10TH STANDARD
MATHEMATICS
WORK BOOK WITH PREVIOUS QUESTION PAPERS
Guidance
:
Sri S.K.B. Prasad,
Educational Officer, DDPI Office, Chitradurga Ph: 94486 94642
Concept:
Sri K.P. Obaiah, Subject Inspector (Mathematics), DDPI Office, Chitradurga Ph: 94495 12518
Resource Persons
:
(1) Sri K.S. Thippeswamy, Girisha Jr. College, Hiriyur Ph: 98866 07469
(2) Sri B.V. Guruprakash, Govt. Girls Jr. College, Chitradurga Ph: 99643 48058
(3) Sri M.J. Rudramuni, B.M.G.H.S., Challakere Ph: 98458 02338
(4) Sri B. Shivakumar, Govt. High School, Thimmappayyanahalli, Challakere Tq. Ph: 98459 05534 Over all Supervision :
Sri H. Manjunath Deputy Director for Public Instruction Chitradurga
Sri N M Ramesh Education Officer Deputy Director for Public Instruction Chitradurga
For copies : Contact the Resource Persons or Phone Numbers given above
CONTENTS S.No.
CHAPTERS
PAGE No.
1. NUMBER SYSTEM
1  80
2. PERMUTATIONS AND COMBINATIONS 3. STATISTICS
81  104 105  122
4. FACTORS AND FACTORISATION
123  180
5. QUADRATIC EQUATIONS
181  222
6. MODULAR ARITHMETIC
223  232
7. PRACTICAL GEOMETRY
233  244
8. THEOREMS ON TRIANGLES AND CIRCLES 9. MENSURATION 10. POLYHEDRA AND NETWORKS
245  290
291  316 317  335
1
NUMBER SYSTEM SET THEORY • Set: A Set is a collection of well defined and well distinguished objects. • Elements: The objects which constitute the set are said to be elements of the set. • Representation of Set: (a)
Roster / Tabulation Method: In this method the set is represented by listing all its elements by commas and enclosing them in flower brackets. Ex: A = {1, 2, 3, ................}
(b) Rule Method: Defining the set is called rule method. P={Set of all natural numbers less than 6}
• Finite Set: A set is said to be finite if it has finite number of elements. • Infinite Set: A set is said to be infinite if it has an infinite number of elemements. • Order of a Set: Is the number of elements it contains. • Empty Set: A set having no elements is said to be an empty set. It is also called null set or void set. It is denoted by {} or Φ. Order of null set A is n(A) = 0.
• Singleton Set: A set having only one element is said to be a singleton set. • Subset: Let A and B be two sets them the set A is said to be a subset of the set B if each element of A is also an elements of B symbolically we write it as A ⊆ B.
• Equal Sets: Two sets are said to be equal if A ⊂ B and B ⊂ A. Two sets A and B are said to be equal if they have exactly same elements.
• Equivalent Sets: Two sets are said to be equivalent if they have same number of elements.
• Universal Sets: The main set under discussion or the set containing all possible values in the given frame of reference is said to be universal set and is denoted by U. It is the Union of all the sets.
• Operation on Sets: (a) Union of Sets: Let A and B be two sets. The Union of A and B is the set of al lthose elements which are either in A or in B or in Both. It is denoted symbolically by A ∪ B = {x : x A or x B} A
B
5
(b) Intersection of Sets: Let A and B be two sets. Then intersection of A and B is the set of all those elements which are in both A and B. If it denoted symbolically by A ∩ B A ∩ B = {x : x A and x B} A
B
• Disjoint Sets: Two sets are said to be disjoint if and only if they have no common elements. A∩ B = Φ n (A ∩ B) = 0 A
B
• Difference of Sets: Let A and B be two sets then (A  B) is the set of those elements of the Set A which are not in the set B (A  B) = {x : x A and x (B  A) = {x : x B and x A
B
B} A}
A
(A  B)
B
(B  A)
• Complement of Sets: Let U be the Universal set and A be any set then the complement of the set A is the set of all those elements of U which are not in the set A. This is denoted by A1 or (U  A). A1 = {x : x U
U and x
A}
A
2
Fundamental Resluts A∪ Φ = A A∩Φ=Φ
(1)
Identity law
:
(2)
Commutative Law
:
i) (A ∪ B) = (Β ∪ Α) ii) (A ∩ B) = (B ∩ A)
(3)
Associative Law
:
i) A ∪ (B ∪ C) = (A ∪ B) ∪ C ii) A ∩ (B ∩ C) = (A ∩ B) ∩ C
(4)
Distributive Property :
i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
(5)
DeMorgan’s Law
i) (A ∪ B)1 = A1 ∩ B1 ii) (A ∩ B)1 = A1 ∪ B1
(6)
If A1 = Φ then A = U
(7)
(A1)1 = A
(8)
A ⊆ B then A1 ⊆ B1
(9)
A ∪ A1 = U and A ∩ A1 = Φ
(10)
If A ∩ B = Φ then A1 ∪ B1 = U
(11)
A  (A  B) = A ∩ B
(12)
If (A  B) = A then (A ∩ B) = Φ
(13)
If B is subset of A then A ∩ B = B and A ∪ B = A
(14)
If A, B, C are finite sets then
:
(a)
n(A ∪ B) = n(A) + n(B) (If A and B are disjoint)
(b)
n(A ∪ B) + n(A ∩ B) = n(A) + n(B)
(c)
n(A ∩ B1) = n(A)  n(A  B)
(d)
n(A ∩ A1) = n(B)  n(A ∩ B)
(e)
n(A ∪ B) = n(A ∩ B1) + n(B ∩ A1) + n(A ∩ B)
(f)
n(A − B) + n(A ∩ B) + n(B − A) = n(A ∪ B)
3
I
Fill the box with correct answer A
B
1
{a, b, c, d}
{b, d, e, f}
2
{1, 2, 3, 4, 5}
{1, 3, 5, 8, 9}
3
{c, a, t}
{r, a, t}
4
{x / x
5
{1, 3, 5, 7}
N&1≤x≤5}
{0, 1, 2}
{2, 4, 6, 8}
II
Write the given sets in Roster method.
(1)
A = {x : x N and 1 ≤ x ≤ 5} A = { 1, 2, 3, 4, 5}
(2)
B = {x : x is a perfect square less than 10}
(3)
C = {x : x is a multiples of 3 less than 10}
(4)
D = {x : x
(5)
E = {x : x is even digit less than 10}
(6)
F = {x : x is an odd numbers less than 10}
(7)
G = {x : x is a prime numbers less than 15}
Z and 2 ≤ x ≤ 2}
4
A ∪B {a, b, c, d, e, f}
A∩ B {b, d}
III
Match the Following. (for Practice purpose)
(a)
1. De Morgan’s Law
(a) (A∪B) = (B∪A)
1. ...................
2. Commutative Property
(b) A∪(B∩C)=(A∪B)∩(A∪C)
2. ...................
3. Associative Property
(c) (A∪B)1 = A1∩B1
3. ...................
4. Compliment of A
(d) A∩(B∪C)=(A∩B)∪(A∩C)
4. ...................
(e) U  A = A1
5. ...................
(f) (A∪B)∪C=A∪(B∪C)
6. ...................
5. Union is distributive over intersection 6. Intersection is Distributive over Union (b)
IV
1. A ∪ B
(a) Φ
1. .................................
2. A ∩ (B ∩ C)
(b) U
2. .................................
3. A ∪ (B ∪ C)
(c) B ∪ Α
3. .................................
4. (A ∩ B)1
(d) (A ∩ B) ∩ C
4. .................................
5. (A ∪ B)1
(e) A1 ∪ B1
5. .................................
6. (A1)1
(f) n(A∪B) + n(A∩B)
6. .................................
7. n(A) + n(B)
(g) A
7. ..................................
8. A ∪ A1
(h) (A∪B)∪C
8. ..................................
9. A ∩ A1
(i) A1 ∩ B1
9. ..................................
Write the shaded portion represents in the given figure.
(1)
(2)
(A  B) ................................ (4)
................................
(3)
................................
(5)
................................
(6)
................................
5
................................
(7)
A
B
(8)
A
A
B
C
C
C
................................
(9)
B
................................
................................
V
Draw a Venn Diagram for the follwoing sets and find A∪B and A∩B.
(1)
If P = {x : x is an odd number less than 11} and Q = {x : x is prime number less than 15}
P = {1, 3, 5, 7, 9} Q = {2, 3, 5, 7, 11, 13} P ∪ Q = {1, 2, 3, 5, 7, 9, 11, 13} P ∩ Q = {3, 5, 7} (2)
3 5 7
If A = {0, 2, 4, 6, 8} and B = {x : x is even digit less than 5}
If M = {x : x N and 5 < x < 15} and N = {odd natural numbers less than 10} M= N=
(4)
Q 1 9
A= B= AB = BA =
(3)
P
n(A) = 65, n(A∩B) = 25 and n(A∪B) = 75 n(B) =
6
2 11 13
(5)
In a school 60% of students invest in NDF and 70% of them invest in NSS. n(NDF) = n(NSS) = n(NDF ∪ NSS) = n(NDF ∩ NSS) =
VI
See the Venn’s diagram and write the following. U A
B 1 5 0
3 4 6
2 8 9
(a) U = {..........................................} (b) A = {...........................................} (c) B = {..........................................} (d) A∪B = {......................................} (e) A∩B = {......................................} (f) AB = {.......................................} (g) BA={........................................} (h) B1 = {..........................................} (i) A1 = {.........................................} (j) (A∪B)1 = {......................................} (k) (A∩B)1 = {......................................}
VII
If U ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} A = {1, 2, 3, 7, 9} and B = {2, 4, 6, 7, 9} write down the following. A1 = B1 = AB= B A= A∪B= A∩B= (A ∪ B)1 = (A ∩ B)1 = 7
VIII
Fill up the blanks with suitable answer.
(1)
If n(A  B) = 5, n(B  A) = 4, n(A ∪ B) = 15 the n (A ∩ B) is .................................
(2)
In the Venn Diagram
(3)
If A and B are two sets, (A ∩ B) = Φ then A and B are ..................................
(4)
If A and B are sets, such that n(A) = 76, n(B) = 24, n(A ∩ B)=10 then n(UB) is
A
unshaded portion represents ................................
..................................... (5)
A, B and C are the sets. Intersection of sets is distributive over Union of Sets is represented as ....................................
(6)
If U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (A ∪ B)1 = {1, 5, 7} then (A ∪ B) is ..................
(7)
If A is subset of B and B is subset of A. Then the relation between A and B is ........ ...................................
(8)
If A and B are the subsets of U and A1 ∪ B1 = {2, 3, 5,} U = {1, 2, 3, 4, 5, 6} then A ∩ B is ........................................
(9)
A and B are the sets. If AB = A then A ∩ B = ...................................
(10)
If B ⊂ A then the value of A ∩ B is ........................................
(11)
The value of A ∩ A is ..........................................
(12)
Compliment of the Universal Set is ........................................
(13)
In an Hotel 30 people speak only Hindi, 20 speak only English, 10 speak both Hindi and English. The number of people in the Hotel is ....................................
(14)
If P and Q are non empty sets and P  Q = P then P ∩ Q is ...................................
IX
Choose the correct answer.
(1)
In a distributive set P ∪ (Q ∩ R) = ................................. (B) (P∪Q) ∩ (P∪R) (A) (P∪Q) ∪ (P∪R) (D) (P∩Q) ∩ (P∩R) (C) (P∩Q) ∪ (P∩R)
(M  10)
8
(J  06)
(2)
In a 1000 students 750 pupil plays cricket, 350 pupil plays voly ball and 150 plays
(J  06)
both number of pupils do not play any game is ................................. (A) 200 (B) 500 (C) 250 (D) 50 (3)
A = {a, b, c, d, e} B = {a, c, e, g, h} C = {c, f, g} then A∩(B∪C) is...................... (J  06) (A) {a,c,e} (B) {g,e} (C) {b,d} (D) {c}
(4)
Union of the sets is distributive over intersection this can be represented by............ (M  06) (B) (X∪Y)∪Z = X∪(Y∪Z) (A) X∪Y = Y∪X (C) X∪(Y∪Z) = (X∪Y)∩(X∪Z) (D) X∩(Y∩Z) = (X∩Y)∪(X∩Z)
(5)
(M  06) If A = {1, 2, 3, 4} and B = {3, 4, 5} then A  B is ........................... (A) {3,4} (B) {1,5} (C) {1,2} (D) {0,1,2}
(6)
Set A is subset of U, n(U) = 11 and n(A1) = 8 then the number of elements in A is
(M  06)
.............................. (A) 19 (7)
(B) 11
(C) 8
(D) 3
Set A = {a,b,c,d} B = {b,c,e} then n (A∩B) is ......................... (A) 4 (B) 3 (C) 7
(M  07)
(D) 2
(8)
U = {0, 1, 2, 3, 4} A = {2, 3, 4} B = {0, 2, 3} then (A∩B)1 is ......................... (M  07) (A) {0,1,2,3,4} (B) {0,1,4} (C) {1,4} (D) { }
(9)
In a 9 traveller 5 can speak Kannada, 2 can speak both Kannada and English then the number of Traveller speak only English is ......................... (A) 5 (B) 3 (C) 4
(M  07)
(D) 6
(10)
(J  07) In the set (A∪B)∪C = A∪(B∪C) this means ......................... (A) Union of sets is commutative (B) Union of the sets is associative (C) Union of the sets is distributive over intersection (D) Intersection of the sets is distributive over union
(11)
Universal set U = {2,3,5,6,10} subset A = {5,6} then A1 represents the diagram .............................. (A) U (B) U A 5 6
2
3 10
(J  07)
2
A 5 6
3 10
9
(C) U A 5 6
2 3 10
(D) U A 5 6
2 3 10
(12)
In a class of 60 students 22 member play volyball, 12 members play both volyball and kho kho, 17 members can not play any game. The number of students play
(J  07)
khokho only is ......................... (A) 32 (B) 28
(C) 33
(D) 21
(13)
Set A = {x N / 1≤ x ≤ 4} and B = {3,4,5} then set A  B is equal to ...................... (M  08) (A) {5} (B) {1,2} (C) {3,4,5} (D) {1,2,3,4}
(14)
(M  08) (A∪B)1 = {2,4,6} then which set represents A1∩B1 ...................... (A) {1,2,3,4,5,6} (B) {2,4,6} (C) {1,3,5} (D) { }
(15)
n(A) + n(B) = n(A∪B) then n(A∩B) is equal to .......................... (C) 1 (A) 0 (B) Φ
(16)
(C) A∪B
(B) B  A
(M  09)
(D) A∩B
P, Q and R are three sets then (P∪Q) ∩ (P∪R) is ............................ (A) P∪(Q∪R)
(18)
(D) 2
Set A = {2,3,4,5} and B = {4,5} which of the following is a Null Set ................... (A) A  B
(17)
(M  08)
(B) P∩(Q∩R)
(C) P∪(Q∩R)
(M  09)
(D) P∩(Q∪R)
Set A and B are the subsets of U, A1∪B1 = {2, 3, 5} and U = {1, 2, 3, 4, 5, 6} then (M  09)
A ∩ B=.......................... (A) {2,3,5} (19)
(B) {1,4}
(C) {1,2,3,4,5,6}
(D) {1,4,6}
In a class of 50 students every one should be the members of Science club or Maths club. 29 students are in science club, 11 students are the members in both the clubs. The number of students only in Maths club is ......................... (A) 21
(20)
(C) 11
(D) 10
In a sets A and B, A  B = A then A ∩ B=.......................... (A) A
(21)
(B) 18
(B) B
(C) U
If A = {1,2,4} B = {1,4,5} then A ∩ B is .......................... (A) {4,5}
(B) {1,5}
(C) {1,4}
10
(M  09)
(M  09)
(D) Φ (J  08)
(D) {2,5}
(22)
If U = {1,2,3,4,5,6} and A = {1,2,3} B = {2,4} then which diagram represents (J  08)
A ∩ B ............................ (A)
U A
1 2 4 3
(C)
U A
U A
B
1 2 4 3
5 6
(D)
B
1 2 4 3
(23)
(B)
B
U A
B
1 2 4 3
5 6
5 6
5 6
According to given Venn diagram n(A) is ............................
A
B
(A)
8
(B)
5
(B)
7
(D)
13
(J  08)
8 5 7
(24)
(25)
In a given statement which one is DeMorgan’s rule ............................ (A) n(A)+n(B) = n(A∪B)+n(A∩B)
(B) n(A1) = n(U)  n(A)
(B) A∪(B∩C) = (A∪B)∩(A∪C)
(D) (A∩B)1 = A1∪B1
A and B are the sets n(A) = 11, n(B) = 7, n(A∩B) = 3 then n(A∪B) is ................... (A) 21
(26)
(B) 15
(J  08)
(D) 10
(B) {0,2,4,6,8}
(C) {2,4,6,8}
(B) Φ
(C) {0}
A = {1,2,3} B = {0,1,3,4} and C = {2,3,4} are the sets then A∪(B∩C) is .............. (J  09)
If P = {2,3,4} and Q = {3,5,7} then (PQ) is ......................... (A) {3,7}
(30)
(J  09)
(D) {Φ}
(A) {0,1,2,3} (B) {0,1,3,4} (C) {1,2,3,4} (D) {2,3,4} (29)
(J  09)
(D) {4,6,8,9}
Set A and B are have no common elements then n(A∩B) is ................... (A) 0
(28)
(C) 8
U = {0,1,2,3,4,5,6,7,8,9} and A = {0,1,3,5,7} then A1 is ................... (A) {0,2,3,4,6,8,9}
(27)
(J  08)
(B) {2,4}
(C) {2,3,4,7}
(J  10)
(D) {3}
If A ⊂ B, B ⊂ C and C ⊂ A then which is correct ............................. (A) A = B
(B) B = C
(C) A = C 11
(D) One of them is null set
(J  10)
(31)
5 boys eat apple only. 4 boys eat banana only, the number of boys eat apple out of
(J  10)
12 boys is ............................. (A) 5
(B) 4
(C) 8
(D) 9
(32)
If A∩B = Φ then A1∪B1 is ............................. (A) Null Set (B) Universal Set (C) Disjoint Set (D) Compliment Set
(33)
If n(AB) = 6, n(BA) = 5, n(A∩B) = 3 then n(A∪B) is ............................. (A) 8 (B) 9 (C) 11 (D) 14
(34)
A = {a,b,c,d,e} B = {a,c,e,g,h} and c = {c,f,g} then A∩(B∩C) is ..................... (A) {a,c,e} (B) {g,e} (C) {b,d} (D) {c}
(35)
In a survey of 5000 persons it was found that 2800 read Indian Express and 2300
(J  10)
read Times of India, while 400 read both. Then the number of people who read neither Indian Express nor Times of India ..................... (A) 200 (B) 300 (C) 600
(D) 900
X
Solve the following.
(1)
If A = {3,4,5,6} and B = {4,5,6,7,8} find A∪B and A∩B and draw Venn Diagram (i) A∪B = {
}∪{
={ U
} A
B
(ii) A∩B = {
}∩{
={
}
} }
A
B
12
(2)
If A = {0,2,4,6,8} and B = {x/x is on even digit less than 5} verify and (ii) A ∩ B = B ∩ A (i) A ∪ B= B ∪ A (i) A∪B = {
}∪{
}
={ B∪A = {
} }∪{
}
={ (ii) A∩B = {
} }∩{
}
={ B∩A = {
} }∩{
}
={
}
(3)
If P ={x/x is an odd numbers less than 11} Q = {x/x is prime number less than 15} verify (i) P∩Q = Q∩P and (ii) P∪Q = Q∪P.
(4)
If A = {x/x N and 5 < x < 10} B = {odd natural numbers less than 8} and C = {even natural numbers less than 9}. Verify Union of sets is associative. A={
}
B∪C = {
}
B={
}
A∪B = {
}
C={
} } ∪
{
}
{
}
A∪(B∪C) = { ={
}
(A∪B)∪C = {
} ∪
={
}
∴ A∪(B∪C) = (A∪B)∪C
13
(5)
If A = {3,4,5,9} B = {4,5,6,8} and C = {5,7,8,9} verify that the intersection of sets (M  07) is associative.
(6)
If A = {5,6} B = {6,7,8} and C = {5,6,7} then find (A∪B)∩(A∪C) (A∪B) = {
}∪{
={
}
}
(A∪C) = {
}∪{
={
}
}
(A∪B)∩(A∪C) = { (7)
(J  10)
}
X = {x : x is a prime numbers less than 10} Y = {x : x is even digit less than 10} Z = {x : x is odd digit less than 10} verify that intersection of set is associative. X={
}
(X∩Y) =
Y={
}
(Y∩Z) =
Z={
}
X∩(Y∩Z) = {
}∩{
={
(J  06)
} }
(X∩Y)∩Z = {
}∩{
={
} }
∴ X∩(Y∩Z) = (X∩Y)∩Z
14
(8)
A = {1,3,4,8,9,12} B = {1,4,9} and C = {2,4,8,10} verify A∩(B∪C) = (A∩B) ∪ (A∩C) B∪C = {
}
A∩(B∪C) = {
A∩B = {
}
={
A∩C = {
}
(A∩B)∪(A∩C) = {
}∩{
} }
}∩{
={
} }
∴ A∩(B∪C) = (A∩B) ∪ (A∩C)
(9)
A = {2,4,6,8,10} B = {1,2,3,4,5,6} and C = {1,3,5,7,9,11,13} then verify A∪(B∩C) = (A∪B) ∩ (A∪C) B∩C = {
}
A∪(B∩C) = {
A∪B = {
}
={
A∪C = {
}
(A∪B)∩(A∪C) = { ={
}∪{
} }
}∩{
} }
∴ A∪(B∩C) = (A∪B) ∩ (A∪C)
(10)
If A={1,2} B = {2,3,5} C = {2,3,6,8} then show that A∪(B∩C) = (A∪B)∩(A∪C) (M  07)
15
(11)
Set A={Positive even nos. less than 11} B={Positive odd nos. less than 11} then show that intersection of sets is distributive over union. (M  08)
(12)
U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} A = {x / x is a perfect square less than 10} B = {x / x is a multiple of 3 less than 10} then verify (i) (A∪B)1 = A1∩B1 (ii) (A∩B)1 = A1∪B1 (i) A = {
}
(A∪B)1 = {
}1
B={
}
={
}
A1 = {
}
A1∩B1 = {
B1 = {
}
={
}
A∪ B = {
}
(A∩B)1 = {
}1
={
}
}∩{
}
∴ (A∪B)1 = A1∩B1
(ii) (A∩B) = {
}
A1∪B1 = { ={ ∴ (A∩B) = A ∪B 1
1
1
16
}∪{
} }
(13)
If U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} A = {1,2,4,6,8} and B = {1,3,5,6,9} verify that the compliment of union of sets is the intersection of their compliments. A1 = {
}
(A∪B)1 = {
B1 = {
}
={
A∪B= {
}
A1∩B1 = {
}1 } }∩{
={
} }
∴ (A∪B)1 = A1∩B1
(14)
U = {1, 2, 3, 4, 5, 6, 7, 8, 9} A = {x / x is perfect square less than 10} B = {x / x is even digit less than 10} then show that (A∪B)1 = A1∩B1
(15)
U = {0,1,2,3,4,5,6,7,8,9} A = {1,4,9} B = {2,4,6,8} then show that (A∪B)1 = A1∩B1
17
(M  10)
(J  09)
(16)
In n(AB) = 25  x, n(B  A) = 24x, n(A∩B) = 2x then find n(A∪B) [ n(AB) + n(BA) + n(A∩B) = n(A∪B)]
(18)
There are 60 students in a class. Every student have to learn at least one of the languages Kannada or English. 45 Students select Kannada and 30 English. Find (J  10) how many students select both the languages by using venn diagram.
∴
Total No. of Students = n(K∪E) =
K
E
Kannada Selected students n(K) = English Selected students n(E) = No. of Students to select both n(K∩E) = ? n(K) + n(E) = n(K∪E) + n(K∩E)
∴ n(K∪E) =
(19)
In a group of 25 people 8 people drink only Tea, 7 can drink only Coffee and 4 can drink both Coffee and Tea. Some one can not drink either Coffee or Tea. Find those number of person and draw Venn diagram. (M  10) No. of Persons who drink Coffee only = n(C)  n(C∩T) = 7 No. of Persons drink both Coffe and Tea = n(C∩T) = 4 No. of Persons who drink Tea only = n(T)n(C∩T) = 8 No. of Persons drink either Coffee or Tea = n(C∪T) = ? n(C∪T) = 7 + 4 + 8 = 19 No. of Persons can not drink either Coffee or Tea is = U  n (C∪T) = 25  19 = 6 18
A 7
B 4 6
8
(20)
In a group of 50 persons each one should drink Coffee or Tea, 14 drink Tea but not Coffee, 30 drink Tea. Find how many drink Coffee but not Tea?
(21)
In a group of 110 girls 60 Pass in the First Test, 45 Pass in the Second Test, if 12 Pass in both. How many girls failed in both the test (Ans = 17 Girls)
19
(22)
In a school 60% of students invest in NDF and 70% of them invest in NSS. Find the percentage of students who invest in both the scheme? (Ans: 30%)
(23)
There are 70 cycles in a cylce stand, 50 cycles have carriers, 35 have bells and carriers, while some have bell but no carriers. Find the number of cycles which have bell but no carrier (Ans: 55)
(24)
In a college of 800 students if 500 select History, 250 select both History and Geography. Find the number of students who select Geography only and History only? (Ans: 300, 250)
20
(25)
62 factories in a country manufacture pens only and 48 manufacture pencils and pens. Some of the factories manufacture pencils only. Find their number if the total number of factories in the country manufacturing pens or pencils is 132. (Ans: 22)
21
SEQUENCE • Sequence: It is an ordered arrangement of numbers according to a given rule. The term of a sequence in successive order is denoted by Tn. The nth term Tn is called general term of the sequence
• Finite Sequence: A Sequence having a finite number of terms. T1, T2, T3 ................... Tn is a Sequence of nth term and Tn is the last term.
• Infinite Sequence: A Sequence having infinite number of terms. T1, T2, T3 ........................
• Series: Sum of the terms of the Sequence is called a Series. T1 + T2 + T3 + ...................... + Tn Finite Series T1 + T2 + T3 + ...................... Infinite Series Sn = T1 + T2 + T3 + ...................... + Tn
• Arithmetic Progression (AP): It is a sequence in which defference between a term and its preceding term is a constant. The constant term is called common difference and it is denoted by ‘d’.
• General form of an A.P.: ‘a’ is the first term, ‘d’ is common difference and Tn is nth term then the General form is a, a+d, a+2d, ................... a + (n1)d and n term is Tn = a + (n1)d th
Tn+1 = Tn + d and
Tn1 = Tn  d
To find the common difference ‘d’ T p  Tq d= and Tp = Tq + (p  q) d P Q If Sequence is in ascending order then the ‘d’ is +ve and if it is in descending order then ‘d’ is ve.
• Sum of n terms of an AP: (1) If First term (a) last term ‘l’ and number of terms ‘n’ are given n Sn = (a + l) 2 a+l The middle term = (average) 2 (2) a, d and n are given Sn =
n 2
[2a + (n1)d]
22
(3) Sum of all natural numbers ÎŁn =
n(n+1) 2
(4) Sum of first n odd natural numbers Sn = n2 (5) Sum to first n even natural numbers Sn = n(n+1)
â€˘ If each term of an AP is increased or decreased or multiplied or divided by the same quantity except zero the resulting series is also in AP. I
Is it a Sequence or not
(1)
1, 1, 1, 1, 1, 1, ....................... It is a sequence
(2)
3, 7, 11, 15, ...........................
(3)
3, 7, 15, 23, ......................... It is not a sequence
(4)
5, 10, 15, 20, 25, .........................
(5)
7, 4, 1, 2, 5, ...........................
(6)
1, 4, 9, 16, 25, ...........................
(7)
2, 6, 8, 54, ................................
(8) (9) (10)
II
1 , 2 , 3 , 4 ..................... 4 5 2 3 1,
1 , 1 , 1 .................... 10 100 1000
1 , 3 , 7 , 9 ..................... 4 4 4 4
Find the nth term of the sequence.
(1) 10, 100, 1000, ............ (2)
1 , 2 , 3 , 4 ........... 2 3 4 5
10, 102, 103, .......
Tn = 10n
1 , 2 , 3 ............ 1+1 2+1 3+1
(3) 5, 9, 13, 17, ..................
5+0, 5+4, 5+8, 5+12 ....
(4) 2, 5, 8, 11, ....................
3x11, 3x21, 3x31,.....
(5) 2, 5, 10, 17, .................
12+1, 22+1, 32+1, ......
23
5+4(n1) = 4n+1
III (A) Find the first 4 terms of the given sequence.
Tn
T1
(1) 4n + 3
T2
T4
7
(2) n2+2 (3)
T3
6
n1 n+1
2 4
1
(4) (1)n (5) 4n21
3
III (B) Find number of terms in a given sequence.
(1) Tn = 2n2+1
Tn = 73
(2) Tn = n2+1
Tn = 65
(3) Tn = 2n23+5
Tn = 255
IV
73 = 2n2+1 2n2 = 72 n2 = 36 n=6
Calculate and fill in given box. Tn
(1) 5n+3
Tn+1
Tn1
Tn+2
Tn2
5(n+1)+3 5n+8
5(n1)+3 5n5+3 5n2
5(n+2)+3 5n+10+3 5n+13
5(n2)+3 5n10+3 5n7
(2) 3n2  2
2 (3) n + 2
(4)
n n+1 24
V
Write ‘a’ and ‘d’ for the following AP. AP
a
d
AP
4
74=3
(5)
2 , 1, 4 , 5 ........ 3 3 3
(2) 10, 20, 30, 40, .....
(6)
1 , 1, 3 , 4 ........ 2 2
(3) 7, 3, 1, 5, ......
(7)
3 , 2, 5 , 3 ........ 2 2
(4) 8, 6, 4, 2, .....
(8)
1 , 7 , 3 ........ 2 2
(1) 4, 7, 10, 13, .......
VI
a
d
In an AP, Tn = a+(n1)d then calculate unknown one and fill in the right box. a
d
n
Tn
(1)
2
3
8
...............
(2)
5
...............
10
23
(3)
3
1 2
8
..............
(4)
............
4
15
59
(5)
15
2
...............
5
(6)
............
3
12
31
(7)
1 2
...............
15
15 2
(8)
8
4
...............
28
(9)
4
5
...............
106
(10)
5
...............
7
7
25
VII
Match the following.
(1)
n term of an AP
(a)
n(n+1)
(1).............................
(2)
Sum of n terms of an AP
(b)
n(n+1) 2
(2)..............................
(3)
Common difference d
(c)
Tn
(3)..............................
(4)
Arithmetic mean
(d)
n2
(4)..............................
(5)
Sn  Sn1
(e)
n [2a+(n1)d] 2
(5)..............................
(6)
Sum of natural numbers
(f)
a+(n1)d
(6)..............................
(7)
Sum of odd numbers
(g)
(8)
Sum of even numbers
(h)
VIII
(A)
A
B
th
a+b 2 TpTq pq
a
l
n
Sn / T n
d Tn = a+(n1)d 86 = 1+(n1)5 86 = 1+5n5 5n = 86+4=90
1+6+11+…+86
1
86
18
5
90 = 18 5 n (a + l) Sn = 2 18 = (1 + 86) 2 n=
= 9 x 87 = 782
(2)
(8)..............................
Find the sum of the series. Series
(1)
(7)..............................
3+7+11+….+123
26
Series
a
n
l
d
Sn / T n
(3) 1+4+7+…+103
(B)
Find the number of terms of the following Arithmetic Series. Series
a
d
Sn
n Sn =
(1) 1+2+3+…= 1275
1
1
n 2
[2a + (n1)d]
1275 =
n 2
[2 x 1 + (n–1) 1]
1275 =
n 2
[2 + n – 1]
1275 1275 =
n 2
(n + 1)
n2 + n – 2550 = 0 n2 + 51n – 50n – 2550 = 0 n(n+51)50(n+51) = 0 (n+51) (n50) = 0 n = 51 n = 50
(2) 1+3+5+….= 625
27
Series
(3)
a
d
n
Sn
1+4+7+….= 590
(C)
In an AP find Sn. Term
(1) Tn = 5 – 2n
T1=a
T2
d=T2T1
3
1
2
Sn S15
n
Sn = 2 [2a + (n1) d] 15
= 2 [2 x 3 + (151) 2] 15
= 2 [6 – 28] =
15 x 22 2
= 165 (2) Tn = 4n – 3
S15
(3) Tn = 5n + 2
S20
28
(J  06)
Term (4)
T1=a
T2
d=T2T1
Sn
Tn = 103n
S40
(M  08)
(D)
Find the sum of the arithmetic series which contains n term and middle term. No. of Terms
Middle Term
Sum of Series
Sn = (1)
19
26
n (a + l) 2
a+l 2
is a middle terms
Sn = 19 x 26 = 498
(M  07)
(2)
25
20
(3)
21
17
29
(E)
If the three numbers are in AP. Sum of the numbers and product of the numbers are given. Find the numbers. Sum of the numbers
Product of the numbers
The numbers are a, a+d and ad Sum: a + a + d + a  d = 15 3a = 15 a = 15 = 5 3
Product: a (a+d) (ad) = 105 5 (52d2) = 105 (1)
15
105
25  d 2 = 105 =2121 3
d 2 = 25  21 = 41 d=2 the numbers are a=5 a+d = 5+2 = 7 ad = 52 = 3 They are (3, 5, 7)
(2)
30
120
30
(3) (2)
(3)
(4)
Sum of the numbers
Product of the numbers
36
1620
24
288
The numbers are a, a+d and ad
31
(F)
Find â€˜xâ€™ if the arithmatic mean between a and b is A.
a
A
b
A=
a+b 2
x1= 2x+3+2x (1)
2x+3
x1
2x
2 x+5 x1 = 2
2x2 = x+5 x=7
(2)
5
x
19
(3)
6a
x
6+a
(J  10)
(4)
2
(x1)
4
32
(5)
3x+1
5x1
5x+1
(6)
2x
x+10
3x+2
(7)
2x+3
x1
2x
IX
(A)
If Σn =
(1)
30 Σn = 30 x 31
n(n+1) find the value of 2
2
1
25
(2)
Σn =
(4)
Σn + Σn
1
= 465 (3)
60
30
1
1
Σn  Σn
33
40
30
1
1
Find the value of n
(1)
Σn = 36
(2)
Σn = 210
(3)
Σn = 820
(4)
3Σn = 630
(5)
Σn1 = 78
(6)
Σn1 = 820
X
(1) Find the sum of all naturals between 50 and 150 ( (Notice the use of words betwen, from and to)
∴
(B)
Σ149  Σ50)
(2) Find the sum of all naturals between 100 and 200 which are divisible by 5.
105 + 110 + 115 + ..................... + 195 = 5(21 + 22 + 23 + ................... + 39) = 5(Σ39  Σ20)
34
(3)
Find the sum of all naturals from 101 to 200 which are not divisible by 4. (Hint: Find sum of 101 to 200 and sum of divisible nos. and subtract)
(4)
Find the sum of all odd natural number from 1 to 100.
(5)
Find sum of all even natural numbers from 1 to 100
35
XI
Find an AP if the ratio between the numbers are given.
Ratio (1)
3rd
9th
1 :3
T18 = 90
T3 T9 a+2d a+8d
= =
1
T18 = 90
3
a + 17d = 90
1
18d = 90
3
3a + 6d = a+8d 2a = 2d a=d AP is 5, 10, 15, 20...... (2)
4th
8th
1 :2
T10 = 30
(3)
5th
10th
1 :2
T12 = 36
d=
90 18
=5
a=5
(J  09)
36
XII
(1)
In an AP if T10 = 2T6 then prove that T4 = 2T3.
(2)
7th term of an AP is 15 and 16th term is 30 find the sum of the 1st and 19th terms.
d=
(3)
Tp  Tq P Q
Tn = a + (n1)d
T19 = ?
In a sequence T1 = 3, Tn = 3Tn1 + 2 for all n > 1. Find first 4 terms of a sequence
37
(4)
In an AP the sum of first 10 terms is 175 and the sum of next ten terms is 175 find AP
S10 = 175, Sn= n
S20 = 350 [2a + (n1) d]
2
S10 = 5 (2a + 9d) = 175 2a + 9d = 35
 (1)
ly 2a + 19d = 70
 (2)
equate (1) and (2) (5)
The sum of the 6 terms of an AP is 345 and difference between first and last term is 55 find the terms of an AP
Sn =
n 2
(a + l)
345 = 6 (a + l) 2
a + l = 115
 (1)
l  a = 55
 (2)
equate (1) and (2) (6)
In an AP of 21 terms sum of the middle 3 terms is 129 and sum of the last 3 terms is 237 find the sequence.
T10 + T11 + T12 = 129 a + 9d + a + 10d + a + 11d = 129 a + 10d = 43  (1) T19 + T20 + T21 = 237 a + 19d = 79  (2) equate (1) and (2) (7)
8th term in an AP is double the 13th term, show that 2nd term is double the 10th term.
T8 = 2T13
T2 = 2T10
38
(8)
Three numbers are in AP whose sum is 18 if 2, 4, 11 are added to them respectively the resulting numbers are in GP find those numbers.
(9)
In an AP of 21 terms the first and last terms are 4 and 64 respectively. Find the sum of the series.
(10)
The angles of a quadrilateral are in AP. If the smallest angle is 150. Find the angles of the quagrilateral. (Assume angles are ad, a+d, a+3d, a3d)
39
(11)
Find the sum of the series 3 + 7 + 11 + ............. up to 42 terms.
(12)
3 nos are in the ratio 2 : 5 : 7. If 7 is substracted from 2nd no. the resulting numbers are in AP. Find the numbers. (The numbers are 2x, 5x, 7x and T2 = 5x  7 2x, 5x7, 7x are in AP)
(13)
Three angles of a triangle are in AP. The smallest angle is 400 then find the angles of the triangle. [Assume angles are (ad), a, (a+d)]
40
(14)
A person deposits Rs. 1000 in the first month. Then every month he increases the monthly deposit by Rs. 60. Calulate his total investment at the end of (M  09) 2 years.
(15)
In an AP seven times the 7th term is equal to the eleven times the 11th term. Then show that the 18th term is equal to zero. (M  10) ( 7T7 = 11T11)
(16)
Check whether 301 is a term in the AP 5, 11, 17, 23..................
41
(17)
The sum of four numbers in AP is 16 their product is 105. Find them. (Assume the numbers as (a3d) (a+d) (ad) (a+3d) and cd is 2d)
(18)
Sum of 3 numbers in AP is 27 the sum of their sequence is 293. Find those numbers (Let the numbers be (ad) (a) (a+d))
(19)
Sides of the right angle triangle are in AP. Show that they are in the ratio 3 : 4 : 5 [Assume the sides are a + d, a, a  d, Apply Pythagorous Theorem)
42
XIII
Fill up the blanks with suitable answer
(1)
The sum of first 13 terms of an AP of which 7th term is 40 is ....................
(2)
The sum of the first twenty positive multpiles of 7 is ...............
(3)
S9 = 81, T9 = 27 Value of S8 is ...........................
(4)
If Tn = (1)n the sum of 4 terms is ...........................
(5)
If a is constant then a + 2a + 3a + ....... + na = .....................
(6)
Sum of the first twenty odd numbers is ........................
(7)
Sum of first twenty even natural numbers is .......................
(8)
If first term of an AP is â€˜aâ€™ and its common defference is d. Then the formula to (J  10)
find the nth term of AP is ...................... XIV
Choose the correct answer.
(1)
The common difference of an AP is d then the relation between 5th and 12th term is ..........................
(2)
(3)
(A) T5 = T126d
(B) T12=T5+6d
(C) T5 = T127d
(D) T12 = T5+8d
In an AP the common difference is .......................... (A) d = Tn+Tn+1
(B) d = Tn1 Tn+1
(C) Tn+1  Tn
(D) Tn+1 + Tn
Sn  Sn1 is ...................... (A) Sn
(4)
(B) Tn
(C) d
(D) Sn+1
In an AP 20th and 30th terms are 201 and 301 respectively the common differnece is ............................. (A) 10
(B) 100
(C) 5
43
(D) 2
(5)
Σn  1 is given by the formula is ........................
(A) (6)
n(n+1) 2
(B)
(B) n + 1
n(n2) 2
(D) 2n(n+1)
(C) 45
(D) 42
(C) 1
(D) 10
10
The value of Σ (1)9 is ...................... n=1
(B) 1
Pth term of an AP is q and Qth term is p then (P+Q)th term is ...................... (A) 0
(10)
(C) n(n+1)
(B) 38
(A) 0 (9)
(D)
Which of the following is the term of the sequence 3, 6, 9, 15, ....... is ................. (A) 25
(8)
(C) n(n+2) 2
Sum of 1+3+5+ .............. + n is ............................. (A) n2
(7)
n(n1) 2
(B) (P + Q)
(C) (P  Q)
(D) (P + Q)
Three numbers are in AP the sum of first and third terms is 14, the middle term is ......................... (A) 6
(11)
(13)
(15)
(D) 7
(B) Σ
n 2
(C) 3Σn
(D) Σn
If the terms of an AP are in ascending the common difference is .................... (A) Positive
(B) Negative
(C) Zero
(D) May be +ve or ve
If (5a  x), 6a, (7a + x) are in AP then the 5th term is .................... (A) 3(3a+x)
(14)
(C) 18
2 + 4 + 6 + 8 + ............... + 2n can be written as .................... (A) 2Σn
(12)
(B) 8
(B) 9(a+3x)
(C) 9(a+x)
(D) 3(3ax)
n(n2+1) is nth term of a sequence the sequence is in .................... (A) AP
(B) GP
(C) HP
(D) None of these
In an AP S5 = 35 and S4 = 22 then, 5th term is .................... (A) 35
(B) 10
(C) 13 44
(M  06)
(D) 22
(16)
(A) 4n  1 (17)
(M  06)
3, 7, 11, 15, ........... the nth term is .................... (B) 4n + 1
(C) 4n + 3
(D) 3n + 4
Ramu put a dot in 1st square, 2 in 2nd, 3 in 3rd and so on. Total number of squares he required to put 55 dot is ............................ (A) 55
(18)
(B) 11
(D) 10
In a sequence Tn = n2  1 and Tn = 35 then value of n is .................... (A) 6
(19)
(C) 9
(M  06)
(B) 36
(C) 34
(J  06)
(D) 6
Geetha climbed 15 steps of a building in the first minute after that she climbed 3 steps less than what she had climbed in the previous minute. If she climbed the building in 5 minute the number of steps she climbed is .................... (A) 75
(20)
(B) 105
(B) 9
(C) 5
(M  07)
(D) 7
10
The value of ÎŁn is ........................
(M  07)
1
(A) 10 (22)
(D) 50
In a sequence Tn = 2n1 the 4th term is .................... (A) 23
(21)
(C) 45
(B) 11
(C) 55
(D) 110
A person put 3 marbles in first box, 5 in second box, 7 in third box and so on. Number of marbles can be put in 16th box is .................... (A) 66
(23)
(C) 31
(D) 35
(B) Tn5 = Tn6 + d
(C) Tn5 = Tn + d
(M  08)
(D) Tn5= Tnd
Sum of 15 terms of an AP is 180 then 8th term is .................................... (A) 8
(25)
(B) 33
(J  07)
In an AP which relation is true .................................... (A) Tn5 = Tn4 + d
(24)
(J  06)
(B) 12
(C) 15
(D) 18
2x+1, 4x, 13x are in AP the value of x is .................................... (A) 2
(B) 3
(C) 4 45
(M  08)
(M  08)
(D) 5
(26)
In an AP Tn = 3n1 the common difference is .................................... (A) 1
(27)
(C) 3n
(D) 2n
(B) 12
(C) 10
(J  09)
(D) 11
(B) 29
(C) 30
(D) 28
If Tn = (1)n the correct relation between the terms is ................................. (M  10) (A) S1 = S2
(31)
(B) 2
In an AP common difference is 3, first term is 1 then the 10th term is ........... (J  09) (A) 27
(30)
(D) 4
In a sequence Tn = 2n2 + 1 then S2 is .................................... (A) 9
(29)
(C) 3
In an AP Tn+5 = 35 and Tn+1 = 23 then common difference is ...................... (M  09) (A) 3
(28)
(B) 2
(M  09)
(B) S2 = S3
(C) S3 = S4
(D) S2 = S4
In an arithmatic sequence if T4=8 and a=2, then its common difference is .......... (A) 6
(B) 4
(C) 2
46
(D) 10
(J  10)
GEOMETRIC PROGRESSION (GP) • A Sequence in which the ratio of a term and its preceding term is a constant Example:
2, 6, 18, 54, .............................. T1, T2, T3, T4 ............................. T2 T1
=
T3 T2
T4
=
.............................
T3
The constant ratio is called the ‘common ratio’ and is denoted by ‘r’ and the first term is ‘a’
• The General Form of a G.P. ‘a’ is the first term, ‘r’ is the common ratio and ‘Tn’ is the ‘n’th term, then the general form is a, ar, ar2, ar3 .................. arn1 The nth term is
Tn = arn1
• If ‘Tn’ and ‘r’ are given, then The succeding term
Tn+1 = Tn x r
The preceding term
Tn1 =
• To find common ratio
also we have
r=
Tp Tq
Tn r T2 T1
= rpq
Note: If each term of a GP is multiplied or divided by the same quantity then the resulting terms are also in GP.
( Verify this with a suitable example)
47
• Sum of the terms of a GP is called Geometric Series Example:
2, 4, 8, 16, 32, ....................... is a GP 2 + 4 + 8 + 16 + 32 + ............... is a Gemetric Series.
• Sum of ‘n’ terms of a Geometric Series rn  1
(i) If r > 1 then
Sn = a
(ii) If r < 1 then
Sn = a
(iii) If r = 1 then
Sn = na
r1 1  rn
1r
• Sum of infinite (∞) terms of a geometric series is S∞ =
a 1r
• The geometric mean (G) begween a and b is G=
•
ab
or
G2 = ab
Sn : S2n = 1 : rn + 1 S2n : Sn = rn + 1
48
I
Write any 5 Geometric Progressions.
(1) (2) (3) (4) (5) II
Write ‘a’ and ‘r’ of the following Geometric Progressions.
(1)
2, 6, 18, 54 .......................... a = ......................
r = ..........................
(2)
5, 10, 20, 40 ........................
(3)
1, 1, 1, 1, .......................
(4)
1,
1
,
2
(5)
1,
1
1
1
,
,
3
............................
8
4 1
1
,
............................
27
9
III
Write the Geometric Progressions for the given ‘a’ and ‘r’.
(1)
a = 3, r = 2
(2)
a = 1, r =
(3)
a=
1 2
,
1 3
r=
1 2
49
IV
Which of the following sequences are in GP. Justify your answer.
(1)
1, 2, 4, 8.............................
(2)
3, 6, 9, 12...........................
(3)
8, 4, 2, 1 ............................
(4)
4, 12, 36, 108 .....................
(5)
12, 6, 3,
3
...........................
2
V
Find the nth terms of the following GP.
(1)
3, 12, 48, ......................
(2)
1, 3, 9, 27 ......................
(3)
3, 6, 12, 24, ......................
(4)
1, 2, 4, 8, ......................
VI
Find the first 4 terms of the follwoing GPâ€™s. Tn
T1
T2
2 x 5n1 2 x 3n1 3n1 1n 2n1
50
T3
T4
VII
Fil up the boxes in the following table with suitable answer.
S.No.
T1
T2
1
1
2
2
T3
8
1
3 4
256
.......................
2
.......................
T4
3
.......................
3
2
32 1
6
729
6
.......................
VIII Four alternatives are given to each questions choose the right answer and write it in the space provided for that.
(1)
In a GP if T4 = (A)
1
1 4
and r =
1 2
, then T5 is ....................
(B) 8
8
(C)
1 4
(D)
1 2
Ans (2)
The (n + 1)th term of a GP is ........................ (A) Tn+1 = Tnrn
(B) Tn+1 = Tnrn1
(C) Tn+1 = Tn rn+1
(D) Tn+1= Tnx r
Ans (3)
If x, 1,
1
are in GP then the common ratio is ........................
x
(A) x
(B) 1
(C)
1 x
(D) 1 x2
Ans (4)
In a GP, which of the following is correct ........................ (A) Tn = r x Tn1
(B) Tn1 = r x Tn
Ans
51
(C) Tn1 x Tn = r
(D) Tnxr = Tn1
(5)
In a GP, if Sâˆž = 3, r = (A) 2
1
then the first term is .................... 3 2 (B) 3 (C) 3
(D)
3 2
Ans (6)
In a GP the first term is 3, and the common ratio is 1, then the sum of ten terms is ........................ (A) 3 x 110
(B) 30
(C) 3 x 19
(D) 10
Ans (7)
The nth term of the GP, 10, 100, 1000 .......... is ......................... (A) 10n
(B) n10
(C) 10n
(D)
10 n
Ans (8)
The 20th term of the GP Tn = xn1 is ........................ (A) x20
(B) 20x
(C) 20x
(D) x19
Ans (9)
The missing term in the GP 2, 4, ........., 16, 32 ........ is ......................... (A) 6
(B) 8
(C) 10
(D) 12
Ans (10)
The nth term of the GP 1, 3, 9, 27, .......... is ........................ (A) 3 x 3n2
(B) 3 x 3n1
(C) 3n1
(D) 3n + 1
Ans (11)
The common ratio of the GP Tn = 2 x 5n1 is ........................ (A) 2
(B)
3
(C) 5
2
Ans 56
(D) 2.5
(12)
In a GP, if T8 : T5 = 8 : 1 then the common ratio is .................... (A) 2
(B) 2
(D) x5
(C) ±
Ans (13)
The 9th term of the GP Tn = 2n1 is ........................ (A) 256
(B) 64
(C) 32
(D) 128
Ans (14)
If 16, x, 25 are in GP, the value of x is ......................... (A) 20
(B)
200
(C)
20
(D) 400
Ans (15)
IF ‘b’ is the GM between ‘a’ and ‘c’ then (A)
b a
(B)
a
a b
= ........................
(C)
b
a c
(D)
b c
Ans (16)
ABCD is a square EFGH is the another square obtained by joining the mid points of ABCD, ly PQRS is the another square obtained by joining the midpoints of EFGH and so on. Then the areas of these squares are in ......................... (A) AP
(B) HP
(C) GP
(D) None of these
Ans (17)
The AM and GM of two numbers are 5 & 4 respectively, then the numbers are ...... (A) 2, 8
(B) 4, 16
(C) 3, 7
(D) 1, 4
Ans (18)
The first term of a GP is ‘a’ and common ratio is ‘r’ then the 4th term is ............... (A) ar3
(B) ar4
(C) ar2
Ans 53
(D) ar
(19)
In a GP which of the following is not correct ? (A) Tn+1 = Tn x r
(B) Tn1 = Tn r
T4
(C) r =
(D) Tn+1 = Tn+r
T3
Ans
IX
Four alternatives are given to each questions, choose the right answer and fill in the blanks. (Previous Question Paper Questions)
(1)
The 6th term of the sequence 1, 1
(A)
(2)
81
a 1r
....... is ...................................
(J  06)
1
(C) 
243
1
(D) 
81
243
1r
(C)
(B) arn1
(M  07)
(D) ar0
a
(M  07)
(C) 8
(D) 12
(M  07)
n(n+1) 2
(B) a
1r
n
(C)
1r
n 2
r 1 n
(n2+1)
(D)
a(r1)
In a GP, if T5 : T2 = 1 : 8, then the common ratio is ................................ (M  08) (B)
1
(C) 2
3
(D)
1 2
In a GP if Tn = 2 x 3n1, then T5 is ................................ (A) 486
(7)
1
27
(B) 16
(A) 3 (6)
9
The formula to find the sum of ‘n’ terms a geometric series is ................................ (A)
(5)
3
1
,
The GM between 4 and 16 is ................................... (A) 4
(4)
1
,
As ‘n’ approaches ‘∞’ the value of S∞ is ................................... (A)
(3)
(B)
1
(B) 162
(M  08)
(C) 243
(D) 8
The GM of three numbers is 4, then their product is ................................ (M  09) (A) 16
Clue:
G=
ab
a, G, b
(B) 64
G2 = ab
ab x G
(C) 128
4 = ab
16 x 4 = 64
(D) 256
ab = 16
2
54
(8)
In a GP, if Tn is the nth term, and common ratio is r, then Tn1 term can be deter mined by
................................
(A) Tn x r (9)
(B)
(C)
r
The 4th term of the sequence (A) 9
(10)
Tn
(J  09)
Tn2 r
(D) Tn+1 x r
3 , 3, 3 3 is ................................
(B) 27 3
(C) 21
(J  10)
(D) 9 3
p, 8, q are in GP then pq = ................................. (A)
64
(B) 2
(J  09)
(C) 2
2
(D) 64
X
Solve the following problems. (Previous Question Paper Questions)
(1)
In a GP if a = 3, r = 2 and Tn = 96 then ST Sn = 189. Solution:
Tn = 96 arn1 = 96
Sn = a
(M  06)
rn  1 r1
................................... ................................... (2)
=
..........................................
Find S3 : S6 of the series 4 + 12 + 36 + ................ Solution:
r=
T2
(J  06)
Clue:
T1
Sn : S2n = 1 : rn + 1 S3 : S8 = 1 : r3 + 1
(3)
(M  07)
In a GP S8 : S4 = 97 : 81 find r Solution:
55
(4)
In a GP if T6 = 32 and r = 2 find a Solution:
(5)
Use Tn = arn1
The 10th term of a GP is 8 times of its 13th term if the first term is 3, then find the sum of infinite terms. (M  09) Solution:
T10 = 8 x T13 T10 T13
a=3
=8
r = .................
Sâˆž = ................................. = .................................
................. = 8 = .................................. ............................. = ................................... r = .......................... = ................................... (6)
Find the sum of the series 2 + 4 + 8 + ............. + 256 Solution:
a = ..............
Tn = arn1
(J  09)
Sn = a
r = ...................
= ..................
=
Tn = .................
= .................
=
n = .................
=
n=?
= 56
rn  1 r1
(7)
The first term of a GP is 64 and the common ratio is ‘r’. If the average of 1st and 4th term is 140 find the value of ‘r’. (A  10) Solution:
a + ar3
a = 64
2
r=?
= 140
64 + ar3 = 140 x 2
T1 = 64 T4 = ar3
ar3 = ..............  .............. 64 r3 = ....................
r3 =
64
r=
(8)
Three Nos. are in GP, whose sum is 26 and their product is 216, find the numebrs. (J  10)
Solution:
Let the nos. be a , a, ar r Product:
a r
Sum :
x a x ar = ..............
........................... = .........................
a r
+ a + ar = 26
................................. .................................
a3 = .................
..................................
a = ..................
.................................. r = .......................
The numbers are
.................., ...................., ...................
57
(9)
Three numebrs are in GP, whose sum is 28 and their product is 512. find the Nos. Solution:
(10)
For what value of K, in K, K2, K+1 makes a GP. Solution:
Clue
G=
K,
K2,
K+1
a
G
b
ab
........................................... ........................................... ........................................... (11)
xb
If a, b, c are in AP and x, y, z are in GP, then show that Solution:
a, b, c are in AP
(
a+c 2
x
xc
=
y=
2
x
x
a 2
x
xc
x
(
xz )c xz )a
c 2
x
x
( (
ya
x
xz
za z
a+c 2
x )c x ( x )c x (
z )c z )c
x
za z
= .............................................................
= .............................................................
= .............................................................
58
yc
x, y, z are in GP
a+c
b=
LHS
&
xc
x
a 2
x z
c 2
za zb
=1
(12)
The sum of 2 terms of a GP is 6 and first term is 2. Find the common ratio. Solution:
a = T1 = .................... S2 = T 1 + T 2
(13)
r=
T2 T1
6 = ........... + T2
.........................................
...................................
.........................................
T2 = ..........................
.........................................
S8 : S4 = 97 : 81, T15 = 16, write the GP. Solution:
(14)
A person invests Rs. 5 as initial amount, then after he deposite Rs. 10 at the end of 1st month and Rs.20 at the end of 2nd month and soon, if he deposite Rs.3,27,680 at the end of 16th month, find the total amount that he invested. Solution:
5 + 10 + 20 + ............................+ 3,27,680 T1
T2
T17
a= r= n = 17 Sn = ?
59
(15)
Sn is the sum of â€˜nâ€™ terms of a geometric series, a = 3 and r = 2, then find S8  S3. Solution:
(16)
How many terms of the series 2+4+8+....... should be added to get a sum of 1022. Solution:
a = ...................... r = ....................... Sn = 1022 n=?
60
(17)
Find the number of terms of the following series 3 + 9 + 27 + ........... = 1092 Sn = a
Solution:
n Â r 1
Clue
r1
a=3 r= Sn = 1092 n=?
(18)
Tn1
In a GP show that
=
Tn+1
1 r2 Hint: Tn1 =
Solution:
Tn r
Tn+1 = Tn x r
(19)
Three numbers whose sum is 18 are in AP, if 2, 4 & 11 are added to them respectively, the resulting numbers are in GP. Find the numbers? Solution:
Let the numbers be ad, a, a+d
in AP
Sum is 18 .................................... .................................... a=6 If 2, 4, 11 are added to ad, a, a+d respectively ad+2,
a+4,
a+d+11
T1
T2
T3
T2 T1
=
are in GP
T3 T2
..................................................... 61
(20)
Find the GM between (x + y) and x3 + 3xy (x + y) + y3 Solution:
a=x+y b = x3 + 3xy (x + y) + y3 G=
ab
G = ................................................. G = ................................................. G = ................................................. G = ................................................. G = x2 + y2 + 2xy (21)
If 8, x+1, T2 are in GP find x. Solution:
Hint G=
62
ab
HORMONIC PROGRESSION (HP) • A Hormonic Progression (HP) is a Sequence in which the reciprocal of terms of it form an arithmatic progression. 1
• If Tn is the nth term of a HP then • nth term of a HP is Tn =
2
•
3
is in AP then
1 a + (n1) d 1
• General term of HP is
a 3 2
Tn
will be the nth term of the AP.
,
1
,
a+d
1 a + 2d
,
1 a + 3d
, ...................
1 a + (n1) d
is in HP
• If a, H, b are in HP, then Hormonic mean between a and b is H =
2ab a+b
• If A, G, H are the AM, GM and HM of two positive numbers a and b then A . G . H are in GP (G = AH )
•A >G> H I
State which of the Sequences are in HP.
(a)
1,
1 1 1 , , .......... 2 3 4
(b)
6, 4, 3 ..............
(d)
1,
Reciprocal of the given Sequence 1, 2, 3, 4 .......... T2  T1 = 2  1 = 1 T3  T2 = 3  2 = 1 it is in AP hence the reciprocal of the Sequence is in HP. (c)
1 1 1 1 .......... , , , 5 10 17 2
63
2 1 2 , , .......... 3 2 5
II
(a)
1 , 2
1 , 1 is in HP then find T and 2, 5, 8 is in AP n 8 5
2=2+0=2+3x0 5=2+3x1=2+3x1 8=2+6=2+3x2 Tn = ................................................ (b)
In
, 1 , 1 , 8 5
1 2
a= Tn
1 Find 12th Term and 8, 13, 18, 23 ...... is in AP 8
d=
n = 12
= a + (n1) d =
III
1 , 1 , 1, 1 ................ find 10th term. 3 5
(c)
In
(1)
In a HP T4 =
1 1 and T10 = find T19. 12 42 1 If Tn = a + (n1) d T4 =
1 a+(
)d
a + 3d = 12
 (1)
=
1 12
T12 =
1 a+(
)d
a + 9d = 42
 (2)
a + 3d = 42
a + 9d = 42  a + 3d = 12
a + 3 x 5 = 12
6d = 30
a = 12  15
d=
30 =5 6 64
a=3
=
1 42
T19 =
T19 =
(2)
In a HP T5 =
(3)
In a HP T10 =
(4)
In a HP T7 =
1 +(
)
=
1 87
6 2 and T1 = find T7 and T12. 19 9
6 2 and T3 = then find T11. 19 9
1 and T13 = 1 find a. 20 38
65
(M  08)
(M  10)
IV
(1) Find the HM between 6 and 3.
(2) If A, G, H are AM, GM and HM of the number 9 and 16 then verify A, G, H are in GP. a=9
A=
b = 16
âˆ´
a+b 2
G=
ab
H=
2ab a+b
a = Ax H =
(3) If A, G, H are AM, GM and HM of the number 2 and 8 then show that A > G > H. a=2 b=8
A= a + b 2
66
G=
ab
H=
2ab a+b
(3) If a, b, c are in HP then find the value of a  b bc 1 , 1 , 1 are in AP b a c T2  T 1 = T 3  T 2
(5) If A, G, H are the AM, GM and HM of two positive numbers a and b then, show that A, G and H are in GP
(6) If
b c a , and are in HP then show that a, b, c are also in HP. a+b b+c c+a
(By adding or dividing a constant term to an each term of an HP there is no change in the given HP)
67
V
Previous Exam paper questions.
(1)
If x, y, z are AM, GM and HM between 2 numbers respectively, then which of (M  06)
following relation is true.
(2)
(a)
x<y<z
(b)
x>y>z
(c)
x=y=z
(d)
x<y<z
(a) (3)
(6)
1 14
(c) 1 17
2
(d)
1 1 , T10 = then d is .......................... 12 42 (b) 5 (c) 3
(b)
1 , 2
1, 4
1 20 (J  06)
(d)
4
1 ....... is .......................... (M  07) 6
(c) 1 2
2
(J  06)
(d)
1 (M  09)
If 11, 13, 15, 17, 19 ............ is an AP, then the terms in HP are (a)
1, 2, 3, 4, .............
(c)
1 , 1 , 1 , 1 , 1 ...... (d) 11 13 15 17 19
(b)
1, 3, 5, 7, ............. 1 , 1 , 1 ......... 10 12 14 (M  07)
If 3, x, 7 are in HP, then the value of x is (a)
(7)
6
(b)
The common difference of the sequence (a)
(5)
1 11
In a HP T4 = (a)
(4)
1 1 1 , , ......... is .......................... 2 5 8
The 6th term of the sequence
21 5
(b)
5
(c)
21
(d)
5 21
The AM and GM between two distinct numbers are 5 and 4 respectively, then their (M  07)
HM is ........................................ (a) (8)
4
(b)
4.5
(c)
5
(d)
3.2
The HM between 1 and 4 is ........................................ (a)
7 5
(b)
8 5
(c)
68
6 5
(d)
7 4
(9)
(10)
(a)
2 (P + Q) PQ
(b)
2 PQ P+Q
(b)
2 (P + Q) PQ
(d)
2P+Q PQ
If x, y, z are in HP, then the harmonic mean is ............................. (a)
(11)
(12)
2xz x+z
(b)
2xy x+y
(c)
(J  09)
2yz y+z
(d)
2xz x+y (J  09)
If A, G, H are AM, GM and HM of a and b then ............................. (a)
A, G, H are in AP
(b)
A, G, H are in GP
(c)
A, G, H are in HP
(c)
A, G, H are not in AP, GP & HP
The HM between 1 and 2 is ............................. (a)
(13)
(M  09)
The Harmonic mean of P and Q is
1 1 2
(b)
1
1 4
(c)
(M  10)
1 1 3
(d)
1
2 3
(M  10)
An example for HP among the following is (a)
1, 1 , 2 , 2 3
3 ........... 4
(b)
1, 1 , 1 , 1 3 6 9
(c)
1, 2 , 1 , 3 2
2 ........... 5
(d)
1,
...........
1 , 1 , 1 ........... 4 7 9
(14)
The reciprocal of the term of a HP are in .............................
(15)
The GM of two numbers is 4
(J  10)
3 and HM is 6, then its AM is .............................
(J  11)
69
70
MATRICES • Matrix: It is a rectangular arrangement of numbers written in the form of rows and columns is called Matrix. The arrangement of numbers are enclosed by brackets like [ ] or ( ) or  . If a matrix has ‘m’ rows and ‘n’ columns then it is called a matrix of order m x n and it has ‘mn’ elements.
• Types of Matrix: (a)
Zero Matrix: The matrix in which all its elements are zero.
(b)
Row Matrix: A matrix having only one row is called a row matrix and it’s order is 1 x n.
(c)
Column Matrix: A matrix having only one column is called a column matrix and its order is n x 1
(d)
Rectangular Matrix: A matrix in which the number of rows is not equal to number of columns and its order is m x n.
(e)
Square Matrix: A matrix in which the number of rows equal to the number of columns and its order is n x n
(f)
Diagonal Matrix: In a square matrix except the principle diagonal all other elements are zero. 1 0 0 0 2 0 0 0 3
(g)
3x3
Scalar Matrix: A diagonal matrix in which the principle diagonal elemnts must be same. 3 0 0 0 3 0 0 0 3
(h)
3x3
Unit Matrix (Identity Matrix): A diagonal Matrix in which principle diagonal element must be one. It is denoted by I. Identity Matrix is also a Scalar Matrix. 1 0 0 0 1 0 0 0 1 71
3x3
(i)
Symmetric Matrix: If a Square Matrix is folded along its principle diagonal the elements which coinside are said to be symmetric with respect to the principle diagonal of the matrix. 5 1 3 1 0 4 3 4 2
(j)
3x3
Skew Symmetric Matrix: A Square Matrix is called a Skew Symmetric Matrix if the elements which are symmetric with the principle diagonal are equal but oppiste in sign while diagonal elements are zero. 0
2
1
2
0 3
1
3
0
3x3
•
Equal Matrix : Two matrix are said to be equal if their order are same and the corresponding elements are equal.
•
Addition of two Matrix: Let A and B be two matrix of order m x n. The sum A + B of the matrix A and B is a matrix obtained by adding their corresponding elements order of A + B is also m x n.
•
Scalar Multiplication of the Matrix: Let A be a matrix of order m x n and K be any scalar. The scalar multiplication of the matrix A by a scalar K is matrix denoted by KA and its elements are scalar multiple of the elements of A. The order of K is also m x n.
•
Multiplication of Matrices: If A =
a c
b d
and B =
AB =
a c
b d
e g
f h
e g =
f h
then AB is
ae + bg ce + dg
af + bh cf + dh
If the number of columns of matrix A and the number of rows of matrix B are equal then only AB exists. If A is of order m x n and B is of order n x p. The product AB is of order m x p.
•
Transpose of a Matrix: Let A be a matrix of order m x n. The matrix obtained from A by interchanging the rows in to columns is called the transpose of the matrix A and it is denoted by A1 and its order is n x m.
72
•
If A and B be any two matrices of same order then,
I. (1)
(2)
(3)
(4)
(5)
(a)
(A + B)1 = A1 + B1
(b)
(A1)1 = A
(c)
If A and B are two matrices such that (A•B) is defined them (A•B)1 = B1•A1
(d)
If A = A1 then the matrix A is a Symmetric Matrix.
(e)
If A = A1 then the matrix A is a Skew Symmetric Matrix.
(f)
If A and B are two matrices A  B = B  A and IA = AI = A
(g)
A + A1 is a Symmetric Matrix and A  A1 is Skew Symmetric Matrix.
(h)
AB = BA.
Matrix A 1
0
0
0
1
0
0
0
1
1
2
3
0
1 3
 1 2
0
2
3
1
0
1
2
1
2
3
2
0
4
3
4
2
3
0
0
0 3
0
0 (6)
Type
Order
0 3
[ 2 1
0]
73
Transpose A1
2A
II.
Match the Following:
(1)
Null Matrix
(a)
(2)
Diagonal Matrix
(b)
(3)
(4)
(5)
(6)
(7)
Scalar Matrix
Unit Matrix
Rectangle Matrix
Row Matrix
Column Matrix
0 3 3 0
(c)
(d)
(e)
(f)
(g)
(8)
Symmetric Marix
(h)
(9)
Skew Symmetric Matrix
(1) ...................................................
2
8
0
1
5
2
0
0
0
0
3
6
6
1
1
0
0
1
3
0
0
3
2
0
0
1
(3) ...................................................
(4) ...................................................
(5) ...................................................
(6) ...................................................
(7) ...................................................
[2 1 3] (i)
(2) ...................................................
3
(8) ................................................... (9) ...................................................
0 1 III.
Find the value of x for the given: Symmetric
(1)
1
x2
2x1 1 2x1 = x2 2xx = 2+1 x = 1 (2)
2+x 3 2x5 5
Skew Symmetric 0
x2
2x1 0 2x1 = (x2) 2x1 = x+2 3x = 3 x = 1 0 x+6
5 0
Scalar 2x1
0
0 2+x 2x1 = 2+x 2xx = 2+1 x=3 2x 0
0 x3
Identity 2x+1
0
0 1 2x+1 = 1 2x = 11 x=0 6+x 0
0 1
.......................
.......................
.......................
.......................
.......................
.......................
.......................
.......................
.......................
.......................
.......................
.......................
74
(3)
2 x3 3x1 2
0 6+x
2x 0
3x+4 0 0 2x+5
x+5 0
0 1
.......................
.......................
.......................
.......................
.......................
.......................
.......................
.......................
.......................
.......................
.......................
.......................
IV.
Fill up the blanks with suitable answer:
(1)
The number of elements of the matrix 3 x 4 is .......................
(2)
The number of columns in the matrix given the total number of elements and the number of rows 30 and 10 is .......................
(3)
The principle diagonal elements of the given matrix 1 2 3 2 3 4
is ...........................
5 6 7
(4)
(5)
If A =
1
1
1
2
2
0
then A1 is ...........................
If the order of the given matrix A is (m x n) then the order of itâ€™s transpose A1 is ...............................
(6)
If A is a square matrix and I is a Unit Matrix of the same order then AI = ...........
(7)
A is a matrix then [{(A1)1}1]1 is ......................
(8)
Matrices A, B and C are of order 3 x 3, 3 x 2 and 3 x 2 respectively then the order of A (B + C) is ...............................
(9)
If A is a Square Matrix of order 2 x 2 then A + A1 is ...............................
(10)
If A is a Square Matrix of order 2 x 2 then A  A1 is ............................... 75
V.
Choose correct answer and fill in the blanks:
(1)
If A = A1 then the matrix A is .......................
(2)
(3)
(A)
Scalar Matrix
(B) Square Matrix
(C)
Symmetric Matrix
(D) Skew Symmetric Matrix
If A = A1 then the matrix A is ....................... (A)
Scalar Matrix
(B) Square Matrix
(C)
Symmetric Matrix
(D) Skew Symmetric Matrix
The value of x, y and z is in
3+x 1+y 3z
(A) (4)
(J  07)
(3, 0, 2)
(B) (3, 0, 1)
0 =
1
is .......................
2 (C) (3, 0, 1)
(D) (1, 3, 2)
A is a matrix of order 3 x 4, B is a matrix of order 3 x 2 and C is a matrix of order 2 x 4. Which of the following has the order 3 x 3 = .......................... (A)
(5)
2
0
0
2
0 3 3
(8)
(D) CAB
(C) BCA1
(B)
0
2
2
0
(C)
2
2
2
2
(D)
2
0
2
0
2
3
3
0
Which of the following is a Symmetric Matrix .......................... (A)
(7)
(B) BCA
Which of the following is a Scalar Matrix .......................... (A)
(6)
ABC
(B)
0
2
0
0
5
(C)
0
5
3
0
Transpose of a row matrix is .................................... (A)
Column Matrix
(B) UnitMatrix
(C)
Square Matrix
(D) Null Matrix
Which of the follwoing relation is true .................................... (A)
A B = B A
(B) A + B = A1 + B1
(C)
A+ B = B +A
(D) AB = BA
76
(D)
(9)
If A = (A)
(10)
5
6
7
12 15
and KA =
1
18 21
(B) 2
If A = (A)
(11)
4
1
2
3
4
7
7
3
3
(C) 3
and A+B = (B)
2
1
(B)
7
11
9
11
then the matrix B is .......................... (C)
3
3
7
7
(D)
3
0
0
3
(B)
1
2
0
3
3
2
1
4
4
2
1
5
(A)
3x 1 5
+
4
(A) 0
B=
A+B
A=
3 5
(C)
11 11 9
(M  06)
0
5
5
0
(D)
0 5 5 0
2
0 2
4
1 2
5
(B)
I=
1
0
0
1
2
1
0
= (B)
3
(D)
(M  07)
(C)
AB
(D)
then IA is .......................... 3
2
1
4
8
3
6
1
then which one is possible......................
AB
(B)
5
(C)
(C)
3
1
2
4
BA
(M  07)
(D)
2
3
4
1
then x is equal to ............................. (M  07)
4
0
(C)
3
(D)
1
is a Skew Symmetric Matrix then the value of x is ....................... (J  07)
x+6 0
(A)
9
(J  06)
1
(A)
(16)
6
is a Skew Symmetric Matrix then the valyue of x is ..............
2x1 1
A=
(15)
x2
1
If A =
(A)
(14)
9
9
2 1
(13)
8
(D) 4
Which one is a Skew Symmetric Matrix ....................... (A)
(12)
then the value of K is .......................
6
(B)
0
(C) 77
5
(D)
6
1 (17)
If A = [ 1 2 3 ] B =
then the order of the matrix A x B is .....................
2
(J  07)
3 1x3
(A)
(18)
0 1
x 5
(A) (19)
(20)
(22)
(A)
0
(B)
AB
1
2
6
=
1
(0.0) 1 3
3
(24)
(D)
3x3
5
(J  07)
(D)
6 (M  08)
2x6
1
3
(A)
0
(AB)1 =
2
(D)
3
then which matrix is possible ..........................
6
(M  08)
A1B
(C)
A1B1
(D)
BA (M  08)
2 (B)
23
2
(A)
1x1
then value of x and y are .............................
and Q =
0 5
(C)
5
(B)
(B)
30 (23)
(C)
1
and B =
1 0 5
y3
(A)
0
(B)
2 3 4
2x
P=
(C)
is a Scalar Matrix then the value of x is .............................
x3
(A)
3x1
then the value of x is .............................
6
0
(A) (21)
0 6 1 5
=
2x 0
A=
(B)
(1.2) 5
(C)
(2.3)
(D)
(3.6)
(M  08)
then PQ = .............................................
6
23
30
5
0
(C)
5
18
0
30
(D)
20 30
7is a diagonal matrix then x is equal to ............................. (M  09)
(B) 2 3 5 6
2
5
3
6
1
(C)
2
(D)
3
then B1A1 = ............................................. (B)
2
3
6
5 78
(C)
2
3
5
6
(M  09)
(D)
5
6
2
3
(25)
If A is a Skew Symmetric MAtrix then which one of the following is true ...................... (J  09)
(A)
(26)
I= (A)
(27)
A= (A)
(28)
(B)
6+x 0 0
A = A
1
5
(B)
1 2
(C)
A =(A ) 1
1 1
(D)
6
(C)
5
(D)
0 1 1
2
0
1
(B)
1
4
1
2
1
(J  09)
3 2
5
2
2
1
(B)
5
2
0
2
(D)
(31)
0
2
1
2
then the value of x is ......................... (J  09)
10 6 3
(C)
4 3
(C)
4
(D)
5
Order of the matrix A is m x n and order of the matrix B is n x p then order of the
If A = (A)
<
1
then AA1 is .............................................
(M  10)
matrix AB is ....................................... (30)
A= A1
is an Identity Matrix then value of x is ........................... (J  09)
1
3x 2
(A) (29)
A=A
1
1
x+2
2x+1
3
2
is a Symmetric Matrix then the value of x is .................
(J  10)
(B)
1
(C)
2
(D)
1
If the order of a given matrix is p x q then the order of its transpose matrix is (J  10)
....................................... VI.
Solve the following problems:
(1)
If A = (i) A + B
1
6
4
5
B=
3
2
4
5
and C = (ii) 2B + C
79
5
1
0
3
then find the follwoing: (iii) B1 + 2C
(2)
Prove that A + A1 is a Symmetric for any Matrix A=
A1 =
A + A1 =
+
=
(3)
Prove that AA1 is a Skew Symmetric for any Matrix.
(4)
Prive that A + B = B + A for any Matrix A=
B=
A+B =
+
B+A =
= (5)
=
Prove that A  B = B  A for any Matrix.
80
+
(6)
If A =
(7)
If A =
4 1 3
2
2 5 1
0
show that 1/2 (A+A1) is Symmetric
B=
3 2
and A + 1/2 X = B then find X.
2 4
∴ X=2
A + 1/2 X = B

/2 X = (B  A)
1
X = 2 (B  A) = 2
=
(8)
If A =
1 2 3 4
and B =
A1 =
2 1 1 3
then verify (A+B)1 = A1+B1.
B1 =
A+B =
A1+B1 =
+
=
=
(A+B)1 =
81
+
(9)
Solve for x given that
(10)
If
(11)
If
(12)
If
1+x
0
1
2
2x 1 0
4
x2 3
4
5
=
+
2
3x 2 1
3
3x
0
1
2
3
2
2
1
1
1
0
1 4 2 1
=
find x.
=
x
+
9
3
2
5
=
82
find x.
0
1
6
5
find x.
7 6 3 4
(J  10)
a+1
b2
c+3
d1
x+y
2
1
xy
(13)
If
(14)
If
(15)
If
(16)
Solve for x and y
0 0
=
find a, b, c and d.
0 0
=
4
3
1
8
find x and y
\
1
5
x
2
3
y
2
=
Solve for x and y.
0
x
y
=
2 1
2
3
1
83
4
(17)
Find x and y if
(18)
If A =
1
2
3
4
A2 = A x A =
1
2
x
5 2
y
=
2 3
2
0 1
1
then find A2 and AA1. 1
2
1
2
3
4
3
4
AA1 =
=
(19)
If A =
2
1
3
3
4
2
4
=
2 1 3 0 2
1
6
1 3 and B =
0 1 2 4
84
then verify (AB)1 = B1A1
(20)
Prove that A(B+C) = AB + AC for any matrix.
(21)
Prove that Multiplication of two matrices is not commulative (AB = BA)
(22)
If A =
2 1 2 1
(M  06)
then find A2
85
(23)
If A =
(24)
If A =
(25)
A=
(26)
A=
1 0 2 3
2 3 5 1
1 2 3 0
1 2 0 3
then find AA1
(J  06)
then find AA1
(M  09)
then find AA1
(M  10)
then find A22A
(M  07)
86
(27)
If A =
(28)
If A =
(29)
If A =
(30)
If A =
1 3 1 0
3 1 2 5
3 1 1 2
1 3 5 7
then show that A2A3I = 0
then show that A28A+13I = 0
then show that A25A+7I = 0
and B =
4 5 7 8
87
then verify A2B2=(A+B)(AB)
3 2
(31)
Find x, if y =
(32)
Find X and Y, if X + Y =
(33)
If
1 4
3 2
2x
1 4
1
and 2x + y =
7 0 2 5
+2
4
3 2
and X  Y =
=
5
1 0
8 4y
************** 88
3 0 0 3
find x and y.
2
PERMUTATIONS AND COMBINATIONS • Points to remember n! = n x (n  1) x (n  2) ................ x 3 x 2 x 1 n! = n(n  1)! 0! = 1 Pr = n(n  1) .......................... (n  r + 1)
n
Pn = n(n  1) .......................... 3 x 2 x 1 = n!
n
Pr =
n
n! (nr)!
r≤n
Pr
n
Cr =
n
r!
Cr =
n(n  1) (n  2) ......... (n  r + 1) r!
Cr =
n! r! (nr)!
n
n
Cr = nCnr or
n
r≤n
CX = nCY => X = Y or X + Y = n
n
C0 = nCn = 1
n
C1 = nCn1 = n
n
Cr =
n
Cr +
n
n r
x
Cr1 =
n
Cr1
n1
Cr
n+1
93
FACTORIAL I
Convert the following products into factorial: Model Problems
(1)
1. 2. 3. 4. 5 = 5!
(2)
6. 7. 8. 9. 10 =
(3)
2. 4. 6. 8. 10 = 25 (1. 2. 3. 4. 5) = 25. 5!
II
Calculate the following
1. 2. 3. 4. 5. 6. 7. 8. 9. 10 1. 2. 3. 4. 5
10 x .......... x ............ 8!
=
10! 5!
(1)
10! 8!
(2)
10 x .......... x ............ x .......... x ............ 10! = 6! 4! 6! x .......... x ............ x .......... x ............
=
=
III
Convert the following product into factorials.
(1)
5. 6. 7. 8. 9. 10
(2)
3. 6. 9. 12. 15. 18
(3)
(n+1) (n+2) (n+3) ..................... (2n)
(4)
1, 3, 5, 7, 9............... (2n1)
IV
Find ‘n’ from the following.
(1)
(n + 1)! = 12 x (n1)! (n+1)!
= 12
........... .......... x ............ x ..........
= 12
(n1)! ............ x ............ = 12 ∴ n = ....................
............ x ............ = 3 x 4 90
(2)
(n + 2)! = 60 (n1)! (n+2)! (n1)! = 60 .......... x ............ x .......... x (n1)! (n1)! ............ x ............ x ............ = 5 x 4 x 3 âˆ´ n = ....................
(3)
If (n + 3)! = 56 (n+1)! find n.
91
= 60
FUNDAMENTAL COUNTING PRINCIPLE If one event occurs in ‘m’ different ways and another event occurs independently in ‘n’ different ways then the two events together can be done in (m x n) different ways.
An Illustration: A coin is tossed, afterwards, a die is thrown. Let us find the total number of outcomes of this experiment. Total number of outcomes of the experiments
=
Number of outcomes of tossing the coin
x
Number of outcomes of throwing the die
[h, t] x [1, 2, 3, 4, 5, 6] 12 = 2 x 6
Examples: (1)
A and B are connected by five routes and B and C by four routes. Then the numbber of ways of travelling from A to C enroute B is .......... x ........... = ............
(2)
I throw a die twice. The total number of outcomes is ........... x ........... = ...............
(3)
The total number of 2 digit numbers in which the digits are non repeating is ............. x .............. = ...............
(4)
The total number of two digit numbers is ............. x .............. = ...............
(5)
The total number of three digit numbers is ............. x .............. = ...............
(6)
The total number of 3 digit numbers with no digit repeated ................ x ................. x .............. = ....................
(7)
In a class there are 14 boys and 11 girls. The number of ways one boy and one girl pair can be formed is ............. x .............. = ...............
(8)
In a restaurant, 5 types of sweets, 4 types of snacks and 6 types of drinks are available. The number of ways a customer can order for a sweet, a snack and a drink is ............. x .............. x .............. = ...............
(9)
There are 20 buses playing, between Bangalore and Hyderabad. Number of ways can a person go from Bangalore to Hyderabad is ............. x .............. = ............... 92
PERMUTATION Permutation is the arrangement of objects which can be made by taking some or all objects from a set of given objects. P3 = 4 x 3 x 2
4
Note: (1)
On the LHS 4 is at the top of P and first factor on the RHS is 4.
(2)
On the LHS 3 is the bottom of P and the number of factors on the RHS is 3.
(3)
Successive factors starting with 4 (i.e. the first factor) are got by declaring each factor by 1 generalisation
(4)
The last factor in (1) on the RHS is got as follows n, (n1), (n2) .......... from an AP with first term n, CD = 1. Tn = a + (n1) d rth term = n + (r  1) (1) = n  r + 1
I
Fill up the blanks.
(1)
7
(2)
................. = 10 x (10  1)
(3)
................. = 6 x (6  1) x (6  2)
(4)
8
II
Find the value of the following.
(1)
5
(2)
10
(3)
6
(4)
4
(5)
n
P4 = 7 x ............... x ............... x ...............
P5 = ............... x ............... x ............... x ............... x ...............
P2 = ............... x ............... = P3 = ............... x ............... x ............... =
P6 = ............... P4 + 4P0 + 4P3 = ............... + ............... + ............... = Pn = ............... 93
(6)
n
(7)
0! = ...............
III
Write the following in the form of nPr.
(1)
8(8  1) =
(2)
8(8  1) (8  2) =
(3)
8(8  1) (8  2) (8  3) =
(4)
8(8  1) (8  2) (8  3) (8  4) =
(5)
8(8  1) (8  2) (8  3) (8  4) (8  5) =
IV
Write the following in the form of nPr.
(1)
n(n1) =
(2)
n(n1) (n2) =
(3)
n(n1) (n2) (n3) =
(4)
(n+1) n(n1) =
(5)
(n+3) (n+2) (n+1) n =
(6)
(n1) (n2) (n3) =
P0 = ...............
94
Teaching Tips: (1)
The following method (of expressing the value as the product of consecutive natural numbers) is easier
(2)
It is always possible to express the value of nPr as the product of consecutive natural numbers Their being done, the number of factors in the product is r, while n is the biggest of the factors.
A
V
Finding unknown ‘n’ or ‘r’.
(1)
If nPn = 5040 find n. Pn = n!
n
n! = 5040 n! = 1 x 2 x ........... x ............. x .............
1
5040
2
5040
3
2520
4
840
5
210
6
42 7
n! = ......... ! ∴
n = ...........
Strategy: Express the given value as product of consecutive natural numbers. (2)
If nPn = 120 find n. n! = 120
120
2
120
3
n! = 1 x 2 x ................................. n! = ........... ! n = .............. (3)
1
n! = 720 n! = ........... x ............. x .................................. n! = ............! n = .............. 95
(B) (1)
If 11Pr = 990 find r Pr = 11 x 10 x 9
11
11
990
10
90 9
P r = P3
11
11
∴r = 3
(2)
Pr = 336
8
8
336
7
Pr = 8 x .......... x ...............
8
Pr = 8P.....
8
∴r = ....................
(3)
Pr = 156
13
13 156
Pr = ............. x .............
13
Pr = ...... P.....
13
∴r = ....................
(C) (1)
P2 = 30
n
n(n1) = 6 x 5 n(n1) = 6 x (61) ∴n = 6
(2)
P3 = 210
n
n(n1) (n2) = ............ x ............. x ............. n(n1) (n2) = ............ x (......  1) x (......  2) ∴n = ..........
96
(3)
P2 = 6
n+1
(n+1) x .............. = 3 x 2 (n+1) x ........... = (2+1) x 2 ∴n = ..........
(4)
(n+3)
P2 = 20
................ x .............. = 5 x 4 ................ x .............. = (2+3) x (2+2) ∴n = ..........
(D) (1)
If nP4 = 12 nP2 then find n. n x (n1) x ............. x ................ = 12 x n x ................ ................. x ................ = 12 n2  3n  ......... + 6 = 12 n2  .......... + 6 = 12 n2  .......... + 6  12 = 0 n2  5n  6 = 0 n2  .......n  ........n  6 = 0 n (n ........)  2 (n3) = 0 (.............) (.............) = 0 ..........  .......... = 0
or
.........  ......... =0
∴n = ..........
or
n = ..........
97
(2)
9(n1) P3 = nP4 find n. 9(
)x(
)x(
) = n x ................ x ................ x ................
∴ .................... = ........................ ∴ n = ........................
(E) (1)
Find the number of permutations of the letters of the word “MILK”. The word “MILK” consists of .................. different letters Number of permutaions = 4P4 = 4 x ................ x ................ x 1
\
= .................... (2)
Find the number of permutations of the letters of the word “WORLD”. The word “WORLD” consists of .................. different letters Number of permutaions = ................................... = ................ ................ ................ = ....................
(F) (1)
In how many ways can 5 students be seated on a bench? Ist Method: 5 Students are arranged in a row ∴ Total number of arrangements
= 5P5 = 5! = 120
98
IInd Method: 1st
2nd
3rd
4th
5th > Places
5
4
3
2
1
Let us take 5 places for five students. The first place can be filled with 5 students. After allocating one, the 2nd can be filled with 4 students, 3rd with 3, 4th and 5th with 2 and 1 respectively. \ Total number of ways
=5x4x3x2x1 = 120 Ways.
(2)
Seven athletes are participating in a race. In how many ways can first three prizes be won? 1st
2nd
3rd
The required number of ways = ................ x .................. x ..................
= .......................... (3)
In how many ways 4 people out of 6 people can be seated in a row for a photograph. 1st
2nd
3rd
4th
The required number of ways = ............... x ................. x ................. x ................
= ..........................
99
Model Problems (1)
How many different numbers of 3 digits can be formed with digit 1, 2, 3, 4, 5 (a) Without repeating (b) With repeatation (a) Without repeatation: 100th
10th
5
4
Ist Method
Unit 3
The hundered place can be filled in 5 ways. The tenth place can be filled in 4 ways The unit place can be filled in 23 ways ∴ The total number of 3 digits = 5 x 4 x 3 = 60
IInd Method This is nothing but number of permutations of five digits, taking three at a time. P3 = 5 x 4 x 3 = 60
5
(b) With repeatation: 100th 5
10th 5
Unit 5
The number of 3 digit numbers that can be formed = 5 x 5 x 5 = 125 (2)
A 3digit numbers can be formed using 0, 1, 2, 3 without repeating any digit? 100th
10th
Unit
Any digit except 0 can occupy the 100th place. Therefore, the first place can be filled in 3 ways, after this second place can be filled in ................. ways, the third place can be filled in ................ ways. ∴
The total number of ways of following the places = ................ x ................ x ................ = .................. ways
∴
The total number of 3 digit numbers = .................... 100
(3)
How many numbers between 400 and 1000 can be made with digits 2, 3, 4, 5, 6 and 0. Solution:
H
T
U
100th Place can be filled only with digits 4, 5, 6 = 3 ways. 10th Place can be filled in ................ ways and Units place can be filled in ...................... ways âˆ´
The total number of number formation = ............. x ............. x ............. s = ....................
(4)
How many even numbers can be formed using all of 1, 2, 3, 5, 6 without repetition of the digit in a number. Solution:
10,000th 1,000th
100th
10th
U
The unit place can be filled only in 2 ways (2 or 6) The remaining box can be filled in ............! ways âˆ´
The total number of ways = ...........! x 2 = ............. x ............. = ....................
(5)
How many seven digit numbers can be formed with digits 1, 2, 3, 4, 5, 6, 7 such that they start with 2 and end with 6 (without repetition). 2
Solution:
6
In this case the first and last position are fixed. Therefore the question is about arranging 5 digits in the rest of 5 places which is .............! ways. = ...................... ways 101
(6)
Find the number of arrangements of word “ARUN” if all the word begin with letter A.
The first place can be filled with ............. and remaining in ............... ways. ∴
Number of arrangements
= ..................... = ...............!
(7)
= .................
How many permutations of all the letters of the word CHEMISTRY (a) begin with ‘M’ Solution:
(b) begin with M and end with ‘T’?
(a)
Number of permutation
= .................... = ....................
(b)
Number of permutation
= .................... = ....................
(8)
In how many ways five friends can sit in a row? In how many of there two friends A and B are side by side? Solution:
(a) Number of arrangements = ...................... = .................... = ................... ways
(b) Here we take the two friends (A and B) as one (as a strategy). Then the number of friends is 4 and hence can be permutated in =........................ ways =........................ ways A and B may be permutated among themselves in = ............. ways ∴ Therefore the total number of arrangements when A and B are side by side
= ....................... = ..................... ways 102
COMBINATION Combination is mere a selectio of different objects. Tips to teach: When n is big and r is close to n, you can avoid unnecessary calculators, using the property nCr = nCnr.
A
Cx = nCy
Case 1:
n
Case 2:
n
(1)
=> x = y
Cx = nCny => x = n  y => x + y = n
If nC6 = nC4 find the value of n. C6 = nC4
Solution:
n
C6 = nCn4
n
6=n4 n = 10
(2)
If nC6 = nC16 find ‘n’ C6 = nC16
Solution:
n
................. = ................. ................. = ................. n = ................. (3)
Find r, if (a) Solution:
(4)
Cr = 10C122r
10
r = 12  2r
or
r + 12  2r = 10
3r = 12
12  r = 10
r=4
r=2
Cr = 10C125r
10
Solution:
r = 12  5r
or
r + 12  5r = ................
6r = ................
............... = ...............
r = ...................
r = ................. 103
(5)
If
Cp = 24C2p+3 find P
24
Solution:
B
(1)
................................
.................................
................................
................................
................................
................................
Find n if nC2 = 15 Solution:
P2
Cr =
n
= 15
2!
n x ....... 2x1
= 15
n x ........... = 15 x 2 n x ........... = 30 n x ........... = 6 x 5 ∴ n = .................
Find n if nC2 = 10 Solution:
C2 = 10
n
P2
n
2!
Pr
n
C2 = 15
n
n
(2)
or
= 10
....... x ....... 2x1
= 10
........... x ........... = 10 x 2 ........... x ........... = 20 ............ x ........... = 5 x 4 ∴ n = .................
104
r!
(3)
If 6Pr = 360 and 6Cr = 15, find r Solution:
Pr =
n
r! =
Cr x r!
n
........ ........
r! = ................ r! = .............! âˆ´ r = .................
(4)
If nPr = 720 and nCr = 120, find r Solution:
Pr =
n
Cr x r!
n
720 = 120 x r! r! = ................ r! = ................! r = .............. (5)
If 36nCr = 3nPr find the value of r. Solution:
36nCr = 3nPr Pr
n
36
r! 36
= 3nPr
=3
......... ................ = ................
r = ..............
105
(6)
C3 = 4 x nC2
n+1
(n+1)
Solution:
P3
P2
n
= 4x
3!
2!
(n+1) x ........... x .........
........... x .........
= 4x
.......... x ........... x ......... .............. =
..............
........... x .........
.............. ..............
.............. = ..............
n = .............. (7)
Find the value of (a) 100C2
(b) 100C99
P2
100
(a)
C2
100
= =
(b)
2!
C (1)
10
(d)
52
Cr = nCnr
n
= 100C10099
100 x 99 2x1
= 100C1
C1 = n
n
= 100
= 4950 (c)
C99
100
C10 = 1
n
Cn = 1
C0 = 1
n
C0 = 1
There are 10 noncollinear points. How many (a) Straight Lines (b) Triangles can be drawn by joining these points. Solution:
(a)
A line is a 2 combination of points. Ten points are given, 2 points joined give a line
âˆ´ The number of straight lines formed
= 10C2 = ................... = ................... = ...................
106
Solution:
(b) A triangle is formed by joining any three noncollinear points in pairs ∴ There are 10 non collinear points. ∴ The number of triangles formed
= ................... = ................... = ................... = ...................
(2)
How many diagonals can be drawn in a pentagon? I Method Consider, a Pentagon, vertices are 5 non collinear points. No. of straight lines can be drawn by using 5 non collinear points
C2
5 =
P2
5
=
......... = ................... = ................... No. of straight lines = No. of side + No. of diagonals. ∴ No. of diagonals =
No. of straight lines can be drawn
No. of diagonals

= 10  5 =5 II Method If a polygon has ‘n’ sides then No. of diagonals = nC2  n In a pentagon no. of diagonals
= 5C2  5 =
............
5
......... =
............
 ..............
......... = .................... 107
III Method No. of diagonals =
n(n3) 2
In a pentagon no. of diagonals
5(53)
=
2 5x2
=
2
=5 (3)
A polygon has 90 diagonals. How many vertices are there for the polygon. Solution: Key Idea:
When all straight lines are got by joining the vertices of the polygon, the diagonals are those other than the sides. No. of diagonals
=
n(n3) 2
= 90
n (n  3) = ............. x ............ n (n  3) = 15 x ........... n (n  3) = 15 x (15  ........) âˆ´ n = ...............
(4)
A polygon has 44 diagonals. Find the number of sides. Solution:
No. of diagonals
=
n(n3) 2
44 = ..............................
n (n3) = ............... x ...............
n (n3) = ............... x (..........  ..........)
n = .......................
108
D (1)
From a group of 20 cricket players, a team of 11 players is to be chosen. In how many ways can this be done? Solution:
The total number of forming = 20C11
the team of eleven
= ......................... = ......................... = ......................... (2)
How many groups of 6 persons can be formed from 8 men and 7 women? Solution:
Required number of ways =
C6
15
= ......................... = ......................... = ......................... (3)
Find the number of combination of the word ‘CAKE’ Solution:
The word ‘CAKE’ has 4 different letters. Select all 4 letters at a time ∴ The number of combinations = ..............................
= .............................. (4)
Find the number of combination of the word ‘SWETHA’ Solution:
Required number of ways
= .............................. = .............................. = ..............................
109
(5)
A team of cricket eleven has to be formed 20 players Ajit is one. In how many ways the team can be formed. In how many ways the team can be formed so as to (a) Include Ajit Solution:
(b) Exclude Ajit
(a) If the team is to contain Ajit, the choice has to made from the other 19 players to select 10 players. This can be done in
= ........................ ways = ........................ = ........................
âˆ´ The total number of ways of forming the team including Ajit = ........
(b) If the team does not contain Ajit, such a team is to be formed from 19 players (excluding Ajit) selecting 11 of the team. âˆ´ The total number of ways of forming the team excluding Ajit = ........
= ........................ = ........................ = ........................ Note: 19C10 + 19C11 = 20C11 because any team formed either includes Ajit or does not include Ajit. E
(1)
In a school of 16 teachers there are 10 men and rest women. A committee of 5 teacher to be formed to represent the class. In how many ways this group can be chosen such that at least has 2 women. Solution: No. of Men
The following possiblities are possible. No. of Women
No. of ways C3 x 6C2 =......................
3
2
10
2
3
............. x ............. = .................
1
4
............. x ............. = .................
0
5
............. x ............. = .................
Total number of ways =................ + .................. + ................. + .............. = .................. 110
(2)
There are 5 bowlers and 10 batsmen in a cricket club. In how many ways can a team of 11 be selected so that the team contain exactly 3 bowlers. Solution:
No. of teams selected = 5C3 x 10C8 = ............. x ............. = ............. x ............. = ...............
(3)
There are 6 bowlers and 9 batsmen in a cricket club. In how many ways can a team of 11 be selected so that the team contain at least 4 bowlers Solution: Bowlers (6)
The following possiblities are possible. Batsmen (9)
No. of ways
4
7
6
5
6
............. x ............. = .................
.....................
C4 x 9C7 = ......................
.....................
............. x ............. = .................
Total number of ways =................ + .................. + ................. = .................. F
(1)
A purse contain 3 coins out of which 6 are golden and 4 are silver. 3 coins are removed from the purse. Find the number of combinations to remove 2 gold coins.
(2)
From 7 English men and 4 Americans a committee as 6 is to be formed. In how many ways can this be done (i) When the committee contain exactly 2 Americans. (ii) At least 2 Americans. Solution 1: The number of ways in which the American can be chosen: .................... The number of ways in which the Englishmen can be chosen: .................... The required number of ways: ............. x ...............
111
Solution 2: Americans (4)
Englishmen (7)
No. of ways
2
4
...................................
3
3
...................................
4
2
...................................
âˆ´ Total number of ways: ............. x ............... x ...............
= ...................... (3)
Out of 7 consonants and 4 vowels, how many words can be made each containing 3 consonants and 2 vowels. The number of combined groups, each containing 3 consonants and 2 vowels
= .............. x ...............
Further, each of these groups contian 5 letters, which may be arranged among themselves = ................. ways = ............................. = .............................
****************
112
3
STATISTICS STANDARD DEVIATION FOR AN UNGROUPED DATA POINTS TO REMEMBER X = Mean
• Arithmatic Mean ( X ) =
X = Score
Σx
N = Number of scores
N
Σ = Sigma Σx = Sum of all Scores
Ex:
Find the mean of scores 20, 23, 26, 27, 29 Σx = 20 + 23 + 26 + 27 + 29 = 125 N= 5 ∴
I
X=
Σx N
125
=
5
= 25
Fill in the blanks Σx
N
130
10
240
20
X
14
20
8
12
360
18
480
16
• Deviation : Deviation = Score  Mean D=xX Ex: 1, 2, 3, 4, 5 X=
1+2+3+4+5 5
=
15 5
=3
113
D =xx = 1  3 = 2 2  3 = 1 33=0 43=1 53=2 Find the deviation for the following scores: 5, 8, 12, 13, 17
ΣX = ...................
X=
N = ..................
Score x
xX=D
5 8 12 13 17
To find Variance and standard deviation of ungroupded data: • Deviation :
2
=
ΣD2
Standard Deviation (
)= =
Ex:
= Variance N = Number of Scores D = Deviations 2
N Variance
ΣD2 N
Find the variance and standard deviation for the follwoing scores: 16, 18, 23, 25, 28 ΣX = 110 N=5 X=
X=
ΣX N 110 5
= 22
114
Score x
x  X = dD
D2
16
16  22 = 6
36
18
18  22 = 4
16
23
23  22 = 1
01
25
25  22 = 3
09
28
28  22 = 6
36 ΣD2 = 98
• Variance
2
= =
ΣD2 N 98
= 19.60
5
• Standard Deviation =
Variance
=
19.60
ΣD2
=
N
= 4.3 II
(1) Find the variance and standard deviations for the following scores: 20, 25, 35, 30, 45, 43 Σx = ............................ N = .............................. X= Score x
Σx N
= ...............................
• Variance ( xX=D
)=
2
D2
ΣD2 N
= .............
20
• Standard Deviation = Variance
25 35 30
=
45 43 ΣD2 =
115
ΣD2 N
= .......................
(2) Find variance and standard deviation for the following scores: 55, 50, 35, 45, 40 Σx = ............................ N = .............................. X=
Score x
Σx N
= ...............................
• Variance
xX=D
ΣD2 N
=............. • Standard Deviation = Variance
50 35 45
=
40
If Variance = 1.69 then SD = 1.69 = 1.3 If SD = 3 then Variance = (SD)2 = 32 =9 Fill up the blanks Variance ( 2)
SD (
ΣD2 N
= .......................
= ..........................
ΣD2 =
III
=
D2
55
Ex:
2
)
1.44 64 0.01 2.5 0.18 7 0.02
116
STANDARD DEVIATION FOR GROUPED DATA POINTS TO REMEMBER X = Score
ΣfX
• Mean of Grouped Data ( X ) =
f = Frequency
N
N = Number Scores X = Mean
Ex:
Ex:
Find the mean from the following table Score(X)
12
14
16
18
20
frequency (f)
6
8
4
7
5
Score X
frequency f
fx
12
6
72
14
8
112
=
16
4
64
= 15.8
18
7
126
20
5
100
N = 30
474
Mean X =
ΣfX N 474 30
=
Find the variance and standard deviation from the given data. Score X
20
30
40
50
60
frequency f
5
4
2
4
5
X
f
fx
X X=D
D2
fD2
20
5
100
20  40 = 20
400
2000
30
4
120
30  40 = 10
100
400
40
2
80
40  40 = 0
0
0
50
4
200
50  40 = 10
100
400
60
5
300
60  40 = 20
400
2000
N = 20
ΣfX=800
ΣfD2 = 4800
117
158 10
•X=
ΣfX
=
N
• Variance (
800
= 40
20
ΣfD2
) =
2
N 4800
=
= 240
20
• Standard Deviation (
)=
15.48
ΣfD2 1 1 25 5 304 4 3088
N 240
=
= 15.48
IV
240.0000 1 140 125 1500 1216 28400 24704 3696
(1) Find the variance and standard deviation for the following data. X
15
20
25
30
35
40
f
8
5
3
2
5
7
X
f
fx
15
8
20
5
25
3
30
2
35
5
40
7 N=
•X=
ΣfX N
X X= D
D2
fD2
ΣfD2 =
ΣfX=
= .......................... = .......................... 118
• Variance(
) = ................................
2
= ............................. = ................................ • ..................................... =
Variance
= .......................................... = .......................................... = .......................................... (2) Find the variance and standard deviation for the following data. X
20
23
25
30
32
35
f
1
2
5
5
2
5
X
f
20
1
23
2
25
5
30
5
32
2
35
5
fx
N=
•X
XX=D
fD2
ΣfD2 =
ΣfX=
= .......................... = .......................... = ..........................
• Variance(
D2
) = ................................
2
= ............................. = ................................ 119
• SD (
)
= = .......................................... = ..........................................
(3) Find the variance and standard deviation for the following data. X
32
35
38
39
40
42
f
2
4
10
6
7
1
X
f
32
2
35
4
38
10
39
6
40
7
42
1
fx
N=
• ............
XX=D
ΣfX=
=
ΣfX N
= .......................... =
ΣfD2 N
= ............................. = ................................ • ............
=
fD2
ΣfD2 =
= ..................................................
• ............
D2
Variance
= ............................. = .............................. = .............................. 120
STANDARD DEVIATION FOR GROUPED SCORES WITH EQUAL CLASS INTERVALS POINTS TO REMEMBER
•X=
Ex:
X = Mid point of the class interval
ΣfX
f = frequency
N
N = Number of frequencies
Find the variance and standard deviation for the following data: C I
1 5
6  10
11  15
16  20
f
1
2
3
4
CI
f
Mid Pt X
fX
XX=D
D2
fD2
15
1
3
3
3  13 = 10
100
100
6  10
2
8
16
8  13 = 5
25
50
11  15
3
13
39
13  13 = 0
0
0
16  20
4
18
72
18  13 = 5
25
100
Σfx130 =130
N = 10
•X
= =
ΣfX N 130 10
• Variance (
= 13
)
ΣfD2
=
2
N 250
=
10
= 25
• SD (
)= = =
Variance ΣfD2 N
25
=5 121
ΣfD2 = 250
V
(1) Find the variance and standard deviation for the following data: CI
59
f
1
CI
f
59
1
10  14
2
15  19
4
20  24
3
25  29
5
10  14 15  19 20  24 25  29 2
4
Mid Pt X
\
3
fX
5
X X= D
• Mean X = .............................. = .............................. = ...............................
• Variance (
)
2
=
ΣfD2 N
= .............................. = ..............................
• SD (
)
fD2
ΣfD2 =
ΣfX =
N=
D2
= = .............................. = .............................. = ................................. 122
(2) Find the variance and standard deviation for the following data: CI
30  34 25  29 20  24 15  19 10  14
f
\
1
CI
f
30  34
1
25  29
4
20  24
6
15  19
5
10  14
3
5 9
1
4
6
X
5
fX
5 9
3
X X= D
ΣfX =
N=
X
N
= .............................. = ...............................
• Variance (
)
2
= .............................. = .............................. = ..............................
• .............................. =
D2
fD2
ΣfD2 =
ΣfX
• ................ =
1
Variance
= .............................. = .............................. = .................................
123
(3) Find the variance and standard deviation for the following grouped data: CI
15  19 20  24 25  29 30  34 35  39 40  44
f
\
3
CI
f
15  19
3
20  24
4
25  29
6
30  34
8
35  39
7
40  44
2
4
6
X
fX
7
XX=D
ΣfX =
N=
•X
8
= .............................. = .............................. = ...............................
• Variance (
)
2
= .............................. = .............................. = ..............................
• SD ( )
= ............................... = ................................ = ................................
..
= .................................
124
2
D2
fD2
ΣfD2 =
(4) Find the variance and standard deviation for the following data: CI
10  20 20  30
f
1
CI
f
5
X
30  40
40  50
50  60
6
5
3
fX
X  X1 = D
D2
fD2
10  20 20  30 30  40 40  50 50  60
ΣfX =
N= \
•X
= .............................. = .............................. = ...............................
• Variance (
)
2
= .............................. = .............................. = ..............................
• SD ( )
= ............................... = ................................ = ................................
..
= .................................
125
ΣfD2 =
COEFFICIENT OF VARIATION (C.V.) POINTS TO REMEMBER
∴
• CV =
CV = Coefficient of Variation
X 100
X
= Standard Deviation X = Mean
Coefficient of Variation determines Consistency or Variability.
• The measure of despersion having less coefficient of variation is more Consist or less Variation. • The measure of despersion having more CV is less Consistant or more Variation Ex: The performance of two teams in an examination is given below. Which team is more Consistant. Team
Mean Score (X)
A
13
0.2
B
15
0.3
CV =
X
(
SD )
X 100
CV of A Team =
0.2 13
X 100
20
=
13
= 1.53 CV of A Team =
0.3
=
15
X 100
30 15
=2
∴ CV of A team is less than CV of B team. A team more Consistant.
126
VI
(1) The performance of Rama and Ravi in Mathematics is given below. Which one is more Consistant? X Rama
60
0.3
Ravi
75
0.5
CV = ................................ CV of Rama
= ................................
CV of Ravi
= ................................
=
0.5 (........)
X 100
= ................................
= ................................
= ................................
CV of ............................. is less than CV of ..............................
âˆ´ ................................ is more Consistant (2) The mean and standard devation of heights and weights of 20 persons are given below X Height in cm
160
0.4
Wight in K gs
50
0.3
In which carecteristic are they more variable? CV = ................................ CV of Height = ................................
CV of Weight = ................................
= ................................
= ................................
= ................................
= ................................
CV of ............................. is more than the CV of ..............................
âˆ´ ................................ is more Variable. 127
(3) The performance the yeild of mango trees Ramappa, Seenappa and Rajappa are given below. The yeild of whom tree is more Consistant.
Trees of
X
Ramappa
50
0.6
Seenappa
75
0.3
Rajappa
60
0.5
CV =
X
x 100
CV of Ramappa
= ................................ = ................................ = ................................
CV of Seenappa
=
0.3 ......
x 100
= ................................ = ................................
CV of Seenappa
=
........ 60
x 100
= ................................ = ................................ On comparing CV of ............................. is least. Hence performance of .............................. is more Consistant
128
(4) Find the coefficient of variation for the follwoing data. CI
59
f
1
CI
f
59
1
10  14
2
15  19
3
20  24
4
10  14 15  19 20  24 2 X
fX
•X
4 XX=D
ΣfX =
N=
\
3
= .............................. = .............................. = ...............................
• SD (
)
=
Variance
= .............................. = .............................. = ..............................
• CV =
X
x 100
= ................................ = ................................ ..
= .................................
129
D2
fD2
ΣfD2 =
VII (1)
The mid point of the class interval 25  29 is .......................
(2)
The mean of 4 scores is 20 and three scores among those are 16, 19, 21 then the 4th score is ..............................
(3)
The mid point of the class interval below the class interval 40  44 is .......................
(4)
Formula to find the mean of individual scores is ...........................
(5)
Formula to find the man of grouped score is ............................
(6)
Formula used to find the standard deviation of individual scores is .........................
(7)
Formula used find the standard deviation of grouped scrores is ...........................
(8)
Formula used to find the CV is ...........................
(9)
The measure compared to find the conisistancy is ...........................
(10)
The measure compared to understand the variability is ...........................
(11)
The variance of the scores is 0.0625. Then the standard deviation is .......................
(12)
The SD of scores is 1.6 then variance is ...........................
(13)
The CV is 30 and the mean is 10, then SD is ............................
(14)
The CV = 1.5,
(15)
If
(16)
The CV of two players Rama and Ramesh are 3/5 and 5/7. The more consistant
= 0.3 then X = ............................
= 0.6, X = 30 then CV = ............................
player is ........................... (17)
The CV of two teams A and B are 2/3, 4/5 then the team which varies more is ............................. **************
130
4
FACTORS AND FACTORISATION POINTS TO REMEMBER • a2 + 2ab + b2 = (a+b)2 • a2  2ab + b2 = (ab)2 • a2  b2 = (a+b) (ab) • a3 + b3 = (a+b) (a2  ab + b2) • a3  b3 = (ab) (a2 + ab + b2) • a4 + a2b2 + b4 = (a2 + ab + b2) (a2  ab + b2)
H.C.F. and L.C.M. by factorisation method:
Expression 1
5x – 10 2
5x  20
Factors
Factors
HCF
5x2 – 20
5x – 10
5(x2)
LCM 5(x2)(x+2)
2
= 5(x2)
= 5(x – 4) = 5(x2 – 22) = 5(x+2)(x2)
2
3
3x – 9
3x – 9
5x2  45
= 3(
5x2 – 20 )
)
= 5(
)
= 5(
)(
a2+7a+12
a2+7a+12
5x2 – 20
a2+8a+15
= ...................
= 5(
)
= 5(
)
= 5(
)(
= a(
) +3(
= (a+3) ( 4
= 5(
a2b2(p216)
a2b2 (p216)
ab(p364)
= a2b2 ( 2 2
=a b (
) )
)
................ ....................... )
ab(p3 – 64) = ab(p3  43)
) )(
)
................ .......................
= ab(p  4)(
)
= ab(p  4)(
)
= a3 – b3
................ .......................
131
Expression 5
a2 – b2
Factors
Factors a2 – b2
HCF
a3 – b3
a3– b3
6
................ .......................
p2 – 4
p2 – 4
p2– 5p + 6
p2– 5p + 6
= p2 – 22
= p2– 3p – 2p + 6
=(
)(
)
= p(
)2(
= (p2) (
7
8x32 – 1
8x3 – 1
a– b
m2 – n2
a–b
18x3y4z9
................ .......................
m2 – n2
(m + n)2 ................ .......................
18x3y4z9
15x4y7z9
15x4y7z9
)
b–a
(m + n)2
10
)
................ .......................
b–a
9
................ .......................
4x2+ 2x + 1
4x2+ 2x + 1
8
LCM
................ .......................
132
Expression 11
a+ b
Factors a+ b
Factors
LCM
a–b
a–b
12
HCF
................ .......................
(x + y)2 (x – y)2
................ .......................
x2 – y2
13
ab, bc, ca ................ .......................
14
a– 2
a–2
a3 – 8
a3 – 8
15
a2 – 9
................ .......................
a2 – 9
a2 + 6a + 9
a2 + 6a + 9
16
7(m+n)2(mn)2 7(m+n)2(mn)2
................ .......................
14 (m2 – n2)
14(m2  n2)
................ .......................
133
H.C.F. by Division Method: Note: (1) Arrange the terms of the expressions in decreasing order of the powers in two columns (2) If the last remainder is a constant and not zero, then the HCF of two expressions is 1 (1)
Find the HCF of 2m2 + 2m + m3 + 1 and 2m + 1 + m2 [Hint: Arrange the terms of the expressions in decreasing order of the powers] m
m3 + 2m2 + 2m + 1
m2 + 2m + 1
m3 + ...........+.............
............+ m
+m+1
m+1
(2)
x3  86x + 35  6x2
x3 + 40  5x2  99x
(3)
x3 + 2x2 + 2x + 1
m
HCF = (m + 1)
(4)
x3  x2  x  2
Find the HCF of expressions x3  7x2 + 14x  8 and x3  6x2 + 11x  6 x3  7x2 + 14x  8
x3  6x2 + 11x  6
134
(5)
Find the HCF of expressions x3  9x2 + 26x  24 and x3  6x2 + 11x  6 x3  9x2 + 26x  24
(6)
x3  6x2 + 11x  6
Find the HCF of expressions x4 + 3x3  x  3 and x3 + x2  5x + 3 x4 + 3x3  x  3
(7)
x3 + x2  5x + 3
Find the HCF of 3x4 + 6x3  12x2  24x and 4x4 + 14x3 + 8x2  8x 3x4 + 6x3  12x2  24x 3x(x3 + 2x2  4x  8)
[Taking 3x as C.F.]
4x4 + 14x3 + 8x2  8x 2x(2x3 + 7x2 + 4x  4) x3 + 2x2  4x  8
[Taking 2x as C.F.] 2x3 + 7x2 + 4x  4
HCF = HCF of 3x and 2x is ............................ HCF of the expressions = ( 135
)(
)
I
Relation Between Two Expressions and their HCF and LCM The product of the HCF and LCM of two expressions is equal to the product of the expressions. AxB=HxL (1)
If A and B are two expressions and their HCF is H, then their LCM can be calculated using the formula is ......................................
(2)
(A)
L=
H xA B
(B)
L=
A HxB
(C)
L=
B AxB
(D)
L=
AxB H
If H and L are the HCF and LCM of the algebraic expression, A and B respectively. Then which of the following is correct? ................................... (A) (C)
(3)
AxH=BxL A H
=
L B
(B)
A B
=
H L
(D)
A L
=
B H
HCF and LCM of two algebriac expressions are ax and 12ax2b3y respectively. If one of the expressions is 4axy, then the other is ........................................ (A) 3xa2b2
(4)
(B) 3x2a
(C) 3x2ab3
(D) 3xab2
If H = (a3), L = (a3) (a2) (a1) and A = a25a+6 then the other experssion ‘B’ is ....................................... (A) a2+3a+3
(5)
(B) a24a+3
(C) a2+2a3
(D) a23a1
The product of HCF and LCM of two expressions is 6a3b4c2. If one expression is 2a3b3c2, then the other is ....................................... (A) 3abc
(6)
(B) 6bc
(C) 3b
(D) 3bc
HCF and LCM of two expressions are 5x2y2 and 10x3y3 respectively. If one of the expressions is 5x2y3 then the other is ....................................... (A) 10x3y3
(B) 10x2y2 136
(C) 10x3y2
(D) 5x3y2
LCM by Division Method II
(A)
(1) L=
Find the LCM of x3 + 2x2 + 2x + 1 and x3  x2  x  2 A x B H
A = x3 + 2x2 + 2x + 1
B = x3  x2  x  2
First find the HCF of two expressions A and B x3 + 2x2 + 2x + 1
1
x3  x2  x  2
x3
HCF = ∴LCM of the two given expressions can be find by using the relation
L=
L=
A H
xB x3 + 2x2 + 2x + 1
x x3  x2  x  2
...................... ............................ ..................
x3 + 2x2 + 2x + 1
∴L = .................................... x (x3  x2  x  2)
137
(2)
Find the LCM of x3  7x2 + 14x  8 and x3  6x2 + 11x  6
(3)
Find the LCM of m4 + 3m3  m  3 and m3 + m2  5m + 3
138
B
(4)
x3  3x2y  18xy2 + 40y3 and x3  4x2y  11xy2 +30y3
(1)
HCF and LCM of two expressions are (x3) and x3  5x2  2x + 24 respectively. If one of the expression is (x2  7x + 12). Find the other. B= =
L x H A x3  5x2  2x + 24 x2  7x + 12
x
x3
.............................. x2  7x + 12
x3  5x2  2x + 24
B = ................................. x (x  3)
139
H=x3 L = x3  5x2  2x + 24 A = x2  7x + 12 B=?
(2)
The HCF and LCM of two algebraic expressions are (a3) and a3  5a2  2a + 24 respectively. If one of the expressions is a2  7a + 12. Find the other. B=
L x H A
.............................. a  7a + 12 2
a3  5a2  2a + 24
B = ................................. x (a  3) (3)
The product of two expressions is a4  9a2 + 4a + 12 and their HCF is (a2). Find their LCM. L= =
AxB H a4  9a2 + 4a + 12 a2
.............................. a2
a4  9a2 + 4a + 12
L = ................................. 140
III
(1)
The HCF and LCM of two expressions are a3 and a3 + a2  17a + 15 respectively. Find the two expressions. a2 + 4a  5 a3
a2 + 4a  5
a3 + a2  17a + 15
= a2 + 5a  a  5
a3  3a2 +
= a (a+5)  1 (a+5)
4a2  17a
= (a+5) (a1)
4a2 + 12a
F1
 5a + 15
F2
 5a + 15
+
0 H
0
F1
A = (a  3) (a + 5)
(2)
H
F2
B = (a  3) (a  1)
= ...................................
= ...................................
= ...................................
= ...................................
The HCF and LCM of two expressions are (m3) and m3 + m2  17m + 15 respectively. Find the two expressions. .................................... m3
m3 + m2  17m + 15
A = (m3) x
...................................
(Factor 1)
B = (m3) x
...................................
(Factor 2)
141
IV
(1)
M and N are two prime numbers, then their HCF is .........................................
(2)
Factors of a2  b2 is .........................................
(3)
If one of the factors of a3  b3 is (ab), then the other one is ..................................
(4)
(a+b) and a2 + b2  ab are the factors of .................................. (A) a3 + b3
(5)
(B) a3  b3
(C) (a + b)3
(D) (a  b)3
Algebraic expression which is having the factors (x+1) and (x+2) is ...................... (A) x2 + 2x + 2
(B) x2 + 3x + 2
(C) x2 + x + 2
(D) x2 + 3x + 2
D) (6)
If one of the factor of a4+a2b2+b4 is a2+b2+ab then the other factor is ...................... (A) a2 + b2
(B) a2  b2
(C) a2 ab + b2
(D) a4+b4ab
D) (7)
The expression obtained by s ubtracting x3  7x2 + 14x  8 from x3  6x2 + 11x  6 is ........................................ (A) x2  3x + 2
(B) x2 + 3x  2
142
(C) x2 + 3x + 2
(D) x2  3x + 2
CYCLIC SYMMETRY An expresion in three variables a, b and c is said to posses “Cyclic Symmetry” if we get back the original expression by changing a to b, b to c and c to a in order. Then the expression is called a cyclically symmetric expression in a, b, c Ex:
(1)
Show that x2 (y  z) + y2 (z  x) + z2 (x  y) is cyclically symmetrical expression. x2 (y  z) + y2 (z  x) + z2 (x  y)
 (1)
Changing the variables x to y, y to z and z to x in cyclic order. y2 (z  x) + z2 (x  y) + x2 (y  z)
 (2)
Compare equation (1) and (2) both are same ∴x2 (y  z) + y2 (z  x) + z2 (x  y) is cyclically symmetrical expression.
I
Show That the following expressions are cyclically symmetrical expression. (1)
(2)
(3)
(4)
a (b  c) + b (c  a) + c (a  b)
 (1)
b (c  a) + ............ + a (.........)
 (2)
ab (a + b) + bc (b + c) + ca (c + a)
 (1)
................+ ...............+ ab(a + b)
 (2)
a2b (a  b) + b2c (b  c) + c2a (c  a)
 (1)
............... + c2a (c  a) + ...............
 (2)
xy2 (x  y) + yz2 (y  z) + zx2 (z  x)
 (1)
yz2 (y  z) + ............... + ...............
 (2)
143
II
Write the following using ‘Σ’ Notation Ex: (1)
a+b+c =Σ a abc
(2)
a2 + b2 + c2  ab  bc  ca = a2 + b2 + c2  (ab + bc +ca) = Σ a2  Σ ab abc
abc
(3)
ab (a + b) + bc (b + c) + ca (c + a)
(4)
x (y2  z2) + y (z2  x2) + z (x2  y2)
(5)
ab2  ac2  a2b + bc2  cb2 +ca2 = ab2 + bc2 + ca2  a2b  cb2  ac2 = ab2 + bc2 + ca2  a2b  b2c  c2a = ab2 + bc2 + ca2  [a2b + b2c + c2a]
III
(1)
The missing term of cyclically symmetrical experssion ab2(a  b) + .............. + ca2 (c  a) is (a) ba2 (b  a)
(2)
(b) bc2 (b  c)
(c) bc2 (a  b)
(d) bc2 (c  b)
Which of the following is a cyclically symmetrical? ....................................... (a) (a+b) (b+c) (c+a)
(b) xy (xy) + yz (yz) + zx (zx)
(c) (ab) (bc) (ca)
(d) 3 (a2+b2) + 2ab + a + b 144
(3)
(4)
(5)
(6)
Which of the following is cyclically symmetrical? ........................................ (a) (ab) + (bc) + (c+a)
(b) a2 (bc) + b2 (c+a) + c2 (cb)
(c) ab(ab) + bc(bc) + ca(c+a)
(d) a2 (b+c) + b2 (c+a) + c2 (c+b)
Which of the following is not cyclically symmetrical? ........................................ (a) a+b+c
(b) a2 + b2 + c2 + 2ab + 2bc + 2ca
(c) a(ab) + b(bc)  c(ca)
(d) ab2 (ab) + bc2 (bc) + ca2 (ca)
Which of the following has no cyclic symmetry? .......................................... (a) x + y + z
(b) xy + yz + zx
(c) xy  yz  zx
(d) x2  y2  z2
Which of the following is be added to ab2+a2c to make it cyclic symmetry....................... (a) a2b
(7)
(b) bc2
(c) b2c
(d) ac2
Which of the following has to be added to a3 + b3 + c3 + 3ab (a + b) to make it cyclically symmetric? ....................................
(8)
(a) bc ( b + c) + ca (c + a)
(b) 3bc (b + c) + ca (c + a)
(c) 3 [bc (b+c) + ca (c+a)]
(d) 3abc (bc+c) + (c+a)
Which of the following expression is in cyclic symmetry if a, b and c are variables................................. (a) a + b + c)
(9)
(b) a2  b2  c2
(c) a  a2  a2
Which of the following is in cyclic with x, y, z variables? (a) x2 + xy
(b) x2 + z2 + xz + yz
(c) x2 + y2 + z2
(d) x2 + y2 + xy + yz
145
(d) a  b  c
IV
(1)
a + b + c in Σ notation is ................................................... (a) Σ a abc
(2)
abc
abc
abc
(d) Σa2ab abc
(b) Σabc
(c) Σa
abc
abc
(d) Σa+b+c abc
(b) Σx (y2+z2) abc
(c) Σ xy(zx) abc
(d) abc Σ xy(z+x)
(b) 2Σc(a+c) abc
(c) 2Σa(b + c) abc
(d) 2Σa(a+b) abc
(b) Σc2(a+b) abc
2 (c) Σa (b+c) abc
2 (d) Σb (a+b) abc
(b) Σab (ba) abc
(c) Σa (b2c2) abc
(d) Σb(ab) abc
(b) Σa(a2b) abc
2 (c) (Σa) abc
2 (d) Σ(a+b+c) abc
(c) Σca(ac)
(d) Σab2+Σabc
Σab (bc) is same as ..................... abc (a) Σab(ab) abc
(10)
abc
a2 + b2 + c2 + 2ab + 2bc + 2ca can be expressed using Σ notation as .................... (a) abc Σa2  abc Σ2ab
(9)
(c) Σa2Σab
ab2  ac2  a2b + bc2  cb2 + ca2 can be written as follows..................... (a) abc Σab (a+b)
(8)
abc
The one which is equal to ab(a + b) + bc(b + c) +ca(c + a) is..................... (a) abc Σab(a+b)
(7)
(b) Σa2+ab
ab + ac + bc + ab + ca + bc can be written by using Σ notation as ..................... (a) 2Σb(a+b)
(6)
(d) Σ ac abc
Write xy2  xz2 + yz2  yx2 + zx2  zy using Σ notation by ........................................... (a) abc Σ x(y2z2)
(5)
abc
ab + bc + ca can be written by using Σ as ........................................... (a) Σab
(4)
(c) Σabc
abc
a2 + b2 + c2  ab  bc  ca is represented in Σ notation is ........................................... (a) Σ a2
(3)
(b) Σab
(b) Σbc(ca) abc
abc
Which of the following cannot be written using Σ notation. (a) a2+b2+c2
(b) ab2+bc2+ca2+3abc
(c) a3b3c3
(d) (ab) + (bc) + (ca) 146
abc
abc
(11)
When Σ notation is used, the expression x2 + y2 + z2  x  y  z becomes ..................... (a) xyz Σ (x2+x)
(12)
c+
1 + b+ b2
(b) xyz Σ(xx2) 1 +c + c2
(a) abc Σ a2 V
(1)
abc
(4)
(b) xyz Σx2
(7)
(c) xyz Σx3
(d) Σxyz xyz
Σa2 + Σ2ab is equal to ........................................... abc
abc
(b) (a+b)3
(c) (a2+b2+c2)
(d) (a+b+c)2
2 If a, b, c are variables, then Σa + b is ..................................... abc
(a) a2 + b + b2 + c + c2 + a
(b) a2  b + b2  c + c2  a
(c) a2 + b  c + b2 + c  a
(d) a2 + b2 + c2
Σx(yz) is expanded and simplified to obtain .....................................
xyz
(b) 0
(c) xz  xy
(d) y(zx)
The expression of Σp2 is .................................... pqr
(a) p2 + q2 + r2 (6)
(d) abc Σa2ab
Σx = 0, then 3xyz is equal to ...........................................
(a) xy + xz (5)
(c) abc Σa2Σab
xyz
(a) (a+b)2 (3)
(d) xyz Σ(x2+x)
1 can ne expressed using Σ notation is ................. a2
(b) Σa2+ab
(a) xyz Σxy (2)
(c) xyz Σx2+Σx
(b) p2
(c) q2
(d) pqr
The expanded form of Σabca(ab) is ........................................... (a) (ab) + (bc) + (ca)
(b) a(bc) + b(ca) + c(ab)
(c) a  b  c
(d) a(ab) + b(bc) + c(ca)
Σa2 (b2  c2) when expanded and simpliefied ........................................... abc
(a) a2b2  a2c2
(b) a2b2  a2c2 + b2c2
(c) a2b2 + b2c2 + c2a2
(d) 0
147
(8)
(9)
(10)
(11)
2 is ............................ Σa2  Σb abc
abc
(a) 1
(b) 0
(c) a2 + b2 + c2
(d) 2a2 + 2b2 + 2c2
Σ (a+b)  Σabc(c+a) is ..............................
abc
(a) 0
(b) a+b+c
(c) 2a+2b+2c
(d) None of the above
2 Σa (b2c2) when expanded and simplified ........................................... abc
(a) a2b2  a2c2
(b) a2b2  a2c2 + b2c2
(c) a2b2 + b2c2 + c2a2
(d) 0
When xyz Σ (x+y) is expanded, we get ........................................... (a) x + y + z
(12)
(14)
(1)
(d) 3xyz
abc
(b) a + b + c
(c) 1
(d) 0
When Σx(y  z) expanded and simplified, its value is ..................................... xyz (a) 0
(b) xy  yz  zx
(c) xy  xz
(d) xy + yz + zx
Σa  Σb is .................................... abc abc (a) a + b + c
VI
(c) 3x + 3y + 3z
The value of Σ (ab) is equal to ..................................... (a) a  b  c
(13)
(b) 2x + 2y + 2z
(b) 2a + 2b + 2c
(c) 0
(d) a  b
If the sum of three numbers is 0 and the sum of the cubes of the same numbers is 99, then the product of these numbers is ........................................... (a) 9
(b) 33
(c) 24
148
(d) 30
(2)
(3)
If a + b + c = 0 the ÎŁ
abc
a+b is ............................ 3
(a) 3
(b) 0
(c) 3
(d) None of the above
The sum and the product of three numbers are 0 and 30 respectively. The sum of their cubes is .............................. (a) 0
(b) 90
(c) 160
149
(d) 900
CONDITIONAL IDENTITY Points to be remember (1)
(a+b)2 = a2 + b2 + 2ab
(2)
(ab)2 = a2 + b2  2ab
(3)
a2  b2 = (a+b) (ab)
(4)
(x + a) (x + b) = x2 + x (a+b) + ab
(5)
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(6)
(a + b)3 = a3 + b3 + 3ab (a+b)
(7)
(a  b)3 = a3  b3  3ab (ab)
(8)
a3 + b3 = (a+b) (a2 + b2  ab)
(9)
a3  b3 = (ab) (a2 + b2 + ab)
(10)
a3 + b3 + c3  3abc = (a + b + c) (a2 + b2 + c2  ab  bc  ca)
(11)
a4 + a2b2 + b4 = (a2 + b2 + ab) (a2 + b2  ab)
(12)
(x + a) (x + b) (x + c) = x3 + x2 (a+b+c) + x (ab + bc + ca) + abc ************
I
(1)
What is an Identity? Give an example.
(2)
What is meant by conditional identity? Give an example.
150
II
(1)
If a + b + c = 0, prove that a3 + b3 + c3 = 3abc. a+b+c=0 a + b = .................. Cubing both sides (a + b)3 = ................................... a3 + .......... + 3ab (a + b) = .............................. a3 + .......... + 3ab ( c) = .............................. a3 + b3 + c3 = ...........................
(2)
If a + b + c = 0, show that (b + c) (b + a) = ac a+b+c=0 b + c = .................. b + a = .................. LHS
= (b + c) (b + a) = .................. x .................. = ........................................
(3)
If a + b + c = 0, prove that (b + c) (b  c) + a (a + 2b) = 0 LHS
= (b + c) (b  c) + a (a + 2b) = b2 .......... ( ................... ) + a2 + .......... = a2 + b2 + 2ab  ( ................... ) = (a + b)2 .......... ( ....................) = (a + b + c) (......................) = 0 (...............................)
LHS = 0 LHS = RHS 151
(4)
If a + b + c = 0, prove that a (a2  bc) + b (b2  ca) + c (c2  ab) = 0 LHS
= a (a2  bc) + b (b2  ca) + c (c2  ab) = a3 ......... (.................) + b3 ......... (.................) + c3 ......... (.................) = a3 + b3 + c3  ............................................. = 3abc  ..............................
(5)
LHS
= ..............................
LHS
= ..............................
If a + b + c = 0, prove that a (b2 + c2) + b (c2 + a2) + c (a2 + b2) = 3abc LHS
= a (b2 + c2) + b (c2 + a2) + c (a2 + b2) = ab2 + ac2 + bc2 + ba2 + ca2 + cb2 = ab2 + ba2 + bc2 + cb2 + ac2 + ca2 = ab (.................) + bc (.................) + ac (.................) = ..................... + ..................... + ......................
(6)
LHS
= ........................................
LHS
=
RHS
If a + b + c = 0, prove that
LHS
=
=
a2 + bc
b2 + ca
a2 bc
+
b2 ca
c2 ab
.......... + ............ + ........... abc
=
....................... abc
=
....................... abc
= .................... 152
+
c2 =3 ab
III
(1)
If a + b + c = 0, then the value of (b + c) (b + a) = .....................................
(2)
If a + 2b + 3c = 0, then the value of a3 + 8b3 + 27c3 = .....................................
(3)
If x + y + z = 0, then
(4)
If a + b + c = 0, then which is equal to (b + c) (c + a)?
2 x2 + y + yz xz
153
z2 = .......................... xy
(5)
If x2 + y2 = 5 and xy = 2 then x + y = .....................................
(6)
(7)
If a + b + c = 0, then (a + b) (b + c) (c + a) = .....................................
(8)
If a  2b  3c = 0 then a3  8b3  27c3 = .....................................
x+
1 x
2
 x
1 x
2
is equal to .....................................
154
(9)
If a + b + c = 0 then,
[ a+bc + b+ca + c+ab ][ a+bc + b+ca + c+ab ]
is equal to .....................................
a b
+
b c
+
c a
(10)
If
= 0, then
(11)
If x2  3x + 1 = 0 then x +
โด
x2  3x + x x x3+
x+
1 x
1 x
+
c3 b3 + a3 c3
1 = ........................ x
x2  3x + 1 = 0 รทx
a3 b3
=0
=0
1 = ............. x
155
= ........................
(12)
1 1 = 2 then p2 + 2 is ................................ p p
If p + p+
1 p
=2
Squaring both sides Â p+
1 p
2
= 22
..................................... ..................................... .....................................
(13)
If a + b + c = 0, then the value of a + b  2c is .....................................
(14)
If x + y = 7 and xy = 10 then the value of
x+y is ..................................... x2  y2
(x  y)2 = (x + y)2  4xy = ..................................... = .....................................
=
x+y x2  y2 x+y (x + y) (x  y)
= .....................................
= ..................................... = .....................................
156
(15)
The value of (a + b)2  (a  b)2 is .................................
(16)
If a + b + c = 0, show that b2  4ac is a perfect square. a+b+c=0 a + c = b Squaring both sides (a + c)2 = (b)2 ............................. = ............................. Subtract 4ac both sides .............................  4ac = .............................  4ac ............................. = ............................. âˆ´b2  4ac = ............................. âˆ´b2  4ac is a perfect square
(17)
x+ LHS
1 x
= 3, then prove that x3 + 2x + = x3 + 2x + 1 = x + x3 3
2 x + + 2
1 = 30 x3 x+
1 x
2 x
+
1 x3 x+
= .........................................
157
1 =3 x
Cubing both sides x+
= ..................... + 2 (.............)
= 30
1 x
3
= 33
x3 +
1 1 +3xxx = 27 x3 x
x3 +
1 = 27  (...........) x3
x3 +
1 = ..................... x3
(18)
2 a2 + b2 + c2 + 3 3 3 3 a +b +c
If a + b + c = 0, show that LHS
=
a2 + b2 + c2 + a3 + b3 + c3
2 3
=
a2 + b2 + c2 + a3 + b3 + c3
2 3
=
a2 + b2 + c2 + a3 + b3 + c3
2ab + 2bc + 2ca 3abc
=
a2 + b2 + c2 + a3 + b3 + c3
2ab + 2bc + 2ca 3abc
=
1 + a
1 + b
1 a 1 c
+
1 b
+
1 c
= 0
bc+ca+ab abc
.......................................
[Substitute a3+b3+c3 = 3abc]
= ...........................
....................................... =
........................
=
........................
........................ ........................
= .................................... (19)
If a + b + c = 0, then prove that a2  bc = b2  ca = c2  ab =  (ab+bc+ca) a2  bc
a+b+c=0
= a x a  bc
b = b  c
= a (b  c)  bc = ab  ca  bc = .............................. (20)
= 0
= ..............................
If a + b + c = 0, then prove that Σa (a2  bc) = 0 abc
Σa (a2  bc)
abc
= a (a2  bc) + ....................... + ....................... = a3  abc + ....................... + ....................... = a3 + ............. + .................  3abc = ............................................. = ............................................. 158
(21)
If 3x  4y = 5 then P.T. 27x3  64y3  180xy = 125 LHS
= 27x3  64y3  180xy = (3x)3 ......... (.............)3 ......... 3 x (.........) x (.........) x (...............) = (..............) ......... (..............) ......... (....................) = ........................................................... = ...........................................................
(22)
If x + y + z = 0, P.T. (xy + yz + zx)2 = x2y2 + y2z2 + z2x2 LHS
= (xy + yz + zx)2
Hint: (a+b+c)2=a2+b2+c2+2ab+2bc+2ca
= (x + y + z) = 0
(23)
[ b+ca
If a + b + c = 0 then P.T. LHS
=
[ b+ca + c+ab +
+
c a+b
][ b+ca + c+ab c+a a+b + =9 b c ]
b c + c+a a+b
][ b+ca +
+
a+b c
]= 9
Hint: a+b+c = 0 a+b = c b+c = a c+a = b
159
(24)
If ab + bc + ca = 1, then S.T. LHS
=
a+b 1ab
=
a+b ab+bc+caab
a+b 1 = 1ab c
[Substitute 1 = ab + bc + ca]
(25)
If 2(a2 + b2) = (a+b)2 show that a = b
(26)
If x + y = 5 and xy = 10, P.T. LHS
= =
x3  y3 x2  y2
= 3
x3  y3 x2  y2
Hint: (xy)2 = (x+y)2  4xy
(xy)3 + 3xy (xy) (x+y) (xy)
= ................................................
= ......................... = ......................... = .........................
= ................................................ = ................................................
160
(27)
If x + y = 8 and xy = 12, Find the value of
[ xy
2
(28)
+
y2 x
] [ yx ] [ yx ] ÷
+1
+1
If b2 = c2 + a2, P.T. (a + b + c) (a  b + c) = 2ac LHS
= (a + b + c) (a  b + c) = [(a+c)+b] [(a+c)b] = .................................................. = .................................................. = ..................................................
(29)
If xy (x + y) = 1, P.T. LHS
1 x3y3
 x3  y3 = 3
1  x3  y3 x3y3 1 3 =  [x3 + y3] xy
= ............................
Hint: xy (x + y) = 1 1 (x + y) = xy Cubing Both Sides 1 3 (x + y)3 = xy
= ............................
............................
=
............................ 161
(30)
If x + y = a, xy = b, S.T. (1 + x2) (1 + y2) = a2 + (1b)2 RHS
= a2 + (1b)2 = (x + y)2 + (1  xy)2
(31)
If x +
1 4 1 = 4, S.T. x3 + 4x2 + 2  3 = 148 x x x
(32)
If x2 +
1 1 3 2 = 7, P.T. x + x x3
= 18
162
(33)
VI
If x2 +
1 1 = 27, then P.T. x + x x2
= ± 29
If a + b + c = 2S, then (a)
a+bc=a+bc+cc = a + b + c  2c = 2S  2c = 2 (S  c)
(b)
a+cb=
(c)
b+ca=
(d)
abc=
163
[Add and Subtract c]
(e)
a2 + b2  c2 + 2ab = a2 + b2 + 2ab  c2 = (.................................)2  c2 = (a + b + c) (a + b  c) = 2S ( ......................... )
(f)
b2 + c2  a2 + 2bc =
(g)
c2 + a2  b2 + 2ca =
(h)
a2  b2  c2 + 2bc
= .......................................
= a2  (b2 + c2  2bc) = a2  (b  c)2 = [a+ (bc)] [a(bc)]
V
(i)
b2  c2  a2 + 2ac =
(j)
c2  a2  b2 + 2ab =
If a + b + c = 2S then prove the following identities. (1)
S2 + (S  a)2 + (S  b)2 + (S  c)2 = a2 + b2 + c2 LHS
= S2 + (S  a)2 + (S  b)2 + (S  c)2 = S2 + S2 + a2  2as + ......... + .........  ......... + ......... + .........  ..........3 = 4S2 + a2 + b2 + c2  (................) ......... (................) ......... (................) = 4S2 + a2 + b2 + c2  2a (......................................) = 4S2 + a2 + b2 + c2  2a (.............) = 4S2 + a2 + b2 + c2  (.............)
LHS
= a2 + b2 + c2
âˆ´ LHS = RHS
164
(2)
a2 + b2  c2 + 2ab = 4S (S  C) LHS
= a2 + b2  c2 + 2ab = a2 + b2 + 2ab  c2 = (..............................)  c2
a2  b2 = (a+b) (ab)
= (..............................) (..............................) = (a + b + c) (a + b  c + C  C)
Add and subtract ‘C’
= (a + b + c) (a + b + c 2C) = 2S (2S  2C) = 2S x 2(S  C) = 4S (S  C) (3)
b2 + c2  a2 + 2bc = 4S (S  a) LHS
= b2 + c2  a2 + 2bc = ..............................  a2 = (..............................) x (..............................) = (..............................) x (..............................) = .............................. = .............................. = ..............................
(4)
c2 + a2  b2 + 2ca = 4S (S  b)
165
(5)
a2  b2  c2 + 2bc = 4(S  b) (S  c) LHS
= a2  b2  c2 + 2bc = a2  [b2 + c2  2bc] = a2  (b  c)2
a2  b2 = (a + b) (a  b)
= [a + (b  c)] [a  (b  c)] = [a + b  c] [a  b + c] = [a + b + c  2c] [a + b + c  2b] = (2S  2C) (2S  2b)
(6)
LHS
= 4 (S  b) (S  C)
LHS
=
RHS
b2  c2  a2 + 2ca = 4(S  C) (S  a) LHS
= b2  c2  a2 + 2ca = b2  [..........................] = b2  (.................)2 = [...........................] [..........................] = [...........................] [..........................] = [...........................] [..........................] = (...................) (...................)
LHS
= (...................) (...................)
LHS
=
RHS
166
(7)
(2bc + a2  b2  c2) (2bc  a2 + b2 + c2) = 16S (Sa) (Sb) (Sc) LHS
= (2bc + a2  b2  c2) (2bc  a2 + b2 + c2) = [a2  b2  c2 + 2bc] [b2 + c2 + 2bc  a2] = [a2  (b2 + c2  2bc)] [b2 + c2 + 2bc  a2] = [a2  (....................)2] [(....................)2  a2] = (..................)(..................) (..................)(..................) = ............................................................. = ............................................................. = ............................................................. = ............................................................. = ............................................................. = .............................................................
(8)
(2ab + c2  a2  b2) (2a  c2 + a2 + b2) = 16S (Sa) (Sb) (Sc)
167
(9)
[2ca + b2  a2  c2] [2ca  b2 + a2 + c2] = 16S (Sa) (Sb) (Sc)
(10)
Sc a2 + b2  c2 + 2ab = 2 2 2 Sb a  b + c + 2ac LHS
=
a2 + b2  c2 + 2ab a2  b2 + c2 + 2ac
=
a2 + b2 + 2ab  c2 a2 + b2 + 2ac  b2
=
168
(11)
b2 + c2  a2 + 2bc Sa = 2 2 2 b  c + a + 2ab Sb
(12)
c2 + a2  b2 + 2ca Sb = c2  a2 + b2 + 2bc Sa
169
(13)
If a + b + c = 2S, then show that LHS
=
a2 + b2  c2 + 2ab a2  b2  c2  2bc
=
a2 + b2 + 2ab  c2 a2  [b2 + c2 + 2bc]
=
a2 + b2  c2 + 2ab S = 2 2 2 a  b  c  2bc Sb
(..................)2  c2 a2  (..................)
=
(a+b+c) (a+bc) (a+b+c) [a(b+c)]
=
a + b + c  2C abc
=
(14)
If a + b + c = 2S, prove that, S2 + (S  a)2 + (S  b)2 + (S  c)2 = a2 + b2 + c2 LHS
= S2 + (S  a)2 + (S  b)2 + (S  c)2 = S2 + S2 + a2  2as + (..........)+(.........)(.........)+(.........)+(.........)(..........) = 4S2 + a2 + b2 +c2  2aS  2bS  2cS = 4S2 + a2 + b2 +c2  2S (a + b + c) = ........................................................ = ........................................................ = ........................................................
170
(15)
c2  a2  b2 + 2ab = 4 (s  a) (s  b)
171
SURDS • The irrational root of a rational number is called a Surd. Example:
2 ,
3 ,
5
4 ,
• The Standard form of a Surd is •
n
Think !
n
9 ,
3
8
are
not Surds, why
a
a > ‘a’ radicand > ‘n’ order of the surd
• Pure Surd: The surds having 1 as their rational coefficients are called Pure Surds. Ex:
3
2 ,
4 ,
5
7
etc.
• Mixed Surd: The surds having their rational coefficients other than 1 are called Mixed Surds. Ex: 2
3 , 7
3 , 8
3
3
etc.
• Like Surd: A Group of Surds of the same order, having the same radicand in their reduced form are caled Like Surds. Ex: (1)
3
(2)
{
2 , 23 2 , 4 8 ,
3
18 ,
2 50 ,
32 } > { 2
2 , 3
2 ,5
2, 4
2}
• Unlike Surd: A Group of surds of different order, diffferent radicand or both in their reduced form are called Unlike Surds. Ex: (1)
2 ,
3
3 ,
3
4
(2)
3 ,
3
5 ,
4
7
(3)
5 ,
3
5 ,
4
5
3
etc
• Binomial Surd: The algebraic sum of two distinct monomial surds or the algebraic sum of a rational number and a monomial surd is called Binomial Surd. Ex: (1)
2 +
3 , 2
7 , p+2
172
q
etc
â€˘ Only like surds can be added or subtracted. â€˘ Multiplication and Division of Surds having same order. (1)
Rule:
n
Ex:
(2)
a x
n
3 x
Rule:
Ex:
n
a
n
b
3
14
3
7
b
=
n
2
=
3x2
=
n
=
3
axb =
6
a b 14 7
=
3
2
Rationalisation of Surds and Rationalising Factors The process of multiplying a surd, by another surd to get a rational number is called Rationalisation. Then the each surd is the rationalising factor of the other. 3 x Here
3
=
9
= 3
3 is the R.F. of
3
173
I
Write the respective orders and radicands of the surds in the following table.
S.No.
Surd
Order
Radicand
S.No.
1 Ex
3
2
3
5
8
4
5
6
8
19
50
7
5
4
3
2 3
5
4
II
Surd
Order
Radicand
Surd
17
19
7
Write the surds having the following order and radicand.
S.No.
Order
Radicand
Surd
S.No.
Order
Radicand
Ex
2
8
8
3
2
14
1
3
6
4
3
7
2
4
18
5
6
15
III
Express the following surds in the index form.
Surd Ex: 3
3½
3
Surd 5
8
Index Form
19
142
9
7 4
Index Form
5 17
4
6
3
7
IV
Express the following in Surd form.
Index Form Ex: 201/3
Surd Form
Index Form
20
5(2)1/3
3
5½
3(27)1/8
81/5
7(x2)1/3
174
Surd Form
Reduction of Surds 75 can be simplified as
Ex:
75 = Here
52 x 3 = 5 3
75 is a pure surd 5
V
25 x 3 =
3 is a mixed surd
Reduce the following surds. Pure Surd
Simplification
Mixed Surd
48 3
32
200 405
4
300
VI
Conversion of mixed surd into pure surd Ex:
11
2
>
112 x 2
=
=
Express the following surds into their pure form Mixed Surd 8 3
3
Pure form
4 2
34 5
121 x 2
10
3
5
6
175
242 Mixed form
Classification of Like Surds Ex:
VII
5
3 , 4
3
2 , 7
3 , 10
3
Set A: 5 3 , 7
3 , 11 3
Set B: 4
3
3
2 , 10
2 , 8
3
2 , 11
3,8
3
2
2
Classify the following surds into set of like surds (reduce them wherever necessary) (1) 5
2 , 6
(3) 125 ,
3
5 , 3
45 ,
12 ,
2, 8
2 , 10
20 ,
27,
5
(2)
18 , 3
300
(4)
a3, 3 a , 4 2a , 6 16a , 10 4a
3
2 , 43 3,
32 , 2 3 24
Addition and Subtraction of Surds (Only like surds can be added or subtracted) Ex:
(1) Addition of
5 = 5
a and 2 a + 2
= (5+2) = 7 Index form
a
a
= 7(a)1/2 176
a a
Sum or difference of the reduced surds can be obtained by adding or subtracting the coefficient.
VIII Simplify:
Ex:
(1)
10
(2)
Find the sum of 3
3
Simplify:
x  8
3
18 + 5
2 +
a and 4
a and express in the index form.
200
2 + 100 x 2
3 2 +5
2 + 10
(3+5+10)
2
(1) Simplify: 7 3 2 +
(3) Simplify:
a ,5
9x2 +5
= 18 IX
x
2
2 3
16 
3
45  3 20 + 4
54
(2) Simplify: 4
5
3 3
12 + 2
75
(4) Simplify: 2 2a + 4 8a  3 2a
177
Addition of Binomial Surds Ex 1: Find the sum of (2
x +3
y ) and (5
x 
Soln: 2
x +3
y + 5
x 
y
2
x +5
x + 3
y 
y
(2+5) 7
x + (31)
x +2
Ex 2: Subtract (5
a +3
b ) from
(8
a +5
b )
y ) Soln:
y
y
(8
8 + 12 , 3
(2) 2+4
3 , 27 
8
Multiplication of Surds (having same order) Rule
:
Ex
:
n
a x
b =
n
2 x
ab
n
3 =
2x3 =
6
Muliply the following surds: (1)
6 x
3 =
(2) 2
7 x
(3) 5
a x 3
(4) (5) (
a ( 3 +
2 = b =
b +
c ) =
2 )( 5 
7 )= 178
a +3
b )
a +5
b 5
a 3
b
8
a 5
a +5
b 3
b
(85)
(1) Find the sum of
b )  (5
8
3
X
a +5
a + (53)
a +2
Add :
b
2 + 2
b
3 +5
2 4
3
3 
2 +
Multiplication of Surds (having different order) Ex
: Multiply
3 x
3
2
The orders are 2 & 3, LCM of 2 & 3 is 6
3 = (3)1/2 = 31/2 x 6/6 = 33/6 = (33)1/6 = 3
2 = (2)1/3 = 21/3 x 6/6 = 22/6 = (22)1/6 =
âˆ´ 6
XI
3 x 26 x
=
6
=
6
3
6
6 6
33 =
22 =
108 5 and
4
2
2 and
3
3
(3) Find the product of
3
3 x
(4) Simplify
4
(Now they have reduced to same order)
27 x 4
(2) Multiply
6
27
2
4
(1) Multiply
6
3
5 x
4
4
2
3
179
(5) Find the product of
(6) Multiply
3
3
x
4
3 and
3
6
5
Rationalisation of Surds and Rationalising Factor Ex:
5 is a surd. When 5 x ∴
5 =
5
25 = 5 is a rational number.
5 is the Rationalising factor of
5
Write the rationalising factors of the following surds. S.No.
5 is multiplied by
Surd
1
3
a
2
x+y
3
m
4
5
x
5
7
y
6
4
p+q
7
6
ab
180
RF
The RF of a binomial surd (Conjugate) Ex:
( 3 +
2 ) is a binomial surd, if this is multiplied by (
( 3 +
2 )( 3 
( 3 )2  (
2 )
>
3 
2 )
[(a+b)(ab) = a2  b2 form]
2 )2
=32 =1 ∴
3 
2
is the RF of
3 +
2
3 +
2
is the RF of
3
2

Write the RF of the following binomial Surds S.No.
Binomial Surd
1
5
a + 3
2
m +
RF (Conjugate)
b n
3
4
x  2
4
10
a 
y b
5
3 +
b
6
7 
3
Simplification by rationalising the denominator. Ex
3
(1)
x 3
Rationalise the denominator and simplify. x
(
(Multiply and divide by RF of the denominator)
x
x 3
x
x x )
2
=
3
x x
181
XII
Rationalise the denominators and simplify the following. (1)
(3)
(5)
2
(2)
5
7
(4)
x
3
(6)
2x
a b
5 3
3a 5
Ex (2) : Rationalise the denominator and simplify 5 3 +
The RF of denominator is
2
5 3 + =
=
5(
x
2 3 
3

2
3 
2
3 
2 )
1
2 )
=
2
[(a+b)(ab) = a2  b2]
( 3 ) 2  ( 2 )2 5(
3 
5(
3 182
2 )
XIII
Rationalise the denominator and simplify the following. (1)
(3)
(5)
XIV
2 x 
(2)
y
m m +
3 6 +
(4)
n
2
(6)
3
10 7 +
2
6 3 +
7 10 
2
3 3
Rationalise the denominators and simplify the following. 5 + 3 (1) 5  3
183
(2)
(3)
(4)
(5)
2
3 +
2
3
2 +
5
2 3 
2
3 2 +
3
3 
2
3 +
2
+

3 3 +
2
2 2 
3
184
(6)
(7)
(8)
6 3 +
7
+
2
3
10 

3
4 5

+
3
3
2
6 +
2
3
5
6 + 2
2 5 +
3
Think (1)
The RF of
(2)
S.T. the RF of x1/3 + y1/3 is x2/3  (xy)1/3 + y2/3
3
x2 is
3
x
why ?
(Clue: Use a3 + b3 formula)
185
(9)
Find a and b if
(10)
Find a and b if
(11)
Simplify:
3  1 3 +1
4+
2
2+
2
4+
5
4
5
+
=a+b
3
=a
b
4
5
4+
5
186
XV
EXAM QUESTIONS (PREVIOUS) (1)
The answer obtained by the subtraction of (5
a  3
b ) from (8
a + 5
b )
is .......................... (2)
The RF of x1/2  y1/2 is ..........................
(3)
The product of 2 and
(4)
Rationalise the denominator and simplify
(5)
The sum of 3
(6)
The value of (2
(7)
The conjugate of xy is ..........................
(8)
Simplify:
(9)
The pure surd of 2
(10)
The RF of a
(11)
The answer obtained by rationalising the denominator of
(12)
Find the product of
(13)
Which of the following is a pair of like surds (a)
2, 3 
3
a ,  13
3 is ..........................
a, 6
2
3
3 3 +
by rationalising the denominator.
2
5 is ..........................
b + c is ..........................
3
3 and
(b)
3
2,
3
6
2
2 is subtracted from
(c)
If
(15)
The product of 3
(16)
The RF of p
(17)
The index form of a
(18)
The product of 2 and
(19)
When
(20)
The area of a rectangle with length (2 units is ..........................
(21)
Simplify by rationalising the denominator.
5 and (
q  q
8 +
3
2 
2,
(14)
2 
5 
a is ..........................
+
2 3
3
3 )2 is ..........................
2 + 3
3 
5 +
8
3 5
is ....................
(d) 4
3, 3
4
3 , then the result is ..........................
3  2) is ..........................
p is .......................... b is .......................... 3
5 is ..........................
50 is simplified we get ..........................
187
5 +
3 ) unit and breadth (2
7 10 
3 3
5 
3)
(22)
The sum of 9a and
(23)
When
(24)
When
(25)
32 +
8
25a is ..........................
is simplified by rationalising the denominator, the result is ............
2
10  2 is multiplied by its RF, the result is ................... 50 = ..........................
(26)
The RF of 5
p  q is ..........................
(27)
When 2
(28)
The order and radicandâ€™s of a n x are .................. and ..................
(29)
The sum of
(30)
The product of
(31)
Rationalise the denominator and simplify
x 
y is subtracted from 5
50 + 6
2 3 
2 and
128 is ..................
5 is ..................
3 + x + b
2
(32)
Conjugate of a
(33)
Find the product of
(34)
Rationalise the denominator and simplify 3
2 + 2
3
3
2  2
3
x =
y , the result is ....................
3
+
2
2 +
x + 2
4 and
3
3
(35)
If 9
(36)
The order and radicands of 5 3m are ..........................
(37)
Rationalise the denominator and simplify 2 3 +
(38)
If 10 +
12 +
4
y is ..........................
3 2 84 = a + 2
(a) 84 (39)
147 , the value of x is ..........................
b then the value of b ..........................
(b) 21
If x =
7  4
Clue (
7 4
(c) 10
3 then ST 7x + 3 =
4 
(d) 2 1 x
= 4.
3 )
************** 188
5
QUADRATIC EQUATIONS • An equation involving a variable whose highest degree is one is a linear equation, linear equation has only one root. General form is
mx + c = 0
• An equation involving a variable whose highest degree is two is a quadratic equation. The quadratic equation has only two roots.
• Quadratic equation involving a variable only in second degree is a pure quadratic equation. General form is
ax2 + c = 0
• Quadratic equation involving a variable in second degree as well as in first degree is an adfected quadratic equation. Standard form of quadratic equation where a, b and c are real numbers and a ≠ 0 is ax2 + bx + c = 0
• A quadratic equation can be solved by factors method or by using the formula x=
 b ± b2  4ac 2a
• The nature of the roots of ax2 + bx + c = 0 depends on the value of (b2  4ac) which is called the discriminant and denoted by Δ (delta)
• If b2  4ac = 0 the roots are real and equal. • If b2  4ac > 0 the roots are real and distinct. • If b2  4ac < 0 the roots are complex or imaginary. • If b = 0 the roots are equal but opposite in sign. • If a = c the roots are reciprocal to each other. • If c = 0 one root is zero. • If a and c have the same sign but ‘b’ opposite both roots are positive. • If a, b and c have the same sign both roots aer negative. • If a and c have opposite signs the roots have opposite sign. • The sum of the roots of ax2 + bx + c = 0 is given by
b a
• The product of the roots of ax2 + bx + c = 0 is given by
c a
• When the roots are given we can form the quadratic equation by x2  (sum) x + product = 0 189
I
(1)
In a perticular quadratic equation a stright line cuts the parabola at (2, 4) and (1, 1) (M  06)
then the roots of the quadratic equation are .................................. (a) (2, 4) (2)
(b) (1, 1)
(4)
(5)
(6)
(d) (4, 1)
Select the pure quadratic equation in the following.................................. (a) 2x2 + x + 5 = 0
(3)
(c) (2, 1)
(b) 3x2 + 1 = 28
(c) x2  x  7 = 0
(d) x  x2 = 0
Nature of the roots of the equation x2  6x + 9 = 0 is .................................. (M  06) (a) Real and Rational
(b) Real and Irrational
(c) Equal to each other
(d) Imaginary
Sum of the roots of x2  5x = 9 is .................................. (a) 5 (b) 5 (c) 9 5 The quadratic equation whose roots are 3 +
2 and 3 
(M  06)
(d) 
9 5
2 is .........................
(a) x2 + 6x  7 = 0
(b) x2  6x + 7 = 0
(c) x2  6x  7 = 0
(d) x2 + 6x + 7 = 0
(M  06)
If the product of the roots of the equation x2 + 3x + q = 0 is zero then the value of q is ................................
(7)
(M  06)
(M  06)
In which of the following equation the parabola obtained? ................................ (a) Equation of Straight Line
(b) Cubic Equation
(b) Quadratic Equation
(d) Linear Equation
(M  06)
(8)
(M  06) Positive root of the equation (2x  1) (x  3) = 0 is ................................ 1 (a) 3 (b) 3 (c) (d) 1 2 2
(9)
If V = πr2h then value of r is ................................ (a) ±
(10)
Vh π
(b) ±
πh V
(c) ±
(M  06)
V πh
(d) ±
The value of x in the equation ax2 + bx + c = 0 is ................................ (a) (c)
+ b ± b2  4ac
(b)
2a  b ‐ b2 + 4ac
(d)
2a 190
 b ± b2  4ac 2a  b ± b2 + 4ac 2a
πV h
(M  06)
(11)
(a) (0, 3) (12)
(13)
(c) 2x2 + x = 105
(d) 2x2 + x + 105 =0
(M  06)
If 3a2  27 = 0 then the value of a is ...................... (b) ±3
(J  06)
(c) ±27
(d) ±1
If (a+8)2  5 = 31 then the value of a is ...................... (b) +2, 14
(J  06)
(c) 2, +14
(d) 2, 14
Roots of the equation x2  x = 6 ...................... (b) +3, 2
(J  06)
(c) 3, 2
(d) 1, +6
Sum of the number and its square is 20 then the number is ...................... (a) 5
(17)
(d) 2
(b) 2x2  x = 105
(a) +1, 6 (16)
(c) 3
(a) x2 + 2x = 15
(a) 2, 14 (15)
(b) (0, 3)
The sum of a number and twice its square is 105. The equation form is ......................
(a) ±9 (14)
(M  06)
The roots of the equation x2  3x = 0 are ................................
(b) 10
(c) 4
(d) 2
m and n are the roots of the equation x2  6x + 2 = 0 then the value of
1 + m
(18)
(19)
(d) 2
(b) ax2 + bx + c = 0
(J  06)
(c) ax2 + c = 0
(d) ax2 + bx = 0
(b) b2  ac
(c) b2 + 4ac
(b) Distinct
The quadratic equation whose roots are 2 +
(J  06)
(c) Equal 3 and 2 
(d) Zero 3 is ......................
(a) x + 4x + 1 = 0
(b) x  4x + 1 = 0
(c) x2 + 2x + 3 = 0
(d) x2  2x  3 = 0
2
2
191
(J  06)
(d) b2 + ac
Roots of the equation x2  2x + 1 = 0 is ...................... (a) Not a real no.s
(21)
(c) 3
The roots of the quadratic equation depends on ...................... (a) b2  4ac
(20)
(b) 1.5
General form of pure quadratic equation is ...................... (a) (bx + c)2 = 0
1 n
(J  06)
is ...................... (a) 6
(J  06)
(J  06)
(J  06)
(22)
Product of the roots of the equation 6k2  3k = 0 is ................................ (a) 2 (b)  1 (c) 1 (d) 0 2 2
(23)
The value of m for which the equation x2 + mx + 4 = 0 has equal roots is ...................... (a) ±4
(24)
(26)
(c) 0
36 the value of a is ...................... a (b) + 3 (c) ‐ 3
(d) ±1
If we solve the equation 4a = (a) ±9
(25)
(b) ± 2
(M  07)
(d) ± 3
Standard form of the equation 2x = 5  x2 is ...................... (a) 2x  5 + x2 = 0
(b) x2 + 2x  5 = 0
(c) x2  2x + 5 = 0
(d) 2x  5  x2 = 0
(M  07)
(M  07)
The quadratic equation whose roots are 3 ± 5 is ...................... (a) x2  6x + 4 = 0
(b) x2  3x + 5 = 0
(c) x2 + 3x  5 = 0
(d) x2 + 6x + 4 = 0
(J  06)
0
(27)
If the roots of the equation are real and distinct which one is correct in the given (M  07)
below ...................... (a) Δ > 0
(b) Δ < 0
(c) Δ = 0
(d) Δ ≤ 0
(28)
(M  07) Sum of the roots of the equation 2x2  5x + 6 = 0 is ...................... 5 (a) (b) 3 (c) 5 (d) 2 2 2 5
(29)
The value of m for which the equation mx2 + 6x + 1 = 0 has equal roots is .................... (a) 6
(30)
(32)
(c) 9
(d) 5
(M  07)
One root of the equation x2  5x = 0 is ‘0’ then other one is ...................... (M  07) (a) 0
(31)
(b) 1
(b)  5
(c) + 5
(d) ± 5
The quadratic equation whose roots are 5 and 6 is ..................... (a) x2  30x  1 = 0
(b) x2  x  30 = 0
(c) x2 + x  30 = 0
(d) x2  x + 30 = 0
The product of the roots of the equation 3x2  7x + 9 = 0 is ...................... (a) 7 (b) 3 (c) 1 (d) 3 3 3 192
(J  07)
(J  07)
(33)
(34)
1 2 gt after solve the equation t is ................................ 2 2S 2S 2g (a) t = ± (b) t = (c) t = g g S If S =
2g S
(d) t = ±
Product of the roots of the equation x2  5x + 6 = 0 is ...................... (a) Real and Disticnt (b) Real and Equal
(35)
(J  07)
(c) Imaginary
(J  07)
(d) Equal (J  07)
A stright line graph Y = 3 is ...................... (a) Passes through the origin (b) Perpendicular to X axis (c) Parallel to X axis and passes through Y axis at 3 (d) Parallel to Y axis and passes through X axis at 3
(36)
(37)
If b = 0 in the equation ax2 + bx + c = 0 then the equation is ...................... (J  07) (a) Pure quadratic equation
(b) Adfected quadratic equation
(c) Equation of the straight line
(d) Linear equation
The value of m for which the equation x2  mx + 16 = 0 has equal roots is ...................... (J  07)
(a) ±4
(b) ±16
(c) ±2
(d) ±8
0
(38)
‘m’ and ‘n’ are the roots of the equation x2  6x + 2 = 0 ...................... (a) 12
(39)
(b) 6
(c) 2
(d) 3
If graphically solving the equation the roots are ..........................
X1
(J  07)
(J  07)
Y
1O 0 1
1
2
X
2
Y1
(40)
Which one of the following is pure quadratic equation .................... (a) 2x + 5 = 13
(41)
(b) x2 + 5 = 26x
(c) x2 = 5x
(d) x2 + 2x = 3
The quadratic equation whose roots are +2 and 2 is ...................... (a) Adfected quadratic equation
(b) Linear equation
(c) Simple linear equation
(d) Pure quadratic equation 193
(M  08)
(M  08)
(42)
(M  08) One of the root of the equation 2x (x + 3) = 0 is ................................ 2 (a) 0 (b) (c) +3 (d) 5 3
(43)
The base of a triangle is 4cm longer than its altitude. If its area is 30 Sqcm. Which of the equation is represented it? ...................... (a) x(x+4) = 30
(44)
(b) ±15
(d) x(x+4) = 60
(c) ±20
(d) ±25
Discriminent of the equation 2x2 = 5x is ...................... (a) 27
(46)
(c) x(x+4)=5
Product of the number and twice of it is 200 then the number is ............... (M  08) (a) ±10
(45)
(b) 2x(x+4) = 40
(M  08)
(b) 25
(M  08)
(c) 23
(d) 10
The value of m for which the equation x2 + mx + 36 = 0 has equal roots is ...................... (a) ±6
(b) ±8
(c) ±12
(d) ±18
(M  08)
0
(47)
Sum and product of the roots of the equation 2x2 = 3x is ...................... (M  08) 15 (a) 3 and 0 (b) 0 and 3 (c) 15 and 0 (d) 0 and 2 2 2 2
(48)
The quadratic equation whose roots are 3 + 2 2 and 3  2 (a) x + 6 + 1 = 0 2
(49)
x
(52)
(M  08)
(d) x  6x  1 =0 2
(J  08)
(b) 1
(c) 9
(d) 5
If a2 = b2 + c2, then c is equal to ................................... (a) ±
(51)
(c) x  6x + 1 = 0 2
If the roots of the equation mx2 + 6x + 1 = 0 are equal then the value of ‘m’ is ............ (a) 6
(50)
(b) x + 6x  1 = 0 2
2 is ..........................
b2 + a2
(b) ±
a2 + b2
(c) ±
a b
IF A = 4πr2, then r is ................................... A (a) ± (b) A (c) 4Aπ 4π 4π
(J  08)
(d) ±
a2  b2 (J  08)
(d)
4π A
“Twice of the square of a number is added to thrice of that number, the sum is 65” the equation for this statement is ................................... (a) 3x2 + 2x = 65
(b) 2x2 + 3x = 65
(c) 2x2  3x = 65
(d) 3x2  2x = 65 194
(J  08)
(53)
In a quadratic equation b2  4ac = 7, then nature of the roots is ................................ (J  08)
(a) real and equal
(b) real and distict
(c) imaginary
(d) ve number (J  08)
(54)
The descriminant of the equation ax2 + bx + c = 0 is ................................. (a) b (b) b2  4ac a + b ± b2  4ac (c) c (d) a 2a
(55)
The roots of the equation x2 + 4x + c = 0 are equal then the value of c is ...................... (a) 3
(b) 4
(c) 5
(d) 12
(J  08)
(56)
(J  08) The sum and product of the equation 4x2 + 1 = 0 are ...................... 1 1 (a) 1 and 4 (b) 0 and 1 (c) 0 and (d) 0 and 4 4
(57)
In a right angled triangle the hypotenous is 13cm one of the remaining sides is 5 cm more than the other, then the relation between them is ...................... (a) x + (x + 5) = 13
(b) x2 + (x2 + 5) = 13
(c) x2 + (x + 5)2 = 132
(d) x2 + (5  x)2 = 132
(J  08)
0
(58)
The roots of a quadratic equation are 0 and (a) 2x2 + x = 0
(59)
(61)
(62)
1 =0 2
(b) x2 +
(c) 2x2 + 1 = 0
(d) 2x2  x = 0
Which one of the following is pure quadratic equation .......................... (a) 2x2  x = 0
(60)
1 , then the equation is ...................... 2 (J  08)
(b) 5x = 3
(c) 4x = 9x2
(d) 2x2 = 16
The graph of a parabola is by ...............................
(J  08)
(a) a linear equation
(b) Simultaneous equation
(c) a quadratic equation
(d) a polynomial equation (M  09)
A equation has only one root those equation is ............................... (a) Quadratic equation
(b) Equation of Stright line
(c) Cubic Equation
(d) Linear Equation
If F =
mv2 then v is ................................... r
(a)
Fm r
(b)
mr F 195
(M  09)
(M  09)
(c)
Fr m
(d)
F rm
(63)
A positive root of the equation (2x  7) (3x  5) = 0 is ................................ (M  09) (a) 7 (b) 2 (c) 3 (d) 5 7 5 7 2
(64)
The value of x in the equation px2 + qx + r = 0 is ................................. (a)
(c)
(65)
 p ± p2  4pq
(b)
2p  p ± r2  4pq
(d)
2r
(M  09)
 q ± q2  4pr 2p  p ± p2  4pq 2q
Length of the rectangle is 4 cm longer than its breadth. If its area is 60 Sq.cm. The (M  09)
equation represented for it is ......................
(66)
(a) x + (x + 4) = 60
(b) x + (x + 4)  60 = 0
(c) (x + 4) x + 60 = 0
(d) (x + 4) x  60 = 0
The nature of the roots of the equation depends upon the value of ......................
(M  09)
(a) b2  4ac (67)
(68)
(70)
(M  09)
(d)
1 2 (M  09)
(b) 10
(c) 15
(d) 5
Sum and product of the roots of the equation are 5 and 4 then the equation is .......... (a) x2 + 5x + 4 = 0
(b) x2  5x + 4 = 0
(c) x2 + x  20 = 0
(d) x2  x  20 = 0
(M  09)
a and b are the roots of the equation x2  5x + 7 = 0 then ab (a+b) ............................... (b) 25
(c) 35
(d) 49
(M  09)
Product of the roots of the equation x2 + 5x + (K+4) = 0 is zero then K = ....................... (M  09)
(a) 5 (72)
(d) b + 4ac
The value of m for which the equation x2  mx + 25 = 0 has equal roots is ......................
(a) 5 (71)
(c) b  4ac
Product of the roots of the equation 2x2 = 3x is ...................... (a) 2 (b) 3 (c) 0 2 3
(a) 20 (69)
(b) b2 + 4ac
(b) 4
(c) 4
(d) 5
If a = 0 in the quadratic equation ax2 + bx + c = 0 then the equation is ............................. (J  09)
(a) Pure quadratic equation
(b) Adfected quadratic equation
(c) Linear equation
(d) Simultaneous equation 196
(73)
Roots of the equation 3x2  3x = 0 are ................................ (a) 0 and 1
(74)
(75)
(b) 0 and 3
(77)
(c) 1 and 3
(d) 0 and 3
If we solve the equation 7y =
35 then the value of y is ............................... (J  09) y
(a) ±
3
7
(b) ±
(c) ±
5 7
(d) ±
(b) x2 + x  42 = 0
(c) 2x2 + x  42 = 0
(d) x2x 42 = 0 (J  09)
The standard form of 2m2 = 2  m is ...................... (a) 2m2 + m  2 = 0
(b) 2m2  m  2 = 0
(c) 2m2  m + 2 = 0
(d) 2m2 + m + 2 = 0
The coefficients a, b, c of the equation 2k2  2k  5 = 0 are substituted the roots (J  09)
obtained are ...................... (a) k =
 (2) ± (2)2  4(2)(5)
(c) k = (78)
(79)
5
Sum of a number and its square is 42 it can be represented as ...................... (J  09) (a) x2 + x + 42 = 0
(76)
(J  09)
2 2 ± (2)2  4(2) (5) 2 (2)
(b) k =
 2 ± (2)2  4(2) (5)
(d) k =
2 (2)  2 ± (2)2  4(2) (5) 2 (2)
In a quadratic equation b2 = 4ac then the roots are ......................
(J  09)
(a) Real and equal
(b) Real and distinct
(c) Imaginary
(d) Imaginary and equal
‘m’ and ‘n’ are the roots of the equation then the general form of the quadratic (J  09)
equation is ............................... (a) x2 + (m + n) x + mn = 0
(b) x2  (m + n) x  mn = 0
(c) x2 + (m  n) x + mn = 0
(d) x2  (m + n) x + mn = 0
(80)
If m and n are the roots of the equation 2x2  6x + 1 = 0 then m2n + mn2 is ................... (J  09) (a) 3 (b) 2 (c) 3 (d) 1 2 2 2 3
(81)
The graphs y = x2 and y = 2  x are intersects at the points (1, 1) and (2, 4) then the roots of the expected equation are ............................... (a) 2 and 2
(b) 1 and 2 197
(c) 0 and 2
(J  09)
(d) 0 and 4
(82)
If v2 = u2 + 2aS then the value of u is ................................ (a) v2 = 2aS
(83)
(85)
(87)
(88)
(89)
(b) ax2  1 = 0
(91)
v2  2aS
(d) 2aS  v2
(c) x2 = 1
(d) x2 + 1 = 0 (M  10)
(d) 5  x2 = x (J  10)
If l2 = r2 + h2 then value of h is ...................... l2  r2
(b) ±
r 2  l2
(c) ±
l2 + r 2
(d) ±
lr
If b = 0 in the equation ax2 + bx + c = 0 the equation becomes ......................(J  10) (a) Adfected quadratic equation
(b) Pure quadratic equation
(c) Linear equation
(d) Simultaneous equation
Sum of the roots of the equation 2x2 = 6x  5 is ...................... (a) 1 (b) 5 (c)  5 2 3 2
(J  10)
(d) 3
Sum of the number and twice of its square is 78. It can be represented as ........................ (a) x + 2x2 = 78
(b) x + (2x)2 = 78
(c) x2 + 2x = 78
(d) x2 + (2x)2 = 78
1 = 2 is same as ................... x 1 2 (a) x + = 22 x
(J  10)
x+
(c) x2  2x + 1 = 0 (90)
(c) ±
The pure quadratic equation in the following is ...................... (a) 4x = 81 (b) x + 1 = 5 (c) (x + 2)2 = 3x x x
(a) ± (86)
v2 + 2aS
The quadratic equation whose roots are 1 and 1 is ................................. (M  10) (a) ax2  x  1 = 0
(84)
(b) ±
(M  10)
(b) x2 + 2x + 1 = 0 (d) x2 + 2x = 0
If in ax2 + bx + c = 0, a = c the roots are ............................... (a) Additive inverses
(b) Multiplicative inverses
(c) Equal
(d) Zero
One root of 4x2  8mx  9 = 0 is the negative of the other root. So the value of m is ............................... (a) 0
(b) 9 8
(c)  9 8 198
(d)  9 4
(92)
The equation 5x2  26x + p = 0 is has reciprocal roots, so the value of p is ................................ (a) 25
(93)
(94)
(95)
(d) 1 5
(c) 5
The graph of x2  6x + 9 = 0 is ......................................... (a) Cuts the X  axis at two points
(b) Cuts the Y  axis at two points
(c) Touches the X  axis at one point
(d) Has no contact with the X  axis
The equation (bc)x2 + (ca)x + (ab) = 0 has ...................... (a) equal roots
(b) irrational roots
(c) rational roots
(d) none
If the roots of the equation ax2 + b = 0 are real and distinct then ...................... (a) ab > 0
(96)
(b) 1
(b) a = 0
(c) ab < 0
(d) a>0, b>0
If the product of the roots of the equation m2 + 6x + (2m  1) = 0 is 1 then the value of m is ...................... (a) 1
(97)
(b) 1
(c)
1 3
(d)  1 3
If α, β are the roots of x2  2x + 2 = 0 then α2 + β2 = ...................... (a) 2
(b) 0
(c) 1
(d) 4
(98)
If the roots of the equation 12x2 + mx + 5 = 0 are in the ratio 3 : 2 then m is ..................... 1 (a) (b) 5 (c) 5 10 (d) 5 10 12 12 12
(99)
In a quadratic equation with a = 1, a student reads the coefficient 16 of x wrongly as 19 and obtains the roots as 15 and 4 the correct roots are .......................... (a) 6, 10
(b) 6, 10
(c) 8, 8
(d) 8, 8
(100) If (x  4) (2x + 5) = 0 the incorrect statement is ............................... 5 5 (a) x = 4 (b) x = (c) x = (d) 2x = 5 2 2 (101) If α and β are the roots of x2  px + 36 = 0 and α2 + β2 = 9 then p = ............................... (a) ± 3
(b) ± 6
(c) ± 8 199
(d) ± 9
(102) The nature of the graph of y = x2 + 5x  24 is ............................... (a)
(b)
Y
Y
X1
X
Y1
(c)
Y1
(d)
Y
X1
X
1 =2 x
1)
x2 – 6x + 4
5)
x+
2)
x2 + 2x = 4
6)
5(x2)(x+3) = 0
3)
x2 +
7)
x + 3x = 5
8)
k=
x+5 =x 2
X1
Y1
II Which of these are quadratic equations
4)
Y
X
Y1
1 =2 x2
X1
X
1 mv2 2
III Classify as pure and adfected quadratic equation.
1)
x2 + x = 5
5)
x2 + 2x = 5
2)
p (p – 3) = 1
6)
k2 – k = 0
3)
x+
1 =2 x
7)
7y =
4)
2m2 = 32
8)
x2 + 2 = 6
200
35 y
IV (A)
Solve the pure quadratic equation
1) 5x2 = 125
x2 =
81 a
2) m2 – 1 = 143
3) 4a =
2 2 5) (x  4) = 9 18
2 6) x  3 = 7 1 2 4 4
125 5
x2 = 25 x=±
25
x = ±5 4) (2m – 5)2 = 81
B
(1) If A = 2πr2 then solve for r and find the value of r, when A = 77 and π =
(2) If B =
3 a2 then solve for ‘a’ and find the value of a when B = 16 4
22 7
3
(3) If V = πr2h then solve for r and find the value of r, when v = 176 and h = 14
201
(M  07)
(4) If r2 = l2 + d2 then solve for d and find the value of d, when r = 5 and l = 4.
(5) If k = 1 mv2 then solve for v and find the value of v, when k = 100 and m = 2. 2
(6) If v2 = u2 + 2aS solve for v, if u = 0, a = 2 and S = 100, find the value of v.
V
Find the roots of the following equations.
1) x (x – 3) = 0
2) (2m+1)(3m2) = 0
3) m2 – 4m = 0
4) (y+6)(y+9) = 0
5) (b3)(b5) = 0
6) (5z2)(7z+3) = 0
202
VI
Solve the quadratic equations. (By factorisation method)
(1) x2 + 15x + 50 = 0
(2) a2 + 5a + 6 = 0
(3) y2 = y + 2
(4) 6  p2 = p
(5) 13m = 6 (m2 + 1)
(6) 0.2t2  0.04t = 0.03 (Multiply by 100)
(7)
7 x2  6x  13
7 =0
203
VII
Solve the following equation by using formula.
(1) x2  7x + 12 = 0
(M  07)
(2) a2 = 4a + 6
(3) x2  8x + 1 = 0
(J  09)
(4) 2x2  2x = 5
(M  09)
(6) x2 + 1 = 8x
(M  08)
(5) 3x2 = 48
(7) x2  2x  4 = 0
(8) 2y2 + 6y = 0
(J  10)
204
VIII
Solve the following equations..
(1) (x+4)(x4) = 6x
IX
(J  07)
(2) 2(a21) = a(1a)
(3) (n3)2 + (n+1)2 = 16
(4) 3(b5) (b7) = 4(b+3)
(5) 8(S1) (S+1) + 2 (S+3) = 1
(6) 11(m+1) (m+2) = 38 (m+1) + 9m
(A) Solve (we first reduce to Standard form and solve by any method)
(1)
5x  2 3x  8 = x+5 x2
(2)
205
12m+1 11m+3 = 10m+11 9m+13
y+1 y 25 + y = y+1 12
(4)
2 1 + = 2 x1 x2 x
(5)
n+2 n+1 + = 2n+13 n2 n1 n+1
(6)
5 2 + = 6 m+6 2(m+4) m+2
(1) The product of two consecutive integers is 182. Find them. (Let the integers be x and (x+1))
(2) The product of two consectutive even numbers is 48. Find them ( Let the number be x and x+2) âˆ´
B
(3)
206
(3) The sum of the squares of three consecutive integers is 194. Find them. (Let the integers are x2, (x+1)2 and (x+2)2)
(4) The perimeter of a rectangular plot is 32m and its area is 60 Sq.m. Find the dimensions. (M  10) ( l+b = 16, lb = 60) âˆ´
(5) Sum of the two numbers is 18 and sum of the squares of those number is 290 find the numbers. ( x + y = 18 and x2 + y2 = 290) (J  07) âˆ´
(6) The length of a rectangular field is three times its breadth. The area of the field is 147 Sq.m. Find the length of the field.
207
(7) The hypotenuse of a right tirangle is 20m. If the other two sides one is 4m longer then the other. Find the length of those two sides.
(8) The perimeter of a right triangle is 30cm. Its hypotenuse is 13cm. Find the other two sides.
(9) The base of a triangle is 4cm longer than its altitude. If its area is 48 Sq.cm. Find the base (M  07) and the altitude of the triangle.
(10) An aeroplane takes one hour less for a journey of 1200 km if its speed is increased by 60km ph. Find its usual speed.
208
(11) A train required one hour less to return from a city 180km away, by travelling 6 km/h faster. How fast did it travel each way?
(12) Sailor Raju takes one hour forty minutes to go 8km downstream and return. If the speed of the current is 2 km/h. Find the speed of the boat in still water. (J  06)
(13) Yashu bought some books for Rs. 60. Had he bought 5 books more for the same amount each book would have cost him one rupee less. Find the number of books bought by (M  06) Yashu.
209
(14) A dealer sells an article for Rs. 24 and gains as much percent as the cost of the article in rupees. Find the cost of the article.
(15) A man bought a certain number of glass vessels for Rs. 600. Two of them were broken. He sold the remaining at a profit of Rs. 10 each gaining Rs. 50 on the whole. How many glass vessels did he buy?
(16) Varsha takes 6 days less than Usha does, to do a piece of work. Together they can do it in 4 days. Find the number of days each will take to do it separately.
210
(17) Two pipes together can fill a tank of water in 6 hours one of them takes 5 hours less than the other to fill it working alone. Find the time each will take to fill the tank.
(18) What number increased by its reciprocal equals: 65 8
(19) In an auditorium the number of seats in each row is 8 fewer than the number of rows. How many seats are in each row if the auditorium seats 609?
211
(20) The head master of a school distributed Rs. 1500 equally among the rank holder of X standard class. If 5 more of them have secured the rank each would have got Rs. 25 less. Find the number of rank holders.
(21) The altitude of a triangle is 5 cm smaller than its base, if its area is 150 Sq.cm., find the base of the triangle. (M  09)
212
X Discuss the nature of the roots. 2
Δ = b – 4ac
Q.E. 1
3d2 – 2d + 1 = 0
2
2n2 – 9n + 8 = 0
3
x2 – 2x + 1 = 0
4
x2 – 2x – 5 = 0
5
2x2 + 5x – 1 = 0
6
x2 + 7x + 12 = 0
a=3
Δ = (2)2 – 4(3)(1)
b = 2
= 412
c=1
= 8 < 0
213
Conclusion Roots are Imaginary
XI
For what positive value of m or p are the roots of the equations are equal, distinct and imaginary.
Quadratic Equation 1
x2 – mx + 9 = 0
2
mk2 – 3k + 1 =0
3
px2 + 2(p4)x + 2= 0
4
x2 – (m+1)x + 9 = 0
5
3x2 + 2mx + 4 = 0
2
Δ = b – 4ac
Roots Roots are are Equal
a=1
= (m)2 – 4(1)(9)
m236=0
b = m = m2  36
m2 = 36
c=9
m = 6±6
214
Imaginary Distinct Imaginary
m>6
m<6
XII Find the sum and product of the roots of the following equations. Quadratic Equation
a
b
c
1
2 2 4x++95==00 x3x – mx
3
4
5
2
x2 – 5 = 4x
3
6x2 – 5x = 0
4
3x2 + 5 = 0
5
2ax2 + bx + ab = 0
6
x2 + mnx + m + n=0
7
2(a2+ b2)x2 + 2(a+b)x + 1 = 0
8
3a2x2 +8abx +4b2=0
215
Sum m + n
b a (4) = 3 4 = 3
m+n =
Product mn mn =
c = a
5 3
XIII
Find the quadratic equations whose roots are given.
1
Roots 5 and 3 5 3
5 3
m+n
mn
+
3 5 x 5 3
3 5
25+9 34 = = 15 15 2
3
3 and 2 5 3
3 and 2 5
5
3 + 5 and 3  5
6
2 + 1 and 2  1
2 ± 3
x2(m+n)x + mn = 0 x2 34 x + 1 = 0 15 15x2 34x + 15 = 0
5 and 3
4
7
=1
Equation
5
216
XIV
(1) x = 2 is one root of 3x2  5x + c = 0. Find the other root.
(2) x = 7 is one root of x2  bx  28 = 0. Find the other root.
(3) Find the value of k, so that the equation x2 + 4x + k + 2 = 0 has one root equal to zero.
(4) The equation 4x2  8mx + 9 = 0 has one root as the negative of the other. Find the value of m.
217
(5) The equation 5x2  26x + p = 0 has reciprocal roots. Find p.
(6) Find the value of q so that the equation 2x2  3qx + 5q = 0 has one root which is twice the other.
(7) Find the value of â€˜pâ€™ so that the equation 4x2  8px + 9 = 0 has roots whose difference is 4.
218
(8) If one root of the equation x2 + px + q = 0 is twice the other prove 3p2 = 16q
(9) If one root of the equation px2 + 3x + 2 = 0 is reciprocal of the other then find the vlaue (M  10) of p.
(10) If the roots of the equation (bc) x2 + (ca) x + (ab) = 0 are equal. Then prove that (J  10) 2b = a + c
219
XV
(1) If ‘a’ and ‘b’ are the roots of 3m2 = 6m + 5 find a 1 b 1 1 (a) + (b) + (c) 2 + 12 b a a b a b
(2) If p and q are the roots of 2x2 + 3x + 6 = 0 find the value of (a) p2 + q2
(b) (p+2q) (2p+q)
220
(d)
1 1 + 3 a3 b
(3) If m and n are the roots of 3x2  6x + 4 = 0 find the value of (a) m2n + mn2
(b)
1 1 + 3 m3 n
(c)
1 1 n m + + 2 n + m +3mn m n
(4) If p and q are the roots of the equation 2x2 + 3x + 6 = 0, find the value of (a) p2 + q2
(b) p3 + q3
221
(c) p2q + pq2
(5) If m and n are the roots of the equation x2 + 6x + 9 = 0 find the value of (a) m2n + mn2
(b)
(c) 13 + 13 m n
(d)
1 + 12 m2 n 1 1 n2 + m +2 m + n n
Â m2
+ 3mn
(6) From the equation whose roots are the squares of the roots of the equation x2 2x+ 4 = 0 (J  07)
(7) Find the value of m for which the equation 2x2 + 3x + m = 0 has equal roots.
222
(J  10)
XVI
(1) Draw the graph of y = x2 and find the value of
(J  06)
7
(2) Draw the graph of y = 2x2 and verify that the value of
223
4 = ±2
(3) Draw the graph of y = x2 and find the value of
3
Â
(4) Draw the graph of y = x2 and y = x + 2 and hence solve x2  x  2 = 0. Â
224
(M  08)
(5) Draw the graph of y = x2 and y = 2x + 3 and hence solve the equation x2  2x  3 = 0
(J  09)
Â
(6) Solve graphically x2 + x  2 = 0.
(M  07)
Â
225
(7) Solve graphically 2x2 + 3x  5 = 0
(8) Draw the graph of 1 x2 2
226
(9) Draw the graph of 3 x2 2
(M  06)
(10) Draw the graph of y = x2
227
(11) Draw the graph of y = 2x2
(M  09)
(12) Draw the graph of y = x2 and y = 6  x and hence solve the equation x2 + x  6 = 0
(M  10)
228
(13) Draw the graph of y = x2 and y = 2  x and hence solve the equation x2 + x  2 = 0
(J  10)
Â
(14) Solve Graphically: 2x2  3x  5 = 0 Â
229
(15) Solve Graphically: 2x2 + 3x  5 = 0 Â
***************
230
6
MODULAR ARITHMETIC • The face of the clock shows only 12 hours. Ex
Ex
(1) :
(2) :
13th hour of the day is equivalent to 1st hour. The relation is expressed as 13 congruent 1 (mod 12) or
13 ≡ 1 (mod 12)
ly
17 ≡ 5 (mod 12)
In an English calendar, if the 1st day is monday then 8th day, 15th day, 22nd day these are also monday. This is an example of modulo 7. i.e.
8 ≡ 1 (mod 7) 15 ≡ 1 (mod 7)
In general
a ≡ b (mod m) > a  b ≡ 0 (mod m) that means (ab) is a multiple of m or ‘m’ is a divisor of (ab) ** If a ≡ b (mod m) then (ab) is always a multiple of ‘m’. Set of residues: If any positive integer is divided by ‘m’ then the reminder will be {0, 1, 2, 3, ............ (m1)} i.e.
Zm = {0, 1, 2, 3, 4, ............ (m1)}
Complete the following
(1) Z3 = {
}
(2) Z4 = {
}
(3) Z5 = {
}
(4) Z6 = {
}
231
Modular Addition
When ‘a’ and ‘b’ are any two integers and ‘m’ is a positive integer then a ⊕m b = r means, ‘r’ is the remainder obtained when (a+b) is divided by ‘m’. Ex:
2 ⊕4 3 = 1
i.e.
2+3 = 5, when 5 is divided by 4 then reminder is 1.
Modular Multiplication
‘a’ and ‘b’ are any 2 integers, m is a positive integer then a ⊗m b = r means, ‘r’ is th remainder obtained when (a x b) is divided by ‘m’. Ex:
2 ⊗4 3 = 2
i.e.
2 x 3 = 6, if 6 is divided by 4, then reminder is 2.
Check whether the following relations are true or false
Ex:
17 ≡ 5 (mod 3)
Solution: 17  5 = 12 12 is a multiple of 3 ∴ relation (congruence) is true.
(1) 75 ≡ 10 (mod 15)
(2) 121 ≡ (8) (mod 17)
(3) 12 ≡ (4) (mod 4)
232
(4) 9 ≡ 23 (mod 12)
(5) 15 ≡ 29 (mod 7)
(6) 175 ≡ 5 (mod 17)
(7) 12 ≡ 8 (mod 6)
Find the sum of the following
3 ⊕4 2
Ex:
Solution 3 + 2 = 5 When 5 is divided by 4, the reminder is 1 ∴ 3 ⊕4 2 = 1
(1) 8 ⊕10 9
(2) 5 ⊕7 4
(3) 4 ⊕5 3
(4) (3 ⊕7 6) ⊕7 4
233
(5) (10 ⊕12 2) ⊕12 3
(6) 5 ⊕7 (6 ⊕7 8)
Find the value of ‘x’
(1) 3 ⊕5 4 = x
(2) 5 ⊕6 x = 5
(3) 3 ⊕4 x = 1
(4) 2 ⊕3 x = 0
Find the product of the following
Ex:
5 ⊗11 10
Solution: 5 x 10 = 50 dividing 50, by 11, the reminder is 6 ∴
5 ⊗11 10 = 6
(1) 5 ⊗6 5
(2) 4 ⊗5 3
234
(3) 5 ⊗7 8
(4) (4 ⊗11 3) ⊗11 7
(5) (5 ⊗12 2) ⊗12 6
(6) (3 ⊗4 5) ⊗4 6
Choose the value of ‘x’ or ‘y’ from the following.
Ex:
x + 2 ≡ 4 (mod 5) (a) 4
Solution: x + 2  4 = 5
(b) 7
x2=5
(c) 5
x=5+2
(d) 3
x=7
(1) 2x + 1 ≡ 1 (mod 10) (a) 2 (b) 9 (c) 1 (d) 5 (2) 4 ⊕ y ≡ 0 (mod 5) (a) 0 (b) 1 (c) 4 (d) 5
235
(3) 3 ⊕ y ≡ 2 (mod 6)
(a) 4 (b) 2 (c) 5 (d) 6 (4) 2y ≡ 1 (mod 5)
(a) 2 (b) 5 (c) 6 (d) 3 (5) y ⊗ y ≡ 1 (mod 8) (a) 6 (b) 3 (c) 8 (d) 4
CAYLEY’S TABLE Cayley’s table represents the modular arithmetic system. Ex (1) : Construction of Cayley’s table under addition modulo 3 i.e. (Z3, +) or ⊕3 (1)
Z3 = {0, 1, 2} ⊕3
0
1
2
0
0
1
2
0 ⊕30 = 0
0 ⊕31 = 1
0 ⊕32 = 2
1
1
2
0
1 ⊕30 = 1
1 ⊕31 = 2
1 ⊕32 = 0
2
2
0
1
2 ⊕30 = 2
2 ⊕31 = 0
2 ⊕32 = 1
236
(2)
Similarly ⊗3 [Multiplication table of Z3] ⊗3
0
1
2
0
0
0
0
0 ⊗30 = 0
0 ⊗31 = 1
0 ⊗32 = 2
1
0
1
2
1 ⊗30 = 1
1 ⊗31 = 2
1 ⊗32 = 0
2
0
2
1
0 ⊗30 = 2
2 ⊗31 = 0
2 ⊗32 = 1
Construct the Cayley’s table for the following. (1)
Z5 under modulo 5 addition.
(2)
⊗6 , Z6
(3)
Z4, ⊕4
237
(4)
Z4, ⊗4
(5)
Q = {0, 2, 4, 6, 8} under ⊕ mod 10.
(6)
S = {1, 5, 7, 11} under ⊗ mod 12
Previous Exam Questions (1)
(2)
Which one of the following is true?
(M  06)
(a)
2 ≡ 3 (mod 3)
(b)
7 ≡ 13 (mod 5)
(c)
3 ≡ 5 (mod 5)
(d)
7 ≡ 12 (mod 5)
If a ≡ b (mod m) then (ab) is always
(M  06)
(a)
greater than ‘m’
(b)
multiple of ‘m’
(c)
lesser than ‘m’
(d)
m=0
238
(3)
The value of 4 ⊗7 8 is ............................
(M  06)
(4)
If 2y ≡ 1 (mod 5) the value of y is ............................
(M  06)
(5)
The value of (3 ⊕7 6) ⊕7 4 is ............................
(J  06)
(6)
The value of a is a ⊗ a ≡ 2 (mod 14) is ............................
(J  06)\
(7)
The value of (4 ⊗11 5) ⊗11 7 is ............................
(J  06)
(8)
Modulo sustem which satisfies 15 ≡ 1 is
(J  06)
(a) mod 7 (9)
(c) mod 3
(d) mod 6
If 2y ≡ 1 (mod 5) then y is ............................ (a) 2
(10)
(b) mod 5
(b) 5
(M  07)
(c) 6
(d) 3
17th hour of a day is equivalent to 5th hour this relation can be written as (a) 17 ≡ 5 (mod 12)
(b) 12 ≡ 5 (mod 17)
(c) 17 ≡ 12 (mod 5)
(d) 17 ≡ 5 (mod 24)
(M  07)
(11)
If y ⊗ y ≡ 1 (mod 8) the value of y is ............................
(M  07)
(12)
The set of residues of modulo 4 is ............................
(M  07)
(13)
The departure time of a bus from Mangalore to Bangalore is 21 : 00 hours, this is (M  08) equal to (a) 9:00 AM
(14)
(16)
(c) 10:00 AM
(d) 10:00 PM
If 3y ≡ 2 (mod 2) the value of y is equal to (a) 1
(15)
(b) 9:00 PM
(b) 3
(M  08)
(c) 4
(d) 5 (J  08)
Which one of the following is a correct relation (a) 8 ≡ 7 (mod 3)
(b) 7 ≡ (2) (mod 4)
(c) 10 ≡ 6 (mod 3)
(d) 18 ≡ 10 (mod 5)
The value of x in 3x ≡ 2 (mod 4) is ............................ (a) 1
(b) 2
(c) 3 239
(J  08)
(d) 4
(17)
The highest residue obtained by the division of a positive integer by (m+1) is (M  09)
(a) 0
(b) 1
(c) m1
(18)
If x ⊕10 x ≡ 2, the value of x is ............................
(19)
If y ⊗ y ≡ 1 (mod 3), the value of y is ............................ (a) 1
(b) 3
(c) 6
(20)
The sum of (3 ⊕7 6) ⊕7 5 is ............................
(21)
Construct Cayley’s table under ⊗10 on S = {2, 4, 6, 8}
(22)
Construct Cayley’s table under addition mod 3, on Z3. **************
240
(d) m ( M

0 9 )
( J

0 9 )
( J

0 9 )

1 0 )

1 0 )
(d) 2
( A
( J
7 I
PRACTICAL GEOMETRY (1)
Two circles having radius 4cm and 3cm touching externally. Then the distance between their centres is ..............................................
(2)
Two circles having radius 4cm and 3cm touching internally. Then the distance between their centres is ..............................................
(3)
In figure, AB = CD = 8cm and OX = 3cm then OC is ..............................................
(4)
C
x X
B P
O2
O1
B
Q
A
(5)
Y
.
.
tangent is ................................................
D
O
A
In the given figure, the transverse common
y
If two circles are intersecting each other, then the number of common tangents drawn is .................................................
(6)
E
In the figure common tangent is ................................
F
D . A
(7)
In the figure common tangent is ................................
C
B A
P
B O C
(8)
Two circles having the radius 3cm and 4.5 cm, the distance between their centers is 5cm then the cirlces are ................................
(9)
In the figure, the segment which makes right angle in a semi circle is ................................
D A
B
O
E F
241
C
G
(10)
Number of common tangents to be drawin to two concentric circles is ...........................
(11)
If two circles having radius 4cm and 2cm, the distance between their centres is 7cms. Then the circles are ................................
(12)
AB and CD are two equal and parallel chords in a circle. If the distance from the centre of the circle to the chord AB = 2x units, then the distance between the chords is .......................................
(13) (14)
ABC is an angle in a major arc. Then ABC is ................................... Radii of two circles are 5cm and 3cm respectively and the distance between the centres is 6cm. Then they are ...................................
(15)
â€˜Oâ€™ is the center of a circle. AB is a chord.
C
From the figure ACB is ...................................
O. A
(16)
B
Two circles when touch externally, then number of transverse common tangents that can be drawn is ...................................
(17)
The angle in a semi circle is ...................................
(18)
If two circles are touching internally, the distance between their centres is equal to difference of their ...................................
(19)
As the length of the chord increases, the perpendicular distance ...................................
(20)
As the length of the chord decreases, the perpendicular distance ...................................
(21)
The biggest chord in a circle is ...................................
(22)
The perpendicular distance between diameter and center of the circle is .......................
(23)
Minor arc subtends ...................................
(24)
Major arc subtends ................................... 242
(25)
What are concentric circles?
(26)
What are congruent circles?
(27)
What is a secant?
(28)
What is a tangent?
(29)
How many common tangents can be drawn to the externally touching circles?
(30)
In the figure POB and PBO are complementary
P O
angle then PB is a ....................................... (31)
In the figure OP ⊥ AB and OQ ⊥ AC.
A
If OP = OQ then the relation between AB and AC is ....................................... (32)
B
P B
Q O C
In the figure AB is a tangen ‘A’ is a point of contact. I OBA = 450 which of the following is true ....................................... (A) OA > AB
(B) AB > OA
(C) AB > OB
(D) OB = AB 243
O
A
B
(33)
No common tangents can be drawn to the following circles ................................... (A) Externally touching circles
(B) Internally touching circles
(C) Intersecting circles
(D) Concentric Circles
(34)
The tangents drawn to a circle at the ends of a diameter are ...................................
(35)
The length of the transverse common tangent drawn to two circles is calculated using the formula...................................
(36)
The length of the direct common tangent drawn to two circles is caculated using the forumla ...................................
(37)
The number of tangents that can be drawn to internally touching circles at their point of contact is ...................................
(38)
In the figure AB = 4cm OP = OQ then CQ is ............................
C
D
Q
O A
II
(1)
Construct a tangent to a circle of radius 3cm at any point â€˜pâ€™ on it.
244
P
B
(2)
Given PQ = 4cm with PQ as diameter draw a circle. Draw two tangents to the circle at ‘P’ and ‘Q’
(3)
In a circle of radius 3cm, draw two radii such that the angle between them is 600. Draw two tangents at the ends of the radii.
(4)
In a circle of radius 3cm, draw two tangents such that the angle between them is 1100.
245
(5)
Construct two tangents to a circle of diameter 4cm from a point 6cm away from the centre.
(6)
Construct two tangents to a circle of radius 3cm from a point 2.5cm away from the circle.
246
(7)
In a circle of radius 2.5 cm, draw a chord of length 3.5cm. Draw two tangents at the ends of chord.
247
III
(1)
What is meant by a direct common tangent?
(2)
What is meant by a transverse common tangent?
Common Tangents
S.No.
No. of Common tangents
Figure A
B
1 C
D
2
3
4
5
248
No. of DCT
No. of TCT
Direct Common Tangent (1)
Draw a direct common tangent to two congruent circles of radii 3cm, whose centres are 8cm apart.
(2)
Construct a direct common tangent to two congruent circles of radii 2.5 cm each and whose centres are 5cm apart.
249
(3)
Two circles of radii 3cm and 2cm have their centres 7cm apart. Draw direct common tangent to the circles.
(4)
Construct a direct common tangent to two circles of radii 3.5 cm and 2cm and whose centres are 1.5 cm apart.
(5)
Draw two direct common tangents to two circles of radii 4.5cm and 2.5cm and their centres are 7cm apart.
250
(6)
Two circles of radii 3cm and 2cm have their centres 3.5cm apart. Draw direct com mon tangents to the circles.
251
Transverse Common Tangent (1)
Draw two congruent circles of radii 3cm having their centres 8cm apart. Draw a transverse common tangent.
(2)
Construct two circles of radii 2cm and 3cm where centres are 8cm apart. Construct a transverse common tangents and measurethe length of trasverse common tangent and verify by calculation.
*************** 252
8
THEOREMS ON TRIANGLES AND CIRCLES • Two polygons having the same number of sides are Similar if and only if (i)
The angles of one triangle are equal to the corresponding angles of the other and
(ii)
The sides of one triangle are proportional to the corresponding sides of the other.
• Congruent triangles and squares always similar. • The symbol ‘’ means similar to • Basic proportionality theorem (Thales Theorem): A stright line drawn parallel to a side of a triangle divides the other two sides proportionately.
• Conversely if a line divides two sides of a triangle proportion the line is parallel to the third side of the triangle.
• Corollary of BPT: If a line is drawn parallel to a side of a triangle then the sides of new triangle formed are proportional to the sides of the given triangle.
THEOREM  1 • If two triangles are equiangular then their corresponding sides are proportional. • Converse of Theorem 1: If the corresponding sides of two triangle are proportional then the triangles are equiangular.
•
^ = F^ then ^ ^ ^ and C (i) If ^ A = D, B=E
A
^
B
Δ ABC  Δ DEF
D
Y
^
X
(ii) If XY  BC then C
(a)
AC AB = DF DE
(b)
AX AY = BX CY
AY (c) AX = AC AB
E
=
BC EF
=
XY BC
F
253
Δ AXY  Δ ABC
THEOREM  2 • Areas of similar triangles are proportional to the square on the corresponding sides. • Areas of similar triangles are proportional to the square on the corresponding medians. • Areas of similar triangles are proportional to the square on the corresponding altitude. • Areas of similar triangles are proportional to the square of the radii of their circum circles.
THEOREM  3 • In a right angle triangle the square on the hypotenuse is equal to the sum of the square on the other two sides.
• Conversely if the sum of the squares on any two sides is equal to the square on the third side those two sides include a right angle.
• If the square on the longest side of a triangle is greater than the sum of the squares on the other two sides the triangle is obtuse angled.
• If the square on the longest side of a triangle is less than the sum of the squares on the other two sides the triangle is Acute angled.
• If a set of three numbers the square of one equal the sum of the squares of the other two the three numbers form a pythagorean triplet.
• According to Boudhayana Shulba Sutra the diagonal of a rectangle produces both area which its length and breadth produce separately.
• Bhaskaracharya (1114 AD) has given a proof for the property of the right angle triangle. .
I
(1)
.
In the figure MN  PR, if QN : NR = 2 : 3 and PQ = 11.5 cm then QM is ........................ P M
Q
(2)
N
(A) 10 cm
(B) 2.3 cm
(C) 4 cm
(D) 4.6 cm
(M  06)
R
Sides of a triangle are of lengths 2cm, 3cm and 4cm respectively. Which of the sets of numbers are sides of a triangle similar to the above triangle. ...............................
(M  06)
(A) 4, 5, 6 (3)
(B) 5, 6, 7
(C) 12, 13, 14
(D) 6, 9, 12
The area of ΔABC = 144 Sq.cm and area of ΔPQR = 25 Sq.cm. Altitude of ΔABC=6cm. If ΔABC  ΔPQR then the correspoinding altitude of ΔPQR is .................
(M  06)
(A) 2.5cm
(B) 5cm
(C) 12cm 254
(D) 6cm
In the figure ABCD is a trapezium AB  DC, which of the
B
^
(4) A
following is equal to AB ? ............................... CD
O C
^
D
(5)
Area of ΔPQR Area of ΔPMN
In the figure MN  QR, P
M
N
R
Q
(B) BC AD
(A) AD BC
(C) AO AB
(M  06)
(D) OB OD
is equal to ....................... (M  06)
(A) QR MN
(B)
2 (C) PQ2 PR
2 (D) MN 2 QR
PQ2 PM2
^ = ACB ^ and AQP ^ = ABC ^ then AP . AB is equal to ............................... In the figure APQ
(6)
(M  06)
A P
x
Q x
B
(A) AQ . AC
(B) AP . AB
(C) AC . BC
(D) BC . AB
C
(M  06) The corresponding sides of two equiangular triangles are ............................ (J  07)
(7)
(A) equal (8)
(B) parallel
(C) proportional
(D) unequal
A vertical pole of 10m casts a shadow of 8m at certain time of the day. The length of the shadow cast by a tower standing next to the pole of height 110m is (M  06)
............................... (A) 80m
(B) 88m
(C) 100m
(D) 18m
In a parallelogram ABCD, P is point on BC. In ΔDCP and ΔBLP, DP : PL ......................
(9)
(J  06)
C
^ ^
D
(A) DC : BL
(B) DC : BP
(C) PC : BL
(D) PC : PL
P
(10)
^ ^
A
L
B
The name of the mathematician who proposed the basic proportionality theorem is (J  06)
...................................... (A) Euclid
(B) Thales 255
(C) Pythagoras
(D) Shreedhar
(11)
In the figure XY  AB, AX = 9cm, XC = 7cm, BC = 20cm,
C
Y
^
X
(A) 11.25 cm
(12)
(B) 10.25 cm
B (C) 10 cm
^
A
(J  06)
BY = ...................................... (D) 15 cm
If the two given triangles are similar then the ratio between their sides is ........................ A
(J  06)
• P
•
x
B
x
C Q C
^
(13) D
AC BC AB = = QR PR PQ
(B)
BC AC AB = = QR PQ PR
(C)
BC AC AB = = RP PQ QR
(D)
BC AC AB = = PQ RP QR
In the trapezium ABCD, AB  CD and the diagonal intersect (M  07) at O, then OD is equal to ........................... OC AB B (A) OB (B) (C) OC (D) AC CD OA OD DB
O
^
A
R
(A)
In the given figure value of PQ is ........................... (M  07)
(14) P Y 1.5m X
Q 8m
(15)
2m
(A) 10 m
(B) 7.5 m
(C) 9.5 m
(D) 3.5 m
R
Select the set of numbers in the following which can form similar triangles ..................... (M  07)
(16)
(A) 9, 12, 18 and 3, 4, 6
(B) 3, 4, 6 and 9, 10, 12
(C) 8, 6, 12 and 2, 6, 3
(D) 3, 4, 5 and 2, 4, 10
Two similar triangles have areas 120 Sq.cm and 480 Sq.cm respectively. Then the ratio of any pair of corresponding sides is ........................... (A) 1 : 4
(17)
(B) 1 : 2
(C) 4 : 1 (D) 2 : 3 ^ ^ In the given figure ABC = AYX then the ratio of the
A
^
B
^
X
(M  07)
Y
corresponding sides is ...................................... (A) AX = AB = CB AY XY AC C (C)
AC BC AB = = AY XY AX 256
(M  07)
(B) AB = BC = AX XY AC AY (D)
AY XY AX = = AB CB AC
(18)
A ladder 13 m long rests against a wall at a height 12m from the ground. Then the distance of the foot of the ladder from the wall is ...................................... (M  07) (A) 1m
(19)
(B) 25m
(C) 5m
(D) 12.5m
In the following figure DE  AB. If AD = 7cm, CD = 5cm, CE = 10cm then the length of BE is ......................................
(J  07)
C
^
(21)
(B) AO . OB = OC . OD
C (C) AB . DC = OB . OD
(D) AO . AB = OC . DC
^ = 900. If AD âŠĽ BC then In a right angled triangle ABC, CAB
C
^ is ...................................... (J  07) the angle equal to ACD
A
B
D
^ (A) ABD ^ (C) CAD
^ (B) DAB ^ (D) ADB
In the figure OA = AB = BC = CD = 1 unit. The unit of OD is
C 900 900
B
900 O
(23)
(J  07)
............................. (A) 1
(B) 2
(C) 3
(D) 4
A
In a Rhambus ABCD diagonals intersect at O. The sum of
A
AC2 + BD2 is ...................................... B
(J  07)
(A) AO . OD = OB . OC
D
(22)
(D) 20cm
correct statement ......................................
O D
(C) 12cm
In the trapezium ABCD, AB  DC which of the following is a
B
^
A
(B) 14cm
B
^
A
(20)
E
^
D
(A) 17cm
O
D
(A) 4AB2
(B) 4AC2
(C) 4BD2
(D) 4AO2
C
257
(J  07)
(24)
In a right angled triangle ABC if CAB = 900 which of the following is correct................. (J  07)
2
(A) BC = AC + AB
(B) AC = AB + BC
(C) AB2 = BC2 + AC2
(D) BC2 = AB2  AC2
2
(25)
2
2
2
2
Which of the following are the sides of the right angled triangle? ............................. (J  07)
(A) 6, 7, 8 (26)
(B) 20, 30, 10
(C) 24, 26, 10
(D) 16, 17, 18
Which of the following polygons are always similar?............................. (A) Isosceles triangles
(B) Right angled triangle
(C) Rhombuses
(D) Squares
(27)
In the figure ΔABC, BE ⊥ AC and CF ⊥ AB. Then which of
A
F
(M  08)
the followig relation is correct?.............................
E
C
B
(M  08)
(A) AE . EC = AF . AC
(B) AE . FC = AF . EB
(C) AB . BC = AC . EB
(D) AE . BC = AB . CF
In ΔXYZ if XY2  YZ2 = XZ2 then the hypotenuse and right angled vertex are.................
(28)
(M  08)
^ (A) XY and X (29)
^ (B) XY and Z
^ (C) YZ and X
^ (D) YZ and Y
Perimeter of a square is 20 cm. Then the length of the diagonal is ............................. (M  08)
(A) 10 (30)
2 cm
(B) 10 cm D
(C) 5
2 cm
(D) 5 cm
In the figure B = C = 900. If AB = 5cm, BC = 6cm and CD = 3cm then AD is equal to ............................
(M  08)
C
B
(A) 8 cm
(B) 9 cm
(C) 10 cm
(D) 12 cm
A
(31)
The dimensions of a rectangula plot is 12m and 16m. The length of the longest line (M  08)
that can be drawn in it is ............................. (A) 16m
(B) 20m
(C) 24m
258
(D) 28m
(32)
In the figure DE  BC, AD : AB = 1 : 2, BC = 6cm then DE is
A
^
B
(A) 1cm
(B) 2cm
C (C) 3cm
(D) 4cm
BC In ΔABC and ΔDEF if AB = = EF DE
(33)
(J  08)
................................
E
^
D
AC then the correct pair of corresponding DF (J  08)
equal angles is .................................
^ ^ and E (A) A (34)
(B) ^ C and ^ F
^ ^ and D (C) B
^ ^ and F (D) A
The two corresponding sides of similar triangles are 3cm and 4cm. The area of largest triangle is 48 Sq.cm. The area of smallest triangle is ..................................... (A) 80 Sq.cm
(35)
(B) 64 Sq.cm
(J  08)
(C) 36 Sq.cm
(D) 27 Sq.cm
Which of the following is a correct statement...........................................
(M 09)
(A) All the rectangles are similar (B) All the rhombuses are similar (C) All the right angled triangles are similar (D) All the equilateral triangles are similar (36)
In ΔABC, PQ  AB the correct relation is ...........................
A
(M 09)
P
B
Q
BQ AP = QC PC
(A)
CP BQ = CA BA
(B)
C (C)
PQ = AB BC BQ
(D) PQ = AB AP QC
In the figure ABC = AQP = 900 then
(37) A Q
(A) BC PQ
(B)
QP BC
(D)
AQ = ................ AB (M 09) AC PQ
P
(C) B
(38)
C
AP AB
If the perimeter of two similar triangles aer in the ratio of 4 : 1 then the ratio between (M  09)
their areas will be ............................. (A) 16 : 1
(B) 4 : 1
(C) 2 : 1 259
(D)
2 :1
(39)
ΔABC  ΔDEF the area of ΔABC is 45cm2 and the area of ΔDEF is 20 cm2 one side
of ΔABC is 3.6 cm then the length of corresponding side of ΔDEF is .............................
(J  09)
(A) 3.4 cm (40)
(B) 2.4 cm
(C) 1.4 cm
(D) 4.4 cm
“If the square on one side of a triangle is equal to the sum of the squares on the other two sides then those two sides contain a right angle” this statement refers to (J  09)
..........................................
(41)
(A) Pythagoras Theorem
(B) Thales Theorem
(C) Converse of Thales Thorem
(D) Converse of Pythagorus Theorem
The length of a diagonal of a square of side 5 cm is ............................. (A) 5
(42)
2 cm
12cm
^
B
^
^ ^
12cm
(C) 10 cm
(D) 10
(A) 12 cm
(B) 14 cm
D (C) 11 cm
(D) 13 cm
C
(B) 8, 15, 18
(C) 8, 15, 17
The corresponding sides of two similar triangles are in the ratio 4 : 9. The ratio
(A) 2 : 3
(A  10)
(47)
(B) 16 : 81
(C) 81 : 16
(D) 14 : 19
The diagonal of a square is 10 2 cm then the length of its sides is .............................
(A  10)
(A) 2cm (46)
(J  09)
(D) 8, 15, 19
between their areas is .............................
(45)
2 cm
^ 6cm^
Which one of the following is Pythagorian Triplet?............................. (A) 8, 15, 16
(44)
5 cm
From the figure the length of AD is ........................... (J  09)
A
11cm E
(43)
(B) 2
(J  09)
(B) 10cm
(C) 8cm
(D) 20cm
The Mathematician who proposed Basic Proportionality Theorem is .......................
(A  10)
^ ^ ^ In ΔABC and ΔDEF, A = D, C^ = F, we can conclude ............................. ^ ^ (B) B is not equal to E (A) ΔABC is similar to ΔDEF (C) ΔABC is not similar to ΔDEF
260
(D) ΔABC is congruent to ΔDEF
(48)
ΔABC has sides of 5, 6 and 7 units while ΔPQR has a perimeter of 360 units. ΔABC  ΔPQR when the sides of Δ PQR are .............................
(A) 2 times the sides of ΔABC
(B) 10 times the sides of ΔABC
(C) 20 times the sides of ΔABC
(D) 18 times the sides of ΔABC
^ AD = BE the incorrect statement ^ = B, In the figure A
C
(49) D
is ..........................................
E
B
A
(50)
(A) AC = BC
(B) CD = CE
(C) DE  AB
(D) DE = AB/2
The areas of similar triangles are not proportional to the square on the corresponding .......................................... (A) sides
(51)
(B) medians
A
(53)
(A) AD2 = BD . CD
(B) AD2 = BC . CD
(C) AD2 = BD + DC
(D) AD2 = BC . BD
B
The relation 392 = 362 + 152 is mentioned as back as in 2000 BC in ............................. (A) Taittiriya Samhita
(B) Bhaskara’s Work
(C) Apasthamba Sutras
(D) Kathyayana Sutras
An example for a pythagorean triplet is ............................. (A) 9, 10, 14
(54)
(D) angular bisectors
^ = 900 AD ⊥ BC so we can write ............................. In ΔABC, A
C D
(52)
(C) altitudes
(B) 7, 25, 24
(C) 5, 5, 5
(D) 6, 6, 8
The sides of a triangle are (a+b), (ab) and 2 ab units it is ............................. (A) A right triangle with hypotenuse (a+b) (B) A right triangle with altitude (a+b) (C) A right triangle with hypotenuse 2 ab (D) A right triangle with hypotenuse (ab)
(55)
ABCD is a rhombus. AC2 + BD2 is equal to ............................. (A) 2AB2
(B) 3AD2 261
(C) 4BC2
(D) CD2
A
(56)
In ΔABC, A = 900, AD ⊥ BC which of the following is correct .............................
B
C
D
(57)
(B) ΔABC  ΔADB
(C) ΔADC  ΔADB
(D) Both A, B and C
In figure
C c
a
E
D
B
A
(58)
D A
a+b = .......................................... a
(A)
a a+b
(B)
b+d b
(C)
a+c a
(D)
c+d c
d
b
^ ^ In the figure ABD = BDC, CD = 4AB then BD = ..........................................
E
B
C
(59)
(A) ΔABC  ΔADC
(A) BE
(B) 4BE
(C) BD = AC
(D) 5BE
In ΔABC, BAC = 900 AD ⊥ BC then the value of x is
B 4 D
.............................
5 x
(A) C (C) 3
A
A
B
^
(60)
5 5
(B) 2
5
(D) 5
5
In a trapizium ABCD, AB  DC and AB = 2CD, if the diagonals intersect at ‘O’ the triangle AOB is
O
.............................
(61)
C
(A) 4 ΔAOD
(B) 4 ΔCOB
(C) 4 ΔCOD
(D) 4 ΔADC
D, E, F are the midpoints of a triangle ΔABC then ΔDEF is equal to ............................. (A)
(62)
^
D
1 ΔABC 2
(B)
1 ΔABC 4
(C)
1 ΔABC 8
(D) ΔABC
In an equilateral triangle ABC, AN ⊥ BC then AN2 = ............................. (A) 3BN2
(B) BN2
(C) 2BN2 262
(D) 4BN2
(63)
The length of the diagonal of a square of side x unit is ............................. (A) x
(64)
x
2
(D)
x 2
(B) 1 : 2
(C) 2 :
3
(D) 3 : 2
In Δ ABC, AB = AC and BD ⊥ AC then BD2 + CD2 = .......................................... (A) 2 AC . CD
(66)
(C) x
The ratio of the length of a side of an equilateral tiangle and its height is ...................... (A) 2 : 1
(65)
(B) 2x
(B) AC . CD
(C)
AC CD
(D) 2 (ACCD)
E = 830 then ^ C is If ABC and DEF are similar triangles in which ^A = 470 and ^ ............................. (A) 600
(B) 700
(C) 500
263
(D) 400
I
II
(1)
In figure PQ  BC fill the blanks in the following A
Q
P
(1)
AB = AP
(2)
AB = PB
(3)
QC = AQ
(4)
AQ = AP
C
B
(2)
In the figure XY  QR write the ratios in terms of a, b and c P c
a x
d
b
(2)
PY PX = PR PQ
===>
(3)
PX = PY PR PQ
R
^
Q
===>
Y
^
X
PY (1) PX = YR XQ
y
= XY QR
===>
XY AY In the figure XY  BC and XZ  AC, show that AX = = BC AC AB
(3)
A
Y
^
X
d C
^
B
Z
In Î”ABC, XY  BC, AB = 4.5 cm, AC = 3.5 cm and AX = 7.2 cm. Find XY.
(4)
A
Q
P
C
B
(5)
In the adjoining figure AD = 11.2 cm, AB = 16.8 cm, AE = 7.6 cm and AC = 11.4 cm. Show that BC  DE C E
A
D
B
264
(6)
In Î”ABC, DE  BC, AD = 4x3, AE = 8x7, BD = 3x1 and CE = 5x3. Find the value of x
(7)
Prove that any two medians of a triangle divide each in the ratio 2 : 1 A E
D G C
B
(8)
2 In the figure XY  BC, AX = a2 and BX = 2a3 and AY = find a. 5 CY A
(9)
Y
C
^
B
^
X
ABCD is a parallelogram. P is any point on BC . DE meets AB produced at L. Prove that DP : PL = DC : BL. D
C P
A
B
L
265
(10)
Show that in a trapezium the line joining the mid points of non parallel sides is parallel to the parallel sides. P
^
D
C Y
X
B
^
A
^ ^ In the figure ABD = BDC, CD = 4AB. Show that BD = 5BE
(11)
D A E
B
C
(12)
In Î”ABC, AB = AC, D is a point on AC. Such that BC2 = AC x CD.
(13)
In the figure ^A = ^ B, AD = BE, show that DE  AB. C
D A
E B
266
(14)
Fill with suitable words. If two triangles are equiangular then their corresponding sides are ................................. M A V
Y
J
P
R
N
Data: In ΔMJR and ΔAPN ^ = PAN ^ JMR ^ MJR = ..................... ^ MRJ = ..................... To Prove: MJ = JR = ......... PN ......... AP Construction: Mark V on MJ and Y on MR such that MV=AP and MY=...........and join VY Proof: In ΔMVY and ΔAPN ^ M = .................
(Data)
MV = AP
(...........................................)
MY = AN
(Construction)
∴
Δ ............. ≅ Δ .............
∴
^ ^ VY = PN and MVY = APN
(S.A.S).
^ ^ ^ MYN = APN = MJR ∴
^ ^ MVY = MJR
∴
VY  .................
(Corresponding Angles)
∴
JR MR MJ = = ........ ......... MV
∴
JR MR MJ = = .......... ........ .........
(BPT and Corollary)
267
(15)
A man whose height is 1.5 m standing 8 m from a lamp post observes that his shadow cast by the light is 2 m in length. How high is the lamp above the ground.
(16)
If one diagonal of a trapezium divides the other in the ratio 2 : 1 then prove that one (M  09) of the parallel sides is twice the other.
268
(17)
Prove that the areas of similar triangles are proportional to the squares on the corresponding sides. P X
Q
R
A
Y
Z
B
Data: Let PQR and XYZ are similar triangles in which QR and YZ are corresponding sides. To Prove:
Area of a ΔPQR Area of ΔXYZ
..................
=
..................
Construction: Draw PA ⊥ QR and XB ⊥YZ Proof:
Area of a ΔPQR = 1/2 x .................. Area of ΔXYZ 1/2 x .................. = QR x ............. YZ .............
(1)
In ΔPAQ and ΔXBY
^ = ........................... PQA ^ = ................ = 90 PAQ ^ ^ QPA = YXB ∴
ΔPAQ  ........................
(Equiangular)
PQ PA = XY XB PA ........
=
(Corresponding Sides)
PQ ........
=
QR ........
QR PA = YZ XB
(2)
Area of a ΔPQR Area of ΔXYZ ∴
Area of a ΔPQR ............................
=
=
QR = YZ
............ YZ2 269
QR (Substitute (2) in (1)) YZ
(18)
Two corresponding sides in two similar triangles are 3.6 cm and 2.4 cm respectively. If the area of the bigger triangle is 45 cm2 find the area of the smaller one.
(19)
Two similar triangle have areas 392 cm2 and 200 cm2 respectively. Find the ratio of any pair of corresponding sides.
(20)
The area of Î”ABC is 25.6 cm2 XY  BC.It divides AB in the ratio 5 : 3. Find the area of Î”AXY.
270
BF2 ΔBOF = CE2 ΔCOE
(21)
In ΔABC, BE ⊥ AC, CF ⊥ AB, BE and CF meet O. Show that
(22)
XY is drawn parallel to base BC of ΔABC. If ΔAXY : trapezium XBCY = 4 : 5, show that AX : XB = 2 : 1
(23)
Prove that the areas of similar triangles have the same ratio as the squares of the corresponding altitudes.
271
(24)
Prove that the areas of similar triangles have the same ratio as the squares of the corresponding medians.
(25)
Prove that the areas of similar triangles have the same ratio as the square on their circum radii.
272
AB2 ΔABD = AC2 ΔACD
(26)
ΔABC is right angled at A, AD ⊥ BC, Show that
(27)
^ = 900, AD ⊥ BC. Show that AD2 = BD . CD In ΔABC, A
(28)
D, E and F are the mid point of ΔABC, show that ΔDEF =
273
1 ΔABC 4
(29)
Prove that in a right angled triangle the square on the hypothenuse is equal to the sum of the squares on the other two sides. M
J
R
P
Data: In ΔMJR, JMR = 900 To Prove:
JR2 = ..................+ ..................
Construction: Draw MP ⊥ JR Proof: In ΔMJR and ΔPJM ^ JMP = .................. = 900 (Data and Construction) ^ ^ MJR = MJP ∴ ∴
∴
(Common)
ΔMJR  ........................
MJ JR = JP ..........
(Equiangular) (Raio of corresponding sides)
JP . JR = .......................
(1)
In ΔMJR and ΔPMR
^ = MPR ^ = 900 JMR ^
ΔMRJ = ............... ∴
(Data and Construction) (Common)
ΔMJR  Δ................ (Equiangular)
JR MR = MR ......... PR x JR = .......................
(2)
JP x JR + PR x JR = ..........................
(Adding i and ii)
JR (JP + PR) = MJ2 + MR2 ∴
JR . JR = MJ2 + MR2
∴
274
.............................. = ..............................
(30)
Test whether the following are pythagoren triplet or not.
i) 5, 3, 4
ii) 8, 15, 17
iii) 9, 10, 14
iv) 7, 24, 25
v) 12, 13, 5
vi) (a+b), (ab), 2 ab
vii) 2.5, 3.5, 4.5
viii) 2.5, 6, 6.5
ix) 3, 3, 5
5 =4 +3 2
2
2
25 = 16 + 9 25 = 25 They form pythagoren triplet
(31)
Find the length of the diagonal of a squares of side (a) 12 cm
(b) 5 unit
275
(32)
The diagonal of square is given find each side of the squaer and perimeter (a) 10 cm
(33)
(b) 8 cm
Calculate the altitude of an equilateral triangle of side (a) 8 cm
(34)
(b) 5 cm
Two poles 10m and 18m high stand vertically 15 m away from each other. Find the distance between their top
276
(35)
The man walks 8km due north then 5 km due east and from there 4km to north. How far is he from the starting point.
(36)
The base of a right circular cone has a diameter of 15 cm and its slant height is 8.5cm find its vertical height.
(37)
A ladder is placed in such a way that its foot is at a distance of 5m from the wall and its top reaches a window 12m above the ground. What is the length of the ladder.
277
(38)
ABCD is a trapezium in which AB  CD and BC ⊥ AB, AB = 7.5 cm, AD = 13cm and CD = 12.5 cm, find BC.
(39)
ABC is a triangle right angled at A, AD ⊥ BC. Find the sides of the triangle ABC if BD = 8cm and CD = 2cm.
(40)
In ΔABC, AD ⊥ BC, prove that AB2  BD2 = AC2  CD2.
278
(41)
In an equilateral ΔABC, AN ⊥ BC, show that AN2 = 3BN2.
(42)
In ΔABC, AB = AC, BD ⊥ AC, Prove that BD2 + CD2 = 2AC . CD
( 4
I n
3
)
ΔABC, AD ⊥ BC, DB : CD = 3 : 1, Prove that BC2 = 2 (AB2  AC2)
279
(44)
ABCD is a rhombus, prove that AC2 + BD2 = 4AB2.
(45)
ΔABC is right angled at C, P is a point on AC, Q is a point on BC prove that
AQ2 + BP2 = AB2 + PQ2
(46)
P is a point of trisection of the base BC of an equilateral ΔABC, prove that 9AP2=7BC2
(47)
In ΔABC, B = 900 P is the midpoint of BC, PN ⊥ AC, prove that AN2  NC2 = AB2.
280
(48)
A ladder of length 2.6m is leaned against a wall when it is at a distance of 2.4 m from the foot of the wall. The top of the ladder touches the bottom edge of the window in the wall. If the foot of the ladder is moved 1.4 m towards the wall is touches the top (A  10) edge of the window. Find the height of the window.
281
TOUCHING CIRCLES
If two circles touch each other the point of contact and the centre of the circles are collinear.
If the two circles touch externally then the distance between their centre is d = R + r
If the two circles touch internally then the distance between their centre is d = R  r
Three circles of radii r1, r2 and r3 touches externally then perimeter of the triangle formed by joining the centres of the circles is 2(r1 + r2 + r3)
The tangents drawn to a circle formed an external point are equal.
If the sides of a quadrilateral are tangents to a circle then the sum of opposite sides equal to the sum of the other pair of sides.
Total Direct Common Total Tangent
Nature
1 d>R+r
2 d=R+r
3 d<R+r
A
B
A
B
A
B
4 d=R–r
A
5 d<R–r
A
6 d=0
A
B
B
Common Tangents Common Tangents DCT TCT
Circles are
4
2
2
Separated
3
2
1
External Touch
2
2
0
Intersect
1
1
0
Internal Touch
0
0
0
One within other
0
0
0
Concentric
282
Mention the kinds of circles in each case.
S.No.
d
R