numbers ending in 1 or 9 is 1. ii) The units digit of 272 is 2. So, the units digit of its square will also be 4, as the units digit of squares of numbers ending in 2 or 8 is 4. iii) The units digit of 3853 is 3. So, the units digit of its square will also be 9, as the units digit of squares of numbers ending in 3 or 7 is 9. iv) The units digit of 52698 is 8. So, the units digit of its square will also be 4, as the units digit of squares of numbers ending in 2 or 8 is 4. v) The units digit of 99680 is 0. So, the units digit of its square will also be 0, as the units digit of squares of all numbers ending in 0 is 0.

Example 3: Observe the following pattern and find the missing numbers.

112 = 121 1012 = 10201 101012 = 102030201 10101012 = ........................ .................. = 10203040504030201 Solution: By the above pattern we can write: 10101012 = 1020304030201 1010101012 = 10203040504030201 Therefore, the required missing numbers are 1020304030201 and 101010101, respectively.

Example 4: Why are the following numbers not square numbers? i) 64000

ii) 89722

Solution: i) 64000 is not a perfect square number, because the number of zeros ending in 64000 is 3, which is an odd number, and the number of zeros at the end of a perfect square is always even. ii) 89722 is not a perfect square because this number ends in 2.

Example 5: Using a suitable pattern, complete the following:

i)

3332 = 12321

ii)

6666662 = 12345654321

125

SQUARES AND SQUARE ROOTS

Solution: Consider the following patterns: 3332 (3)2 × (111)2 9 × 12321 = = = 9 i) 12321 12321 12321 6666662 62 × 111112 36 × 12345654321 = 36 ii) = = 1234564321 1234567321 12345654321 Hence,

666666 = 36 12345654321

5.3 ESTIMATION OF SQUARE ROOT Estimation and approximation refer to an acceptable approximation of the actual number. These methods help in the calculation and estimation of a number's square root.

1. If the number of digits in a perfect square 'n' is even, then the number of digits in n .

2 2. If the number of digits in a perfect square 'n' is odd, then the number of digits in n +1 . its square root is 2 Example: Estimate the value of 80 to the nearest whole number. its square root is

Solution: We know that 64 < 80 < 81 and = 64 8 and = 81 9.

64 < 80 < 81 So, 8 < 80 < 9 Since 80 is much closer to 81 than 64.

So,

80 is approximately equal to 9.

5.4 FINDING SQUARE ROOT BY REPEATED SUBTRACTION 5.4.1 Square roots by repeated subtraction 1. Let a be the given perfect square whose square root is to be calculated.

126

IL Foundation Series Class 8

2. Subtract 1, 3, 5, 7... from a successively until the result is zero.

3. Count the number of steps performed to arrive at zero. Let this number be ' n ', then

a = n.

Example: Find the square root of 16 using repeated subtraction.

Solution: We have: 16 − 1 = 15 15 − 3 = 12 12 − 5 = 7 7 −7 = 0 Here, subtraction is performed 4 times. ∴ 16 = 4.

However, this method is not suitable for calculating the square roots of large numbers. So, we shall discuss other methods of calculating the square roots of large numbers.

5.5 FINDING SQUARE ROOT BY PRIME FACTORISATION We have studied in the previous section that when we multiply a natural number by itself, we get the square of the given number.

For example: if we multiply 2 by itself, i.e., 2 × 2, we get 4; 2 × 2 = 4 or 22 = 4.

Here, we say that the square of 2 is 4 or 2 is the square root of 4 .

If we write 16 as the product of prime factors, then we have : 16 = 2 × 2 × 2 × 2 Grouping these factors into pairs, we have : 16 = (2 × 2) × (2 × 2) Thus, we can say that to find the square root of a number (perfect square), we follow the following steps: 1. Write down the number as the product of its prime factors.

2. As the number is a perfect square, it will be possible to group the factors into pairs.

3. Write one factor from each pair and find their product.

4. The product obtained is the required square root of the given perfect square.

Example: Find the square root of 400 by the prime factorisation method.

127

SQUARES AND SQUARE ROOTS

Solution: The prime factorisation of 400: 2

400

2

200

2

100

2

50

5

25

5

5 1

∴

400 = 2 × 2 × 2 × 2 × 5 × 5

∴

400 = 2 × 2 × 5 = 20

5.6

FINDING SQUARE ROOT BY LONG DIVISION

5.6.1 Square root by long division method We have learnt to find the square root of perfect squares by the prime factorisation method. But, when the square numbers are very large, finding the square root using the prime factorisation method proves lengthy as well as difficult. In such cases, we use the method of long division.

This method begins by pairing the perfect square the way we have already done. Dividing the number 225 into periods, we have 225 = 2 25. So, 225 has two periods.

The following steps are explained with the help of an example.

Example: Find the square root of 225.

Solution: Step 1: In the given number, form the pairs and place bars (as shown above) on them starting from the units digit.

Step 2: Think of the largest square number less than or equal to the first period (from the left). Write it just below the first period and its square root as quotient and divisor. (put the quotient above the period) Step 3: Subtract the product of quotient and divisor.

128

1

1

5

2

25

-1 25

1

25

-1

25

0

0

IL Foundation Series Class 8

Step 4: Bring down the second period on the right of the remainder. This is the new dividend.

Step 5: Double the quotient and write this number on the left of the remainder with a blank on the right for the next digit as the next possible divisor.

Step 6: Enter a new digit to fill in the blank and as the new digit in the quotient such that the product of this new digit in the quotient with the new divisor is less than or equal to the new dividend.

Step 7: Subtract and bring down the next period (if any).

Step 8: Repeat steps 5, 6, and 7, and fill all bars that have been considered. The final quotient is the required square root.

Example: Find the square root of 363609.

Solution: The given number is 363609.

Applying the long division method, we have : 6

0

3

6 36

36

09

36 120 0

36 0

1203

36

09

36

09 0

∴ 363609 = 603 SOLVED EXAMPLES Example 1: Find the least number which must be added to 306452 to make it a perfect square.

Solution: Applying the long division method, we have : 5

5

3

6 30

64

52

25 105 5

64

5

25

1103

39

52

33

09

6

43 129

SQUARES AND SQUARE ROOTS

From above, the given number 306452 is greater than (553)2. i.e., 5532 < 306452 < 306916 (i.e.,5542 ).

Since we have to find the least number that should be added to 306452 to make it a perfect square, if we add 306916 - 306452 = 464 to 306452, then the sum 306452 + 464 will be a perfect square.

So, the required least number is 464.

Example 2: Find the square root of 169 by the method of successive subtractions.

Solution: Here, 169-1 = 168 160-7 = 153 133-13 = 120 88-19 = 69 25-25 = 0

168-3 = 165 153-9 = 144 120-15 = 105 69-21 = 48

165-5 = 160 144-11 = 133 105-17 = 88 48-23 = 25

Thus, in the case of 169, we need to take 13 steps to get zero by successive subtractions.

Therefore, the square root of 169 is 13.

Example 3: A teacher wants to arrange the maximum possible number of 6000 students in a field such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after the arrangement.

Solution: Since 71 students were left, so the remaining students = 6000 - 71 = 5929 ∴ The prime factorisation of 5929 is : 7

5929

7

847

11

121

11

11 1

∴ 5929 = 7 × 7 × 11 × 11 So,

5929 =7 × 11 =77

Hence, the required number of rows is 77.

130

IL Foundation Series Class 8

Example 4: Find the least perfect square, which is exactly divisible by each of the numbers 6, 9, 15 and 20.

Solution: The least number divisible by each of the numbers 6, 9, 15 and 20 is their LCM, which is :

(2 × 2 × 3 × 3 × 5) i.e., 180.

Now, 180 = 2 × 2 × 3 × 3 × 5 To make it a perfect square, it must be multiplied by 5. ∴ The required number = 180 × 5 = 900 Example 5: The length of the rectangular field is 7 times its breadth. Its area is 1792 sq.m. Find the cost of fencing it at the rate ₹1.50 per metre.

Solution: Let the breadth of the field = x and the length = 7x Area of the rectangular field = 1792 sq. m ⇒ ⇒ lllbblblbb= 179999922222 ====17 ⇒ 17 17 ⇒ ⇒ 17 ⇒ 1792 ⇒77777xxxxx× = 1792 ⇒ ××××xxxxx= = 1792 ⇒ = 1792 ⇒ = 1792 ⇒ 1792 ⇒ 77777xxxxx222222= = 1792 ⇒ = 1792 ⇒ = 1792 ⇒ = 1792 1792 1792 1792 1792 1792 ⇒ ⇒ xxxxx222222= = ⇒ = ⇒ = ⇒ = 77777 22 2 ⇒ 256 ⇒ xxxxx222 = = 256 ⇒ = 256 ⇒ = 256 ⇒ = 256 ⇒ 11116666 ⇒ xxxx= = ⇒ = ⇒ = Length = 7x = 112 m Breadth = x = 16 m Perimeter = 2(l + b) = 2(112+16) = 2(128) = 256 m The cost of fencing per 1 m = ₹1.50 The cost of fencing per 256 m = 256 × 1.50 = ₹384

5.7

PYTHAGOREAN TRIPLETS

For any natural number m > 1, we have (2m)2 + (m2 - 1)2 = (m2 + 1)2.

So, m2, m2 -1, and m2 + 1 form a Pythagorean triplet.

Consider the following: 32 + 42 = 9 + 16 = 25 = 52

131

SQUARES AND SQUARE ROOTS

The collection of numbers 3, 4 and 5 is known as a Pythagorean triplet. 6, 8, and 10 is also a Pythagorean triplet, since: 62 + 82 = 36 + 64 = 100 = 102 Again, observe that 52 +122 = 25 + 144 = 169 = 132.

The numbers 5, 12 and 13 form another such triplet.

Example: Write the Pythagorean triplet whose smallest member is 8.

Solution: We can get Pythagorean triplets by using the general form 2m, m2 - 1, m2 + 1.

Let us first take m2 - 1 = 8 So, m2 = 8 + 1 = 9 which gives m = 3 Therefore, 2m = 6 and m2 + 1 = 10 The triplet is 6, 8, 10. But 8 is not the smallest member of this.

So, let us try, 2m = 8 Then, m = 4 We get, m2 - 1 = 16 - 1 = 15

And m2 + 1 = 16 + 1 = 17 The triplet consists of 8, 15, and 17, with 8 as the smallest member.

Note: All Pythagorean triplets may not be obtained using this form. For example, another triplet 5, 12 and 13 also have 12 as a member.

5.8 SQUARE ROOT OF A DECIMAL NUMBER We may find the square root of a decimal number without converting it into a rational number. We do it as follows.

1. Place bars on an integral part of the number in the usual manner.

2. Place bars on the decimal part for every pair of digits beginning with the first decimal place.

3. Start finding the square root using the division process as usual.

4. Place the decimal point in the quotient as soon as the integral part is exhausted.

5. Stop when the remainder becomes zero. The quotient at this stage is the square root.

Example: Find the square root of 477.4225.

Solution: Here, the number of decimal places is already even.

132

IL Foundation Series Class 8

So, place bars on the integral and decimal parts and proceed as given below: 2 1 . 8 5 2 4

77 . 42

25

4 41

77 41

428

36

42

34

24

2

18

25

2

18

25

4365

0

Thus, we have

477.4225 = 21.85 .

SOLVED EXAMPLES Example 1: A decimal fraction is multiplied by itself. If the product is 251953.8025, find the fraction.

Solution: Let the required fraction be 'x', then: = x × x 251953.8025 = ⇒ x 2 251953.8025 = ⇒x

251953.8025

Now, we mark off the periods and find the square root of 251953.8025 as below.

5 0 1 . 9 5 5 25

19

53 . 80

25

25 100

19 0

1001 10029 100385

19

53

10

01

9

52

80

9

02

61

50

19

25

50

19

25 0

133

SQUARES AND SQUARE ROOTS

Example 2: The area of a square playground is 291.0436 square metres. Find the length of each side of the playground.

Solution: Area of a square playground = 291.0436 m 2 Side of the square playground = 291.0436 1

7 . 0

6

1 2

91 . 04

36

1 27 1

91

1

89

340

2

04 0

3406

2

04

36

2

04

36 0

We find that 291.0436 = 17.06 Thus, each side of the square playground is 17.06 m Example 3: Find a Pythagorean triplet in which one member is 12.

Solution: If we take, m 2 − 1 = 12 Then, m 2 = 12 + 1 = 13 Then, the value of m will not be an integer.

So, we try to take m 2 + 1 = 12 Again, m 2 = 11 will not give an integer value for m.

So, let us take: 2m = 12 m=6 Thus, m 2 − 1 = 36 − 1 = 35 and m 2 + 1 = 36 + 1 = 37 .

Therefore, the required triplet is 12, 35, 37.

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IL Foundation Series Class 8

QUICK REVIEW 2

• If a natural number m can be expressed as n , where n is also a natural number, then m is a square number.

• All square numbers end with 0, 1, 4, 5, 6 or 9 at the units place.

• Square numbers can only have an even number of zeros at the end.

• The square root is the inverse operation of the square.

• There are two integral square roots of a perfect square number. The positive square root of a number is denoted by the symbol √.

For example: 32 = 9 gives 9 = 3.

• For any natural number m > 1, we have (2m)2 + ( m 2 − 1) = ( m 2 + 1) .

2

2

So, 2m, m 2 − 1 , and m 2 + 1 forms a Pythagorean triplet.

• If the number of digits in a perfect square 'n' is even, then the number of digits in its square root is

n .

2

• If the number of digits in a perfect square 'n' is odd, then the number of digits in its square root is

n +1 .

2

WORKSHEET - 1 I. SQUARE NUMBERS AND ITS PROPERTIES 1.

2.

Which of the following numbers are perfect squares? i) 16 ii) 32 iii) 1296 iv) 373756

v) 4375686

Observe the following patterns and find the missing numbers: 12 + 22 + 22 = 32 22 + 32 + 62 = 72 32 + 42 + 122 = 132 42 + 52 + 202 = ___ 52 + 62 + ___ = 312

3.

By just examining the units digit, verify which of the following are not perfect squares. i) 1024 ii) 1025 iii) 189 iv) 1022

v) 1234 135

SQUARES AND SQUARE ROOTS

4.

Find the squares of 331 and 339.

5.

Show that the following numbers are not perfect squares: i) 7927 ii) 1058 iii) 33453

6.

Which of the following numbers would have an odd square number? i) 432 ii) 2826 iii) 7779 iv) 82004

7.

Consider the following pattern: 352 =3 × (3 + 1) hundred + 25 =1225 452 =4 × (4 + 1) hundred + 25 =2025 1152 =11 × (11 + 1) hundred + 25 =13225 Using the above pattern, find the squares of: i) 25 ii) 85 iii) 105

8.

iv) 305

Consider the following pattern : 532 = ( 52 + 3 ) hundred + 32 = 2809 57 2 = ( 52 + 7 ) hundred + 7 2 = 3249 Using the above pattern, find the squares of : i) 51 ii) 54 iii) 56

9.

10.

11.

12.

iv) 59

Find the square of 789 without actual multiplication.

Find the squares of the following numbers using the column method. Verify the result by finding the square using the usual method : i) 24 ii) 37 iii) 71 iv) 96 Find the squares of: i) 127

ii) 235

iii) 443

iv) 251

Find the squares of the following numbers using the diagonal method: i) 89

ii) 275

iii) 293

iv) 346

II. PYTHAGOREAN TRIPLETS AND ESTIMATION OF SQUARE ROOT 1.

i) (1, 2, 3)

ii) (3, 4, 5)

iii) (6, 7, 8)

iv) (10, 24, 26)

v) (1, 1, 1)

vi) (12, 35, 37)

2.

3.

136

Which of the following triplets are Pythagorean?

Find the Pythagorean triplets, one of whose number is:

250 iii) 16 iv) 14 1000 250 Estimate the value of the250 following to the nearest whole number.

350 1000

i) 6

i)

ii) 10

250

ii) 1000

1000

350

350

500

500

iii)

350 500

iv)

500

IL Foundation Series Class 8

4.

5.

What could be the possible ones digits of the square root of each of the following numbers? i) 9801

ii) 99856

iii) 998001

iv) 657666025

Without doing any calculation, find the numbers which are surely not perfect squares. i) 153

ii) 257

iii) 408

iv) 441

III. FINDING SQUARE ROOT BY PRIME FACTORISATION, REPEATED SUBTRACTION AND LONG DIVISION 1.

Find the square roots of the following using the prime factorisation method.

i) 16

ii) 441

iii) 4096

iv) 8281

v) 47089

vi) 190969

2.

Find the smallest number by which 125 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained.

3.

Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained.

4.

The product of the two numbers is 1296. If one number is 16 times the other, find the numbers.

5.

23716 students are sitting in a stadium in such a manner that there are as many students in a row as there are rows in the stadium. Find the number of rows.

6.

Find the square roots of 121 and 64 by the method of successive subtractions.

7.

Write the prime factorisation of the following numbers to find the square roots.

8.

i) 9604

ii) 7056

Check if the following numbers are perfect squares. If yes, find their square roots.

i) 1936

9.

For each of the following numbers, find the smallest number by which it should be multiplied to get a perfect square. Also, find the square root of the square number so obtained.

i) 1200

10.

For each of the following numbers, find the smallest number by which it should be divided to get a perfect square. Also, find the square root of the square number so obtained.

i) 2800

11.

ii) 8281

ii) 1008

iii) 2028

ii) 45056

Find the square root of the following numbers by the long division method :

i) 44100

ii) 27225

12.

Find the least numbers which must be subtracted from each of the following numbers to make them perfect squares.

i) 4931

ii) 18265

iii) 1234321 iv) 4937284

iii) 26535 137

SQUARES AND SQUARE ROOTS

13.

Find the least number which must be added to each of the following numbers to make them perfect squares.

i) 4931

ii) 2361

iii) 5607

14.

Find the least number of four digits which is a perfect square.

15.

Find the greatest number of four digits that is a perfect square.

16.

The students of class VIII of a school donated `2401 in all for the prime minister's National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

17.

2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

18.

Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

19.

Find the greatest number of 6 digits that is a perfect square.

20.

Find the least number of 6 digits that is a perfect square.

21.

The area of square field is 60025 m2. A man cycles along its boundary at 18 km/h. In how much time will he return to the starting point?

22.

Find the least number which must be added to 506900 to make it a perfect square. Find this perfect square and its square root.

23.

24.

Find the missing digit in 2220x so that the number becomes a perfect square.

What least number must be added to 5607 to make it a perfect square? Find this perfect square and its square root.

IV. FINDING SQUARE ROOT OF DECIMAL NUMBERS 1.

2.

5 7 v) 12 8 Find the value of each of the following up to three places decimal.

i) 5

ii) 237.615

iii) 0.00064

iv)

vi) 2

1 12

4.

2 11 iv) 2 5 18 244 2 m . Find the length of each side of the field.

The area of a square field is 80 729 Find the square root of 12.0068 up to three decimal places.

5.

Find the square root of the following decimal numbers :

3.

6.

138

Find the square root of each of the following up to three places of decimal.

i)

3

5 12

i) 2.56

ii) 1

2 7

ii) 42.25

iii)

2

iii) 84.8241

iv) 0.813604 v) 0.00038809

Which fraction, when multiplied by itself, gives 227.798649?

IL Foundation Series Class 8

7.

The area of a square playground is 256.6404 m 2 . Find the length of one side of the playground.

WORKSHEET - 2 I. MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 36 = = and 5 2.236,= then ?

1.

If 2 1.414 5

a) 2.6632 b) 2.6832 c) 1.6832 d) 2.5832

2.

3.

1.7 = ? (up to correct to three places of decimal) a) 0.305

b) 0.304

c) 1.204

d) 1.304

b) 0.232

c) 1.231

d) 1.232

0.053361 = ?

a) 0.231

4.

A decimal fraction is multiplied by itself. If the product is 0.7225, then the decimal fraction is:

a) 0.75

5.

6.

7.

8.

9.

10.

11.

12.

b) 0.85

c) 0.95

d) 0.65

72 as the sum of two consecutive integers is __________. a) 40 + 9

b) 24 + 25

c) 36 + 13

d) 32 + 17

Among the following, which one is not a Pythagorean triplet? a) 3, 4, 5

b) 6, 8, 10

c) 5, 12, 13

d) 2, 3, 4

The number of non-square numbers that lie between the pair of numbers 5002 and 5012 is: a) 999

b) 1002

c) 1000

d) 1001

The units digit in the square of the number 78 is : a) 4

b) 3

c) 2

d) 5

The units digit in the square of the number 1111 is : a) 2

b) 1

c) 3

d) 4

The smallest 3-digit perfect square is: a) 961

b) 999

c) 100

d) 125

The number of digits in the square root of 62500 is : a) 1

b) 2

c) 4

d) 3

The possible ones digit of the square root of 676 is: a) 4, 6

b) 5, 7

c) 1, 8

d) 2, 9 139

SQUARES AND SQUARE ROOTS

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

The number of natural numbers lie between 82 and 92 are: a) 17

b) 16

c) 18

d) 19

A perfect square number between 30 and 40 is: a) 36

b) 32

c) 33

d) 39

Between 50 and 60, the perfect square number is: a) 56

b) 55

c) 54

d) none

Among the following, a perfect square number is: a) 10000

b) 2222

c) 32543

d) 888

Among the following, which would end with digit 1? a) 1322

b) 87 2

c) 2092

d) 722

Among the following, which would have 6 at the units place? a) 17 2

b) 342

c) 252

d) 492

The greatest number of 5 digits, which is a perfect square is: a) 99856

b) 88577

c) 77777

d) 33333

The square of which of the following numbers will be odd? a) 42

b) 54

c) 66

d) 81

The number of zeroes in the square of the number 50 is: a) 1

b) 2

c) 3

d) 4

The smallest number by which 32 should be multiplied to get a perfect square is: a) 2

b) 3

c) 4

d) 8

II. FILL IN THE BLANKS 1.

42.25 =

2.

1132 − 1122 =

.

3.

The number of perfect squares less than 100 is

4.

Square numbers can only have

5.

If

6.

7.

140

.

49 a = , then a is .

64 512 1 1 The value of . + is 16 9 The sum of the first n odd natural numbers is

. number of zeroes at the end.

.

IL Foundation Series Class 8

0.9 × 1.6 =

8.

9.

.

The square root of a proper fraction is

than the given fraction.

III. SUBJECTIVE QUESTIONS 1.

Find the value of a 2 + 2 ( b 2 + c 2 ) , if b =1, c =3 and a =2.

2.

Find the least square number exactly divisible by 8, 12, 15 and 20.

3.

Find the value of x if

4.

Find the length of the side of a square whose area is 441 m 2 .

5.

Express 212 as the sum of two consecutive integers.

6.

What is/are the remainder(s) when a perfect square is divided by 3?

7.

Give the reason to prove that 640 and 81000 are not perfect squares.

8.

If a number ends with 3 zeroes, how many zeroes will its square have?

9.

Is 196 a perfect square? Explain.

10.

If 784 = 28 , then find the value of 7.84 + 0.0784 .

11.

Without calculating the square roots, find the number of digits in the square root of: ii) 10201

2401 = 7 x .

i) 225

iii) 654481

iv) 39702601

12.

The cost of levelling a square lawn at ` 250 per m2 is ` 13322, find the cost of fencing it at ` 5 per metre.

13.

Find the square root of the following using the prime factorisation method:

i) 1024

14.

In an auditorium, the number of rows is equal to the number of chairs in each row. If the capacity of the auditorium is 2025, find the number of chairs in each row.

15.

The students of a class arranged a picnic. Each student contributed as many rupees as the number of students in the class. If the total contribution is `1156, find the strength of the class.

16.

17.

ii) 9604

Find the square root of the following by using the long-division method: i) 729

ii) 1024

Find the square roots of the following fractions:

64 400 ii) 225 169 18. Find the value of 5.29 × 1.69 .

i)

141

6

CUBES AND CUBE ROOTS

6.1 INTRODUCTION TO CUBES AND CUBE ROOTS In this chapter, we will extend the idea of squares and square roots to cubes and cube roots. As we have seen, we can only calculate the square roots of positive real numbers. However, cube roots are defined for positive as well as negative numbers. In fact, the cube roots of negative numbers are negatives of the cube roots of their absolute values.

6.1.1 Cube of a number Cube: The cube of a number is the number itself raised to the power 3.

Thus, if a is a number, then the cube of a is a3, that is, a 3 = a × a × a .

Example: 23 = 2 × 2 × 2 = 8. The cube of 2 is 8.

(1.2)3 = 1.2 ×1.2 ×1.2 = 1.728. The cube of 1.2 is 1.728.

3

2 8 2 2 2 2 8 . The cube of is .   = × × = 3 27  3  3 3 3 27

Hardy-Ramanujan number: 1729 is the smallest Hardy-Ramanujan number. There are infinitely many such numbers. 1729 can be expressed as a sum of two cubes in two different ways.

3 3 1. 1729= 1 + 12

1729 = 93 + 103 = 23 + 163 2. 4104 4104 = 93 + 153 3. 13832 = 183 + 203 13832 = 23 + 243 Cube of a natural number: The cube of a natural number is the natural number itself raised to the power 3.

Thus, if m is a natural number, then m3 is the cube of m .

That is, m3 = m × m × m.

Perfect cube: A natural number is said to be a perfect cube if it is the cube of some natural number.

Example: 8 is a perfect cube because there is a natural number 2 such that 8 = 2 × 2 × 2 = 23.

12 is not a perfect cube because there is no natural number whose cube is 12.

142

IL Foundation Series Class 8

6.2 PROPERTIES OF CUBES 1. The cube of a negative rational number is always negative.

Example: (−1)3 =( −1) × ( −1) × ( −1) =−1 (−3)3 =( −3 ) × ( −3 ) × ( −3 ) =−27 3

a a a a 2. The cube of a rational number is   = × × b b b b 3. The cube of an even number is even.

Example: 43 = 64 (4 and 64 both are even.)

4. The cube of an odd number is odd.

Example: 53 = 125 (5 and 125 both are odd.)

5. The sum of the cubes of the first n natural numbers is equal to the square of their sum.

That is, 13 + 23 + 33 +….. + n 3= (1 + 2 + 3 +…. + n)2 .

SOLVED EXAMPLES Example 1: Show that 189 is not a perfect cube.

Solution: Resolving 189 into prime factors: 189 = 3 × 3 × 3 × 7 = 33 × 7 So, we can’t express 189 as a product of triplets. ∴189 is not a perfect cube.

Example 2: Show that 216 is a perfect cube.

Solution:

216 = 2 × 2 × 2 × 3 × 3 × 3 216 = 6 × 6 × 6 = (6)3 216 can be expressed as a product of triplets.

So, 216 is a perfect cube.

The cube of 6 is 216.

Example 3: What is the smallest number by which 3087 must be multiplied so that the product is a perfect cube?

Solution:

3087 = 3 × 3 × 7 × 7 × 7 143

CUBES AND CUBE ROOTS

To make 3087 a perfect cube, it must be multiplied by 3.

Example 4: What is the smallest natural number by which 392 must be divided so that the quotient is a perfect cube?

Solution: 392 = 2 × 2 × 2 × 7 × 7.

To make 392 a perfect cube, it must be divided by 7 × 7 = 49.

6.3 CUBE ROOTS Cube root: A number m is the cube root of a number n , if n = m3 .

In the other words, the cube root of a number n is the number m whose cube gives n.

The cube root of a number n is denoted by 3 n .

3

n is also called the radical, n is called the radicand, and 3 is called the index of the radical.

Example: i) 8 = 23

∴38 = 2 (−5)3 ii) −125 = ∴ 3 −125 =−5 iii) 0.008 = (0.2)3

∴ 3 0.008 = 0.2 Cube root of a natural number: A natural number m is the cube root of a natural number n if n = m3 , and we write 3 n = m . Thus, 3 n = m ⇔ n = m3 .

Example: 3 27 =3 ⇔ 27 =33

144

IL Foundation Series Class 8

6.4 FINDING CUBE ROOT BY PRIME FACTORISATION Steps to find the cube root of a number by prime factorisation: Step 1: Express the given number as a product of primes.

Step 2: Make triplets of the same prime.

Step 3: Find the product of primes, choosing one from each triplet.

Step 4: This product is the required cube root of the given number.

Example: 216 = 2 × 2 × 2 × 3 × 3 × 3 3

216 = 2 × 3 = 6

6.5 PROPERTIES OF CUBE ROOTS 1. 3 −a =− 3 a − 3 27 = −3 Example: 3 −27 =

= 3a×3b 2. 3 ab Example: 3 8 × 27 = 3 8 × 3 27 = 2 × 3 = 6 3. 3

a 3a , where b ≠ 0 . = b 3b

125 Example: 3= 64

125 5 = 3 64 4

3

6.6 FINDING CUBE ROOT BY ESTIMATION We can use the following steps to find the cube root of a large number that is a perfect cube.

Example: Find the cube root of 857375 by using the estimation method.

Step 1: Start with making groups of three digits starting from the right most digit of the cube number 857375.

857 Second group

375 First group

We get 375 and 857 as two groups of three digits each.

145

CUBES AND CUBE ROOTS

Step 2: The first group, i.e., 375 will give you the ones (or units) digit of the required cube root. The number 375 ends with 5.

We know that 5 comes at the ones place of a number only when its cube root ends in 5.

So, we get 5 at the ones place of the cube root.

Step 3: Now, take the second group, i.e., 857. We know that 93 = 729 and 103 = 1000 . Also, 729 < 857 < 1000 . So, taking the smaller number's cube root, i.e., 9 as the tens place of the required cube root, we get 3 857375 = 95.

SOLVED EXAMPLES Example 1: The volume of a cubical box is 32.768 cubic metres. Find the length of the side of the box.

Solution: Let the length of the side of the box be x metres. The volume is x3 cubic metres. ∴ x3 = 32.768 = x

3

= 32.768

32768 = 1000

3

3

3

32768 1000

3 = = 32768 32 and 3 1000 10

Hence,= x

3 3

32768 32 = = 3.2 m.

1000 10

Example 2: Find the volume of a cube whose surface area is 150 m 2 .

Solution: Let the length of each edge of the given cube be x metres. Then, the surface area = 6 x 2 .  Surface area = 150 m2 Þ 6 x 2 = 150 150 Þ x2 = = 25 6 Þ x = 25 = 5´5 = 5 \ Volume of a cube = x 3 cubic metres = 53 = 5´5´5 = 125 cubic metres The volume of the required cube is 125 cubic metres.

146

IL Foundation Series Class 8

Example 3: Evaluate 3 2744 .

Solution: 2744 = 2 × 2 × 2 × 7 × 7 × 7 3

2744 = 2 × 7 = 14

Example 4: Find the cube root of ( −1000 ) .

Solution:

− 3 1000 We know that 3 −1000 = 1000

= 2×2×2×5×5×5

3

1000 = 2 × 5 = 10

3

−1000 = − 3 1000 = −10

Example 5: Evaluate 3 125 × 64 .

Solution: 3

=

125 × 64 3

125 × 3 64

= 3 5×5×5 × 3 4× 4× 4 = 5 × 4 = 20 Example 6: What is the smallest number by which 1323 must be multiplied so that the product is a perfect cube?

Solution: 3

1323

3

441

3

147

7

49

7

7 1

1323 = 3 × 3 × 3 × 7 × 7 We should multiply 1323 by 7 to form a perfect cube.

147

CUBES AND CUBE ROOTS

Example 7: What is the smallest number by which 392 must be multiplied so that the product is a perfect cube?

Solution: 2

392

2

196

2

98

7

49

7

7 1

Resolving 392 = {2 × 2 × 2} × 7 × 7 We find that 2 occurs as a prime factor of 392 thrice but 7 occurs as a prime factor only twice. Thus, if we multiply 392 by 7, then it will be 2 × 2 × 2 × 7 × 7 × 7 which is a perfect cube. Hence, we must multiply 392 by 7 so that the product becomes a perfect cube.

Example 8: What is the smallest number by which 704 must be divided so that the quotient is a perfect cube?

Solution: 2

704

2

352

2

176

2

88

2

44

2

22 11

Resolving 704 into prime factors, and grouping the factors in triplets of equal factors, we get:

704 = {2 × 2 × 2} × {2 × 2 × 2} × 11 So, if we divided 704 by 11, then the quotient would be {2 × 2 × 2} × {2 × 2 × 2}. ∴We must divide 704 by 11 so that the quotient is a perfect cube.

QUICK REVIEW

148

•

The numbers obtained when a number is multiplied by itself three times are known as cube numbers. For example: 1, 8, 27, etc.

•

Numbers like 1729, 4104, 13832, are known as Hardy-Ramanujan numbers. They can be expressed as a sum of two cubes in two different ways.

IL Foundation Series Class 8

•

During the prime factorisation of any number, if each factor appears three times, then the number is a perfect cube.

•

For a given number x , we define its cube as, cube of x = x × x × x , and denote it by x 3 .

•

A natural number is a perfect cube if it can be expressed as a product of triplets of equal factors.

•

Properties of cube of a number:

1. The cube of a negative rational number is always negative.

3

a a a a 2. The cube of a rational number is   =   ×   ×   . b b b b 3. The cube of an even number is even.

4. The cube of an odd number is odd.

•

The cube root of a number x is the number whose cube gives x . The cube root of x is denoted by 3 x .

•

Properties of cube roots: 1.

2.

3.

3

−a =− 3 a

3

ab =

3

a 3a , where b ≠ 0 = b 3b

3

a×3b

WORKSHEET - 1 I.

CUBE OF A NUMBER AND ITS PROPERTIES 1. Find the smallest number by which 3087 must be multiplied so that the product is a perfect cube.

2. Find the smallest number by which 392 must be divided so that the quotient is a perfect cube.

3. Evaluate (0.8)3.

4. Write the cubes of the first three multiples of 3.

5. Using prime factorisation, determine which of the following are perfect cubes. i)128

ii) 343

iii) 729

iv) 1331

6. Is 9720 a perfect cube? If not, find the smallest number by which it should be divided to get a perfect cube.

149

CUBES AND CUBE ROOTS

7. What is the smallest number by which 3600 should be multiplied so that the quotient is a perfect cube?

8. If one side of a cube is 15 m in length, find its volume.

9. Find the length of each side of the cube if its volume is 512 cm3.

10. Three numbers are in the ratio 1 : 2 : 3, and the sum of their cubes is 4500. Find the numbers.

3

1    11. Solve:  52 + 122 2     

( )

3

1    2 2 12. Solve:  6 + 8 2     

( )

13. Check whether 1728 is a perfect cube using prime factorisation.

14. Write cubes of 5 natural numbers which are multiples of 3, and verify the following: 'The cube of natural number, which is a multiple of 3 is a multiple of 27'.

15. Write cubes of 5 natural numbers of the form 3n + 1 (e.g. 4, 7, 10...), and verify the following: The cube of a natural number of the form 3n + 1 is a natural number of the same form.

16. What is the smallest number by which 8640 must be divided so that the quotient is a perfect cube?

17. By what number would you multiply 231525 to make it a perfect cube?

18. Parikshit made a cuboid of plasticine with sides 5 cm, 2 cm, and 5 cm. Find the number of such cuboids needed to form a cube.

19. Find the smallest number by which 128 must be divided to get a perfect cube.

20. The cube of a number is 8 times the cube of another number. If the sum of the cubes of the numbers is 243, then find the difference between the two numbers.

21. Find the sum of the digits of the smallest number possible, which, when multiplied by 1800, gives a perfect cube.

II.

CUBE ROOT OF A NUMBER AND ITS PROPERTIES 1. Evaluate 3 2744 .

2. Find the cube root of ( −1000 ) .

3. Evaluate 3 125 × 64 .

4. Evaluate 3 27 + 3 0.008 + 3 0.064 .

150

IL Foundation Series Class 8

5. The difference between two perfect cubes is 189. If the cube root of the smaller number is 3, then find the cube root of the larger number.

6. Using prime factorisation, find the cube roots of: (i) 512 (ii) 2197.

7. Find the cube roots of: (i) -125 (ii) -5832 (iii) -17576.

8. What is the smallest number by which the number 26244 can be divided so that the quotient is a perfect cube? Also, find the cube root of the quotient.

9. Find out the cube root of 13824 by the prime factorisation method.

10. Evaluate: 3 729 .

2197 11. Evaluate: 3 −512 .

1331 12. Find the value of 13 + 23 + 33 + … + 103 .

48 16, then what is the value of x ?

13. If 3 x + 3 = x

14. Solve: 3 0.125 + 3 0.729 .

15. What should be added to 2714 to make the sum a perfect cube?

16. Evaluate:3 3

3.43 .

10

17. Evaluate: 3 16 × 1372 ÷ 8 × 288 .

18. Find the value of

3

−2744 × 3 −216 .

64 3 729

19. If the cube root of 175616 is 56, then find the value of 3 175.616 + 3 0.175616 + 3 0.000175616 ?

20. The four-fifths of one-eighth of three-fourths of A is 64. What is the cube root of the threefifth of A?

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. Among the following, the cube of 6 is: a) 16 b) 36 c) 126

d) 216

151

CUBES AND CUBE ROOTS

2. Among the following, which is not a perfect cube? a) 125 b) 144 c) 1000 d) 216 3. If the volume of a cubical box is 474.552 cubic metres, then the length of each side of the box is: a) 7.2 m b) 7.4 m c) 7.6 m d) 7.8 m 4. The least number by which 3087 should be multiplied so that the product is a perfect cube, is: a) 7 b) 9 c) 3 d) 49 5. The ones digit in the cube root of 1728 is: a) 1 b) 2 c) 3

d) 9

6. Among the following, the number that is not a perfect cube is: a) 1331 b) 512 c) 343 d) 100 7. Harshith makes a plastic cuboid with sides 5 cm, 4 cm, and 2 cm . The number of cubes that can be formed from this cuboid if each edge of the cube is 2 cm is: a) 20 b) 5 c) 10 d) 16 8. Among the following, the false statement is: a) The cube of any odd number is odd. b) A perfect cube does not end with two zeroes. c) The cube of a single-digit number may be a single-digit number. d) There is no perfect cube that ends with 7.

9. Among the following, the smallest number by which 2401 must be divided to obtain a perfect cube is: a) 7 b) 6 c) 5 d) 9 10. If the volume of a cube is 64 cm3 , then the edge of the cube is: a)4 cm b)8 cm c)16 cm d) 6 cm 11. 3 125 × 64 is: a) 10

b) 20

c) 30

d) 40

24389 3 m , then the edge of the cube is: 216 23 26 29 32 m m m m a) b) c) d) 6 6 6 6 13. The ones digit in the cube root of the cube number 4096 is: a) 2 b) 6 c) 4 d) 9 12. If the volume of a cube is

14. The number by which 10000 must be divided in order to get a perfect cube is: a) 2 b) 5 c) 10 d) 100

152

IL Foundation Series Class 8

15. The cube of an odd natural number is: a) Even b) Odd c) Either even or odd

d) Prime number

16. The cube of an even natural number is: a) Even b) Odd c) Either even or odd

d) Prime number

17. The cube of which of the following negative integers ends with 3? a) -11 b) -20 c) -14 d) -17

512 18. The cube of which of the following rational numbers is ?

2197

5 4 8 9 a) b) c)

d) 6 7 13 11 1 19. Which of the following decimals have the cube as ?

8 a) 0.7

b) 0.5

c) 1.0

d) 1.2

20. Which of the following statements is false? a) -216 is a perfect cube.

2 b) The cube of is 8 .

5 125 c) If a and b are integers such that a 2 > b 2 , then a 3 > b3 . d) The cube of all odd natural numbers is odd.

21. Which of the following is the perfect cube of a number? a) 289 b) 537 c) 414

d) 729

22. For which of the following can we conclude that the cube root of 343 is 7? a) 343 = 1+ 7 +19 + 37 + 61+127 b) 343 = 1+ 7 +19 + 37 + 91+127 +169 c) 343 = 1+ 7 +19 + 37 + 61+ 91+127 d) 343 = 1+ 7 +19 + 37 + 61+127 +169 23. The units and tens digit of the cube root of 4913 is: a) 1 and 7 b) 9 and 1 c) 7 and 1

d) 2 and 7

24. The cube root of a number gives 12 as the answer through prime factorisation. The number is: a) 1728 b) 3375 c) 9261 d) 4096

2 25. The cube of 1 is: 3 12 17 2 a) 27 b) 4 27

c) 1

17

29

d) 1

4 9

153

CUBES AND CUBE ROOTS

26. Find the difference between the square root and the cube root of the number ( 729 × 64 ) . a) 646

b) 252

c) 144

d) None of these

II. FILL IN THE BLANKS 1. The cube root of 512 is x, then 3 x is _________.

2.

3

−512 = _____________ .

343

3. If 3 x = y then x = _______________ .

3

 2  2  4.    = _________________ .  5  

5.

3

125 = ______________ .

216

6.

3

0.027 = ____________ .

7.

3

−1 =_____________ .

8. Cube of a positive number is __________.

9. Cube root of ______ is equal to 9.

10. 3 43 × 63 = __________.

III. SUBJECTIVE QUESTIONS 1. The volume of a cubical box is 216 m3 . Find the length of the longest rod that can be placed in the box.

2. Is 72k a perfect cube, if k = 2? If yes, then find cube root of 72k .

3. The side of a cube is 11 cm . Find its volume.

4. Find the cube root of the predecessor of the Hardy-Ramanujan number.

5. Is 8000 a cube of an odd or an even number?

6. The volume of a metallic cube is 42875 cm3 , find the side of the cube.

7. Find the value of 3 0.1 × 0.1 × 0.1 × 10 × 10 × 10 .

154

IL Foundation Series Class 8

8. Which is the smallest number by which 243 must be divided to get a perfect cube?

9. What will be the units digit of the cube of 532?

10. I am the cube of a number, and I am also double the square of the same number, who am I?

155

7

COMPARING QUANTITIES

7.1 RATIO AND PROPORTION 7.1.1 Ratio Suppose we compare the weights of two girls, Rashmi and Tanya, weighing 16 kg and 23 kg, respectively. We can do it in two ways: i) The difference in their weights is (23 - 16) kg or 7 kg. We say that Tanya is heavier than Rashmi by 7 kg.

This is known as comparison by difference. ii) We can write,

Quantity A Quantity B

=

16 23

16 of Tanya’s weight.

23 This is known as comparison by division. i.e., Rashmi’s weight =

When we compare two quantities of the same kind by division, we say that we have formed a ratio of the two quantities. We use the symbol ’:’ to express a ratio.

Ratio means comparing two quantities.

Ratio formula: A ratio is a quantitative relationship between two or more quantities. The general formula for a ratio is expressed as

Ratio =

Quantity A Quantity B

Example: If the number of apples (Quantity A) is 5 and the number of oranges (Quantity B) 5 is 3, the ratio of apples to oranges is 3 .

This can also be written as 5:3.

7.1.2 Proportion When we have four quantities, and the ratio of the first to the second matches the ratio of the third to the fourth, we say that the four quantities are in proportion. This relationship is denoted a c as a:b :: c:d or = . b d The terms ’a’ and ’d’ are referred to as the extremes, while ’b’ and ’c’ are identified as the means.

156

IL Foundation Series Class 8

Example: The ratio of length to breadth of a rectangle is 3:4. If the length of the rectangle is 15 cm, find its breadth.

Solution: Let the length be represented by 3x and the breadth by 4x, where x is a constant.

Given that the length is 15 cm, we can set up a proportion:

15 3x = 4x breadth

Therefore, 3x × breadth = 4x × 15 breadth = 20 cm So, the breadth of the rectangle is 20 cm.

7.2 PERCENTAGE The word per cent is an abbreviation of the Latin phrase ’per centum’, which means per hundred or hundredths.

Thus, the term 'per cent' means per hundred (or) for every hundred. The symbol % represents per cent.

Percentage formula: Percentages express a proportion out of 100 and are calculated using the following formula: Part × 100 Whole This formula represents the relationship between a part and the whole in terms of a percentage.

Percentage =

Example: If you have 25 blue marbles out of a total of 100 marbles, the percentage of blue marbles is calculated as The Percentage of blue marbles =

25 × 100 = 25% 100

So, 25% of the marbles are blue.

35 .

100 Thus, a fraction with its denominator 100 is equal to that per cent, with the numerator being the same as the percentage.

Percent as a fraction: We have 35% = 35 hundredths =

8 12 = 8%, = 12% etc.

100 100 To convert a fraction into a per cent, we multiply the fraction by 100 and put the per cent sign (%).

So,

4 4 = × 100 % = 80% 5 5 To convert a per cent into a fraction, we divide it by 100 and remove the per cent sign (%).

Thus,

157

COMPARING QUANTITIES

25 13 , 13% = etc.

100 100 Percent as a ratio: A per cent can be expressed as a ratio where the first term is the given per cent and the second term is 100. ∴ 25% =

For example, 8% = 36% =

8 2 = = 2 : 25 100 25

36 9 = = 9 : 25 100 25

Percent in decimal form: To convert a given per cent into decimal form, we express it as a fraction with a denominator of 100, and then we represent this fraction as a decimal.

For example: i)

65% =

65 = 0.65 100

7.4 = 0.074 100 The percentage increase or decrease of a quantity is calculated as below: ii) 7.4% =

Percentage increase or decrease =

Increase or Decrease ×100 Initial value

SOLVED EXAMPLES Example 1: Find 12% of ` 1200.

Solution: 12% of ` 1200 = `

12 × 1200 = ` 144 100

Example 2: If 23% of a is 46, then find a.

Solution: We have 23% of a =

23 100 × a = 46 ⇒ a = × 46 ⇒ a = 200 100 23

Example 3: 72% of 25 students are good at mathematics. How many are not good at it?

Solution: We have, Number of students who are good in mathematics = 72% of 25 =

72 × 25 = 18 100

∴ Number of students who are not good at mathematics = 25 - 18 = 7 158

IL Foundation Series Class 8

Example 4: If Chameli had `600 left after spending 75% of her money, how much did she have in the beginning?

Solution: Suppose Chameli had ` x in the beginning then, money spent by Chameli = 75% of x = `

75 3x ×x=` 100 4

∴ Money left with Chameli = ` x `

3x 4

4x-3x x =` 4 4

But it is given that she had ` 600 left after spending 75% of her money. ∴

x = 600 ⇒ x = 600 × 4 = 2400. Hence, Chameli had ` 2400.

4

Example 5: Find the percentage of pure gold in 22-carat gold, if 24-carat gold is one hundred per cent pure gold.

Solution: In 22-carat gold, 22 parts out of 24 parts is pure. ∴ Percentage of pure gold in 22 carat gold =

22 2 × 100 % = 91 %.

24 3

Example 6: The salary of an officer has been increased by 50%. By what percentage must the new salary be reduced to restore the original salary?

Solution: Let the original salary be `100.

Then, increase in the salary = 50% of `100 = `50.

Salary after increment = `150.

Now, in order to restore the original salary, a reduction of `50 should be made on `150.

Thus, reduction on `150 = `50 50 50 1 ⇒ Reduction `100 = × 100 = 33 150 150 3 1 Hence, reduction on new salary = 33 %.

3 Reduction on ` 1 = `

159

COMPARING QUANTITIES

7.3

PROFIT, LOSS AND DISCOUNT

Cost price: The amount paid to purchase an article or the cost of making an article is known as its cost price. The cost price is abbreviated as CP. Selling price: The price at which an article is sold is known as its selling price. The selling price is abbreviated as SP. 7.3.1 Profit Profit: If the selling price (SP) of an article is greater than the cost price (CP), the difference between the selling price and cost price is called profit.

Thus, if SP > CP, then Profit = SP - CP ⇒ SP = CP + profit ⇒ CP = SP - profit Profit percentage: The profit percentage is the profit that would be obtained for a CP of ` 100.

Profit × 100 CP Thus, in case of profit or gain (i.e. if SP > CP), i.e. Profit per cent =

i) Profit = SP - CP ii) SP = Profit + CP iii) CP = SP - Profit iv) Profit per cent = v) Profit =

Profit ×100 CP

CP × Profit % 100

vi) SP = CP + Profit ⇒ SP = CP +

160

Profit % × CP 100

⇒ SP =

100 + Profit% 100

vii) CP =

100 × SP 100 + Profit%

× CP

IL Foundation Series Class 8

7.3.2 Loss If the selling price (SP) of an article is less than the cost price (CP), the difference between the cost price (CP) and the selling price (SP) is called a loss.

Thus, if SP < CP, then Loss = CP - SP ⇒ CP = SP + Loss ⇒ SP = CP - Loss Loss percentage: The loss percentage is the loss that would be made for a CP of ` 100.

Loss × 100 CP Thus, in the case of a loss (i.e., when SP < CP).

That is, the loss percentage =

We have, i) Loss = CP - SP ii) SP = CP - Loss iii) CP = SP + Loss iv) Loss % =

Loss × 100 CP

CP × Loss % 100 vi) SP = CP - Loss v) Loss =

CP × Loss % 100 100 - Loss % ⇒ SP = × CP 100 100 × SP vii)CP = .

(100 - Loss %) ⇒ SP = CP -

7.3.3 Discounts You might have noticed that when buying goods, there is a price marked on every article. This price is known as the marked price (MP) of the article. To clear stocks or increase sales, shopkeepers sometimes offer a certain percentage of rebate on the marked price for cash payments. This rebate is known as a discount.

Thus, SP = MP - Discount Also, rate of discount = Discount % =

Discount × 100 MP

161

COMPARING QUANTITIES

Discount % × MP 100 (100 - Discount %) × MP ⇒ SP = 100 Now SP = MP - Discount = MP -

⇒ MP =

100 × SP and Discount %= (100 - Discount%)

MP - SP MP

× 100

Example: Suresh went shopping for clothes at a store that is currently offering a 20% discount on selected T-shirts. The original price of a T-shirt is `500. Calculate the discounted price of the T-shirt after the 20% discount.

Solution: To find the discounted price or selling price, you can use the formula: SP or Discounted Price = MP - Discount SP or Discounted Price = MP -

Discount Percentage × MP 100

For the T-shirt with an original price of `500 and a 20% discount: 20 × 500 100 Discounted Price = 500 - ( 0.2 × 500) Discounted Price = 500 -

Discounted Price = 500 - 100 Discounted Price = `400

SOLVED EXAMPLES Example 1: A shopkeeper marks his goods at a price such that, after allowing a discount of 12.5% for cash payment, he still makes a profit of 10%. Find the marked price of an article which costs him `245.

Solution: We have, CP of the article = ` 245 , Gain = 10% SP = 100 + gain % × CP 100 110 SP = ` 100 + 10 × 245 = ` × 245 = ` 269.50 100 100 Now, SP = ` 269.50 , Discount %=12.5

162

IL Foundation Series Class 8

MP =

26950 100 × 269.50 100 × SP =` =` = ` 308 87.5 100 - 12.5 100 - discount %

Example 2: By selling a stool for ` 67.50, a carpenter loses 10%. How much per cent would he gain or loss by selling it for ` 82.50?

Solution: We have SP = ` 67.50 and loss % = 10 CP =

CP = `

100 × SP 100 - loss % 100 × 67.50 = ` 75 100 - 10

If SP = ` 82.50 , then SP > CP So gain = SP - CP = ` 82.50 - ` 75 = ` 7.50 Gain % =

7.4

7.50 Gain × 100 = × 100 = 10% 75 CP

TAXES

Sales tax: Sales tax is a tax imposed on the final sale of goods and services at the retail level. It is usually collected by the seller and remitted to the government.

Value Added Tax (VAT): VAT is a consumption tax that is assessed at each stage of the production and distribution chain. It is based on the value added at each stage.

Formula: VAT Rate × Original Cost 100 Goods and Services Tax (GST): GST is a comprehensive indirect tax that subsumes various taxes like central excise, service tax, and state-level VAT. It is levied on the supply of goods and services.

Total cost with VAT = Original Cost +

SOLVED EXAMPLES Example 1: The price of a T.V set inclusive of VAT is `13,530. If the rate of VAT is 10%. Find it basic price.

Solution: Let the basic price of T.V set be ` x, then VAT at the rate of 10% on ` x = `

10 ×x 100 163

COMPARING QUANTITIES

Thus, the sale price of the T.V set = ` x + x = ` 11x 10 10 It is given that the sale price of the T.V set is `13,530 ∴ 11x = 13530 10 13530 × 10 ⇒x= = 12300 11 Hence, the basic price of the T.V set is `12,300.

Example 2: If a washing machine is sold at `18000 but the basic price is `15000. Find the rate of VAT for this purchase.

Solution: Let the rate of VAT be x%.

We know that, VAT % =

(Selling price - Base price) × 100 Base Price

(18000-15000) × 100 15000 3000 = × 100 15000 =

= 20% Hence, the rate of VAT is 20%.

7.5

COMPOUND INTEREST

7.5.1 Simple interest If Principal = P, Rate = R% per annum, and Time = T (in years) then the simple interest (S.I) P×R×T is given by S.I = .

100 7.5.2 Compound interest If the borrower and the lender agree to fix a certain interval of time (such as a year, half-year or quarter-year), during which the amount (equal to the Principal + Interest) at the end of each interval becomes the principal for the next interval, then the total interest over all the intervals, calculated in this manner, is called compound interest and abbreviated as C.I. C.I. = Amount - Principal

164

IL Foundation Series Class 8

Computation of compound interest by using formulae: Let P be the principal and the rate of interest be R% per annum. If the interest is compounded annually, then the amount, A, and the Compound Interest, C.I., at the end of n years are given by, R n R n -1 and C.I = A - P = P 1 + 100 100 Computation of compound interest when the interest is compounded half-yearly.

A=P 1+

2n A= P 1 + R and C.I = A - P 200

Compound interests can also be used to find an increase or decrease in population.

Let P be the population of a city or town at the beginning of a certain year. If the population grows at a rate of R1% during the first year and R2% during the second year, then the R1 R2 population after 2 years = P 1 + × 1+ 100 100 This formula may also be extended for more than 2 years.

Let P be the population of a city or town at the beginning of a certain year. If the population R n decreases at a rate of R% per annum, then population after n years = P 1 100 The difference between CI and SI after 2 years can be calculated by (shortcut) Difference = P

R 2 100

SOLVED EXAMPLES Example 1: Find the compound interest on ` 1000 for two years at 4% per annum.

Solution: Principal for the first year = ` 1000 PRT 1000 × 4 × 1 = ` 40 I = 100 100 Amount at the end of first year = ` 1000 + ` 40 = ` 1040 Interest for the first year = `

1040 × 4 × 1 = ` 41.60 100 Principal for the second year = ` 1040 + ` 41.60 = ` 1081.60 Interest for the second year = `

Compound Interest = (1081.60 - 1000) = 81.60 165

COMPARING QUANTITIES

Example 2: Maria invests 93750 at 9.6% per annum for 3 years, and the interest is compounded annually. Calculate: i) the amount standing to her credit at the end of the second year ii) the interest for the third year Solution: i) We have, principal for the first year = 93750 Rate of interest = 9.6% per annum 93750 × 9.6 × 1 = 9000 100 Amount at the end of the first year = 93750 + 9000 = 102750 Interest for the first year =

102750 × 9.6 × 1 = 9864 100 Amount at the end of second year = (102750 + 9864)= 112614 Interest for the second year =

ii)Principal for the third year = 112614 112614 × 9.6 × 1 = 10810.94 100 Example 3: Find the compound interest on 12000 for 3 years at 10% per annum compounded annually.

Interest for the third year =

Solution: We know that the amount A at the end of n years at the rate of R% per annum when the interest is compounded annually is given by A=P 1+

R n 100

Here, P = 12000 R =10% per annum and n = 3 Amount after 3 years = P 1 + A=

12000 × 1 +

10 3 = 100

R 3 100

12000 × 1 +

1 3 11 11 11 = 12000 × × × = 15972 10 10 10 10

Now, compound interest = A - P = 15972 - 12000 = 3972

166

IL Foundation Series Class 8

Example 4: Compute the compound interest on 12000 for 2 years at 20% per annum when compounded half-yearly.

Solution: We know amount A at the end of n years at the rate of R % per annum when the interest is compounded half-yearly is given by A=P 1+

R 2n 200

Here, principal, P = 12000, R = 20% per annum and n = 2 years Amount after 2 years = P 1 +

R 2n 200

12000 1 +

20 2×2 200

=

= 12000 × 1 +

= 12000 ×

11 4 10

= 12000 ×

14641 = 17569.20 10000

1 4 10

Compound interest = 17569.20 - 12000 = 5569.20 Example 5: Find the compound interest at the rate of 10% per annum for four years on the principal, which in four years gives ` 1600 as simple interest at the rate of 4% per annum.

Solution: Let ` P be the principal This principal gives ` 1600 as S.I in four years at the rate of 4% per annum.

P=

S.I × 100 1600 × 100 =` = `10000 RT 4×4

Now, we have P = ` 10,000, R = 10% and n = 4 R n 10 4 = `10000 × 1 + 100 100 4 11 4 = ` 10000 1 + 1 = ` 10000 × = ` 14641 10 10 Compound Interest = ` 14641 - ` 10000 = ` 4641 Amount after 4 years = P 1 +

167

COMPARING QUANTITIES

QUICK REVIEW • • •

Quantity A = a = a: b b Quantity B a c Proportion⇒ = = a:b : : c:d b d Part Percentage = × 100 Whole Ratio =

• Gain or loss is always based on the CP Gain = SP - CP Gain Gain % = × 100 % CP 100 + Gain% SP = × CP 100 •

Gain or loss is always based on the CP Loss = CP - SP Loss × 100 % CP SP = 100 - Loss % × CP 100 Loss % =

•

The discount is allowed on the marked price SP = Marked price - Discount

•

n Amount after n years when interest is compounded annually is given by A = P 1 + R 100 Compound Interest = Amount - Principal

If the present population of a place is P and it decreases at R% per annum, then population after R n n years = P 1 100

WORKSHEET - 1 I.

RATIO, PROPORTION AND PERCENTAGE 1.

168

Write each of the following as per cent. i)

7 25

ii)

14 625

iii) 0.8

iv) 0.005

IL Foundation Series Class 8

2.

7 125 Convert the following percentages to fractions and ratios. v) 15:16

vi) 111:125

vii)

i) 25%

ii) 0.25%

iii) 0.3%

4.

(5x + 3y) . (x–2y) If 4P = 7Q = 21R, then what is P:Q:R ?

5.

A B C If 3 = 2 = 5 , then what is the value of ratio (C + A)2 : (A + B)2 : (B + C)2 ?

3.

If (x + y):(x - y) = 11:1, find value of

1 1 1 1 6. A sum of ` x was divided between A, B, C and D in the ratio 3 : 5 : 6 : 9 . If the difference between the shares of B and D is ` 832, then find the value of x.

7.

If A exceeds B by 40%, B is less than C by 20%, then A:C is:

8. A is 120% of B and B is 65% of C. If the sum of A, B and C is 121.5, then what is the value of C - 2B + A? 9. The price of sugar is increased by 20%. By what per cent must the consumption of sugar be decreased so that the expenditure on sugar may remain the same?

10. The ratio of the number of cans of orange, pineapple, and mixed fruit juices kept in a store is 8:9:15. If the store sells 25%, 33.33% and 20% of orange, pineapple and mixed fruit juices cans, respectively, then what is the ratio of number of cans of these juices in the remaining stock?

11. In an examination, B obtained 20% more marks than those obtained by A, and A obtained 10% less marks than those obtained by C. D obtained 20% more marks than those obtained

by C. By what percentage are the marks obtained by D more than those obtained by A? 12. Two numbers are 40% and 80% lesser than a third number. By how much per cent is the second number to be enhanced to make it equal to the first number?

13. The price of sugar is increased by 18%. A person wants to increase the expenditure by 12% only. By what per cent, correct to one decimal place, should he decrease his consumption?

14.

Radha earns 22% of her investment. If she earns `187, then how much did she invest?

15.

x is 5% of y, y is 24% of z. If x = 480, find the values of y and z.

16. Deepti attended school for 216 days in a full year. If her attendance is 90%, find the number of days the school was open.

17.

Find: i) 16.5% of 5000 metre

ii) 25% of 10 kg

iii) 25% of `1000

18. A cricketer scored a total of 62 runs in 96 balls. He hit 3 sixes, 8 fours, 2 twos and 8 singles.

What percentage of the total runs came in

i) Sixes

ii) fours

iii) two

iv) singles 169

COMPARING QUANTITIES

II. PROFIT, LOSS, DISCOUNT AND TAXES 1. A retailer buys a radio for `225. His overhead expenses are `15. If he sells the radio for `300, determine his profit per cent.

2.

A student buys a pen for `90 and sells it for `100. Find his gain and gain per cent.

3. Ramesh bought two boxes for `1300. He sold one box at a profit of 20% and the other box at a loss of 12%. If the selling price of both boxes is the same, find the cost price of each box.

4. Rohit buys an item at a 25% discount on the marked price. He sells it for `660, making a profit of 10%. What is the marked price of the item?

5. Vikram bought a watch for `825. If this amount includes 10% VAT on the list price, what was the list price of the watch?

6. Sunita purchases a bicycle for `660. She has paid a VAT of 10%. Find the list price of the bicycle.

7. A shopkeeper gives an 11% discount on a television set, and the cost price of it is `22,250.

Then, find the marked price of the television set.

8. The marked price of an almirah is `3000. The shopkeeper gave a 12% discount on it. Find the total discount and selling price of the almirah.

1 9. After allowing a discount of 7 % on the marked price, an article is sold for ` 555. Find its 2 marked price.

10. Jasmine allows a 4% discount on the marked price of her goods and still earns a profit of 20%. What is the cost price of a if the marked price is `850?

11. The cost of furniture inclusive of VAT is `7150. If the rate of VAT is 10%, find the original cost of the furniture.

12. A colour TV is available for `13440 inclusive of VAT. If the original cost of the TV is `12000, find the rate of VAT. III. COMPOUND INTEREST 1. Find the compound interest on `64000 for 1 year at the rate of 10% per annum compounded quarterly.

1 2. Find the amount and the compound interest on ` 8000 for 1 years at 10% per annum, 2 compounded half-yearly.

3. What will `125000 amount to at the rate of 6% if the interest is calculated after every four months?

170

IL Foundation Series Class 8

4. At what rate per cent compound interest per annum will ` 640 amount to ` 774.40 in 2 years?

5. The value of a refrigerator, which was purchased 2 years ago, depreciates at 12% per annum. If its present value is ` 9680, how much was it purchased for?

6. The population of a town increases at the rate of 50 per thousand. If its population after 2 years will be 22050, find its present population.

7. The population of a town was decreasing every year due to migration, poverty, and unemployment. The present population of the town is 6,31,680. Last year, the migration was 4%, and the year before last, it was 6%. What was the population two years ago?

8. Abha purchased a house from Avas Parishad on credit. If the cost of the house is `64000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one and a half-years.

9. Find the amount of ` 50,000 after 2 years, compounded annually. The rate of interest is 8% p.a. during the first year, and 9% p.a. during the second year.

10. Find the amount to be paid at the end of 17 months on ` 1,800 at 8% per annum compounded annually.

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1.

2.

3.

4.

5.

3 as a per cent is 5 a) 30%

b) 40%

c) 45%

d) 60%

6:5 when expressed as a percentage is 1 % b) 90% c) 120% 3 If 5% of a number is 9, then the number is a) 83

a) 45

b) 90

c) 135

d) 6.5%

d) 180

What per cent of 90 is 120?

1 % 3 What per cent of 10 kg is 250 g?

1 % 3

a) 75%

b) 33

c) 133

a) 25%

b) 5%

c) 10%

d) None of these

d) 2.5%

171

COMPARING QUANTITIES

6.

7.

8.

9.

If 40% of x = 240, then x = a) 60

b) 600

c) 6000

d) 960

b) 480

c) 600

d) 560

b) 100x

c)

2 1 is ?

7 35 b) 10%

c) 20%

If (180% of x) ÷ 2 = 504, then x = a) 400 x% of y is y% of a) x

x

100

d)

y 100

What per cent of a) 2.5%

d) 25%

10. Raju buys a toy for ` 75 and sells it for ` 100. His gain per cent is 1 1 % d) 37 % 3 2 11. A bat is bought for ` 120 and sold for ` 105. Then the loss per cent is

a) 25%

b) 20%

c) 33

a) 15%

b) 12

1 % 2

c) 16

2 % 3

d) 14

1 % 5

12. Bananas are bought at 3 for ` 2 and sold at 2 for ` 3. Then the gain per cent is

a) 25%

b) 50%

c) 75%

d) 125%

13. On selling 100 pencils, a man gains the selling price of 20 pencils. His gain % is

a) 20%

b) 25%

c) 22

1 % 2

d) 16

2 % 3

6 14. The selling price of an article is 5 of the cost price. Then, the gain per cent is

a) 20%

b) 25%

c) 30%

d) 120%

15. The price of a watch, including 10% VAT, is ` 825. Then the basic price is

a) `540

b)`500

c) `750

d)`800

16. On selling a chair for `720, a man loses 25%. To gain 25%, it must be sold for

a)`900

b) `1200

c)`1000

d) `1400

17. The compound interest on ` 5000 at 8% per annum for 2 years, compounded annually is

172

a) `800

b) `825

c) `832

d) `850

IL Foundation Series Class 8

18. The compound interest on ` 6250 at 8% per annum for 1 year, compounded half-yearly is

a) `500

b) `510

c) `550

d) `560

1 19. The sum that amounts to ` 4913 in 3 years at 6 4 % per annum compounded annually is

a)`4096

b) `4000

c) `4085

d) `4075

20. Oranges are bought at 5 for ` 10 and sold at 6 for `15. The gain per cent is

a) 50%

b) 40%

c) 35%

d) 25%

21. By selling a radio for ` 950, a man loses 5%. What per cent shall he gain by selling it for ` 1040?

a) 4%

b) 4.5%

c) 5%

d) 9%

22. A number is first increased by 10% and then reduced by 10%. Then the number

a) does not change b) decreases by 1% c) increases by 1%

d) none of these

23. A period of 4 hours 30 minutes is what per cent of a day?

3 2 % b) 20% c) 16 % 4 3 24. One-third of 1206 is what per cent of 134?

a) 18

a) 3%

b) 30%

c) 20%

d) 19%

d) 300%

25. A number exceeds 20% of itself by 40. Then the number is

a) 50

b) 60

c) 80

d) 320

II. FILL IN THE BLANKS 1.

0.9 per cent can be expressed as _____.

2.

If the discount is 10%, an item bought for `90 is priced at ______.

3.

75% of 480 = x × 15, then x is ______.

4.

Discount per cent is always calculated on _______.

5.

C.I. and S.I for one year are ______.

6.

When CP. > SP., then there is always a ______.

7.

In case of C.I., the principal ______ every year.

173

COMPARING QUANTITIES

8.

The number of quarters in 2

1 years is ________.

2

R T - 1 =_____.

100 10. A city contains 40% males, 30% females, and the rest are children. The percentage of children is ________.

9.

P 1+

III. SUBJECTIVE QUESTIONS 1. An item marked at `1500 is offered in a sale at a discount of 10%. How much does the customer pay to buy it?

2. A fan is sold at `1615 after allowing a discount of 5% on the marked price. Find the marked price.

3.

A refrigerator is marked for `12000 and is sold for `10,200. Find the discount per cent.

4.

What single discount is equal to successive discounts of 20% and 10%?

5.

What discount per cent is offered in a sale if the banner reads buy 3 get 2 free?

6. Marked price of a pair of shoes is ` 900. A customer paid `54 as S.T. for it. What is the rate of S.T.?

7.

After giving a discount of 5% an item is sold for `190. What was its marked price?

8. If interest is compounded half-yearly for 18 months, what is the number of conversion periods?

9. What is the difference between simple interest and compound interest compound annually on a sum of `10000 with a rate of interest of 10% for 2 years?

121 10. A sum compounded annually becomes times itself in 2 years. Determine the rate of 100 interest per annum.

174

8

8.1

ALGEBRAIC EXPRESSIONS AND IDENTITIES

I NTRODUCTION TO ALGEBRAIC EXPRESSIONS AND POLYNOMIALS

8.1.1 Introduction to algebraic expressions Variable

A symbol which can take various numerical values is called a variable or literal.

Example: p, q, r, a, b, c, etc.

Constant

A symbol with a fixed value is called a constant.

Example: In 3 + x, 3 is a constant, and x is a variable.

If we say 'a' is a constant in a + x, then 'a' takes a fixed value.

Term

Constants alone or variables alone or their combinations by operation of multiplication or division are called terms. y 2 Example: 5, y, 4 z, 5 x 2 yz, , , etc. z x Constant term

A term of an expression having no variable is called a constant term.

3 5 Example: 2, , , 3 , etc.

4 6 Algebraic Expression

The combination of terms obtained by the fundamental operations ( +, −, ×, ÷ ) is called an algebraic expression.

Example: 5 x − 3, 3 − 2y, 4a, 5 ÷ y .

Terms 5x

‒

3

Coefficient Variable

Constant

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ALGEBRAIC EXPRESSIONS AND IDENTITIES

In the above expression 5 x − 3, ' x ' is a variable whose value is unknown, and it can take any value.

'5’ is known as the coefficient of x.

'3’ is the constant term that has a definite value.

Equality of two algebraic expressions

Two algebraic expressions are said to be equal when they have the same terms irrespective of the order.

Example: 2 x 2 + y 3 and y 3 + 2 x 2 are equal expressions.

8.1.2 Introduction to polynomials •

n expression containing only one term in which powers of variables are non-negative integers A is called a monomial.

Example: 4 xyz, 2l 2 m 2 , 8 pq , etc.

Note: Every monomial is a term, but every term need not be a monomial.

Example: 2x is a monomial and also a term.

2 2 But is only a term, not a monomial, because = 2 x −1, where the power of 'x' is a negative x x integer.

•

An expression containing two monomials is called a binomial.

Example: 2 + x, 3y + 4 z , etc.

•

An expression containing three monomials is called a trinomial.

Example: x + y − z, 3 xz − 4 xy + 2 zy , etc.

•

An expression containing one or more monomials is called a polynomial.

Example: 2a − 4 b, 5 x + y + z , etc.

•

An expression containing one or more terms is called a multinomial.

4 Example: 2 + , 3 x + y − z , etc. x Note: All polynomials are multinomials but every multinomial need not be a polynomial.

Factors

Each term of an algebraic expression is a product of one or more numbers and/or literals. These numbers or literals are known as factors of the term.

Example: 6 = 2 × 3, where 2, 3 are called factors of 6, 5 xy = 5 × x × y , where 5, x, y are factors of 5xy.

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Coefficient

In a product containing two or more factors, each factor is called the coefficient of the product of the other factors.

Example: In 6x, 6 is the numerical coefficient of 'x' and 'x' is the literal coefficient of 6.

Note: When the numerical coefficient of a term is +1 or -1, there is no need to mention 1.

Example: The coefficient of x 2 in x 2 + 3 x + 5 is 1, the coefficient of xy in 5 x 2 + 7 xy + 10 y 2 is 7.

Note: The degree of zero polynomial is not defined.

Degree of a monomial

The degree of a monomial is the sum of the powers of variables involved in it.

Example: Degree of 5 x 2 y is 3.

Note: Every non-zero number is considered a monomial with degree zero.

Degree of polynomial

The greatest degree of terms in a polynomial is called the degree of polynomial.

Example: 1) Degree of 5 x 2 + 6 x 3 + 7 x + 2 is 3.

(

2) Degree of x 3 + x 4

) is 4 × 2 = 8.

2

Note: The degree of a non-zero constant polynomial is zero.

Example: Consider the constant '6'. It can be written as 6x° The degree of 6x° is 0 because the highest power of x in this expression is 0.

Like terms

Terms which contain the same literal factors are called like terms or similar terms.

In like terms, the numerical coefficient may be different.

Example : x, 7 x, 9 x.

Example : 4 x 2 yz, -5x 2 yz, Unlike terms

3 2 x yz.

4

The terms which do not have the same literal factors are called unlike terms.

Example: 3 x, 4y; 6 x 2 , 5 xy.

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ALGEBRAIC EXPRESSIONS AND IDENTITIES

SOLVED EXAMPLES Example 1: Find the number of terms in each of the following expressions and mention the name of the polynomial. i) 3 x 2 − 4y + 2

ii) 4abc2 − 7 xy

Solution: i) The expression 3 x 2 − 4y + 2 has 3 terms 3 x 2, −4y, and 2. This is a trinomial because it contains three terms. ii) The expression 4abc2 − 7 xy has 2 terms 4abc2 and −7xy . This is a binomial because it contains two terms.

Example 2: Write the numerical coefficient of the following. −3 i) 4 x 2 y ii) −6a 2b 2 c2 iii) abc 4 Solution: i) The numerical coefficient of 4 x 2 y is 4. ii) The numerical coefficient of −6a 2b 2 c2 is −6.

3 3 iii) The numerical coefficient of − abc is − .

4 4 Example 3: Find the coefficient of x in each of the following. i) 6x

ii) 10lmn

iii) −x

Solution: i) The coefficient of x in 6x is 6. ii) The expression 10lmn does not contain x, so the coefficient of x is 0. iii) The coefficient of x in -x is -1.

Example 4: W rite the degree of the following expressions.

2 i) 2 x 3 − 5 x 2 + 3 x + 2 ii) a + b + c − 3abc 3 Solution: 8 i) The degree of the expression x 3 − 5 x 2 + 3 x + 2 is 3, which is the highest power of x.

3 ii) The degree of the expression a + b + c − 3abc is 3, which is the total number of variables in the term with the most variables.

Example 5: State the coefficient and the degrees of the following polynomials. i) 10 x5

ii) −8

Solution: i) The polynomial 10 x5 has a coefficient of 10 and has a degree of 5.

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ii) The polynomial −8 has a coefficient of −8 and a degree of 0 because it is a constant term.

8.2 ADDITION OF ALGEBRAIC EXPRESSIONS The addition of algebraic expressions means adding the like terms of the expressions.

8.2.1 Methods of addition of algebraic expressions Horizontal method

In this method, like terms should be added and unlike terms should be written separately by using associative law of addition.

Example : Add 3 x + 4y and x − 2y .

Solution: 3 x + 4y + x − 2y = 3 x + x + 4y − 2y = 4 x + 2y Example : Add 9 x 2 − 4 x + 5 and −3 x 2 + 2 x − 1 Solution:

( 9x − 4x + 5 ) + ( −3x + 2x − 1 ) 2

2

= 9x2 − 4x + 5 − 3x2 + 2x − 1 = x 2 ( 9 − 3 ) + x ( −4 + 2 ) + 4 = x 2 ( 6 ) + x ( −2 ) + 4 = 6x2 − 2x + 4 Vertical Method

i) In this method, the expressions to be added are written one below the other. ii) The like terms of each type are placed in separate columns. iii) The sum will be written below that column. iv) If a particular like term is absent in an expression, the place is left vacant.

Examples: 1.

4a

−

6b

+

0

0

+

2b

+

4c

5a

+

0

−

2c

9a

−

4b

+

2c

2.

x2 y

−

4x2

−

xy 2

0

+

5x2

− 5 xy 2

7 x2 y

+

2x2

+

8x2 y

+

3x2

− 6 xy 2

0

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ALGEBRAIC EXPRESSIONS AND IDENTITIES

8.3 SUBTRACTION OF ALGEBRAIC EXPRESSIONS The difference of two like terms is the difference in the numerical coefficient of the two like terms.

8.3.1 Additive inverse of a number The additive inverse of any number is obtained by simply changing its sign, so the additive inverse of a number is also called the negative of that number.

Example: The additive inverse of 10 is ( −10 ).

8.3.2 Additive inverse of an expression The additive inverse or the negative of an expression is obtained by replacing each term of the expression by its additive inverse.

Example: The additive inverse of 2 x − 1 is −2 x + 1.

To subtract 1st expression from the 2nd expression, the additive inverse of the 1st expression should be added to the 2nd expression.

If A and B are two algebraic expressions, then A − B = A + ( − B ).

Example: Subtract 3a + 5b from 8a + 7 b.

Solution: ( 8a + 7b ) − ( 3a + 5b ) = ( 8a + 7b ) + ( −3a − 5b ) = 8a + 7 b − 3a − 5b = 8a − 3a + 7 b − 5b = 5a + 2b 8.3.3 Methods of subtraction of algebraic expressions Like additions, subtraction can also be done in two ways: i) Horizontal method ii) Vertical method Example: Subtract 3a + 4b; 2a + b .

Solution: Horizontal method 3 a + 4 b − ( 2a + b ) = 3a + 4b − 2a − b = 3a − 2a + 4b − b = a + 3b 180

IL Foundation Series Class 8

Vertical method 3a

+

4b

2a

+

b

(-)

(-) a

+

3b

SOLVED EXAMPLES Example 1: Add 9 x 2 − 4 x + 5 and 3 x 2 + 2 x − 1 .

Solution: 9x2 4x 5

3x2 2x 1

9x2 4x 5 3x2 2x 1 9x2 3x2 x2 9 3

4x 2x x

4 2

5 1 ∵grouping like terms

4

12 x 2 2 x 4 Example 2: Subtract 2 x − x 2 + 5 from −4 x − 3 + 7 x 2.

Solution:

( −4x − 3 + 7 x ) − ( 2x − x + 5 ) = −4x − 3 + 7 x − 2x + x − 5 = ( 7 x + x ) + ( −4 x − 2 x ) + ( −3 − 5 ) 2

2

2

2

2

2

= x 2 ( 7 + 1 ) + x ( −4 − 2 ) + ( −8 ) = x 2 ( 8 ) + x ( −6 ) + ( −8 ) = 8x2 − 6x − 8

(

) (

)

∴ −4 x − 3 + 7 x 2 − 2 x − x 2 + 5 = 8 x 2 − 6 x − 8 Example 3: Add 7 x 2 − 4 x + 5, − 3 x 2 + 2 x − 1 and 5 x 2 − x + 9.

Solution:

( 7 x − 4x + 5 ) + ( −3x + 2x − 1 ) + ( 5x − x + 9 ) 2

2

2

= 7 x2 − 4x + 5 − 3x2 + 2x − 1 + 5x2 − x + 9

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ALGEBRAIC EXPRESSIONS AND IDENTITIES

(

)

= 7 x 2 − 3 x 2 + 5 x 2 + ( −4 x + 2 x − x ) + ( 5 − 1 + 9 )(grouping like terms ) = x 2 ( 7 − 3 + 5 ) + x ( −4 + 2 − 1 ) + ( 13 )(adding like terms ) = x 2 ( 12 − 3 ) + x ( 2 − 5 ) + 13 = 9 x 2 − 3 x + 13

1 1 5 −1 1 1 1 Example 4: Add 5 x 2 − x + , x 2 + x − and −2 x 2 + x − .

5 6 3 2 2 2 3 Solution: 1 5  1 2 1 1   2 1 2 1  5 x − x +  +  − x + x −  +  −2 x + x −  3 2  2 2 3  5 6  1 1 1  5 1 1    1 =  5x2 − x2 − 2x2  +  − x + x + x  +  − −  2 2 5  2 3 6    3 1    1 1 1 5 1 1 = x2  5 − − 2  + x  − + +  +  − −  2    3 2 5 2 3 6  −10 + 15 + 6   15 − 2 − 1   10 − 1 − 4  = x2  +   + x 2 30 6       5  11   12  = x2   + x  +  2  30   6  11 5 = x2 + x + 2 2 30 3 4 1 12 3 2 Example 5: Subtract x 2 y + y − x 2 yz from x 2 yz − xyz + x 2 y .

2 5 3 5 5 3 Solution: 3 2 4 1  12  3   x 2 yz − xyz + x 2 y  −  x 2 y + y − x 2 yz  5 3 5 3  5  2  12 2 3 2 2 3 2 4 1 2 = x yz − xyz + x y − x y − y + x yz 5 5 3 2 5 3

1 4  12   −3 2 2 2  3 =  x 2 yz + x 2 yz  +  x y + x y  − xyz − y 3 3 5  5   2  5 4  −3 2  3  12 1  +  − xyz − y = x 2 yz  +  + x 2 y  5  2 3 5  5 3 4  36 + 5  2  −9 + 4  3 = x 2 yz   + x y  − xyz − y 5  15   6  5

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IL Foundation Series Class 8

4  41  2  −5  3 = x 2 yz   + x y  − xyz − y 5  15   6  5 =

41 2 5 3 4 x yz − x 2 y − xyz − y 15 6 5 5

8.4 MULTIPLICATION OF ALGEBRAIC EXPRESSIONS 8.4.1 Multiplication of monomials We have the '× ' sign for multiplication. It need not be written between the product of a numeral and a literal number.

Example: 5 x = 5 × x Rules to be followed to multiply monomials

i) Numerical coefficient in the product = product of numerical coefficients in the monomials ii) Literal coefficient in the product = product of literal coefficients in the monomials These rules may be extended for the product of three or more monomials.

Example: 6ab × 4b = ( 6 × 4 ) ( ab × b ) = 24ab 2 1  5  Example: Multiply:  p2q  × ( −12 pq ) ×  p3  3  6  Solution: 1 2  5 3  p q  × ( −12 pq ) ×  p  3  6  1 5 = × ( −12 ) × p6q 2 3 6 =

( −5 × 2 ) p6q 2 = −10 p6q 2 3

3

 −4 2   −15 2 2  Example: Multiply: ( −12cd ) ×  c d × c d   3   2  Solution: −4 −15 2 2  c d  ( −12cd ) ×  c2 d  ×   3   2  = −12 ×

−4 −15 × × cd × c2 d × c2 d 2 3 2

= −120 c5 d 4 183

ALGEBRAIC EXPRESSIONS AND IDENTITIES

8.4.2 Multiplication of a monomial and a binomial We know the distributive property a ( b + c ) = ab + ac or ( b + c ) a = ba + ca .

This property gives us the method of multiplication of a monomial and a binomial. In order to multiply a binomial by a monomial, we must multiply each term of the binomial by the monomial.

This can be done in two ways: i)Horizontal method Example: 2 x ( 3 x + 4y ) = 2 x × 3 x + 2 x × 4y = 6 x 2 + 8 xy ii) Column method 3x

6x2

+

4y

×

2x

+

8xy

Example: If A = x + 2y, B = y − 2 z , and C = z + 2 x , then find the values of the following. i) A + 2 B

ii) 2 A − 3C

iii) A − 2 B + 3C

Solution: Given A = x + 2y, B = y − 2 z, C = z + 2 x ; i) A + 2 B = x + 2y + 2 ( y − 2 z ) = x + 2y + 2y − 4 z = x − 4 z + 4y ii) 2 A − 3C = 2 ( x + 2y ) − 3 ( z + 2 x ) = 2 x + 4y − 3 z − 6 x = −4 x + 4y − 3 z iii) A − 2B + 3C = x + 2y − 2 ( y − 2 z ) + 3 ( z + 2 x ) = x + 2y − 2y + 4 z + 3 z + 6 x = 7 x + 7 z iv) 2A + 3B − 5C = 2 ( x + 2y ) + 3 ( y − 2 z ) − 5 ( z + 2 x ) = 2 x + 4y + 3y − 6 z − 5 z − 10 x = −8 x + 7 y − 11z

184

iv) 2A + 3B − 5C

IL Foundation Series Class 8

8.4.3 Multiplication of a monomial and a polynomial Multiply each term of the polynomial by the monomial.

Example: Simplify: 2 x ( 3 x + 4y + 6 z ) Solution: 2 x ( 3 x + 4y + 6 z ) = 2 x × 3 x + 2 x × 4y + 2 x × 6 z = 6 x 2 + 8 xy + 12 xz 5 1 7  Example: Multiply  xy 2 − xy + x 2 y  and (xy).

4 6 2  Solution: 1 2  7 2 5  xy − xy + x y  ( xy ) 4 6 2  7 2 5 1 2 xy xy xy xy x y xy 2 4 6 7 2 3 x y 2

5 2 2 x y 4

1 3 2 x y 6

8.4.4 Multiplication of two binomials Multiply each term of the first binomial with each term of the second binomial and add the like terms in the product. Suppose ( a + b ) and ( c + d ) are two binomials. By using the distributive law of multiplication over addition, we can find their product as given below.

( a + b )×( c + d ) = a ×( c + d ) + b ×( c + d ) = (a×c )+(a×d )+(b×c )+(b×d ) = ac + ad + bc + bd.

Note: The product of two factors with the same sign is positive, and the product of two factors with the opposite signs is negative.

Example: 3 x × 2y = 6 xy −5 x × −2 x = 10 x 2

3 x × −2y = −6 xy Example: Find the product of ( 2 x + 1 ) and ( 3y − 2 ).

Solution:

( 2 x + 1 ) ( 3y − 2 ) = 3y ( 2 x + 1 ) − 2 ( 2 x + 1 ) = 6 xy + 3y − 4 x − 2 185

ALGEBRAIC EXPRESSIONS AND IDENTITIES

Example: Find the product of ( 2 x − 3 ) and ( 5 x + 4 ).

Solution:

( 2x − 3 ) × ( 5x + 4 ) = 5x ( 2x − 3 ) + 4 ( 2x − 3 ) = 10 x 2 − 15 x + 8 x − 12 = 10 x 2 − 7 x − 12

SOLVED EXAMPLES Example 1: If the length of a square is increased by 5 units and the breadth is decreased by 3 units, then find the area of the rectangle formed.

Solution: Let the side of the square be x.

The length of the square is increased by 5 units.

The length of the rectangle formed = x + 5 The breadth of the square is decreased by 3 units.

The breadth of the rectangle formed = x − 3 The area of the formed = length × breadth A = l × b = ( x + 5 )( x − 3 ) = x 2 − 3 x + 5 x − 15 = x 2 + 2 x − 15 Example 2: Find the volume of a rectangular box with given length = 2ax, breadth = 3by and height = 5cz.

Solution: Length = 2ax, breadth = 3by , height = 5cz The volume of rectangular box ( V ) = l × b × h V = 2ax × 3by × 5cz = ( 2 × 3 × 5 )( ax )( by )( cz ) = 30abc × xyz

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Example 3: Solve 4 xy × 5 x 2 y 2 × 6 x 3 y 3.

Solution: 4 xy × 5 x 2 y 2 × 6 x 3 y 3 =

( 4xy × 5x y ) × 6x y 2 2

3 3

= 20 x 3 y 3 × 6 x 3 y 3 = 120 × x 3 y 3 × x 3 y 3 = 120 x 6 y 6 Example 4: Multiply ( 7 xy + 5y ) by 3xy.

Solution:

( 7 xy + 5y ) × 3 xy = 7 xy × 3 xy + 5y × 3 xy = 21x 2 y 2 + 15 xy 2 Example 5: Multiply ( 2 x + 3y ) and ( 4 x − 5y ).

Solution:

( 2 x + 3y ) × ( 4 x − 5y ) = 2 x × ( 4 x − 5y ) + 3y ( 4 x − 5y ) = ( 2 x × 4 x − 2 x × 5y ) + ( 3y × 4 x − 3y × 5y )

(

) (

= 8 x 2 − 10 xy + 12 xy − 15y 2

)

= 8 x 2 − 10 xy + 12 xy − 15y 2 = 8 x 2 + 2 xy − 15y 2

(

)

1  1 Example 6: Multiply  x − y  and 5 x 2 − 4y 2 .

4  5 Solution:

(

(

)

)

(

1  1 1 1 2 2 2 2 2 2  x − y  × 5 x − 4y = x 5 x − 4y − y 5 x − 4y 4  5 4 5

(

)

1 1 1 1 = x × 5 x 2 − x × 4y 2 − y × 5 x 2 + y × 4y 2 5 4 4 5 4 5 = x 3 − xy 2 − x 2 y + y 3 5 4

)

1  4 5 1 ∴  x − y  × 5 x 2 − 4y 2 = x 3 − xy 2 − x 2 y + y 3 4  5 4 5

187

ALGEBRAIC EXPRESSIONS AND IDENTITIES

(

) (

)

Example 7: Multiply 3 x 2 + y 2 by x 2 + 2y 2 .

Solution: Column method 3x 2 + y 2 × x 2 + 2y 2 3x 4 + x 2 y 2

Multiplying 3x 2 + y 2 by x 2

+ 6 x 2 y 2 + 2y 4

Mulltiplying 3x 2 + y 2 by 2y 2

3x 4 + 7 x 2 y 2 + 2y 4

Adding the like terms

Horizontal method

( 3 x + y ) ( x + 2y ) = 3 x × ( x + 2y ) + y × ( x + 2y ) 2

2

2

2

2

2

2

2

2

2

= 3 x 2 × x 2 + 3 x 2 × 2y 2 + y 2 × x 2 + y 2 × 2y 2 = 3 x 4 + 6 x 2 y 2 + x 2 y 2 + 2y 4 = 3 x 4 + 7 x 2 y 2 + 2y 4 Example 8: Multiply ( 0.5 x − y ) by ( 0.5 x + y ).

Solution:

( 0.5 x − y ) × ( 0.5 x + y ) = 0.5 x ( 0.5 x + y ) − y ( 0.5 x + y ) = 0.5 x × 0.5 x + 0.5 x × y − y × 0.5 x + y × ( −y ) = 0.25 x 2 + 0.5 xy − 0.5 xy − y 2 = 0.25 x 2 − y 2

(

) (

)

Example 9: Multiply 2 x 2 − 4 x + 5 by x 2 + 3 x − 7 .

Solution: Horizontal Method

( 2x − 4x + 5 ) × ( x + 3x − 7 ) = 2x ( x + 3x − 7 ) − 4x ( x + 3x − 7 ) + 5 ( x + 3x − 7 ) 2

2

2

2

2

2

= 2 x 4 + 6 x 3 − 14 x 2 − 4 x 3 − 12 x 2 + 28 x + 5 x 2 + 15 x − 35 = 2 x 4 + 6 x 3 − 4 x 3 − 14 x 2 − 12 x 2 + 5 x 2 + 28 x + 15 x − 35 = 2 x 4 + 2 x 3 − 21x 2 + 43 x − 35

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IL Foundation Series Class 8

Column Method 2x2 - 4x + 5 x2 + 3x - 7 2x4 - 4x3+ 5x2 + 6x3 - 12x2 + 15x - 14x2 + 28x - 35 2x4 + 2x3 - 21x2 + 43x - 35

FINDING THE VALUE OF AN EXPRESSION FOR A GIVEN VALUE OF THE VARIABLE

8.5

We know that the value of an algebraic expression depends on the values of the variables within the expression. To find the value of an expression, we substitute the values of the variables in the expression and then simplify.

Consider the expression 2 x + 5y − 1. Let’s find the values for x = 4 and y = 2.

Step 1: Substitute the given values for x and y into the expression.

2 x + 5y − 1 = 2 ( 4 ) + 5 ( 2 ) − 1 Step 2: Evaluate the expression using the order of operations. = ( 2× 4 ) + ( 5×2 ) −1 = ( 8 ) + ( 10 ) − 1 = 18 − 1 = 17 Therefore, the value of 2 x + 5y − 1 for x = 4 and y = 2 is 17.

SOLVED EXAMPLES Example 1: Find the value of the following expressions when n = −2. (i) 3n − 2

(ii) 3n 2 + n − 2

Solution: (i) 3n − 2, substituting n = −2 into 3n − 2 , we get 3 ( −2 ) − 2 = −6 − 2 = −8 (ii) 3n 2 + n − 2, substituting n = −2 into 3n 2 + n − 2, we get 3(−2)2 + ( −2 ) − 2 = 3 ( 4 ) − 2 − 2 = 12 − 2 − 2 = 8 189

ALGEBRAIC EXPRESSIONS AND IDENTITIES

Example 2: Find the value of the following products ( x + 2y ) ( x − 2y ) at = x 1= ,y 0 .

Solution:

( x + 2y ) ( x − 2y ) = x ( x − 2y ) + 2y ( x − 2y ) = x × x − x × 2y + 2y × x − 2y × 2y = x 2 − 2 xy + 2 xy − 4y 2 = x 2 − 4y 2 When= x 1= , y 0, we get,

( x + 2y ) ( x − 2y ) = x 2 − 4y 2 = 12 − 4(0)2 =1-0=1

8.6 ALGEBRAIC IDENTITIES 8.6.1 Identity An identity is an equality which is true for all values of the variable. Clearly, an identity implies that the expressions on either side of the equality sign (=) are identical.

8.6.2 Standard identities We shall now study three standard identities. These identities are very useful in factorisation and simplification of algebraic expressions.

Identity 1

(a + b)2 = a 2 + 2ab + b 2 (or) a 2 + b 2 + 2ab i.e., Square of the sum of two terms = ( Square of the first term ) + ( Square of the second term ) + 2 × ( First term ) × ( Second term ) Example: (2 x + 3)2 = (2 x)2 + 2 × 2 x × 3 + 32 = 4 x 2 + 12 x + 9 Identity 2

(a − b)2 = a 2 − 2ab + b 2 (or) a 2 + b 2 − 2ab i.e., the square of the difference of two terms = ( Square of the first term ) + ( Square of the second term ) − 2 × ( First term ) × ( Second term ) Example: (x − 1)2 = x 2 − 2 × x × 1 + 12 = x 2 − 2 x + 1

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IL Foundation Series Class 8

Identity 3

( a + b ) ( a − b ) = a2 − b2 i.e., ( First term + Second term )( First term - Second term ) = ( First term )2 − ( Second term )2 Example: ( 2 x + 3y ) ( 2 x − 3y ) = (2 x)2 − (3y)2 = 4 x 2 − 9y 2

SOLVED EXAMPLES Use a suitable identity to find the following.

Example 1: ( 2 x + 3y)2 Solution: ( 2 x + 3y)2 = (2 x)2 + 2 ( 2 x )( 3y ) + (3y)2 = 4 x 2 + 12 xy + 9y 2  ( a + b )2 = a 2 + 2ab + b 2 

Example 2: ( 2 x − 3y)2 Solution: (2 x − 3y)2 = (2 x)2 − 2 ( 2 x )( 3y ) + (3y)2 = 4 x 2 − 12 xy + 9y 2  ( a − b )2 = a 2 − 2ab + b 2  Example 3: ( 2 x − 3y ) ( 2 x + 3y ) Solution: ( 2 x + 3y ) ( 2 x − 3y ) = ( 2 x )2 − ( 3y )2   ( a + b ) ( a − b ) = a 2 − b 2    2 2 = 4 x − 9y 4  Example 4:  x 2 + 3   3 Solution: 2

2

2

4 2  4 2 4 2 2  x + 3  =  x  + 2  x  ( 3 ) + (3) 3  3  3  =

16 4 x + 8 x 2 + 9  ( a + b )2 = a 2 + 2ab + b 2  9

191

ALGEBRAIC EXPRESSIONS AND IDENTITIES

5  3 5  3 Example 5:  x + y   x − y  6  4 6  4 Solution: 2

5  3 5  3  5  3  x + y   x − y  =  x  −  y  6  4 6  4  6  4

=

2

9 2 25 2  x − y   ( a + b ) ( a − b ) = a 2 − b 2  16 36

Example 6: (93)2 Solution: ( 93)2 = (90 + 3)2 = (90)2 + 2 ( 90 )( 3 ) + (3)2 = 8100 + 540 + 9 ( a + b )2 = a 2 + 2ab + b 2  = 8649 Example 7: Simplify the following by using ( a + b ) ( a − b ) = a 2 − b 2. ii) 1282 − 77 2

i) 68 × 72 Solution:

i) 68 × 72 = ( 70 − 2 ) ( 70 + 2 ) = (70)2 − (2)2  ( a + b ) ( a − b ) = a 2 − b 2  = 4900 − 4 = 4896 ii) 1282 − 77 2 = ( 128 + 77 ) ( 128 − 77 ) = 205 × 51 a 2 − b 2 = ( a + b ) ( a − b )  = 10455

QUICK REVIEW •

192

hile adding (or subtracting) polynomials, first look for like terms and add (or subtract) them; W then arrange the unlike terms.

IL Foundation Series Class 8

•

I n carrying out the multiplications of a polynomial by a binomial (or trinomial), we multiply term by term, i.e., every term of the polynomial is multiplied by every term in the binomial (or trinomial). Note that in such multiplication, we may get terms in the product which are like and must be combined.

•

n identity is an equality, which is true for all values of the variables in the equality. On the A other hand, an equation is true only for certain values of its variables. An equation is not an identity.

•

The following are the standard identities: (a + b)2 = a 2 + 2ab + b 2 (a − b)2 = a 2 − 2ab + b 2 (a + b)(a − b) = a 2 − b 2

WORKSHEET - 1 I.

INTRODUCTION TO ALGEBRAIC EXPRESSIONS AND POLYNOMIALS 1. Identify the terms and their coefficients for each of the following expressions.

i) 8 x 2 yz − 6 xy

ii) y 2 + y + 1

iv) 7 − ab + bc − ca

v)

x y + − xy 2 2

iii) 4 x 2 y 2 − 7 x 2 y 2 z 2 + z 2 vi) 0.3 x − 0.2 xy + 0.7

2. Classify the following polynomials as monomials, binomials, and trinomials. Which polynomials do not fit under any category? i) a + b

ii) 125

iii) z 4 + z 3 + z 2 + z

iv) 6 + a + 4b

v) 2a − 3a 2

vi) 3 x 2 + 4 x 3 + 2 x

vii) 3 x + 4y − 5 x

viii) 3a − 14a 2

ix) xy + yz + zt + tx

x) xyz

xi) x 2 y + xy 2

xii) 2 p + 2q

3. Identify the like terms in the following and write them in groups. i) 2 x, 3y, − 5 x, − 7 y, 3 xy iii) 3 x, 4 xy, − yz,

1 zy 2

ii) 3 xy, − 4 p2q 2 , 9ba, 7 p2q, 20 yx, 10ab iv) abc, ab 2 c, acb 2 , c2ab, b 2ac, a 2bc, cab 2

II. ADDITION OF ALGEBRAIC EXPRESSIONS 1. Simplify: i) p + 7 p − 2 p

ii) 6a + 5a − 3a

iii) 3 x − 2y − 3 z + 7 x − 5y + 9 z

193

ALGEBRAIC EXPRESSIONS AND IDENTITIES

2. Add the following expressions. i)

3 2 2 2 a b, a b 4 8

ii) −6abc, 2abc, − abc iii) 8a 2b 2 , 5a 2b 2 , − 6a 2b 2 , 2a 2b 2

3. From the following figure, find the lengths of PR, QS and PS in terms of x. If x = 2 cm, find the lengths of the above line segments.

P

4x

Q

R

3x

2x

S

4. Add the following expressions. i) u + v, u − v

ii) a 2 + b 2 , a 2 − b 2

iii) 8 p − 9q, 5q + 6 p

iv) 2 x 3 + 3 x 2 − x − 5, x 3 − 3 x 2 + 6, x 2 + 2 x − 5

5. Add the following expressions. i) a + 5b + 7c 2a

+

10b

+

c

ii)

2x2

−

3xy

+

y2

−7 x 2

−

5xy

−

2y 2

4x2

+

xy

−

6y 2

6. Add the following expressions.

2

2

2

2

i) 4 xy − 7 x y, 12 x y − 6 xy and 3 x 2 y + 5 xy 2 ii) 7 x 2 − 4 x + 5, − 3 x 2 + 2 x − 2 and 7 x 2 − 2 x − 9 iii)

2 3 −6 a, a, a 3 5 5

11 12 13 −11 12 13 xy + y + x, y − x − xy 2 5 7 2 5 5 7 3 1 2 5 3 3 7 2 1 3 5 v) x − x + , x + x − x + and x 2 − x − 2 2 2 3 2 4 3 2 2 III. SUBTRACTION OF ALGEBRAIC EXPRESSIONS iv)

1. Simplify the following. i) 7 p − ( −3 p )

ii) 8y − ( 3y )

iii) ( −9t ) − ( 5t )

ii) −3a from 7a

iii) −6xy from −2xy

2. Subtract. i) 4x from 10x

3. In the following problems, subtract the second expression from the first. i) 2a + 3b, a + b

194

ii) 3 x + 4y, x − 2y

iii) −3 x 2 + y 2 , 4 x 2 + 7 y 2

IL Foundation Series Class 8

4. In the following problems, subtract the third expression from the sum of the first two. i) 3a − 2 b + 4c, − 2a + b − 5c, 3a − 2 b + c ii) 4a 3 + a 2 + 5, 2a 2 − 5a + 6, 2a 3 − 3a 2 + 5a − 6 5. If x = 2a 2 − 5a + 7 and y = −3a 2 + 4a + 6, then find: ii) ( x + y ) − ( x − y )

i) x − y

6. Subtract 2a + 3b − c from 3a + 4b + 6c using vertical method.

7. If A = x 2 + x + 1, B = x 2 − x + 1, then find the values of the following. i) A + B

ii) A − 2 B .

8. How much is x 2 + 2 x less than x 2 − 8 x ?

9. Subtract the following algebraic expression: i) - 5xy from 12xy ii) 2a − b from 3a − 5b iii) 2 x 3 − 4 x 2 + 3 x + 5 from 4 x 3 + x 2 + x + 6

(

)

(

)

iv) The sum of 8a − 6a 2 + 9 and −10a + 8a 2 − 8 from −3 4 4 2 3 1 v) x 2 y − xy 2 + xy from x 2 y + xy 2 − xy 5 3 3 2 3 vi)

ab 35 3 4 6 − bc + ac from bc − ac 5 5 7 3 5

IV. MULTIPLICATION OF ALGEBRAIC EXPRESSIONS 1. Find the following products in their simplest form. ii) ( −3 x ) × ( 4y )

i) 2 x × 4

 4   −3 2 2  iv)  l 2  ×  l m  5   2  2. Find the products of the following.

(

iii) ( −12 xy ) × 6 x 3

)

i) 5a ( 6a − 3b )

ii) 8a 2 ( 2a + 5b )

iii) 9 x 2 ( 5 x + 7 )

iv) 2 x 2 3 x − 4 x 2

(

)

3. Multiply the following.

2

2

2

i) 3 x − 4 xy + 2 xy and 2 x y

 x3 2x 1  ii)  − +  and (15x) 5 5   3 195

ALGEBRAIC EXPRESSIONS AND IDENTITIES

4. Simplify the following. i) ( 3 p − 5 ) ( 5 p − 6 )

ii) ( 4 f − 3 g ) ( 5 f + 6 g )

5. Simplify:

(

)

i) x 3 − 2 x 2 + 5 x − 7 ( 2 x − 3 ) ii) ( 5 x + 3 ) ( x − 1 ) ( 3 x − 2 )

(

iii) 2 x 2 + 3 x − 5

)( 3x − 5x + 4 ) 2

iv) ( 5 − x ) ( 6 − 5 x ) ( 2 − x ) v) ( 3 x + 2y ) ( 4 x + 3y ) − ( 2 x − y ) ( 7 x − 3y )

( ) ( ) vii) ( x − 2 x + 3 x − 4 ) ( x − 1 ) − ( 2 x − 3 ) ( x − x + 1 ) viii) ( x − 2y ) ( x + 4y ) x y vi) x 2 − 3 x + 2 ( 5 x − 2 ) − 3 x 2 + 4 x − 5 ( 2 x − 1 ) 3

2

2

2

2

2 2

ix) ( 1.5 x − 4y ) ( 1.5 x + 4y + 3 ) − 4.5 x + 12y V. FINDING THE VALUE OF AN EXPRESSION FOR A GIVEN VALUE OF THE VARIABLE 1. If a = 4, b = −4 , find the value of: i) a 2 + b 2

ii) a 2 + ab + b 2

iii) a 2 − b 2

2. When a = −1, b = 0 , find the value of the given expressions: ii) 3a 2 + 2b 2 + 10

i) 2a + 3b

iii) 5a 2b + ab 2 + 2ab

3. Simplify the expressions and find the value if x is equal to 5. i) x + 5 + 3 ( x − 1 )

ii) 2 ( x + 1 ) + 6 x − 7

iii) 2 x + 3 ( x − 5 )

iv) 7 ( 2 x − 1 ) + 5 x + 1

4. Simplify these expressions and find their values if x = 2, a = −3, b = −1. i) 4 x − 6 − 2 x + 7

ii) 12 − x + 6 x + 2

iii) 5a + 6 − 2a + 8

iv) 12 − 2b − 6 − 3b +a

5. i) If z = 12, find the value of 2 z 3 − 4 ( 2 z − 10 ). ii) If p = −10, find the value of p2 − 3 p − 100.

6. Simplify the expression and find its value when a = 4 and b = −2.

(

)

2 a 2 + ab + 4 − 2ab

196

IL Foundation Series Class 8

VI. ALGEBRAIC IDENTITIES 1. Simplify the following. i) ( p + 3q ) ( p − 3q )

ii) ( l + m ) ( l − m )

iii) ( 2 x + 4 ) ( 2 x − 4 )

2. Expand each of the following.

1  ii)  2 x −  x 

2

i) (a − b)

2

iii) (5 x − 3y)2

3. Express each of the following as a perfect square. i) 9 x 2 + 12 xy + 4y 2

ii) 81x 2 + 90 xy + 25y 2

4. Simplify the following.

(

) −(l −m ) 2

i) (x + 4)2 + (x + 2)2

ii) l 2 + m2

iii) (a + b)2 − (a − b)2

iv)(a + b)2 + (a − b)2

2

2

2

5. Use an appropriate formula to find the values of the following. i) (102)2

iii) ( 20.5 )( 19.5 )

ii) (96)2

6. If x + y = 7, xy = 12 , then find the value of x 2 + y 2.

7. Expand. i) (a + 2b + 3c)2

ii) (a − 2b − 3c)2

8. Verify the following. i) ( ab + bc ) ( ab − bc ) + ( bc + ca ) ( bc − ca ) + ( ca + ab ) ( ca − ab ) = 0

(

)

ii) ( a + b + c ) a 2 + b 2 + c2 − ab − bc − ca = a 3 + b 3 + c3 − 3abc

( ) iv) ( m + n ) ( m − mn + n ) = m + n iii) ( p − q ) p2 + pq + q 2 = p3 − q 3 2

2

3

3

v) ( a + b ) ( a + b ) ( a + b ) = a 3 + 3a 2b + 3ab 2 + b 3 vi) ( a − b ) ( a − b ) ( a − b ) = a 3 − 3a 2b + 3ab 2 − b 3 9. Simplify. i)

8.47 × 8.47 − 1.53 × 1.53 6.94

ii) 16a 2 + 40ab 2 + 25b 4

10. Find the value of x, if 10000 x = (9982)2 − (18)2 11. If ab = 5, then find the value of (a + b)2 − (a − b)2.

(

)

1  1  12. If 49 x 2 − b =  7 x +   7 x − , then find the value of b .

2  2  197

ALGEBRAIC EXPRESSIONS AND IDENTITIES

13. What is the value of

1 + 2008 1 + 2009 1 + 2010 1 + 2011×2013 ?

14. Three real numbers x, y, z are such that x 2 + 6y = −17, y 2 + 4 z = 1 and z 2 + 2 x = 2 .

What is the value of x 2 + y 2 + z 2 ?

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. Which of the following is a constant term?

a) x

b) 4x

c) 6

d) 3 x 2

c) x 2 + z 2

d) 3z

2 3 + 2 2 x y

d) 4 +

2. Which of the following is a monomial? a) x + y

b) a + b + c

3. Among the following, identify the trinomial. a) 2 +

4 n

b) x + y + z

c)

5 x2

4. The number of terms in ax 3 + bx 2 + cx + d is a) 1

b) 2

c) 3

d) 4

c) 23

d) 32

c) 1

d) not defined

c) 2

d) 0

5. The coefficient of x in 23x is a) 2

b) 3

6. The degree of zero polynomial is a) 0

b) 2

7. The degree of 5 x 3 + 7 x 2 + 6 x + 2 is a) 1

b) 3

8. The terms which do not have the same literal terms are called a) like terms

b) unlike terms

c) variables

d) constants

b) like terms

c) literals

d) constants

b) 10 p

c) 9p

d) 11p

b) 10y

c) 4y

d) 6y

9. x, 2 x, 4 x, 6 x are called a) unlike terms 10. 3 p + 5 p + 3 p = a) 12p 11. 5y + 3y − 2y = a) 8y

198

IL Foundation Series Class 8

12. Sum of −6abc, 2abc, − abc is a) −5abc

b) 6abc

c) 5abc

d) 8abc

c) 4 x + 2y

d) 4 x + 4y

13. Sum of 3 x + 4y and x − 2y is a) 4 x − 2y

b) 4 x + 6y

14. When we add a 2 + b 2 and additive inverse of a 2 − b 2, the result is a) 6a 2

b) 2a 2

c) 3a 2

d) 2b 2

15. While adding −b + c and 2a + 3 b + 2c, the result is a) 2a + 2b + 3c

b) 3a + 2b + 5c

c) −a − 2b − 3c

d) 3a + 4b + 5c

16. The additive inverse of 23 is 1 a) -23 b) 23 17. The additive inverse of 2 x − 1 is

c) 0

d) 1

a) −2 x + 1

c) −2 x − 1

d) − ( 2 x + 1 )

b) A − ( − B )

c) A + B

d) B - A

b) −14p

c) 10 p

d) −6p

b) −3a − b

c) 3a + b

d) a + 3b

c) 2 x 2

d) 2 x − 2

b) 2 x + 1

18. If A and B are two expressions, then A − B = a) A + ( − B ) 19. −2 p − ( −12 p ) = a) 14p 20. ( 3a + 4b ) − ( 2a + b ) = a) b − 3a

21. If A = x 2 + x + 1, B = x 2 − x − 1 , then A + B = a) 2 x 2 + 2

b) 2 x + 2

22. When we subtract 3a + 5b from 8a + 7 b, the result is a) 5a − 2b

b) 5a + 2 b

c) 2a + 5b

d) 2a − 5b

23. When we subtract ( 2a + 3b ) from a + b, the result is a) 3a + 4b

b) −a − 2b

c) a + 2b

d) 0

24. If A = a + b + c and B = a − b + c, then the value of A − B = a) 2b

(

b) 2c

c) 2a

d) −2b

b) −72 x 4 y

c) −108 x 3 y 2

d) −801x 3 y 2

c) 2 x + 2 x 2 y

d) − 2 x 2 + 2 xy

)

25. ( −12 xy ) 6 x 3 = 4

a) +72 x y

26. The product of ( 2x ) ( x + y ) is a) 2 x 2 + 2y

b) 2 x 2 + 2 xy

(

) 199

ALGEBRAIC EXPRESSIONS AND IDENTITIES

27. Simplify: 6 p − 8q = 2. a) 3 p − 4q = 1

b) 3 p − 5q = 0

c) 4 p − 3q = 0

d) 5 p − 4q = 1

28. In the multiplication of a monomial and a binomial, we use the ____________ property a) Commutative

b) Distributive

c) Associative

d) Inverse

b) x 2 − 6

c) x 2 − 5 x + 6

d) − x 2 + 5 x

b) a 2 − b 2

c) a 2 + 2ab + b 2

d) a 2 − 2ab + b 2

b) a 2 + b 2

c) a 2 − 2ab + b 2

d) a 2 − b 2

b) a 2 + b 2

c) a 2 − 2ab + b 2

d) a 2 − b 2

29. ( x + 2 ) ( x + 3 ) = a) x 2 + 5 x + 6 30. ( a + b ) ( a − b ) = a) a 2 + b 2 31. (a − b)2 = a) a 2 + 2ab + b 2 32. (a + b)2 = a) a 2 + 2ab + b 2

(

)

33. If a + b = 6, a − b = 2 , then the value of 2 a 2 + b 2 is ______. a) 20 34. If x 4 + a) 7

b) 30

c) 40

d) 10

1 1  = 23 , then the value of x −   will be x x4  b) −7 c) −3

d) 3

2

x2 + 3x + 1 1 is 35. If x + = 1 then the value of 2 x + 7x + 1 x a) 1

b)

3 7

c)

1 2

d)2

1  36. If x 2 − 3 x + 1 = 0 and x > 1, then the value of  x −  x  a)

5

b) 1

c)

2

37. Assertion (A): ( −5 x ) ( −2 x ) = 10 x 2.

Reason (R): The product of two factors with the same sign is positive. a) Both A and R are correct, and R is the correct explanation of A b) Both A and R are correct, but R is not the correct explanation of A c) A is correct, R is incorrect d) A is incorrect, R is correct

200

d) ± 5

IL Foundation Series Class 8

38. Assertion (A): ( 2a + 1 ) ( 2a − 1 ) = 4a 2 − 1.

Reason (R) : ( a + b ) ( a − b ) = a − b .

2

2

a) Both A and R are correct, and R is the correct explanation of A b) Both A and R are correct, but R is not the correct explanation of A c) A is correct, R is incorrect d) A is incorrect, R is correct 39. Assertion (A): ( a + 3b ) ( a − 3b ) = a 2 − 9b 2.

Reason (R): (a + b)2 = a 2 + 2ab + b 2. a) Both A and R are correct, and R is the correct explanation of A b) Both A and R are correct, but R is not the correct explanation of A c) A is correct, R is incorrect d) A is incorrect, R is correct II. MATCH THE FOLLOWING 1. Match the following.

Polynomial

Degree

i. 5 x 2 − 2 x + 7 x5

p. 5

ii. x

q.10

( ) iv. ( x + 2 ) iii. x + x 2 2

3

5

r. 1 s. 6

a) i-r, ii-s, iii-q, iv-p

b) i-q, ii-p, iii-s, iv-r

c) i-s, ii-p, iii-q, iv-r

d) i-p, ii-r, iii-s, iv-q

2. Match the following.

Column - I

Column - II

i.3 x + 2y + 5 x + 4y =

p. 8 x + 6y

ii. 5 x − 2y + 4 x − 3y =

q. 0

iii. 6 x + y − 2 x − y =

r. 4x

iv. x + y − x − y =

s. 9 x − 5y

a) i-p, ii-s, iii-r, iv-q

b) i-p, ii-q, iii-s, iv-r

c) i-s, ii-p, iii-r, iv-q

d) i-s, ii-r, iii-p, iv-q

201

ALGEBRAIC EXPRESSIONS AND IDENTITIES

3. Match the following.

Column - I

Column - II

i. ( 5 x − 3y ) − ( 3 x + 2y )

p. 4x

ii. ( x + y + z ) + ( x − y + z ) − ( − x + 2 z )

q. 3x

iii. 8x is subtracted from 10x

r. 2 x − 5y

iv. 12x is added to −8x

s. 2x

a) i-p, ii-s, iii-q, iv-r

b) i-r, ii-q, iii-s, iv-r

c) i-p, ii-s, iii-r, iv-q

d) i-r, ii-q, iii-s, iv-p

III. SUBJECTIVE QUESTIONS 1. Add:

3 5 2 2 7 7 5 5 5 a − b + c, a − b + c and a + b − c .

3 2 4 2 4 5 3 2 2

4 4 2 3 1 2. Subtract: x 2 y − xy 2 + xy from x 2 y + xy 2 − xy .

5 3 3 2 3 4 2 1  1  2  3. Simplify:  y 2 − y + 11  −  y − 3 + 2y 2  −  y − y 2 + 2 .

7 3 3  7  7  4. Find the area of a rectangular box with length 3mn and breadth 4np.

5. Find the product of

3 2 2 x y , 0.5 xy 2 z 2 , 1.16 x 2 yz 3 , 2 xyz .

4 2

6. Find the product of −5 x y,

−2 2 8 xy z, xyz 2 and − 1 z.

3 15 4

Verify the result when= x 1= , y 2 and z = 3.

7. Find the product: 8. Multiply:

−8 9 3  xyz  xyz 2 − xy 2 z 3 .

27 4 2 

−3 2 3 x y by ( 2x − y ) , and verify the answer for x = 1 and y = 2.

2

 −a a 2   b b 2  + 9. Multiply:   by  −  .

9   2 3   7 10. Multiply: ( 0.8a − 0.5 b ) by ( 1.5a − 3 b ).

(

)

(

)

11. Simplify: x 2 − 3 x + 2 ( 5 x − 2 ) − 3 x 2 + 4 x − 5 ( 2 x − 1 ).

12. Using the formula for squaring a binomial, evaluate the following i) (1001)2 202

ii) (703)2

IL Foundation Series Class 8

13. Simplify using the formula ( a − b ) ( a + b ) = a 2 − b 2. i) 1.8 × 2.2 14. If x +

ii) (467)2 − (33)2

1 1 = 20,then find the value of x 2 + 2 . x x

15. If x 2 +

1 1 1 = 18 then find the value of x + and x − .

2 x x x

203

9

MENSURATION

9.1 PERIMETER AND AREA OF POLYGONS AND CIRCLES 9.1.1 Perimeter and area of polygons A rectilinear figure is a plane figure bounded by line segments. A simple closed rectilinear figure is known as a polygon.

Its perimeter is the sum of all its sides. The unit of measurement of the perimeter is the given unit of length. The area of a closed plane figure is the region enclosed by its boundaries. It is measured in square units.

Let us look at formulas for the area and perimeter of some of the polygons.

Square

a

a Let a denote the sides of the square. Then, i) Perimeter = 4a units ii) Area = a2 sq. units Rectangle

b

l Let l and b denote the length and breadth of the rectangle. Then,

204

IL Foundation Series Class 8

i) Perimeter = 2 (l + b) units ii) Area = lb sq. units Triangle

a

c

h b

Let a, b, and c denote the sides, and h denote the height of the triangle. Then, i) Perimeter = a + b + c units ii) Area =

1 bh sq. units or 2

s(s − a )(s − b )(s − c) sq. units where s is semi-perimeter

Equilateral Triangle

a

a h a

Let a denote the three equal sides, and h denote the height of the triangle. Then, i) Perimeter = 3a units ii) Area =

3 2 1 a = ah sq. units 4 2

Parallelogram

a

h b

Let a and b denote the sides and h denote the height of the parallelogram. Then, i) Perimeter = 2(a + b) units ii) Area = b × h sq. units

205

MENSURATION

9.1.2 Perimeter and area of circle

r The perimeter (or) circumference of a circle is c = 2π r (or ) π d units The area of a circle is π r 2 sq. units.

Area of a semi-circle =

π r2 2

sq. units.

SOLVED EXAMPLES Example 1: Find the area of the shaded region if ABCD is a square with a side length of 14 cm and APD and BPC are semi-circles.

B

C

P

A

D

Solution: Area of shaded region = Area of square ABCD - Area of two semi-circles

 1 22  = (14 × 14) − 2  × × 72  2 7  2 2 =− 196 154 cm = 42 cm

Example 2: Find the area of the shaded region in the given figure, where the radius of the circle is 2 cm. (Take π = 3.14 and

3 = 1.73)

6 cm

6 cm

6 cm

206

IL Foundation Series Class 8

Solution: 3 × 62 36 3 Area of the equilateral triangle = = = 9 3 4 4

= 9 ×1.73 = 15.57 cm 2 2 Area of the circle= π × 2 = 4 × 3.14 = 12.56 cm 2 ∴ Area of the shaded region = 15.57-12.56 = 3.01 cm2 Example 3: In the figure given below, ABCD is a rectangle inscribed in a circle. Find the area of the shaded region. (Take, π = 3.14 ) B

A 3 cm 4 cm D

C

Solution: Given, length = 4 cm, breadth = 3 cm Diagonal of the rectangle = Diameter of the circle

∴ Radius of the circle=

=

42 + 32

=

16 + 9=

25= 5 cm

5 = 2.5 cm 2

∴ Area of the circle = π × (2.5)

2

3.14 × 6.25 = 19.63cm 2

Area of rectangle = 4 × 3 = 12cm 2 Area of the shaded region = Area of the circle - Area of the rectangle

= 19.63 - 12 = 7.63 sq.cm.

207

MENSURATION

9.2 SURFACE AREA OF SOLID OBJECTS Knowing the surface area of objects is important for various practical reasons in everyday life and numerous fields like construction, manufacturing, packaging, art and design, etc. Let us look at the surface area formulas of some solid shapes.

9.2.1 Cube

a

a a

For a cube of side a, Lateral surface area= 2h ( l + b )= 2a ( a + a )= 4a 2 sq. units 2 2 2 Total surface area of a cube =2(lb + bh + lh)= 2 ( a + a + a ) = 6a2 sq. units

Diagonal of a cube =

a2 + a2 + a2 =

3a 2 =

3a units

9.2.2 Cuboid b h l l

h 5 h b l 2

h l 3

b l

4

l

b h

h

b l

6

h

b

b

For a cuboid with length l, breadth b and height h, we have: Lateral surface area = 2( length × height ) + 2( breadth × height ) =2(lh ) + 2(bh ) =2h (1 + b ) sq. units Total surface area of a cuboid = 2( length × height ) + 2( breadth × height ) + 2( length × breadth ) = 2(lh + bh + lb ) sq. units Diagonal of a cuboid =

208

l 2 + b 2 + h 2 units

IL Foundation Series Class 8

9.2.3 Right circular cylinder

h r

For a right circular cylinder radius of the base r and height (or length) h, i) Curved surface area = 2πrh ii) Total surface area = 2πr (h + r ) 9.2.4 Right circular cone

l

h r

For the right circular cone of height h, slant height l, and radius of base r, we have i) Curved surface area = π rl ii) Total surface area = Curved surface area + Area of the base

= π rl + π r 2

= π r (l + r )

9.2.5 Sphere

r

r

For the sphere of radius r, we have i) Surface area = 4π r

2

209

MENSURATION

9.2.6 Hemisphere The hemisphere can either be solid or hollow. a. Solid hemisphere r

For the solid hemisphere of radius r, i) Curved surface area = 2π r 2 ii) Total surface area = 2π r 2 + π r 2 = 3π r 2 b. Hollow hemisphere Inner Radius r R Outer Radius

For the hollow hemisphere of r as the inner radius and R as the outer radius, we have i) Curved surface= area 2π ( R 2 + r 2 ) = 3π R 2 + π r 2 ii) Total surface area 9.2.7 Frustum of a cone r1

l

h r2

If h = vertical height of the frustum of a cone, l = slant height of the frustum, r1 and r2 are radii of the two bases (ends) of the frustum. i) Curved surface area of a frustum of a cone = π l ( r1 + r2 ) where l = ii) Total surface area of a frustum of a cone = π l ( r1 + r2 ) + π ( r1 + r2 ) 2

210

2

h 2 + ( r1 − r2 )

2

IL Foundation Series Class 8

SOLVED EXAMPLES Example 1: Find the total surface area and curved surface area of the cone whose radius is 5 inches 22

and slant height is 4 inches. (Use π = 7 ) Solution: Given, Radius (r) = 5 inches Slant height (l) = 4 inches 22 π= 7 Total surface area (T) = πr(r + l) 22 22 990

=( ) × 5 × (5 + 4) = ( ) × 5 × 9 = square inches 7 7 7 So, the total surface area of the cone is approximately 141.43 square inches.

22 440 )×5×4= square inches.

7 7 Therefore, the curved surface area of the cone is approximately 62.86 square inches.

Curved surface area (S) = πrl = (

Example 2: A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm, and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

7 cm 7 cm 6 cm

7 cm

Solution: Given, Diameter of the hemisphere, (d) = 14 cm 14 cm = 7 cm 2 Height of the cylinder, (h) = Total height of vessel - height of hemisphere

Radius of the hemisphere, (r) =

= 13 cm - 7 cm = 6 cm

211

MENSURATION

The inner surface area of the vessel = Surface area of the cylinder + Surface area of the hemisphere = 2π rh + 2π r 2 = 2ππr (h + r ) 22 =2 × × 7(6 + 7) =572 cm 2 7

Therefore, the inner surface of the vessel is 572 cm2 Example 3: A tent has the form of a cylinder topped with a cone. The cylindrical part has a height of 2.1 m and a diameter of 4 m. The conical top has a slant height of 2.8 m. Determine the area of the canvas required to make the tent. Additionally, calculate the cost of the canvas at a rate of `500 per m2 (Note: The base of the tent will not be covered with canvas.)

2.8 m 2m 2m 2.1 m

2m 2m

Solution: Given, 4 Radius of cylindrical base, (r) = 2 m = 2 m Height of cylindrical portion, (h) = 2.1 m C.S.A of cylindrical portion = 2πrh 22 × 2 × 2.1 7 = 26.4 m 2 = 2×

The radius of the conical base, (r) = 2 m The slant height of the conical portion, (l) = 2.8 m C.S.A of conical portion = πrl

=

22 × 2 × 2.8= 17.6 m 2 7

∴ Area of the canvas used = 26.4 + 17.6 = 44 m2

₹. 22000 ∴ Total cost of the canvas used =500 500 × 44 × 44 = =RsRs . 22000

212

IL Foundation Series Class 8

Example 4: Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth, and height of 15 m, 10 m, and 7 m, respectively. From each can of paint, 100 m2 of area is painted. How many cans of paint will he need to paint the room?

Solution: Given, Length of the hall (l) = 15 m Breadth of the hall (b) = 10 m Height of the hall (h) = 7 m Lateral surface area of four walls = (h × l + h × b + h × l + h × b) = 2(hl + hb)

= 2(7 × 15 + 7 × 10) m2

= 2(105 + 70) m2

= 350 m2

Area of the ceiling = l × b

= 15 × 10 = 150 m2

Area of the hall to be painted = Lateral surface area of walls + Area of the ceiling

= 350 m2 + 150 m2 = 500 m2

Since one can of paint cover a 100 m2 area, 500 The number of cans required = = 5 100 Example 5: Find the side of the cube whose surface area is 600 cm2.

Solution: Let the length of each side be a.

Given that the surface area of the cube = 600 cm2 We know that the surface area of the cube = 6a2 ⇒ 600cm 2 = 6 × (a ) 2

⇒ a2 = 100 cm2 ⇒ a = 10 cm Thus, the side of the cube = 10cm Example 6: Find the total surface area of a hollow hemisphere whose outer and inner radii are 3.2 cm and 2.1 cm, respectively.

Solution: Given, Outer radius ( Router ) = 3.2 cm 213

MENSURATION

Inner radius

( r ) = 2.1 cm inner

2 2 TSA = 3π Router + π rinner

= 3π (3.2) 2 + π (2.1) 2 = 3π (10.24) + π (4.41) = 30.72π + 4.41π = 35.13π = 35.13 × 3.14 = 110.40 sq. cm

9.3 VOLUME OF SOLID OBJECTS In the previous section, we discussed the surface area of different solid shapes. In this section, let us discuss the formulas to find the volume of these solid shapes.

9.3.1 Cube

a

a

a

For a cube of side a, the volume is given by: The volume of a cube = l × b × h = a × a × a = a 3 cu. units 9.3.2 Cuboid h b l

For a cuboid with length l, breadth b, and height h, we have The volume of a cuboid = l × b × h cu. units.

9.3.3 Right circular cylinder

h r 214

IL Foundation Series Class 8

For a right circular cylinder of base radius r and height (or length h), the volume is given by: The volume of a cylinder = π r 2 h 9.3.4 Right circular cone

l

h r

For a right circular cone of height h, slant height l, and radius of base r, The volume of a cone =

1 2 πr h 3

9.3.5 Sphere

r

r

For a sphere of radius r, The volume of a sphere =

4 3 πr 3

9.3.6 Hemisphere r

For a hemisphere of radius r, The volume of a hemisphere =

2 3 πr 3

215

MENSURATION

9.3.7 Frustum of a cone r1

h

l

r2

If h = vertical height of the frustum of a cone, l = slant height of the frustum, r1 and r2 are radii of the two bases (ends) of the frustum.

1 2 2 The volume of a frustum of a cone = π h ( r1 + r2 + r1r2 ) 3

9.4 SURFACE AREA AND VOLUME OF PRISM AND PYRAMID 9.4.1 Surface area and volume of prism The total surface area of the prism is the sum of the areas of its lateral faces and the two bases.

Total surface area of a right prism = (lateral surface area) + 2 (area of the end surface) The volume of a prism is the product of the area of the base and the height.

Volume = (area of the base) × height 9.4.2 Surface area and volume of pyramid The total surface area of a pyramid is the sum of the area of its base and the areas of its triangular lateral faces.

216

IL Foundation Series Class 8

Total surface area of a right pyramid = (lateral surface area) + (base area) The volume of a pyramid is one-third of the product of the area of the base and the height.

Volume =

1 (area of the base) × height 3

SOLVED EXAMPLES Example 1: A closed cylindrical container with a radius of 7 cm and height of 10 cm is to be made out of a metal sheet. Find the volume of the cylinder made.

Solution: Radius = 7 cm Height = 10 cm 22 × 7 × 7 ×10 = 22 × 70 = 1540 cm 3 7 Example 2: The diameter of an ice cream cone is 7 cm, and its height is 12 cm. Find the volume of ice cream that the cone can contain.

Volume = π r h = 2

Solution: The volume of the ice cream cone =

1 2 πr h 3 2

1 22  7  = × ×   ×12 = 22 × 7 = 154 cm 3 3 7 2

Example 3: Find the volume of the frustum of a cone with a height of 15 in, a large base radius of 25 in, and a small base radius of 4 in. Express the answer in terms of in.

217

MENSURATION

Solution: Given: Height (h) = 15 in.

Large base radius (r1) = 18 in.

Small base radius (r2) = 6 in.

Now, substitute the values into the formula for the volume of the frustum of a cone: 1 = V π h ( r12 + r22 + r1r2 ) 3 π ×15 (18) 2 + (18 × 6) + (6) 2  = V 3 = 5π [324 + 108 + 36] V = 5π × 468 V V = 2340π in3 Example 4: Find the volume of a sphere whose radius is 10 cm.

Solution: Radius (r) = 10 cm 4 3 πr 3 4 = × 3.14 × 10 × 10 × 10 3 = 4186.6 cm3

Volume of a sphere =

Example 5: A matchbox measurement is 4 cm by 2.5 cm by 1.5 cm. What will be the volume of a packet containing 12 such matchboxes? How many such packets can be placed in a cardboard box with dimensions of 60 cm × 30 cm × 24 cm ?

Solution: Volume of a matchbox =(4 × 2.5 ×1.5)cm 3 =15 cm 3 Volume of a packet containing 12 matchboxes =× (12 15)cm 3 = 180 cm 3 Now, the volume of the cardboard box = (60 × 30 × 24)cm 3 = 43200 cm3 43200 = 240 180 Example 6: Three cubes of sides 3 cm, 4 cm, and 5 cm are melted, forming the new cube. Find the side of the new cube.

Number of packets that can be put in the cardboard =

Solution: The volume of a cube = a3 cu. units The volume of a cube with side 3 cm = 33 = 27 cm3 218

IL Foundation Series Class 8

The volume of a cube with side 4 cm = 43 = 64 cm3 The volume of a cube with side 5 cm = 53 = 125 cm3 Total volume of a new cube = 27 + 64 + 125 = 216 cm3 The edge of a new cube = a3 = 216 cm3 a3 = 6 × 6 × 6 a = 6 cm Example 7: Find the total surface area and volume of a right triangular prism, which has a base area of 72 square units, the base perimeter of 36 units, and the length of the prism is 8 units.

Solution: Given, base area = 72 square units, base perimeter = 36 units, and length of prism = 8 units Thus, the surface area of the right triangular prism is calculated using the formula: Surface Area = (lateral surface area) + (2 × base area) We know that LSA = (perimeter of the base × length of the prism) Surface Area = (perimeter of the base × length of the prism) + (2 × base area)

SA = (36 × 8) + (2 × 72) = SA (288 + 144) SA = 432 sq. units Volume of the prism = Area of base × height = 72 × 8 = 576 cu. units Example 8: Find the volume of the pyramid with a base area of 36 square units and a height of 6 units.

Solution: For the volume of the pyramid, we use the formula : 1 × base area × height 3 1 = × 36 × 6 3

Volume =

= 72 cubic units

QUICK REVIEW • Square: Perimeter = 4a units and Area = a2 sq. units • Rectangle: Perimeter = 2 ( l + b ) units and Area = lb sq. units 219

MENSURATION

1 bh sq.units or s(s − a )(s − b )(s − c) sq.units 2 • The perimeter of a circle = 2π r (or )π d units, the area of a circle is π r 2 sq. units and the a area of a 2 πr sq. units. semi-circle = 2 r ( hand + r total • Right circular cylinder : Curved surface area = 22π rh ) surface area = 2π r ( h + r ) • Triangle: Perimeter = a + b + c and Area =

l ( rand l ( r1++r) r2 ) • Right circular cone: Curved surface area = π rl ) surface area = π r(l 1 + r2total • Sphere: Surface area = 2π r 2 and surface area of a hemisphere = 2π r 2 • Frustum of cone: Curved surface area of a frustum of a cone = π l ( r1 + r2 ) where l=

h 2 + ( r1 − r2 )

2

and total surface area of a frustum of a cone = π l ( r1 + r2 ) + π ( r12 + r22 )

• Volume of a right circular cylinder = π r 2 h • Volume of a right circular cone = 4 3 πr 3 2 • Volume of a hemisphere = π r 3 3

1 2 πr h 3

• Volume of a sphere =

• Volume of a frustum of a cone =

1 π h ( r12 + r22 + r1r2 ) 3

• Total surface area of a right prism =( lateral surface area) + 2( area of the end surface) • Volume of a prism = (area of the base) × height • Total surface area of a right pyramid = (lateral surface area) + (area of the base) • Volume of a pyramid =

1 (area of the base) × height 3

WORKSHEET - 1 I. PERIMETER AND AREA OF POLYGONS AND CIRCLES 1. Find the area of the shaded region.

G

H 5

I

A B

5

C 5

L

K E

220

J

25

F

D

20

IL Foundation Series Class 8

2. Find out the area of a triangle whose base is 5 cm and altitude is 8 cm.

3. A flooring tile has the shape of a parallelogram whose base is 24 cm, and the corresponding height is 10 cm. How many such tiles are required to cover a floor of an area of 1080 square metres?

4. As shown in the figure, a square and a rectangular field with the given measurements have the same perimeter. Which field has a larger area?

6m

8m

5. Find out the area of an equilateral triangle with sides of 4 cm and an altitude of 8 cm.

6. A circle has a diameter of 10 cm. Find the area of the semi-circle.

7. What is the area of a circular fountain whose diameter is 4 feet?

8. A parallelogram has a base of 10 cm and a height of 8 cm. Calculate its area and perimeter.

9. Find the area of the shaded region in the figure in which four semi-circles are drawn, taking 22  the mid-point of each side of the square as the centre. (Take π =  7 

14 cm

10. A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 6 2 cm , then what is the area of the triangle?

11. In the figure, ABCD is a parallelogram. Find out the area of triangle ECB. D

5m

C

3.5 m A

7m

E

B

221

MENSURATION

II. SURFACE AREA OF SOLID OBJECTS 1. A thin, hollow, hemispherical sailing vessel is made of metal covered by a conical canvas tent.

The radius of the hemisphere is 14 m, and the total height of the vessel (including the height of the tent) is 28 m. Find the area of the metal sheet and the canvas required.

h H = 28 m r = 14 m

2. The curved surface area of a cone is 670 cm, and its radius is 15 cm. Find the total surface area of the cone.

3. A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, its radius at the upper base is 4 cm, and its slant height is 15 cm, find the area of material used for making it.

4. The slant height of the frustum of a cone measures 4 cm, and the perimeters of its circular ends are 18 cm and 6 cm, respectively. Determine the curved surface area of the frustum.

5. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 91 cm and the length is 1.25 m.

6. The radius of a conical tent is 5.6 m, and the slant height is 12 m. Find the length of canvas required to make the tent if the width of the canvas is 4 m.

7. Find the lateral surface area of a cylinder whose circumference is 44 cm and height is 10 cm.

8. Find the length of the largest pole that can be placed in a room of dimensions 12 m × 4 m × 3 m.

9. The three different face diagonals of a cuboid (rectangular parallelopiped) have lengths 39 cm, 40 cm, and 41 cm. Find the length of the main diagonal of the cuboid, which joins a pair of opposite corners.

222

IL Foundation Series Class 8

10. Find the surface areas of the following shapes: 3 cm

12 cm

1 cm

i)

1 cm

ii) 24 cm 12 cm

4 cm 12 cm

1 cm 18 cm 3 cm 3 cm

iii)

5 cm 2 cm

8 cm

5 cm

12 cm

iv) 5 cm

18 cm

4 cm 5 cm

20 cm

III. VOLUME OF SOLID OBJECTS 1. Metallic spheres of radius 6 cm, 8 cm, and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

2. The diameter of cylinder A is 7 cm, and the height is 14 cm. The diameter of a cylinder is 14 cm, and the height is 7 cm. Verify which cylinder has greater volume and surface area.

3. A closed cylindrical container, the radius of which is 7 cm and height 10 cm, is to be made out of a metal sheet. Find i) the area of the metal sheet required. ii) the volume of the cylinder made. iii) the cost of painting the lateral surface of the cylinder at the rate of 4 per cm2.

4. A hollow hemispherical bowl with a thickness of 1 cm has an inner radius of 6 cm. Find the volume of metal required to make the bowl.

5. A drinking glass has the shape of a frustum of a cone with a height of 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Calculate the capacity of the glass.

4 cm 2 cm

2 cm 14 cm

1 cm 1 cm 2 cm

223

MENSURATION

6. A conical cup has a height of 21 cm, and the diameter of its base is 18 cm. Find the amount of water it can hold.

7. The radius and slant height of a cone are in the ratio 8:17. If its curved surface area is 544π cm 2 , then find its volume.

8. A solid cylinder and a cone have the same radius and height. If the volume of a cylinder is 27 cm3, then what will be the volume of a cone?

9. The surface area of the three co-terminus faces of a cuboid is 6,15, and 10 cm2, respectively.

Find the volume of the cuboid.

10. The edges of a cuboid are in the ratio 1: 2 : 3, and its surface area is 88 cm3. Find the volume of the cuboid.

IV. SURFACE AREA AND VOLUME OF PRISM AND PYRAMID 1. A right rectangular prism has a length of 8 cm, a width of 5 cm, and a height of 10 cm.

Calculate the total surface area of the prism, given that the lateral surface area is 140 square centimetres.

2. A triangular prism has a base with a height of 6 cm and a base area of 18 square centimetres.

If the height of the prism is 12 cm, calculate the total volume of the prism.

3. A right square pyramid has a lateral surface area of 48 square units and a base side length of 6 units. Calculate the total surface area of the pyramid.

4. A right triangular pyramid has a lateral surface area of 36 square centimetres. The base of the pyramid is a right-angled triangle with a base length of 4 centimetres and a height of 8 centimetres. Calculate the total surface area of the pyramid.

5. The area of the base of a right circular prism is 50 cm2, and its height is 8 cm. What is its volume?

WORKSHEET - 2 I.

MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. If the perimeter of a square is 4 m, then its area is

a) 4 m2

b) 2 m2

c) 1 m2

d) 8 m2

7 cm and height of 7 cm is 2 c) 77 cm2 d) 147 cm2

2. The curved surface area of a cylinder, whose radius is a) 154 cm2

b) 308 cm2

3. The area of a triangle whose sides are 4 cm, 13 cm, and 15 cm is a) 576 cm2

b) 30 cm2

c) 26 cm2

d) 24 cm2

4. The side of an equilateral triangle of area 64 3 cm 2 is a) 12 cm 224

b) 14 cm

c) 18 cm

d) 16 cm

IL Foundation Series Class 8

5. The length of a rectangle is 1 cm more than its width, and its perimeter is 14 cm. Then, the area of the rectangle is a) 16 cm2

b) 12 cm2

c) 48.75 cm2

d) 10 cm2

6. The shape of a glass (tumbler) is usually in the form of a) A cylinder

b) Frustum of a cone

c) A cone

d) A sphere

7. A cylindrical pencil sharpened at one edge is a combination of a) A cylinder and a cone b) A cylinder and frustum of a cone c) A cylinder and a hemisphere d) Two cylinders 8. The curved surface area and the total surface area of a cylinder with a base radius of 7 m and height of 9 m, respectively, are: a) 396 m2, 704 m2

b) 704 m2, 396 m2

c) 396 m2, 702 m2

d) 704 m2, 399 m2

9. The height of the right circular cylinder is 10 cm, and the radius of the base is 7 cm. Then, the difference between the curved surface area and the total surface area is a) 300 cm2

b) 308 cm2

c) 302 cm2

d) 304 cm2

10. A right circular cone is 84 cm in height. The radius of the base is 350 cm. Then, the curved surface area is a) 385000 cm2

b) 358000 cm2

c) 396000 cm2

d) 345000 cm2

11. The volume of a solid hemisphere of radius 2 m is 352 3 376 3 m b) 576m3 c) m d) 600m3 21 9 12. The curved surface area of a right circular cone of height 15 cm and base diameter of16 cm is

a)

a) 60π cm 2

b) 68π cm 2

c) 120π cm 2

d) 136π cm 2

13. In a right circular cone, the cross-section made by a plane parallel to the base is a a) circle b) frustum of a cone c) sphere d) hemisphere 14. If the surface area of a sphere is 616 cm2, then its diameter is a) 14 cm

b) 7 cm

c) 21 cm

d) 28 cm

15. If the slant height of the frustum of a cone is 6 cm and the perimeters of its circular bases are 24 cm and 12 cm, respectively. Then, the curved surface area of the frustum is a) 96 cm2

b) 112 cm2

c) 108 cm2

d) 118 cm2

16. The ratio of the total surface area to the lateral surface area of a cylinder whose radius is 20 cm and height of 60 cm is a) 2 : 1

b) 4 : 3

c) 6 : 3

d) 3 : 2

225

MENSURATION

17. A road roller 4 m wide and with a diameter of 3.5 m makes 4 revolutions in 1 minute. How long will it take to roll over an area of 6160 m2 ? a) 20 min

b) 35 min

c) 17.5 min

d) 40 min

18. The number of coins, each of radius 0.75 cm and thickness 0.2 cm, to be melted to make a right circular cylinder of height 8 cm and base radius 3 cm is a) 460

b) 500

c) 600

d) 640

19. If the length of a rectangle is 8 cm and each of its diagonals measures 10 cm, then the breadth of the rectangle is a) 5 cm

b) 6 cm

c) 7 cm

d) 9 cm

20. The base of a triangle is four times its height, and its area is 50 cm2. Then, the length of its base is a) 10 cm

b) 15 cm

c) 20 cm

d) 25 cm

21. If the radius of a sphere is increased by 10%, then its volume will be increased by a) 11.1%

b) 22.1%

c) 33.1%

d) 44.1%

22. Twelve solid spheres of the same size are made by melting a solid metallic cylinder with a base diameter of 2 cm and height of 16 cm. The diameter of each sphere is a) 4 cm

b) 3 cm

c) 2 cm

d) 6 cm

23. A cylindrical tank with a diameter of 35 cm is full of water. If 11 litres of water is drawn off, the water level in the tank will drop by 80 90 21 cm b) cm c) 14 cm d) cm 7 7 2 24. A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8 : 27 : 27. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to:

a)

a) 10

b) 50

c) 60

d) 20

25. A room has a floor size of 15 × 6 sq. cm. What is the height of the room if the sum of the areas of the base and roof is equal to the sum of the areas of the four walls? a) 1.12 cm

b) 3.24 cm

c) 4.29 cm

d) 2.5 cm

II. SUBJECTIVE QUESTIONS 1. Find the CSA of a cylinder with a base radius of 7 cm and a height of 12 cm.

2. Find the TSA of a cylinder with a radius of base 5 cm and a height of 9 cm.

3. Find the area of a shaded region.

226

IL Foundation Series Class 8

20 m

25 m

15

25 m

4. The shape of a garden is rectangular in the middle and semi-circular at the ends, as shown in the diagram. Find the area and the perimeter of this garden.

7m

7m

20 m

5. The area of square ABCD is 16 cm2. Find the area of the square joining the mid-points of the sides.

D

R

S

A

C

Q

P

B

6. Mrs. Kaushik has a square plot with the side 25 m with the measurement shown in the figure.

She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ₹ 55 per m2.

[Hint: Area of garden = Area of square - Area of rectangle]

227

MENSURATION

20 m 25 m

15

House

25 m

Garden 25 m

7. A cylindrical pipe, open at both ends, has an outer radius of 8 cm, a thickness of 2 cm, and a length of 20 cm. Find the volume of metal required to manufacture the pipe.

8. A closed cylindrical tank with a radius of 7 m and a height of 3 m is made from a sheet of metal. How much sheet of metal is required?

9. Find the height of the cylinder whose volume is 1.54 m3 and the diameter of the base is 140 cm.

10. A sphere has volume 36π cm 3, find the radius of the sphere.

11. If the radii of the circular ends of a bucket of height 40 cm are 28 cm and 7 cm, respectively, then find the volume of the bucket.

12. The internal and external radii of a spherical shell are 3 cm and 5 cm, respectively. It is melted and recast into a solid cylinder with a diameter of 14 cm. Find the height of the cylinder. Also, find the total surface area of the cylinder.

22 (Take π = ) 7 13. A canal is 300 cm wide and 120 cm deep. The water in the canal is flowing at a speed of 20 km/h. How much area will it irrigate in 20 minutes if 8 cm of standing water is desired?

14. A gold ingot in the shape of a cylinder is melted, and the resulting molten metal is moulded into a few identical conical ingots. If the height of each cone is half the height of the original cylinder and the area of the circular base of each cone is one-fifth that of the circular base of the cylinder, then how many conical ingots can be made?

228

ANSWER KEY 1. RATIONAL NUMBERS Worksheet 1

2.

3.

4.

7.

8.

11

56 103

2 5

72 3

10

−1

36 −31

iii)

60

−46 3

7

4 10 3

9. i) 10. i)

11.

5

iii) −

5. i) 6.

2

2

3 −5 6

iii)

−10 3

2

35

ii)

ii)

ii)

11

iv)

3 −1

iii)

−11 81

6

12. −1 13.

14.

15.

−29

6

II. Properties and comparison of rational numbers 1. NA 2. NA 3. NA 4. NA 5. NA 6. NA 9

iv) – 1

8. NA 9. NA 10. NA 11. NA

ii)

v)

−16

5 3

7

13

2

iii)

5

=

−39

7

−5

iv)

11

−10 7

ii) − ii)

>

6 −7 9

iv) −

−2

2

ii)

7 9

−6

7 6

<

7

5

−7

iv) 0

3

2. √65 − √55 + √91 − √77 3. 52√3 − 8√42 + 91√14 − 196 4. − 196 5. 6 6. 0 IV. Density property and decimal representation of rational numbers 1. i) Non-Terminating Repeating ii) Terminating Decimal 2. i) 3.5 ii) 7.8 3. 0 4. NA

iii)

9 5 11

, ,

2 2

2

6. i) 0.36, Terminating Decimal ii) 0. 90909.., Non-Terminating

5 1

−1

5

5.

255 31

7. i)

14. i)

7

III. Irrational numbers and absolute value 1. −335

81

1

11

15.

99 −35

10 20

13. i)

3

<

9 13 −11 33

iii) −

82

iv)

ii)

1

16 8

iv)

−8

iii)

I. Operations on rational numbers 1. i) −

12. i)

7.

8.

7 11 23

,

,

, and

6 12 18 −6 −9

,

13 26

29

24

V. Rational numbers on the number line 1. NA

3

10

Worksheet 2 I. Multiple choice questions with single correct answer 1. c 2. a 3. d 4. a 5. a 6. b 7. d 8. b 9. a 10. b 11. d 12. b 13. c 14. b 15. d 16. a 17. c 18. b 19. a 20. c 21. a 22. d 23. c II. Fill in the blanks 1. infinite 2. density

229

ANSWER KEY 3.

4.

5.

6.

7.

−64

5

23.

2 24. 8 25. 0 26. −6

125 32

75 −17 20 11 3 1

2. LINEAR EQUATIONS IN ONE VARIABLE

21 7

Worksheet 1 I. Solving linear equations in one variable

8.

5 9. 0 10. 18 11. Positive 12. 1 6 13.

5 14. A 15. Positive 16. −4 III. Subjective questions 1.

2.

3.

4.

5.

6.

7.

1. i)

vii) 2 x) – 0.914 2. i) 8 iv) 36

9 291

315 𝑝𝑝𝑝𝑝

is the additive inverse of

15 −96 8

5

8. 28 m 8 9.

15.

1

24

6 −11

16.

9 17. 7 18. 19 19. 18 −77 20.

168 −3

21. i) 5 22. 250

230

v) −2

viii) 0.6 xi) 0.125 ii) 4 v) -5

viii) 1

10. First prize:

₹10000

35 33

xi) 12

ii)

5

ii)

15

2

71

iii) 3 vi)

2

ix)

3

3

ix) 1 xii) 207 iii) 2.7 vi) 13 2

xii) -8

II. Application of linear equations in one variable 1. 19 2. 15 3. 120,120, 160 4. Number of bees = 4, and flowers = 3 5. 84 6. 72 7. 24 8. ₹96000 9. ₹ 12,50,000 Third prize:

−1 −1

10. , 2 3 11. 0 −5 −5 −5 12. , , 7 4 2 13. 2 7 14. i)

3

10

ii) 4

vii) 2 x)

60 −11

91 55

5

iv) 1

13

𝑞𝑞𝑞𝑞 1

2

9 ₹ 5000 9

Second prize:

₹7500

11. 24 buffaloes, 36 cows and 90 goats 12. 7,35 13. 36 14. 62 15. Shobu’s age=

10 3

9

years Shobu’s mother’s

age = 20 years 16. Granddaughter’s age = 6 years and Grandfather’s age = 60 years 17. 72 18. Aman’s age = 60 years, Son’s age = 20 years 19. 21 years 20. A = 60 years, B = 30 years

ANSWER KEY 21. Father’s age = 30 years, son’s age = 5 years 22. 10 years 23. 39 years 24. 5 years ago 25. 20 years 26.

13

29.

7

21

27. Sobha's initial speed was 80 km/hr, and she increased her speed to 80+10=90 km/hr for the next two hours.

28. Ramaswamy took 1.875 hours to walk to the town and 0.625 hours to come back by bicycle. km

3

30. 1.6 km/hr

Worksheet 2 I. Multiple choice questions with single correct answer 1. a 2. a 3. b 4. b 5. c 6. d 7. b 8. b 9. c 10.c 11. a 12. d 13.c 14. d 15. d 16. d 17. b 18. b 19. b 20. b 21. b 22. c 23. a 24. c 25. a II. Fill in the blanks 1. 18 2. 13,10 3. 29 4. 2 5. 5 cm 6. –

𝑑𝑑𝑑𝑑

𝑐𝑐𝑐𝑐 8

7. −

5

6. −

12

8. 11 III. Subjective questions 1. 220, 352 2. 5 3. 3 (𝑥𝑥𝑥𝑥 + 3) years 4. 4𝑝𝑝𝑝𝑝 𝑝 9 = 11 5. 112 9

7. 𝑥𝑥𝑥𝑥 + 4 = 10 8. 𝑥𝑥𝑥𝑥 + 𝑦𝑦𝑦𝑦 = 180 9. 5 10. 31, 32

3. UNDERSTANDING QUADRILATERALS

Worksheet 1 I. Polygons 1. 74° 2. 12 3. 135° 4. Exterior angle of regular pentagon = 72° Exterior angle of regular decagon = 36° Ratio = 2 : 1 5. i) 14 ii) 44 iii) 90 iv) 170 6. i) 720° ii) 1260° iii) 1440° 7. i) Equilateral triangle ii) Square iii) Regular hexagon 8. i) 110° ii) 50° 9. i) 40° ii) 24° iii) 10° 10. 140° 11. 15 sides 12. 24 sides 13. i) 60° ii) 120° 14. 12 sides II. Quadrilaterals and types of quadrilaterals 1. 68°, 58°, 98°, 136° 2. 36°, 72°, 108°, 144° 3. 120° 4. 80° 5. 180°

6. 3�√55 + 4� cm2 7. 40°, 140°, 140° 8. ∠OAB = 30°, ∠OAD = 60° 9. 130° 10. 75° 11. 84° 12. Trapezium 13. 13 cm 14. 3 15. True 16. 25 cm, 55 cm 17. 45° 18. 60° 19. 18 cm

231

ANSWER KEY 20. 𝑥𝑥𝑥𝑥 = 90°, 𝑦𝑦𝑦𝑦 = 3 units

Worksheet 2 I. Multiple choice questions with single correct answer 1. d 2. c 3. b 4. a 5. b 6. a 7. b 8. a 9. a 10. a 11. a 12. a 13. a 14. b 15. d 16. a 17. a 18. a 19. c 20. c 21. d 22. a 23. c 24. c 25. b 26. d 27. c 28. b 29. b 30. c 31. d 32. d 33. c 34. d 35. b 36. a 37. a 38. c 39. b II. Multiple choice questions with multiple correct answers 1. a, c 2. b, c 3. d 4. a, b, c 5. b, d III. Fill in the blanks 1. 360° 2. 0 3. 24 cm 4. 60° 5. Rectangle 6. 120° 7. Trapezium 8. 180° 9. Hexagon 10. Non-parallel 11. 9 IV. Subjective questions 1. 123° 2. 90° 3. 45°, 75°, 105°, 135° 4. 360° 5. i) 18 sides ii) 8 sides iii) 72 sides iv) 12 sides 6. 72° 7. 2340° 8. i)17 sides ii) 5 sides 9. 60°, 120°, 100°, 120°, 140° 10. 80 232

11. i) 105° ii) 120° 12. 35 13. 85°, 85° 14. 90° 15. 83°, 97°, 83°, 97° 16. 9 cm and 15 cm 17. 60° 18. 50° 19. 4 cm and 8 cm 20. 60°

4. DATA HANDLING

Worksheet 1 I. Representation of data – pictograph, bar graph, double-bar graph 1. i) Class VIII ii) No iii) 12 girls 2. i) Blue ii) 5(Red, green, blue, yellow, orangv) 3.

i) 1998 ii) Yes. The demand for English books increased. Difference in the demand for English books from 1995 to 1998 is 620 − 350 = 270 Whereas for Hindi books the difference is 650 − 500 = 150 4. i) number of students in class VIII of a school ii) In the year 2004 − 05 iii) In the year 2007 − 08 iv) False 5. i) Cats ii) 8 1 1 3 11

6. i) , , ,

2 4 8 20

7. i) 1991 − 92

ii) B ii) 3

ANSWER KEY iii) 32.4% iv) 1995 − 1996 v) 115 II. Representation of data – histogram 1.

2.

3.

i) 0

ii) 70-80

i) 19-23 ii) 5 iii) 39 – 43 iv) 11 v) 20% 4. i) Youngest: 2, eldest: 1 ii) Most: 35-40, Least: 45-50 and 50-55 iii) 5

5.

6. i) 10 ii) 30 iii) 40 iv) 2 7. i) heights (in cms) of girls of Class 8 ii) group 140– 145 iii) 1 girl iv) Number of girls in Group A: 3 girls Group B: 11 girls Group C: 6 girls 8.

III. Representation of data – pie chart 1. ₹2500 2. i) food ii) children’s education and savings iii) 300

233

ANSWER KEY 11. i) Experiment ii) Experiment iii) Event iv) Event 12. i) Certain to happen. ii) May or may not happen iii) Certain to happen iv) Impossible to happen v) Impossible to happen vi) May or may not happen

3.

13. i)

14. i)

4.

15. i)

5. i) ₹540000 6. i) 2:1 iii) 144∘ 7. i) Hindi iii) 225 IV. Probability 1. i) 1 iii)

1 2

2. 20% 5

ii) ₹150000 ii) 96∘ ii) 30 ii) 0

26

4. 0.4 5.

6.

3 5 2 5

7. Red balls are more.

8. i) 9.

1

25 43

ii)

10 2

iii)

10. i)

1 2

1 2

iv)

5

iii)

234

1609 3

11 8

iii)

16. i)

3.

1

365 460

5

10

ii)

9

10

3

10

iv) 0

11 5

36

iii) 1

17.

11

13.

2

ii)

ii)

ii)

364

365 232

1149 3

10 9

iv)

11

ii) 0

36

Worksheet 2 I. Multiple choice questions with single correct answer 1. b 2. c 3. d 4. b 5. a 6. c 7. d 8. b 9. b 10. c 11. a 12. d 13. d 14. c 15. a 16. a 17. b 18. b 19. a 20. c 21. a 22. d 23. b 24. a 25. b 26. d 27. c 28. c 29. b 30. d 31. c 32. d 33. b 34. b 35. b 36. c 37. d II. Fill in the blanks 1. frequency 2. frequency 3. 25 4. 40 − 45 5. 20 − 30 6. its parts 7. probability 8. class mark or mid-point.

9. frequency.

10. 0 and 1 11. 1 12. impossible 5

14. random experiment

ANSWER KEY 15. 𝑛𝑛𝑛𝑛 III. Subjective questions 1. i) Martin ii) 700 iii) Anwar, Martin, Ranjit Singh 2. i) Number of tourists ii) Manali iii) 4000 3.

4. i) 5 iii) 6 v) 7 5. NA 6. a. i) 65 iii)30 b.

6

8. i)

1

7.

5

2

iii)

1 3

ii) 2 iv) 22 ii) 75 iv) 225

ii)

1 6

9.

4 9

10. i)

1

26 1

iii)

2

ii)

1 4

iv)

3

13

5. SQUARES AND SQUARE ROOTS

Worksheet 1 I. Square numbers and its properties 1. 16 and 1296 are perfect squares 2. 212 and 302 3. 1022 4. 109561, 114921 5. NA 6. 431 and 7779 7. i) 625 ii) 7225 iii) 11025 iv) 93025 8. i) 2601 ii) 2916 iii) 3136 iv) 3481 9. 601961 10. i) 576 ii) 1369 iii) 5041 iv) 9261 11. i) 16129 ii) 55225 iii) 196249 iv) 63001 12. i) 7921 ii) 75625 iii) 85849 iv) 119716 II. Pythagorean triplets and estimation of square root 1. ii), iv) and vi) 2. i) (6, 8, 10) ii) (10, 24, 26) iii) (8, 63, 65) iv) (14, 48, 50) 3. i) 15 ii) 31 iii) 18 iv) 22 4. i) 1,9 ii) 6 iii) 1,9 iv) 5 5. i), ii) and iii) III. Finding square root by prime factorisation, repeated subtraction and long division 1. i) 4 ii) 21 iii) 64 iv) 91 v) 217 vi) 437 2. 5, 25 3. 3, 21 4. 9,144 5. 154 6. 11, 8 235

ANSWER KEY 7. i) 98 ii) 84 8. i) 44 ii) 91 9. i) 3,60 ii) 7,84 iii) 3, 78 10. i) 7, 20 ii) 6, 26 11. i) 210 ii) 165 iii) 1111 iv) 2222 12. i) 31 ii) 40 iii) 291 13. i) 110 ii) 40 iii) 18 14. 1024 15. 9801 16. 49 17. 4i 18. 14400 19. 998001 20. 100489 21. 196 seconds 22. 6549 23. 1 24. 18, 75 IV. Finding square root of decimal numbers 1. i) 2.236 ii) 15.414 iii) 0.025 iv) 0.645 v) 0.935 vi) 1.443 2. i) 1.848 ii) 1.133 iii) 1.549 iv) 1.615 3. 242 4. 3.465 5. i) 1.6 ii) 6.5 iii) 9.210 iv) 0.902 v) 0.0197 6. 15.093 7. 16.02 Worksheet 2 I. Multiple choice questions with single correct answer 1. b 2. d 3. a 4. b 5. b 6. d 7. c 8. a 9. b 10. c 11. d 12. a 13. b 14. a 15. d 16. a 17. c 18. b 19. a 20. d 21. b 22. a

236

II. Fill in the blanks 1. 6.5 2. 15 3. 9 4. Even 5. 56 6.

5

12 2

7. 𝑛𝑛𝑛𝑛 8. 1.2 9. Greater III. Subjective questions 1. 24 2. 3600 3. 4 4. 21m 5. 220 + 221 6. either 0 or 1 7. 640 and 81000 ends with odd number of zeroes 8. 6 9. Yes. Explanation:196 = 2 × 2 × 7 × 7 = (2 × 7)2 = 142 clearly, 196 is expressed as the square of another natural number.

So, 196 is a perfect square.

10. 3.08 11. i) 15 ii) 101 iii) 809 iv) 6301 12. 145.98 13. i) 32 ii) 98 14. 45 15. 34 16. i) 27 ii) 32 4 20 ii) 17. i) 15 13 18. 2.99

6. CUBES AND CUBE ROOTS

Worksheet 1 I. Cube of a number and its properties 1. 3 2. 49 3. 0.512 4. NA 5. NA 6. NA 7. NA 8. 3375 m3

ANSWER KEY 9. 8 cm 10. 5, 10, 15 11. 50653 12. 85184 13. NA 14. NA 15. NA 16. 5 17. 105 18. 20 19. 2 20. 3 21. 6 II. Cube root of a number and its properties 1. 14 2. – 10 3. 20 4. 3.6 5. 6 6. i) 8 ii) 13 7. i) – 5 ii) 18 iii) – 26 8. 9 9. 24 10.

9

13

11. −

8

11

12. 55 13. 1728 14. 1.4 15. 30 16. 0.7 17.

7

12

18. 189 19. 6.216 20. 8 Worksheet 2 I. Multiple choice questions with single correct answer 1. d 2. b 3. d 4. c 5. b 6. d 7. b 8. c 9. a 10. a 11. b 12. c 13. b 14. c 15. b 16. a 17. d 18. c 19. b 20. c 21. d 22. c 23. c 24. a 25. B 26. d

II. Fill in the blanks 1. 2

2.

−8 7

3. 𝑦𝑦𝑦𝑦 3

4.

5.

64

15625 5 6

6. 0.3 7. −1 8. positive 9. 729 10. 24 III. Subjective questions

1. 6√3 m 2. No. 72𝑘𝑘𝑘𝑘 is not a perfect cube.

3. 1331cm3 4. 12 5. even number 6. 35 cm 7. 1 8. 9 9. 8 10. 2

7. COMPARING QUANTITIES

Worksheet 1 I. Ratio, proportion and percentage 1. i) 28% ii) 2.24% iii) 80% iv) 0.5% v) 93.75% vi) 88.8% vii) 5.6%

2.

i)

1 4

iii)

3

1000 45

ii)

1

400

3. – 4 4. 21:12:4 5. 64:25:49 6. 7592 7. 28:25 8. 24 9. 16.67% 10. 1:1:2 1 11. 33 % 3 12. 200% 237

ANSWER KEY 13. 5.1% 14. 850 15. 𝑦𝑦𝑦𝑦 = 9600, 𝑧𝑧𝑧𝑧 = 40000 16. 240 17. i) 825m ii) 2.5kg iii) ₹ 250 18. i) 29.03% ii) 51.61% iii) 6.4% iv)12.9% II. Profit, loss, discount and taxes 1. 25% 2. Profit = 10, profit percentage = 11.11% 3. ₹ 550 and ₹ 750 4. ₹ 880 5. ₹ 750 6. ₹ 600 7. ₹ 25,000 8. ₹ 360 9. ₹ 600 10. ₹ 680 11. ₹ 6,500 12. 12% III. Compound interest 1. ₹ 6,644.025 2. ₹ 1,261 3. ₹ 1,32,651 4. 10% 5. ₹ 12,500 6. 1,20,000 7. 7,00,000 8. ₹ 4,921 9. ₹ 8,860 10. ₹ 1,948.37 Worksheet 2 I. Multiple choice questions with single correct answer 1. d 2. c 3. d 4. c 5. d 6. b 7. d 8. a 9. b 10.c 11. b 12. d 13. b 14. a 15. c 16. b 17. c 18. b 19. a 20. d 21. a 22. b 23. a 24. d 25. a II. Fill in the blanks 1. 0.09 2. ₹ 100 3. 24 4. Marked price 5. Equal 6. Loss 7. Changes 8. 10 238

9. CI 10. 30% III. Subjective questions 1. ₹ 1350 2. ₹ 1700 3. 15% 4. 28% 5. 40% 6. 6% 7. ₹ 200 8. 3 9. ₹ 100 10. 10%

8. ALGEBRAIC EXPRESSIONS AND IDENTITIES

Worksheet 1 I. Introduction to algebraic expressions and polynomials 1. i) Terms: 8𝑥𝑥𝑥𝑥 2 𝑦𝑦𝑦𝑦𝑧𝑧𝑧𝑧 and −6𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦.

Coefficients: 8 and -6. ii) Terms: 𝑦𝑦𝑦𝑦 2 , 𝑦𝑦𝑦𝑦, and 1.

Coefficients: 1, 1 (for y), and 1 iii) Terms: 4𝑥𝑥𝑥𝑥 2 𝑦𝑦𝑦𝑦 2 , −7𝑥𝑥𝑥𝑥 2 𝑦𝑦𝑦𝑦 2 𝑧𝑧𝑧𝑧 2 , and 𝑧𝑧𝑧𝑧 2 .

Coefficients: 4, −7, and 1 iv)Terms: 7, −𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎, 𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏, and −𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏.

Coefficients: 1 , −1, 1 ,and −1 𝑥𝑥𝑥𝑥 𝑦𝑦𝑦𝑦 v) Terms: , , and −𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦.

2 2 11

Coefficients: , ,and −1.

22

vi) Terms: 0.3𝑥𝑥𝑥𝑥, −0.2𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦, and 0.7.

Coefficients: 0.3,−0.2, and 0.7.

2. i) Binomial ii) Monomial iii) Polynomial (do not fit in any of these categories) iv) Trinomial v) Binomial vi) Trinomial vii) Trinomial viii) Binomial ix) Polynomial (do not fit in any of these categories) x) Monomial xi) Binomial xii) Binomial 3. i) (2𝑥𝑥𝑥𝑥, −5𝑥𝑥𝑥𝑥), (3𝑦𝑦𝑦𝑦, −7𝑦𝑦𝑦𝑦),3𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦 ii)(3𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦, 20𝑦𝑦𝑦𝑦𝑥𝑥𝑥𝑥), (−4𝑝𝑝𝑝𝑝2 𝑞𝑞𝑞𝑞 2 , 7𝑝𝑝𝑝𝑝2 𝑞𝑞𝑞𝑞 ), (9𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎, 10𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 1

iii) 3𝑥𝑥𝑥𝑥, 4𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦, �−𝑦𝑦𝑦𝑦𝑧𝑧𝑧𝑧, 𝑧𝑧𝑧𝑧𝑦𝑦𝑦𝑦� 2

ANSWER KEY iv) (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏, 𝑏𝑏𝑏𝑏 2 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎, 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑎𝑎𝑎𝑎 2 ), (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 2 𝑏𝑏𝑏𝑏, 𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏𝑎𝑎𝑎𝑎 2 , 𝑎𝑎𝑎𝑎2 𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏, 𝑎𝑎𝑎𝑎 2 𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 ) II. Addition of algebraic expressions 1. i) 6𝑝𝑝𝑝𝑝 ii)8𝑎𝑎𝑎𝑎 iii)10𝑥𝑥𝑥𝑥 − 7𝑦𝑦𝑦𝑦 + 6𝑧𝑧𝑧𝑧 ii)−5𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 2. i)𝑎𝑎𝑎𝑎2 𝑎𝑎𝑎𝑎 iii)9𝑎𝑎𝑎𝑎2 𝑎𝑎𝑎𝑎 2 3. PR = 14 cm, QS = 10 cm, PS = 18 cm 4. i) 2𝑢𝑢𝑢𝑢 ii) 2𝑎𝑎𝑎𝑎2 iii)14𝑝𝑝𝑝𝑝 𝑝 4𝑞𝑞𝑞𝑞 iv) 4𝑥𝑥𝑥𝑥 3 + 3𝑥𝑥𝑥𝑥 𝑥 4 5. i) 3𝑎𝑎𝑎𝑎 + 15𝑎𝑎𝑎𝑎 + 7𝑏𝑏𝑏𝑏 + 𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 ii) −𝑥𝑥𝑥𝑥 2 − 7𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦 − 7𝑦𝑦𝑦𝑦 2 6. i) 8𝑥𝑥𝑥𝑥 2 𝑦𝑦𝑦𝑦 + 3𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 2 ii)11𝑥𝑥𝑥𝑥 2 − 4𝑥𝑥𝑥𝑥 𝑥 6 iii) −

iv)

51

𝑎𝑎𝑎𝑎

𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦 −

14 3

v)5𝑥𝑥𝑥𝑥

1

15

31

𝑦𝑦𝑦𝑦 𝑦

19

𝑥𝑥𝑥𝑥

10 35 11 7 + � � 𝑥𝑥𝑥𝑥 2 − � � 𝑥𝑥𝑥𝑥 4 2

III. Subtraction of algebraic expressions 1. i) 10𝑝𝑝𝑝𝑝 ii) 5𝑦𝑦𝑦𝑦 iii) −4𝑡𝑡𝑡𝑡 2. i)6𝑥𝑥𝑥𝑥 ii)10𝑎𝑎𝑎𝑎 iii) 4𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦 3. i) 𝑎𝑎𝑎𝑎 + 2𝑎𝑎𝑎𝑎 ii)2𝑥𝑥𝑥𝑥 + 6𝑦𝑦𝑦𝑦 2 2 iii) −7𝑥𝑥𝑥𝑥 + 6𝑦𝑦𝑦𝑦 4. i)−2𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎 − 2𝑏𝑏𝑏𝑏. ii)2𝑎𝑎𝑎𝑎3 + 6𝑎𝑎𝑎𝑎2 − 10𝑎𝑎𝑎𝑎 + 17 5. i)5𝑎𝑎𝑎𝑎2 − 9𝑎𝑎𝑎𝑎 + 1 ii) −6𝑎𝑎𝑎𝑎2 + 8𝑎𝑎𝑎𝑎 + 12 6. 𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎 + 7𝑏𝑏𝑏𝑏 7. i) 2𝑥𝑥𝑥𝑥 2 + 1 ii) −𝑥𝑥𝑥𝑥 2 + 3𝑥𝑥𝑥𝑥 𝑥 1 8. 𝑥𝑥𝑥𝑥 2 − 8𝑥𝑥𝑥𝑥 9. i)17𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦 ii) 𝑎𝑎𝑎𝑎 𝑎 4𝑎𝑎𝑎𝑎 iii) 2𝑥𝑥𝑥𝑥 3 + 5𝑥𝑥𝑥𝑥 2 − 2𝑥𝑥𝑥𝑥 + 1 iv) −2𝑎𝑎𝑎𝑎2 + 2𝑎𝑎𝑎𝑎 𝑎 4 1

v)− 𝑥𝑥𝑥𝑥 2 𝑦𝑦𝑦𝑦 + 3

vi)−2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 +

23

15

5

𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 2 − 𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦

10 184𝑏𝑏𝑏𝑏𝑐𝑐𝑐𝑐

−

𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 7

3

IV. Multiplication of algebraic expressions 1. i) 8𝑥𝑥𝑥𝑥 ii) −12𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦 iii) −72𝑥𝑥𝑥𝑥 4 𝑦𝑦𝑦𝑦

6

iv) − 𝑙𝑙𝑙𝑙 4 𝑚𝑚𝑚𝑚2 5

ii) 16𝑎𝑎𝑎𝑎3 + 40𝑎𝑎𝑎𝑎2 𝑎𝑎𝑎𝑎 2. i) 30𝑎𝑎𝑎𝑎2 − 15𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 iii) 45𝑥𝑥𝑥𝑥 3 + 63𝑥𝑥𝑥𝑥 2 iv) 6𝑥𝑥𝑥𝑥 3 − 8𝑥𝑥𝑥𝑥 4 3. i) 6𝑥𝑥𝑥𝑥 4 𝑦𝑦𝑦𝑦 𝑦 8𝑥𝑥𝑥𝑥 3 𝑦𝑦𝑦𝑦 3 + 4𝑥𝑥𝑥𝑥 3 𝑦𝑦𝑦𝑦 2

ii) 5𝑥𝑥𝑥𝑥 4 − 6𝑥𝑥𝑥𝑥 2 + 3𝑥𝑥𝑥𝑥 4. i) 15𝑝𝑝𝑝𝑝2 − 43𝑝𝑝𝑝𝑝 + 30 ii) 20𝑓𝑓𝑓𝑓 2 + 9𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 − 18𝑓𝑓𝑓𝑓2 5. i) 2𝑥𝑥𝑥𝑥 4 − 7𝑥𝑥𝑥𝑥 3 + 16𝑥𝑥𝑥𝑥 2 − 29𝑥𝑥𝑥𝑥 + 21 ii) 15𝑥𝑥𝑥𝑥 3 − 16𝑥𝑥𝑥𝑥 2 − 2𝑥𝑥𝑥𝑥 + 6 iii) 6𝑥𝑥𝑥𝑥 4 − 𝑥𝑥𝑥𝑥 3 − 22𝑥𝑥𝑥𝑥 2 + 36𝑥𝑥𝑥𝑥 𝑥 20 iv) 60 − 92𝑥𝑥𝑥𝑥 + 41𝑥𝑥𝑥𝑥 2 − 5𝑥𝑥𝑥𝑥 3 v) −2𝑥𝑥𝑥𝑥 2 + 30𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦 + 3𝑦𝑦𝑦𝑦 2 vi) −𝑥𝑥𝑥𝑥 3 − 22𝑥𝑥𝑥𝑥 2 + 30𝑥𝑥𝑥𝑥 𝑥 9 vii) 𝑥𝑥𝑥𝑥 4 − 5𝑥𝑥𝑥𝑥 3 + 10𝑥𝑥𝑥𝑥 2 − 12𝑥𝑥𝑥𝑥 + 7 viii) 𝑥𝑥𝑥𝑥 5 𝑦𝑦𝑦𝑦 2 + 4𝑥𝑥𝑥𝑥 3 𝑦𝑦𝑦𝑦 3 − 2𝑥𝑥𝑥𝑥 3 𝑦𝑦𝑦𝑦 4 − 8𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 5 ix)2.25𝑥𝑥𝑥𝑥 2 − 16𝑦𝑦𝑦𝑦 2 V. Finding the value of an expression for a given value of the variable 1. i) 32 ii) 16 iii) 0 2. i) −2 ii) 13 iii) 0 3. i) 22 ii) 35 iii) 10 iv) 89 4. i) 5 ii) 24 iii) 5 iv) 8 5. i) 3400 ii) 30 6. 36 VI. Algebraic identities ii)𝑙𝑙𝑙𝑙 2 − 𝑚𝑚𝑚𝑚2 1. i) 𝑝𝑝𝑝𝑝2 − 9𝑞𝑞𝑞𝑞 2 iii) 4𝑥𝑥𝑥𝑥 2 − 16 2. i) 𝑎𝑎𝑎𝑎2 − 2𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎 2 ii)

4𝑥𝑥𝑥𝑥 2 −4𝑥𝑥𝑥𝑥𝑥𝑥 𝑥𝑥𝑥𝑥 2 2

iii) 25𝑥𝑥𝑥𝑥 − 30𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦 + 9𝑦𝑦𝑦𝑦 2 ii) (9𝑥𝑥𝑥𝑥 + 5𝑦𝑦𝑦𝑦)2 3. i) (3𝑥𝑥𝑥𝑥 + 2𝑦𝑦𝑦𝑦)2 4. i) 2𝑥𝑥𝑥𝑥 2 + 12𝑥𝑥𝑥𝑥 + 20 ii) 4𝑙𝑙𝑙𝑙 2 𝑚𝑚𝑚𝑚2 iii) 4𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 iv) 2𝑎𝑎𝑎𝑎2 + 2𝑎𝑎𝑎𝑎 2 5. i) 10404 ii) 9216 iii) 399.75 6. 25 7. i)𝑎𝑎𝑎𝑎2 + 4𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 + 6𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 + 4𝑎𝑎𝑎𝑎 2 + 12𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 + 9𝑏𝑏𝑏𝑏 2 ii)𝑎𝑎𝑎𝑎2 + 4𝑎𝑎𝑎𝑎 2 + 9𝑏𝑏𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 + 12𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 − 6𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 8. NA 9. i) 10 ii) 4𝑎𝑎𝑎𝑎 + 5𝑎𝑎𝑎𝑎 2 10. 9964 11. 20 12.

1 4

239

ANSWER KEY 13. 2009 14. 14

Worksheet 2 I. Multiple choice questions with single correct answer 1. c 2. d 3. b 4. d 5. c 6. d 7. b 8. b 9. b 10. d 11. d 12. a 13. c 14. d 15. a 16. a 17. a 18. a 19. c 20. d 21.c 22.b 23.b 24.a 25.b 26.b 27.a 28.b 29.a 30.b 31.c 32.a 33.c 34.d 35.c 36.d 37.a 38.a 39.c II. Match the following 1. d 2. a 3. d III. Subjective questions 23

9

53

1. � � 𝑎𝑎𝑎𝑎 𝑎 � � 𝑎𝑎𝑎𝑎 + � � 𝑏𝑏𝑏𝑏 6 1

2

2. − 𝑥𝑥𝑥𝑥 𝑦𝑦𝑦𝑦 + 3

2

4 23 10

2

𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 − 𝑥𝑥𝑥𝑥𝑦𝑦𝑦𝑦

3. −𝑦𝑦𝑦𝑦 − 7𝑦𝑦𝑦𝑦 + 12 4. 12mn2 p sq. units 5. 1.74𝑥𝑥𝑥𝑥 6 𝑦𝑦𝑦𝑦 5 𝑧𝑧𝑧𝑧 8 4

6. − 𝑥𝑥𝑥𝑥 4 𝑦𝑦𝑦𝑦 4 𝑧𝑧𝑧𝑧 5 9 4

20 5 3

2

7. − 𝑥𝑥𝑥𝑥 2 𝑦𝑦𝑦𝑦 2 𝑧𝑧𝑧𝑧 3 + 𝑥𝑥𝑥𝑥 2 𝑦𝑦𝑦𝑦 3 𝑧𝑧𝑧𝑧 4 9

3

3

8. −3𝑥𝑥𝑥𝑥 3 𝑦𝑦𝑦𝑦 3 + 𝑥𝑥𝑥𝑥 2 𝑦𝑦𝑦𝑦 4 9. −

𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 14

+

𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 2 21

2

+

𝑎𝑎𝑎𝑎2 𝑏𝑏𝑏𝑏 18

−

𝑎𝑎𝑎𝑎2 𝑏𝑏𝑏𝑏 2 27

10. 1.2𝑎𝑎𝑎𝑎2 − 3.15𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 + 1.5𝑎𝑎𝑎𝑎 2 11. −𝑥𝑥𝑥𝑥 3 − 22𝑥𝑥𝑥𝑥 2 + 30𝑥𝑥𝑥𝑥 𝑥 9 12. i) 1002001 ii) 494204 13. i) 3.96 ii) 217000 14. 398 15. √20 and 4

9. MENSURATION

Worksheet 1 I. Perimeter and area of polygons and circles 1. 200 m2 2. 20 cm2 3. 45000 tiles 4. Square 5. 4√3 cm2 240

6.

7.

8.

9.

39.27 cm2 12.57 feet2 A= 80 cm2 and P = 36cm 308 cm2

10. 16√3 cm2 11. 3.5 m2 II. Surface area of solid objects 1. Area of metal = 1232m2 Area of canvas required =616√2 𝑚𝑚𝑚𝑚2 2. 1377.14 cm2 2

3. 710 cm2 7

4. 48 cm2 5. 2681.25m2 6. 52.8 m 7. 440 sq.cm 8. 13 m 9. 49 10. i) 2016 cm2 ii) 30 cm2 iii) 2042 cm2 iv) 401.2 cm2 III. Volume of solid objects 1. 12 cm 2. Cylinder B has a greater volume and surface area than Cylinder A. ii) 1540 cm3 3. i) 748 cm2 iii) ₹1760 4.

5588 21

cm3

2

5. 102 cm3 3

6. 1728 cm3 7. 2560π cm3 8. 9 cm3 9. 30 cm3 10. 48 cm3 IV. Surface area and volume of prism and pyramid 1. 340 cm2 2. 108 cm3 3. 84 square units 4. 52 cm2 5. 400 cm3 Worksheet 2 I. Multiple choice questions with single correct answer 1. c 2. a 3. d 4. d 5. b

ANSWER KEY 6. b 7. a 8. a 9. c 10. c 11. a 12. d 13.a 14. a 15. c 16. c 17. b 18. d 19. b 20. c 21. c 22. c 23. d 24. b 25. c II. Subjective questions 1. 528.11 cm2 2. 439.82 cm2 3. 325 m2 4. 129.5 m2 5. 8 cm2 6. 17,875 rupees 7. 1760 cm3 8. 440 m2 9. 1m 10. 3 cm 11. 432120 cm3 12. Height = 2.67 cm (approx.)

Surface area = 430.89 cm2 (approx.)

13. 30 hectares 14. 30

241