3.Sample Question Paper - 1 4.Sample Question Paper - 2 5.Sample Question Paper - 3 6.Sample Question Paper - 4 7.Sample Question Paper - 5 8.Sample Question Paper - 6 9.Sample Question Paper - 7
Question Paper - 9
To obtain HCF & LCM For any two positive integers, a and b HCF (a, b) x LCM (a, b) = a x b For Example f(x) = 3x²y and g(x) = 6xy² HCF = 3xy and LCM = 6x²y²
Every composite number can be expressed as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur Composite Number =
Theorems Real Numbers Prime Factorization method Fundamental Theorem of Arithmetic
1. Let p be a prime number. If p divides a², then p divides a, where a is a positive integer 2. √2, √3 are irrational numbers
Degree
Polynomials
Graphical Representation of Quadratic Polynomial
a ₁ x + b ₁ y + c ₁ = 0, a ₂ x+b ₂ y+c ₂ =0 where a ₁ , b ₁ , c ₁ , a ₂, b ₂, c ₂ are real numbers.
Each solution (x, y), corresponds to a point on the line representing the equation and vice-versa
Algebraic Interpretation
Nature of Roots
Quadratic Equations
If a, b, c are in A.P. Then b = here, b → arithmetic mean a + c 2
Arithmetic mean
General form
• Fixed number in arithmetic progression which provides the to and fro terms by adding/subtracting from the present number. • Can be positive or negative.
• It is denoted by d
Example a, a+d, a+2d, a+3d,...... a+(n-1)d
a n = l(n1) d here, l → last term d →c ommon difference a n → n th term
Definition
Arithmetic Progressions
Common Difference
From the end
Sequence of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
n th term
a n = a + (n1) d here, a → last term d → common difference a n → n th term
Sum (S)
When first & last terms are given:
S n = (a+a n ) or S n = (a+ l ) here, a → first term n → total no. of terms a n → n th term l → last term n 2 n 2
When first term and common difference are given: S n = ( 2a + (n -1) d ) here, a → first term d → common difference n → total no. of terms n 2
From the beginning
Sum of first n positive integers
Let S n = 1 + 2+ 3 + ...... n here, a = 1, last term, l = n S n = = n(a + l ) 2 n(1 +n) 2
3. If in two triangles, corresponding angles are equal, then their corresponding sides are in" the same ratio (or proportion) and hence the two triangles are similar.(AAA criterion)
4. If in two triangles, sides of one triangle are proportional to (ie, in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar. (SSS criterion)
5. If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.(SAS criterion)
Triangles
Theorems
1. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
2. If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Similarity
(i) Corresponding angles are equal (ii) Corresponding sides are in the same ratio
Example
Are the following points vertices of a square: (1,
Since,
All four sides and diagonals are equal Hence, ABCD is a square
Find point of trisection of line segment AB, A(2,-2) and B(-7, 4) Co-ordinates of P = Co-ordinates of Q = (-4 , 2) = (-1 ,
Sine of ∠ A = BC AC Cosine of ∠ A = AB AC Tangent of ∠ A = BC AB Cotangent of ∠ A = AB BC Secant of
A = AC AB Cosecant of ∠ A = AC BC
Study of relationships between the sides & angles of a triangle
Relationships between the Ratios
Trigonometry
Note: How to learn the relation "Some people have" sin θ = "Curly Brown Hair" cos θ = "through proper Brushing"tan
Trigonometric Ratios
Introduction to Trigonometry
Trigonometric Identities
Values of trigonometric ratios between 0 to 90°
P: Perpendicular B: Base H: Hypotenuse
Some Applications of Trigonometry
LineofsightAngleofelevation
AngleofDepression
Q
90
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
radius
In a fixed plane, The locus of a point equidistant from a fixed point is called 'circle'. The fixed point is called the centre & separation of points is the radius of the circle.
Circles
Tangent and tangent point
Only one common point (A) between circle and line PQ. A Here, PQ is tangent at point A lying on the circle.
1. There is no tangent to a circle passing through a point lying inside the circle.
2. There is one and only one tangent to a circle passing through a point lying on the circle.
Two common points (A and B) between line PQ and circle O Q B P r A
3. There are exactly two tangents to a circle through a point lying outside the circle.
Segment
Areas Related to Circles
Portion of the circular region enclosed between a chord and the corresponding arc
Sector of a circle
Formulae
Portion
Example
Given: Inner diameter of the Cylindrical glass = 5 cm Height = 5 cm
Find: Actual capacity of cylindrical glass.
Sum of surface areas of the faces of solid
Mode where, l = lower Limit of median class n = Sum of frequencies h = with of class interval f = frequency of median class c.f. Cumulative frequency of the class preceding to median class Median =
where, l = lower limit of modal class f₁ = frequency of modal class
= frequency of class preceding to modal class
= frequency of class succeeding to modal class
= width of class interval
= class marks.
Direct Method where,
Assumed Mean Method
= frequencies x i = class marks
Frequency obtained by adding the frequencies of all the classes preceding the class A collection,analysis,interpretationofquantitative data
ExperimentalProbability
Sure or Certain Event When event having probability to occur as 1 C omplementary Event For event E, complement event, P(E) = 1-P(E)
Definitions
Elementary Event Event having only one outcome of the experiment
Sum of probabilities of all elementary events is 1. For events A, B, C; P(A) + P(B) + P(C) = 1
Value
Probability
Theoretical Probability
What actually happens in an experiment
P(E) = Number of outcomesfavourable to E
Number of all possible outcomes of the experiment
0 ≤ P(E) ≤ 1
What we expect to happen in an experiment
Number of trials in which the event happened
P(E)=
Total number of trials
Maximum Marks: 80 Time: 3 hours
General Instructions:
Read the following instructions carefully and follow them:
1. This question paper contains 38 questions. All Questions are compulsory.
2. This Question Paper is divided into 5 Sections A, B, C, D and E.
3. In Section A, Question numbers 1-18 are multiple choice questions (MCQs) and questions no. 19 and 20 are Assertion- Reason based questions of 1 mark each.
4. In Section B, Question numbers 21-25 are very short answer (VSA) type questions, carrying 02 marks each.
5. In Section C, Question numbers 26-31 are short answer (SA) type questions, carrying 03 marks each.
6. In Section D, Question numbers 32-35 are long answer (LA) type questions, carrying 05 marks each.
7. In Section E, Question numbers 36-38 are case study-based questions carrying 4 marks each with sub parts of the values of 1, 1 and 2 marks each respectively.
8. There is no overall choice. However, an internal choice in 2 questions of Section B, 2 questions of Section C and 2 questions of Section D has been provided. An internal choice has been provided in all the 2 marks questions of Section E.
9. Draw neat and clean figures wherever required. Take �� = 22 wherever required if not stated. 7
10.Use of calculators is not allowed.
(Section A)
Section A consists of 20 questions of 1 mark each. Q.No. Questions Marks
1. If �� =22 ×3�� ,�� = 22 ×3×5,�� = 22 ×3×7 and LCM (��,��,��) = 3780, then �� is equal to (A)1 (B) 2 (C) 3 (D) 0 1
2. The shortest distance (in units) of the point (2,3) from y-axis is (A) 2 (B) 3 (C) 5 (D) 1
3. If the lines given by 3x +2ky =2 and 2x+5y +1=0 are not parallel, then k has to be (A) 15 (B) ≠15 4 4 (C) any rational number (D) any rational number having 4 as denominator 1
A quadrilateral ABCD is drawn to circumscribe a circle. If BC=7cm, CD=4cm and AD=3cm, then the length of AB is
3cm (B) 4cm (C) 6cm (D) 7cm
5. If �������� + �������� = �� ,then �������� �������� will be
6. Which one of the following is not a quadratic equation?
7. Given below is the picture of the Olympic rings made by taking five congruent circles of radius 1cm each, intersecting in such a way that the chord formed by joining the point of intersection of two circles is also of length 1cm.Total area of all the dotted regions (assuming the thickness of the rings to be negligible) is
For Visually Impaired candidates
The area of the circle that can be inscribed in a square of 6 cm is
8. A pair of dice is tossed. The probability of not getting the sum eight is
9. If 2sin5
10. The sum of two numbers is 1215 and their HCF is 81, then the possible pairs of such numbers are
11. If the area of the base of a right circular cone is 51cm2 and it's volume is 85cm2 , then the height of the cone is given as (A)5 cm (B) 5cm (C) 5cm (D) 5cm 6 3 2
12. If zeroes of the quadratic polynomial a��2 + b�� +c (a, c ≠0) are equal, then (A) c and b must have opposite signs (B) c and a must have opposite signs (C) c and b must have same signs (D) c and a must have same signs
13. The area (in cm2) of a sector of a circle of radius 21cm cut off by an arc of length 22cm is (A)441 (B) 321 (C) 231 (D) 221
14. If ∆ABC ~∆DEF, AB=6cm, DE=9cm, EF=6cm and FD=12cm, then the perimeter of ∆ABC is (A)
15. If the probability of the letter chosen at random from the letters of the word “Mathematics” to be a vowel is 2 2��+1, then x is equal to (A)
16. The points A(9,0), B(9, -6) ,C(-9,0) and D(-9,6) are the vertices of a (A) Square (B) Rectangle (C) Parallelogram (D) Trapezium
17. The median of a set of 9 distinct observation is 20.5. If each of the observations of a set is increased by 2,then the median of a new set
(A) is increased by 2 (B) is decreased by 2 (C) is two times the original number (D) Remains same as that of original observations
18. The length of a tangent drawn to a circle of radius 9 cm from a point at a distance of 41cm from the centre of the circle is (A)40cm (B) 9cm (C) 41cm (D) 50cm
DIRECTIONS: In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R)
Choose the correct option:
(A)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) (B)Both assertion (A) and reason (R) are true and reason (R) is not the explanation of assertion (A) (C)Assertion (A) is true but reason (R) is false. (D)Assertion (A) is false but reason (R) is true.
19. Assertion (A): The number 5n cannot end with the digit 0, where n is a natural number
Reason (R): A number ends with 0, if its prime factorization contains both 2 and 5
20. Assertion (A): If cosA + cos2A=1, then sin2A + sin4A =1 Reason (R): sin2A + cos2A =1
(Section – B)
Section B consists of 5 questions of 2 marks each. 21.(A) (B)
The A.P 8, 10, 12,……. has 60 terms. Find the sum of last 10 terms.
Find the middle term of A.P 6,13, 20, ……., 230
If ������(��+��) =1 and ������(�� ��) = √3 2 ,0° < ��,�� < 90°, find the measure of angles �� and ��
23. If AP and DQ are medians of triangles ABC and DEF respectively, where ∆ABC~
24. (A) (B)
DEF, then prove that ����
A horse, a cow and a goat are tied, each by ropes of length 14m, at the corners A, B and C respectively, of a grassy triangular field ABC with sides of lengths 35m, 40m and 50 m. Find the area of grass field that can be grazed by them.
Find the area of the major segment (in terms of ��) of a circle of radius 5cm, formed by a chord subtending an angle of 90˚ at the centre.
25. A ∆ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 10 cm and 8 cm respectively. Find the lengths of the sides AB and AC, if it is given that ar(∆ABC) = 90cm2
For Visually Impaired candidates:
A circle is inscribed in a right-angled triangle ABC, right angled at B. If BC=7cm and AB=24cm, find the radius of the circle
Section C consists of 6 questions of 3 marks each.
26. In Figure, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B.
Prove that ∠ AOB = 90°
For Visually Impaired candidates:
Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that ∠APB= 2(∠OAB)
27. In a workshop, the number of teachers of English, Hindi and Science are 36, 60 and 84 respectively. Find the minimum number of rooms required, if in each room the same number of teachers are to be seated and all of them being of the same subject.
28. Find the zeroes of the quadratic polynomial 2��2 – (1+2√2)�� + √2 and verify the relationship between the zeroes and coefficents of the polynomial.
29. If ��������+ �������� = √3 ,then prove that �������� + �������� =1 OR
Prove that �������� ��������+1 =cosecA + cotA ��������+
30. On a particular day, Vidhi and Unnati couldn’t decide on who would get to drive the car. They had one coin each and flipped their coin exactly three times. The following was agreed upon:
1. If Vidhi gets two heads in a row, she would drive the car 2. If Unnati gets a head immediately followed by a tail, she would drive the car.
Who has greater probability to drive the car that day? Justify your answer.
The monthly income of Aryan and Babban are in the ratio 3:4 and their monthly expenditures are in ratio 5:7. If each saves ₹ 15,000 per month, find their monthly incomes. OR
Solve the following system of equations graphically:
2�� + �� =6, 2 �� – �� – 2=0. Find the area of the triangle so formed by two lines and ��axis.
For Visually Impaired candidates:
Five years hence, fathers age will be three times the age of son. Five years ago, father was seven times as old as his son. Find their present ages.
Section D consists of 4 questions of 5 marks each
32. A train travels at a certain average speed for a distance of 63km and then travels at a distance of 72km at an average speed of 6km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?
33. Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the sameratio.
Hence in ΔPQR, prove that a line ℓ intersects the sides PQ and PR of a ∆PQR at L and M respectively such that LM II QR. If PL = 5.7cm, PQ=15.2cm and MR=5.5cm, then find the length of PM (in cm)
34.(A)
(B)
From a solid right circular cone, whose height is 6cm and radius of base is 12cm, a right circular cylindrical cavity of height 3cm and radius 4cm is hollowed out such that bases of cone and cylinder form concentric circles. Find the surface area of the remaining solid in terms of ��
An empty cone of radius 3cm and height 12cm is filled with ice-cream such that the lower part of the cone which is (1 6)th of the volume of the cone is unfilled (empty) but a hemisphere is formed on the top. Find the volume of the ice-cream.
35.(A) (B) If the mode of the following distribution is 55, then find the value of �� Hence, find the mean.
A survey regarding heights (in cm) of 51 girls of class X of a school was conducted and the following data was obtained:
Heights (in cm) Number of girls less than 140 04 less than 145 11 less than 150 29 less than 155 40 less than 160 46 less than 165 51
Find the median height of girls. If mode of the above distribution is 148.05, find the mean using empirical formula. 5
(Section – E)
Section E consists of 3 case study-based questions of 4 marks each.
36. In a class, the teacher asks every student to write an example of A.P. Two boys Aryan and Roshan writes the progression as 5, 2,1,4, and 187, 184, 181, respectively. Now the teacher asks his various students the following questions on progression.
Help the students to find answers for the following:
i. Find the sum of the common difference of two progressions. ii. Find the 34th term of progression written by Roshan. iii. (A) Find the sum of first 10 terms of the progression written by Aryan.
(B) Which term of the progressions will have the same value?
37. A group of class X students goes to picnic during winter holidays. The position of three friends Aman, Kirti and Chahat are shown by the points P, Q and R
(i) Find the distance between P and R. 1 (ii) Is
(iii) (A)
(B)Let S be a point which divides the line joining PQ in ratio 2:3. Find the 2 coordinates of S.
For Visually Impaired Candidates:
A group of class X students goes to picnic during winter holidays. Aman, Kirti and Chahat are three friends. The position of three friends Aman, Kirti and Chahat are shown by the points P, Q and R. The co-ordinates of P (2,5), Q (4,4) and R (8,3) are given.
(i) Find the distance between P and R. 1 (ii) Is Q the midpoint of PR? Justify by finding midpoint of PR. 1 (iii) (A) Find the point on x-axis which is equidistant from P and Q. 2
(B) Let S be a point which divides the line joining PQ in ratio 2:3. Find the 2 coordinates of S.
38. India gate (formerly known as All India war memorial) is located near Karthavya path. (formerly Rajpath) at New Delhi. It stands as a memorial to 74187 soldiers of Indian Army, who gave their life in the first world war. This 42m tall structure was designed by Sir Edwin Lutyens in the style of Roman triumphal arches. A student Shreya of height 1 m visited India Gate as a part of her study tour.
i. What is the angle of elevation from Shreya’s eye to the top of India Gate, if she is standing at a distance of 41m away from the India Gate?
ii. If Shreya observes the angle of elevation from her eye to the top of India Gate to be 60o, then how far is the she standing from the base of the India Gate?
iii. (A) If the angle of elevation from Shreya’s eye changes from 45° to 30° , when she moves some distance back from the original position. Find the distance she moves back.
mfrforomm
(B) If Shreya moves to a point which is at a distance of 41m fromthe India Gate, √3 then find the angle of elevation made by her eye to the top of India Gate.
Read the following instructions carefully and follow them:
1. This question paper contains 38 questions.All Questions are compulsory
2. This question paper is divided into 5 SectionsA, B, C, D, and E.
MAX. MARKS: 80
3. In SectionA, Question numbers 1-18 are multiple-choice questions (MCQs) and questions number 19 and 20 areAssertion-Reason based questions of 1 mark each.
4. In Section B, Questions numbers 21-25 are very short answer (VSA) type questions, carrying 02 marks each.
5. In Section C, Questions numbers 26-31 are short answer (SA) type questions, carrying 03 marks each.
6. In Section D, Questions numbers 32-35 are long answer (LA) type questions, carrying 05 marks each.
7. In Section E, Questions numbers 36-38 are case study-based questions carrying 4 marks each with subparts of the values of 1, 1, and 2 marks respectively.
8. Draw neat and clean figures wherever required. Take �� = 22 7 wherever required if not stated.
9. Use of calculators is not allowed.
SectionA
SectionAconsists of 20 questions of 1 mark each.
1. The product of a rational number and an irrational number is A)a rational number only B)an integer C)an irrational number only D) Either rational or irrational 1
2. The graph of �� =��(��) is shown in the figure for some polynomial ��(��).
3. In the given figure, graphs of two linear equations are shown. The pair of these linear equations is:
A) consistent with infinitely many solutions.
B) inconsistent but can be made consistent by extending these lines.
C) consistent with unique solution.
D) inconsistent.
4. The equation of x-axis is of the form
5. In anA.P., if �� = 4,�� =7 and ���� =4, then ' �� ' is
6. Adie is rolled once. The probability that a composite number comes up, is:
7. The perpendicular bisector of the line segment joining the points ��(1,5) and ��(4,6) cuts the ��-axis at
8. We have, ����‖���� and ����‖����. Then,
9. The mid-point of the line segment joining the points ( 1,3) and (8,3 2) is:
12. Two men are on opposite sides of a tower. They observe the angles of elevation of the top of the tower as 30∘ and 45∘ respectively. If the height of the tower is 100 m, then the distance between them is
13. In the given figure,AB is a tangent to a circle with center O. If OA=6cm and ∠OAB= 30∘, then the radius of the circle is:
14. In Figure, PQ and PR are tangents drawn from P to a circle with centre O. If ∠������ = 35∘, then
15. Area of a segment of a circle of radius �� and central angle 90∘ is:
16.
O is the centre of a circle of diameter 4 cm and OABC is a square, if the shaded area is 1 3 area of the square, then the side of the square is
17. Using empirical relationship, the mode of a distribution whose mean is 7.2 and the median 7.1, is:
18. If a digit is chosen at random from the digits 1,2,3,4,5,6,7,8,9; then the probability that this digit is an odd prime number is:
19.
Assertion (A): √3,2√3,3√3,4√3 this series forms anA.P.
Reason (R): Since common difference is same and equal to √3 therefore given series is anAP.
A)BothAand R are true, and R is the correct explanation ofA
B)BothAand R are true, but R is not the correct explanation ofA
C)Ais true, but R is false
D)Ais false, but R is true
20. Assertion (A): If we join two hemispheres of same radius along their bases, then we get a sphere.
Reason (R):Atank is made of the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and radius is 30 cm. The total surface area of the tank is 358m2 .
A)Both A and R are true, and R is the correct explanation of A
B)Both A and R are true, but R is not the correct explanation of A
C) A is true, but R is false
D)Ais false, but R is true
Section B
If the ratio of the sums of first �� terms of twoA.P.'s is (7��+1):(4��+27), find the ratio of their ��th terms.
29. Prove that parallelogram circumscribing a circle is a rhombus.
Atangent PT is drawn parallel to a chordAB as shown in figure. Prove that APB is an isosceles triangle.
Calculate the median from the following data:
Section D
Section D consists of 4 questions of 5 marks each
32. The length of the hypotenuse of a right triangle exceeds the length of its base by 2 cm and exceeds twice the length of altitude by 1 cm. Find the length of each side of the triangle.
Atruck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed which is 20 km per hour more than the first speed. If the truck covers the total distance in 5 hours, then find the first speed of the truck.
33. In the given figure, DEFG is a square and ∠������ =90∘ Prove that
34. Avessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm, and the total height of the vessel is 13 cm. Find the inner surface area and the volume of the vessel. OR
Ahemispherical depression is cut out from one face of a cubical block of side 7 cm, such that the diameter of the hemisphere is equal to the edge of the cube. Find the surface area of the remaining solid.
35. Find the missing frequencies in the following distribution, if the sum of the frequencies is 120 and the mean is 50.
Section E
Section E consists of 3 case study-based questions of 4 marks each.
36. Read the following passage and answer the following questions: Rainbow is an arch of colours that is visible in the sky after rain or when water droplets are present in the atmosphere. The colours of the rainbow are generally, red, orange, yellow, green, blue, indigo and violet. Each colour of the rainbow makes a parabola. We know that any quadratic polynomial ��(��)=����2 +����+��(�� ≠0) represents a parabola on the graph paper.
i.The graph of a rainbow ��=��(��) is shown in the figure. Write the number of zeroes of the curve.
ii.If the graph of a rainbow does not intersect the �� -axis but intersects y -axis at one point, then how many zeroes will it have?
iii.If a rainbow is represented by the quadratic polynomial ��(��)=
+(��+1)��+��, whose zeroes are 2 and 3, find the value of �� and ��
37. Read the following passage and answer the following questions: Priyanshu is very intelligent in maths. He always tries to relate the concept of maths in daily life. One day he is walking away from the base of a lamp post at a speed of 1m/s Lamp is 4.5 m above the ground.
i.If after 2 second, length of shadow is 1 meter, what is the height of Priyanshu?
ii.What is the minimum time after which his shadow will become larger than his original height?
iii.What is the distance of
this point?
38. To raise social awareness about the hazards of smoking, a school decided to start a 'No smoking' campaign. 10 students are asked to prepare campaign banners in the shape of a triangle. The vertices of one of the triangles are P( 3,4),Q(3,4) and R( 2, 1).
i.What are the coordinates of the centroid of △PQR ?
ii.If T be the mid-point of the line joining R and Q, then what are the coordinates of T ?
iii.If �� be the mid-point of line joining �� and ��, then what are the coordinates of �� ?
∴ our assumption is wrong
⇒6+3√2 is an irrational number.
22. In △ABC, it is given that
In △ADC, it is given that
From (i) and (ii), we get
23. Let the sides of the quadrilateral �������� touch the circle at ��,��,�� and ��. Since, the lengths of the tangents from an external point to a given circle are equal.
∴AP=AS
24. We have,
Here, we used the Pythagoras identity: sin2 �� =1
Property used: Factorization and cancellation of common term.
RHS:
Here, we used the reciprocal identity: 1
�� So, LHS = RHS. FinalAnswer: Hence Proved.
26. The greatest number of plants that can be planted in a row =HCF(81,45,63)
∴ hypotenuse of triangle =2��+1 and base of triangle =2�� 1.
Using Pythagoras theorem, (2��+1)2 =��2 +(2�� 1)2
Rejecting �� =0,∴�� =8 hypotenuse of triangle 2×8+1=17cm and base of triangle 2×8 1=15cm
Let the average speed of truck be ��km/h. 150 �� + 200 ��+20 =5 or, 150��+3000+200�� =5��(��+20) or, ��2 50�� 600=0
or, ��2 60��+10�� 600=0
or, ��(�� 60)+10(�� 60)=0
or, (�� 60)(��+10)=0
or, �� =60; or �� = 10
as, speed cannot be negative
Therefore, �� =60km/h
Hence, first speed of the truck =60km/h
33. Given A△������ in which ∠������ =90∘ and DEFG is a square.
Proof
i.In △������ and △������, we have
∠������ =∠������ =90∘
∠������ =∠������ [corresponidng angles]
[∵����‖���� andAB is the transversal]
∴△������ ∼△������
ii.In △������ and △������, we have
∠������ =∠������ =90∘
∠������=∠������ [corresponding angles]
[∵����‖���� andAC is the transversal] ∴△
and
iv. △������ ∼△������
34. Radius =7cm
Height of cylindrical portion =13 7=6cm
Inner surface area of the vessel =2����2 +2����ℎ
=2× 22 7 ×7×7+2× 22 7 ×7×6
=572cm2
Volume of the vessel = 2 3 ����3 +����2ℎ = 2 3 × 22 7 ×7×7×7+ 22 7 ×7×7×6 = 4928 3 or 1642.67cm3 approx.
Therefore, inner surface area and volume of the vessel is 572cm2 and 1642.67cm3 respectively.
Edge of the cube, a=7cm. Radius of the hemisphere, �� = 7 2
Surface area of remaining solid = total surface area of the cube - area of the top of hemispherical part + curved surface area of the hemisphere
given mean = 50
Mean = ∑��������
��
⇒50= 3480+30��1 +70��2 120
⇒6000=3480+30��1 +70��2
⇒3��1 +7��2 =252… (i)
Also, 68+��1 +��2 =120
⇒��1 =52 ��2 2
Substituting in (i), we have 3(52 ��2)+7��2 =252
⇒4��2 =96
⇒��2 = 24
⇒��1 =52 24=28
Hence, ��1 =28 and ��2 =24
SECTION E
36. i.Graph of ��=��(��) intersects X - axis at two distinct points. So we can say that no of zeros of �� =��(��) is 2 . 1 ii.There will not be any zero if graph of ��(��) does not intersect �� - axis. 1
iii. ��2 +(��+1)��+�� is the quadratic polynomial. 2 and -3 are the zeros of the quadratic polynomial.
Thus, 2+( 3)= (��+1) 1
⇒ (��+1) 1 =1
⇒a+1=1
⇒a=0
Also, 2×( 3)=b
⇒b= 6 1 OR
If -4 is zero of given polynomial then, ( 4)2 2( 4) (7��+3)=0
⇒16+8 7p 3=0
⇒7p=21
p=3
37. i.Distance covered in 2sec=2m length of shadow =1m
Total distance from base =2+1=3m
1 3 = height of Rohan 4.5 height of Priyanshu =15m =150cm
ii.When �� >15m distance walked =dm
hence, the time must be 3 sec ∴ minimum time after which his shadow become larger than his original height =3s
After 4 and distance =4m Shadow length = �� m
1.5
∴After 4 sec, the shadow length will be 2 m
38. i.We have, P( 3,4),Q(3,4) and R( 2, 1)
∴ Coordinates of centroid of △
iii. Coordinates of U
The centroid of the triangle formed by joining the mid-points of sides of a given triangle is the same as that of the given triangle.
So, centroid of △
MATHEMATICS STANDARD
SAMPLE QUESTION PAPER - 2
CLASS – X (2025-26)
TIME: 3 Hrs
General Instructions:
Read the following instructions carefully and follow them:
1. This question paper contains 38 questions.All Questions are compulsory
2. This question paper is divided into 5 SectionsA, B, C, D, and E.
MAX. MARKS: 80
3. In SectionA, Question numbers 1-18 are multiple-choice questions (MCQs) and questions number 19 and 20 areAssertion-Reason based questions of 1 mark each.
4. In Section B, Questions numbers 21-25 are very short answer (VSA) type questions, carrying 02 marks each.
5. In Section C, Questions numbers 26-31 are short answer (SA) type questions, carrying 03 marks each.
6. In Section D, Questions numbers 32-35 are long answer (LA) type questions, carrying 05 marks each.
7. In Section E, Questions numbers 36-38 are case study-based questions carrying 4 marks each with subparts of the values of 1, 1, and 2 marks respectively.
8. Draw neat and clean figures wherever required. Take �� = 22 7 wherever required if not stated.
9. Use of calculators is not allowed.
A
SectionAconsists of 20 questions of 1 mark each.
In the given figure, graph of a polynomial ��(��) is given. Number of zeroes of ��(��) is:
Section
3. Two coins are tossed simultaneously. The probability of getting at most one tail is:
Amonth is selected at random in a year. The probability that it is March or October, is
5. The root(s) of the quadratic equation x2
6. The distance between the points (sin ��,cos ��) and (cos ��, sin ��) is
√2 units
In △ABC it is given that
The difference of the areas of a minor sector of angle 120∘ and its corresponding major sector of a circle of radius
cm , is
9. The area of the sector of a circle of radius 12 cm is 60��cm2. The central angle of this sector is:
The product of two consecutive even integers is 528. The quadratic equation corresponding to the above statement, is
(1+��)2�� =528
��(��+2)=528
The abscissa of any point on the ��-axis is
+4)=528
12. If four sides of a quadrilateralABCD are tangential to a circle, then
-1
14. The angle of depressions of the top and bottom of 10 m tall building from the top of a multi-storied building are 30∘ and 60∘ respectively. Find the height of the multistoried building and the distance between the two buildings.
15. In the given figure, PQ is a tangent to the circle with centre O . If ∠OPQ=��,∠POQ= ��, then ��+�� is:
17. If the area of a sector of a circle is 7 20 of the area of the circle, then the angle at the centre is equal to
18. Consider the following frequency distribution:
difference of the upper limit of the median class and the lower limit of the modal class is
19. AssertionA): The 11th term of an AP is 7,9,11,13 is 27.
Reason (R): If s�� is the sum of first �� terms of anAP then its ��th term ���� is given by
=s�� +s�� 1. A) BothAand R are true, and R is the correct explanation ofA
B) BothAand R are true, but R is not the correct explanation ofA
C)Ais true, but R is false
D)Ais false, but R is true
20. AssertionA): If we join two hemispheres of same radius along their bases, then we get a sphere.
Reason (R):Atank is made of the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and radius is 30 cm. The total surface area of the tank is 3.3m2 .
A) Both �� and �� are true, and �� is the correct explanation of ��
B) Both �� and �� are true, but �� is not the correct explanation of ��
C) �� is true, but �� is false
D)Ais false, but R is true
Section B
Section B consists of 5 questions of 2 marks each.
21. Define a prime number and a composite number. Hence
b. The long and short hands of a clock are 6 cm and 4 cm long respectively. Find the sum of distances travelled by their tips in 24 hours, (use �� =3.14 ).
Section C
Section C consists of 6 questions of 3 marks each.
26. Mr. Patil has three classes. Each class has 28, 42 and 56 students respectively. Mr Patil wants to divide each class into groups so that every group in every class has the same number of students and there are no students left over. What is the maximum number of students Mr. Patil can put into each group? 3
27. One zero of the polynomial ��2 2�� (7��+3) is -1, find the value of �� and the other zero. 3
28. a.Alady has only 25 -paisa and 50 -paisa coins in her purse. If she has 50 coins in all totalling ₹ 19.50, how many coins of each kind does she have? OR
b. The sum of two numbers is 8. If their sum is four times their difference, find the numbers. 3
29. a.AtriangleABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 6 cm and 8 cm respectively. Find the lengths of the sidesAB andAC.
b. In the given figure, PT and PS are tangents to a circle with centre O , from a point P , such that PT=4cm and ∠TPS=60∘. Find the length of the chord TS.Also, find the radius of the circle. 3
Find the mean of the following frequency distribution:
Section D consists of 4 questions of 5 marks each
32. a.Agirl is twice as old as her sister. Four years hence, the product of their ages (in years) will be 160. Find their present ages
b. The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages was four times the father's age at that time. Find their present ages.
34. a.Astudent was asked to make a model shaped like a cylinder with two cones attached to its ends by using a thin aluminium sheet. The diameter of the model is 3 cm, and its total length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model.
b. In a cylindrical vessel of radius 10 cm, containing some water, 9000 small spherical balls are dropped which are completely immersed in water which raises the water level. If each spherical ball is of radius 0.5 cm, then find the rise in the level of water in the vessel.
35. The marks obtained by 45 students of a class in a test are given below:
Find the mean and median marks.
Section E
Section E consists of 3 case study-based questions of 4 marks each.
36. Read the following passage and answer the following questions:
Suman is celebrating his birthday. He invited his friends. He bought a packet of toffees/candies which contains 360 candies. He arranges the candies such that in the first row there are 3 candies, in second there are 5 candies, in third there are 7 candies and so on.
i.Find the total number of rows of candies.
ii.How many candies are placed in last row?
iii. a. IfAditya decides to make 15 rows, find the total candies.
b. IfAditya decides to make 21 rows, find the total candies.
37. Read the following passage and answer the following questions:
An observer on the top of a 40 m tall light house (including height of the observer) observes a ship at an angle of depression 30∘ coming towards the base of the light house along straight line joining the ship and the base of the light house. The angle of depression of ship changes to 45∘ after 6 seconds.
i.Find the distance of ship from the base of the light house after 6 seconds from the initial position when angle of depression is 45∘
ii. Find the distance of ship from the base of the light house when the angle of depression is 30∘ .
iii. a. Find the speed of the ship?
iii. b. Find the distance between two positions of ship after 6 seconds?
38. The top of a table is hexagonal in shape.
On the basis of the information given above, answer the following questions:
i.Write the coordinates ofAand B.
ii.Write the coordinates of the mid-point of line segment joining C and D.
iii.a. Find the distance between M and Q.
b.Find the coordinates of the point which divides the line segment joining �� and �� in the ratio 1:3 internally.
Marking Scheme
ITs diagonals intersect each other at the point O .
To prove: ����
Construction:
Through O, draw a line OE parallel toAB or DC intersectingAD at E
23. Let BL=BN=x (tangents from external points are equal)
CL=CM=y
AN=AM=z
∴AB+BC+AC=2x+2y+2z=30
⇒x+y+z=15… (i)
Also x+z=10,x+y=8 and y+z=12
Let the sides of the quadrilateral �������� touch the circle at ��,��,�� and ��. Since, the lengths of the tangents from an external point to a given circle are equal.
∴AP=AS
⇒BP=BQ
CR=CQ
⇒DR=DS
Subtracting from equation (i)
y=5,z=7 and x=3
∴BL=3cm,CM=5cm and AN=7cm.
24. a. LHS =(Sec�� tan ��)2
proved.
25. a. Shaded portion indicates the area which the horse can graze. Clearly, shaded area is the area of a quadrant of a circle of radius r = 21 m .
Length of rectangle = 70 m, breadth = 52 m.
Rope length = 21 m.
Since 21<52 and 21<70, the quarter circle lies completely inside the field.
So no cutting is required.
Area = 1 4 ����2
= 1 4 ×��×(21)2
= 1 4 ×��×441
=11025�� m2
Take �� = 22 7
=110.25× 22 7
= 24255 7 =3465��2
b. Long hand makes 24 rounds in 24 hours
Short hand makes 2 round in 24 hours
radius of the circle formed by long hand =6cm and radius of the circle formed by short hand =4cm.
Distance travelled by long hand in one round = circumference of the circle =2����
=2×6×��
=12��cm
Distance travelled by long hand in 24 rounds =24×12��
=288��
Distance travelled by short hand in a round =2����
=2×4��
=8��cm
Distance travelled by short hand in 2 round
=2×8��
=16��cm
Sum of the distances =288��+16�� =304��
=304×3.14
=954.56cm
Thus, the sum of distances travelled by their tips in 24 hours is 954.56 cm .
SECTION C
26. Given: Mr Patil has three classes. Each class has 28, 42 and 56 students respectively. Mr Patil wants to divide each class into groups so that every group in every class has the same number of students and there are no students left over.
To find: The maximum number of students Mr. Patil can put into each group.
For maximum number of students to put into each group
Mr Patil should have to take H.C.F of 28, 42 and 56 students
H.C.F of 28, 42 and 56 so maximum number of students Mr. Patil can put into each group is 14.
27. Let ��(��)=��2 2�� (7��+3)
Since -1 is a zero of ��(��). Therefore,
��( 1)=0
( 1)2 2( 1) (7��+3)=0
1+2 7�� 3=0
3 7�� 3=0 7��=0 �� =0
For finding zeros of ��(��), we put, ��(��)=0
2 2�� 3=
2 3��+�� 3=0 ��(�� 3)+1(�� 3)=0 (�� 3)(��+1)=0
Put �� 3=0 and ��+1=0, we get, Thus, �� =3, 1
Thus, the other zero is 3. 1
28. a. Suppose the number of 25 - paisa coins be �� and the number of 50 - paisa coins be ��.
Then, ��+�� =50 …………………..(i)
She has a total of ₹19.50, 25��+50�� =19.50×(100) ⇒25��+50�� =1950
Subtracting equation (i) from (ii), ⇒��=28
Substituting ��=28 in (i), we get �� =22. 1
∴ the number of 25 paisa coins =22 the number of 50 paisa coins =28 1 b.
Let the numbers be '��' and '��'.
Given that, sum of two numbers is 8.
��+�� =8 …………….(1)
Also, their sum is four times their difference
��+�� =4(�� ��)
��+�� =4�� 4��
3�� 5�� =0.
Substituting value of 'a' from equation (2) in equation (1) ⇒ 5 3 ��+�� =8 ⇒ 5��+3�� 3 =8 ⇒8�� =24 ⇒��=3
Using this value of '��' in (1), gives
=8 3=5
�� =5 and �� =3
We know that the lengths of tangents drawn from an exterior point to a circle are equal.
a. If the present age of sister be ��, then, by the first condition of the question, we have, present age of the girl =2�� By the second condition of the question,we have,
(2��+4)(��+4)=160
2��2 +8��+4��+16=160
2��2 +12�� 144=0
2��2 +(24 12)�� 144=0
2��(��+12) 12(��+12)=0 (2�� 12)(��+12)=0
Let the age of father =�� years age of son =(45 ��) years (�� 5)(45 �� 5)=4(�� 5) On Solving �� =36
Age of father = 36 years Age of son = 9 years
33. It is given that PQRS is a parallelogram,
To prove OC ‖ SR
In △OPS and △OAB , PS‖AB
∠POS=∠AOB [common angle]
∠OSP=∠OBA [corresponding angles]
∴△OPS∼△OAB [byAAAsimilarity criteria]
Then,
(i) [by basic proportionality theorem]
In △CQR and △CAB, QR‖PS‖AB
∠QCR=∠ACB [common angle]
∠CRQ=∠CBA [corresponding angles]
∴△CQR∼△CAB
Then, by basic proportionality theorem =
[PS≅QR Since, PQRS is a parallelogram, ]
From Equation (i) and (ii),
OS OB = CR CB or OB OS = CB CR
On subtracting from both sides, we get,
OB OS 1= CB CR 1
⇒ OB OS OS = (CB CR) CR
⇒ BS OS = BR CR
By converse of basic proportionality theorem, SR‖OC
Hence proved.
For upper conical portion
Radius of the base (r)=1.5cm
(h1)=2cm
For lower conical portion
For
portion
Hence, the volume of the air contained in the model that Rechel made is 66
b. Step 1: Volume of one small sphere
Step 2: Total volume of 9000 spheres
���� = �� 2[2��+(�� 1)��]
⇒360= �� 2[2×3+(�� 1)×2]
⇒360=��[3+(�� 1)×1]
⇒��2 +2�� 360=0
⇒(��+20)(�� 18)=0
⇒�� = 20 reject
�� =18 accept
ii.Since there are 18 rows number of candies placed in last row ( 18 th row) is
���� =��+(�� 1)��
⇒��18 =3+(18 1)2
⇒��18 =3+17×2
⇒��18 =37 1
iii. a. If there are 15 rows with same arrangement
Sn = n 2[2a+(n 1)d]
⇒S15 = 15 2 [2×3+(15 1)×2]
⇒S15 = 15 2 [6+14×2]
⇒S15 =255
There are 255 candies in 15 rows. 2
iii. b. If there are 21 rows with same arrangement
Sn = n 2[2a+(n 1)d]
⇒S21 = 21 2 [2×3+(21 1)×2]
⇒S21 = 21 2 [6+40]
⇒S21 =483
There are 483 candies in 21 rows. 2
The distance of ship from the base of the light house after 6 seconds from the initial position when angle of depression is 45∘ In △ABC
45∘ = ����
ii.The distance of ship from the base of the light house when angle of depression is 30∘ . In △ABD
b. M(5,11) and N(9,11)
Z = (1×9+3×5 1+3 , 1×11+3×11 1+3 )
Z(6,11)
MATHEMATICS STANDARD
SAMPLE QUESTION PAPER - 3
CLASS – X (2025-26)
TIME: 3 Hrs
General Instructions:
Read the following instructions carefully and follow them:
1. This question paper contains 38 questions.All Questions are compulsory
2. This question paper is divided into 5 SectionsA, B, C, D, and E.
MAX. MARKS: 80
3. In SectionA, Question numbers 1-18 are multiple-choice questions (MCQs) and questions number 19 and 20 areAssertion-Reason based questions of 1 mark each.
4. In Section B, Questions numbers 21-25 are very short answer (VSA) type questions, carrying 02 marks each.
5. In Section C, Questions numbers 26-31 are short answer (SA) type questions, carrying 03 marks each.
6. In Section D, Questions numbers 32-35 are long answer (LA) type questions, carrying 05 marks each.
7. In Section E, Questions numbers 36-38 are case study-based questions carrying 4 marks each with subparts of the values of 1, 1, and 2 marks respectively.
8. Draw neat and clean figures wherever required. Take �� = 22 7 wherever required if not stated.
9. Use of calculators is not allowed.
A
SectionAconsists of 20 questions of 1 mark each.
(HCF × LCM) for the numbers 70 and 40 is:
2800 1 2. The graph of a polynomial is shown in Figure, then the number of its zeroes is:
Section
3. The lines represented by the linear equations �� =�� and �� =4 intersect at ��. The coordinates of the point �� are:
4. If the roots of equation ����2 +����+�� =0,�� ≠0 are real and equal, then which of the following relation is true?
5. The sum of the first 100 even natural numbers is:
2550
10100
6. The coordinates of the point �� dividing the line segment joining the points ��(1,3) and ��(4,6) in the ratio 2:1 are
7. If P(�� 3,4) is the mid-point of the line segment joining the points Q( 6,5) and R( 2,3), then the value of a is
8. In the given figure, if AB‖DC, thenAP is equal to
9. If sin ��+cos �� =�� and sec ��+cosec�� =��, then ��(��2 1)=
10. If O is the centre of a circle,AOC is its diameter and B is a point on the circle such that ∠ACB=50∘. IfAT is the tangent to the circle at the pointA, then ∠BAT= 1
11. Area of a sector of a circle of radius 36 cm is 54��cm2. The length of the corresponding arc of the sector is:
12. In a circle of radius 21 cm, an arc subtends an angle of 60∘ at the centre. The length of the arc is
13. Acard is drawn at random from a well-shuffled deck of 52 playing cards. The probability of getting an ace of spade is:
14. If three coins are tossed simultaneously, then the probability of getting at least two heads, is
15. The shadow of a 5 m long stick is 2 m long.At the same time, the length of the shadow of a 12.5 m high tree is
16. Apendulum swings through an angle of 30∘ and describes an arc 8.8 cm in length. Find the length of the pendulum.
17. In the given figure, if PT is a tangent to a circle with centre O and ∠TPO=35∘, then the measure of ∠�� is:
18. Consider the following frequency distribution of the heights of 60 students of a class:
Height (in cm)
Number of students
150-155 15
155-160 13
160-165 10
165-170 8
170-175 9
175-180 5
The sum of the lower limit of the modal class and upper limit of the median class is A) 330 B) 320 C) 310 D) 315
19. AssertionA): In the given figure, a sphere circumscribes a right cylinder whose height is 8 cm and radius of the base is 3 cm. The ratio of the volumes of the sphere and the cylinder is 125:54
Reason (R): Ratio of their volume = Volume of sphere Volume of cylinder
A) BothAand R are true, and R is the correct explanation ofA
B) BothAand R are true, but R is not the correct explanation ofA
C)Ais true, but R is false
D)Ais false, but R is true 1
20. AssertionA): If ���� is the sum of the first �� terms of anA.P., then its ��th term ���� is given by ���� =���� ���� 1
Reason (R): The 10th term of theA.P. 5, 8, 11, 14, ... is 35.
A) Both �� and �� are true, and �� is the correct explanation of ��
B) Both �� and �� are true, but �� is not the correct explanation of ��
C) �� is true, but �� is false
D)Ais false, but R is true 1
Section B
Section B consists of 5 questions of 2 marks each.
In the given figure, if ∠1=∠2 and △NSQ≅△MTR.Then prove that △PTS≅△PRQ
23. In the given figure, common tangentsAB and CD to two circles intersect at E. Prove that AB=CD.
25. Three horses are tied each with 7 m long rope at three corners of a triangular field having sides 20m,30m and 40 m. Find the area of the plot which can be grazed by the horses.
In the given figure, two concentric circles with centre O are shown. Radii of the circles are 2 cm and 5 cm respectively. Find the area of the shaded region.
Section C
Section C consists of 6 questions of 3 marks each.
26. Ashopkeeper has 120 litres of petrol, 180 litres of diesel and 240 litres of kerosene. He wants to sell oil by filling the three kinds of oils in tins of equal capacity. What should be the greatest capacity of such a tin? 3
27. Find the zeroes of quadratic polynomial ��2 2�� 8 and verify the relationship between the zeroes and their coefficients.
28. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original fraction is 29 20. Find the original fraction.
29. Prove that parallelogram circumscribing a circle is a rhombus.
Atangent PT is drawn parallel to a chordAB as shown in figure. Prove that APB is an isosceles triangle.
The sum of two numbers is 34. If 3 is subtracted from one number and 2 is added to another, the product of these two numbers becomes 260. Find the numbers.
33. In the given figure PA,QB and RC are each perpendicular to AC. If AP=��,BQ=�� and
34. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter 1 of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
OR
Aconical vessel of radius 6 cm and height 8 cm is completely filled with water.Asphere is lowered into the water and its size is such that when it touches the sides, it is just immersed as shown in Figure. What fraction of water over flows?
35. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
Section E
Section E consists of 3 case study-based questions of 4 marks each.
36. Read the following passage and answer the following questions: The students of a school decided to beautify the school on an annual day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 metre. The flags are stored at the position of the middlemost flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time.
i.How much distance did she cover in pacing 6 flags on either side of center point?
ii.Represent above information inArithmetic progression.
iii.How much distance did she cover in completing this job and returning to collect her books?
37. Read the following passage and answer the following questions:
Jagdish has a field which is in the shape of a right angled triangleAQC. He wants to leave a space in the form of a square PQRS inside the field for growing wheat and the remaining for growing vegetables (as shown in the figure). In the field, there is a pole marked as O
i.Taking O as origin, coordinates of P are ( 200,0) and of Q are (200, 0). PQRS being a square, what are the coordinates of �� and ��?
ii.What is the area of square PQRS?
iii.What is the length of diagonal PR in square PQRS?
If �� divides ���� in the ratio ��:1, what is the value of ��, where point �� is (200, 800)?
38. The law of reflection states that when a ray of light reflects off a surface, the angle of incidence is equal to the angle of reflection.
Suresh places a mirror on level ground to determine the height of a pole (with traffic light fired on it). He stands at a certain distance so that he can see the top of the pole reflected
from the mirror. Suresh's eye level is 1.5 m above the ground. The distance of Suresh and the pole from the mirror are 1.8 m and 6 m respectively.
i.Which criterion of similarity is applicable to similar triangles?
ii.What is the height of the pole?
iii.If angle of incidence is ��, find tan ��
OR
Now Suresh move behind such that distance between pole and Suresh is 13 meters. He place mirror between him and pole to see the reflection of light in right position. What is the distance between mirror and Suresh?
21.
Here �� and �� are integers so �� 6�� �� is a rational number. So √2 should be a rational number. But √2 is an irrational number. It is �� contradiction. Hence result is 6+√2 is a irrational number
22. Since △NSQ≅△MTR
∴∠SQN=∠TRM ⇒∠Q=∠R (in △PQR )
∠�� =90∘ 1 2 ∠��
Again 1 =∠2 [given in △PST ](Isosceles property)
∴∠1=∠2= 1 2(180∘ ∠��) =90∘ 1 2 ∠�� 1
Thus, in △PTS and △PRQ
∠1=∠Q [∵ Each =90∘ 1 2∠��]
∠2=∠R,∠P=∠P (Common)
△PTS≅△PRQ
23.
Common tangentsAB and CD to two circles intersect at E.
As we know that, the tangents drawn from an external point to a circle are equal in length.
∴EA=EC (i) and EB=ED (ii) 1
On adding Eqn. (i) and (ii), we get EA+EB=EC+ED
⇒AB=CD 1
Hence proved.
Given, 3cot A=4
⇒cot A= 4 3
We know that,
25. Given 3 horses are tethered with 7 m long ropes at three corners of △ABC
Here radius of sectors, r=7m
Given sides of △ABC are AB=20cm,BC=30m,CA=40m
Area of the plot which can be grazed
Area of sector OABC=
of sector OED = ��×2
SECTION C
26. The required greatest capacity is the HCF of 120, 180 and 240.
240=180×1+60
180=60×3+0 HCF is 60.
Now HCF of 60, 120
120=60×2+0
∴ HCF of 120,180 and 240 is 60.
∴ The required capacity is 60 litres.
27. Let ��(��)=��2 2�� 8
By the method of splitting the middle term, ��2 2�� 8=��2 4��+2�� 8 =��(�� 4)+2(�� 4)=(�� 4)(��+2)
For zeroes of ��(��), p(x)=0
⇒(�� 4)(��+2)=0
⇒�� 4=0 or ��+2=0
⇒�� =4 or �� = 2
⇒�� =4, 2
So, the zeroes of ��(��) are 4 and 2
We observe that, Sum of its zeroes
=4+( 2)=2 = ( 2) 1 = ( Coefficient of ��) Coefficient of ��2
Product of its zeroes =4��( 2)= 8= 8 1 = Constant term Coefficient of ��2
Hence, relation between zeroes and coefficients is verified.
28. Let the base of the right triangle be x cm.
Then altitude =(x 7)cm
Hypotenuse =13cm
By Pythagoras theorem
( Base )2 +( Altitude )2 =( Hypotenuse )2
⟹��2 +(�� 7)2 =(13)2
⟹��2 +��2 14��+49=169
⟹2��2 14�� 120=0
⟹2(��2 7�� 60) =0 or ��2 7�� 60=0
⟹��2 12��+5�� 60=0
⟹��(�� 12)+5(�� 12)=0
⟹(��+5)(�� 12)=0
Either ��+5=0 or �� 12=0
⟹�� = 5,12 2
Since side of the triangle cannot be negative. So, �� =12cm and �� = 5 is rejected.
Hence, length of the other two sides are 12 cm, (12-7) =5cm
OR
Let the denominator be ��, then numerators =�� 3
So the fraction be �� 3 �� By the given condition, new fraction = �� 3+2 ��+2 = �� 1 ��+2 �� 3 �� + �� 1 ��+2 = 29 20 1 (�� 3)(��+2)+��(�� 1) ��(��+2) = 29 20
20[(�� 3)(��+2)+��(�� 1)]=29(��2 +2��)
20[(��2 3��+2�� 6)+(��2 ��)]=29(��2 +2��)
20(��2 �� 6+��2 ��)=29��2 +58��
20(2��2 2�� 6)=29��2 +58��
11��2 98�� 120=0
11��2 110��+12�� 120=0 2
(11��+12)(�� 10)=0
∴�� =10
∴ The fraction is 7 10 29.
GivenABCD is a parallelogram in which all the sides touch a given circle
To prove:-ABCD is a rhombus
Proof :
∵ABCD is a parallelogram
∴AB=DC and AD=BC
Again AP,AQ are tangents to the circle from the pointA
∴AP=AQ
Similarly, BR = BQ
CR = CS
DP=DS
∴(AP+DP)+(BR+CR)=AQ+DS+BQ+CS=(AQ+BQ)+(CS+DS)
⇒AD+BC=AB+DC
⇒BC+BC=AB+AB[∵AB=DC,AD=BC]
⇒2BC=2AB
⇒BC=AB
Hence, parallelogramABCD is a rhombus 1 OR
Given,
Construction: Join PO and produce it to D .
Proof: Here, ���� ⊥����
∠������ =90∘
Also, ����‖����
∴∠TPD+∠ADP=180∘
⇒∠ADP=90∘
OD bisectsAB [Perpendicular from the centre bisects the chord]
In △ADP and △BDP
AD=BD
∠������ =∠������ [Each 90°] ���� =����
∴△������ ≅△������ [SAS]
∠������ =∠������ [C.P.C.T.]
∴△PAB is isosceles triangle.
30. Given, 1+sin2 �� =3sin ��cos ��, then we have to prove that tan �� =1, or 1 2
�� =100⇒ �� 2 =50 Median Class =60 80 �� =60,��.��.=30,�� =23,ℎ =20 we know that, Median =��+ �� 2
Then, according to given condition, we have
��+��=34… (i) and (�� 3)(��+2)=260… (ii)
On substituting the value of �� from equation (i) in equation (ii), we get 1 (�� 3)(34 ��+2)=260 [∵�� =34 ��]
⇒(x 3)(36 x)=260
⇒36x x2 108+3x=260
⇒x2 39x+368=0
⇒x2 (16x+23x)+368=0
⇒x2 16x 23x+368=0
⇒x(x 16) 23(x 16)=0
⇒(x 16)(x 23)=0
⇒x=16 or x=23 3
When x=16, then y=34 16=18
When x=23, then y=34 23=11
So, the numbers are 16,18 or 23,11 1
33. △PAC∼ΔQBC
Adding (i) and (ii)
+
34. According to the question, a hemispherical depression is cut from one face of the cubical block such that the diameter �� of the hemisphere is equal to the edge of the cube.
Let the radius of hemisphere =��
= �� 2
Now, the required surface area = Surface area of cubical block -Area of base of hemisphere + Curved surface area of hemisphere 2
∴ distance of Suresh from mirror =13 �� = 13 – 10 =3m 2
OR
TIME: 3 Hrs
General Instructions:
MATHEMATICS STANDARD
SAMPLE QUESTION PAPER - 4
CLASS – X (2025-26)
Read the following instructions carefully and follow them:
1. This question paper contains 38 questions All Questions are compulsory
2. This question paper is divided into 5 SectionsA, B, C, D, and E.
MAX. MARKS: 80
3. In SectionA, Question numbers 1-18 are multiple-choice questions (MCQs) and questions number 19 and 20 areAssertion-Reason based questions of 1 mark each.
4. In Section B, Questions numbers 21-25 are very short answer (VSA) type questions, carrying 02 marks each.
5. In Section C, Questions numbers 26-31 are short answer (SA) type questions, carrying 03 marks each.
6. In Section D, Questions numbers 32-35 are long answer (LA) type questions, carrying 05 marks each.
7. In Section E, Questions numbers 36-38 are case study-based questions carrying 4 marks each with subparts of the values of 1, 1, and 2 marks respectively.
8. Draw neat and clean figures wherever required. Take �� = 22 7 wherever required if not stated.
9. Use of calculators is not allowed.
SectionA
SectionAconsists of 20 questions of 1 mark each.
1. sec�� when expressed in terms of cot��, is equal to:
2. If PA and PB are tangents to the circle with centre O such that ∠APB= 50∘, then
is equal to
3. Which of the following is not probability of an event?
1 13 %
4. The roots of the quadratic equation ��2 4=0 is/are:
2 only
5. If ��,��,��,��,�� are inAP, then the value of �� 4�� + 6�� 4�� + �� is
0
3
6. Prime factorisation of 882 is:
23 ×3×72
1
7. The volume of a cylinder of radius �� is 1 4 of the volume of a rectangular box with a square base of side length ��. If the cylinder and the box have equal heights, what is �� in terms of ��?
8. (sec2�� 1)(1 cosec2��) is equal to:
9. In △ABC, if ∠C=3∠B=2(∠A+∠B), then ∠C= A) 120° B) 60°
90°
10. The pair of linear equations �� =0 and ��= 6 has: A) no solution B) only solution C) a unique solution D) infinitely many solutions
11. From an external point Q, the length of tangent to a circle is 12 cm and the distance of Q from the centre of circle is 13 cm. The radius of circle (in cm) is A) 7
12. Which of the following is not anA.P.? A) 2,4,8,16,... B) 1.2, 3.2, 5.2, 7.2,... C) 2,5 2,3,7 2,…
��,2��,3��,4��,...
13. If the sum and the product of zeroes of a quadratic polynomial are 2√3 and 3 respectively, then a quadratic polynomial is: A) ��2 2√3�� 3
(�� √3)2 C) ��2 +2√3��+3
��2 +2√3�� 3
14. The diameter of a circle is of length 6 cm. If one end of the diameter is (-4, 0), the other end on x-axis is at:
(2, 0) B) (6, 0)
(4, 0)
(0, 2)
15. If 3 is the least prime factor of number '��' and 7 is the least prime factor of number '��', then the least prime factor of ��+��, is
16. If one zero of the quadratic polynomial ��2 5��+�� is 4, then the value of �� is A) 18 B) 18
36
17. The distance between the points P( 11 3 ,5) and Q( 2 3,5) is: A) 3 units B) 4 units
36 1
18. Asolid consists of a circular cylinder with an exact fitting right circular cone placed at the top. The height of the cone is h. If the total volume of the solid is 3 times the volume of the cone, then the height of the circular cylinder is
2h
19. Assertion (A): The probability of selecting a vowel from the letters of the word MOBILE is 1 3
Reason (R): The word MOBILE has 6 letters, out of which 3 letters are vowels.
A)BothAand R are true, and R is the correct explanation ofA.
B)BothAand R are true, but R is not the correct explanation ofA.
C)Ais true, but R is false.
D)Ais false, but R is true.
20.
Assertion (A):ABCD is a trapezium with DC‖AB.E and F are points onAD and BC respectively, such that EF‖AB. Then AE ED = BF FC
Reason (R):Any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally.
A)BothAand R are true and R is the correct explanation ofA.
B)BothAand R are true but R is not the correct explanation ofA.
C)Ais true but R is false.
D)Ais false but R is true.
Section B
Section B consists of 5 questions of 2 marks each.
21. In the given figure,
22. Abag contains 3 red balls and 5 black balls.Aball is drawn at random from the bag. What is the probability that the ball drawn is i.red? ii.not red?
23. Reeti prepares a Rakhi for her brother Ronit. The Rakhi consists of a rectangle of length 8 cm and breadth 6 cm inscribed in a circle as shown in the figure. Find the area of the shaded region. (Use �� =3.14)
24. Name the type of quadrilateral formed, if any, by the points (4,5),(7,6),(4,3),(1,2), and give a reason for your answer.
Find the coordinates of the point which divides the line segment joining the points (7, 1) and ( 3, 4) internally in the ratio 2: 3.
25. The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is 1 less than twice its 8th term, find theAP.
28. Two concentric circles are of radii 7 cm and �� cm respectively, where �� >7. A chord of the larger circle of length 46 cm, touches the smaller circle. Find the value of ��.
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ=2∠OPQ.
29. The ratio of incomes of two persons is 9:7 and the ratio of their expenditures is 4:3. If each of them manages to save ₹ 2000 per month, then find their monthly incomes.
Draw the graph of the pair of equations 2x+y=4 and 2x y=4. Write the vertices of the triangle formed by these lines and the ��-axis.Also find the area of this triangle.
30. 144 cartons of Coke cans and 90 cartons of Pepsi cans are to be stacked in a canteen. If each stack is of the same height and if it equal contain cartons of the same drink, what would be the greatest number of cartons each stack would have?
31. Find �� so that the point P( 4,6) lies on the line segment joining A(��,10) and B(3, 8). Also, find the ratio in which P dividesAB.
Section D
Section D consists of 4 questions of 5 marks each
32. In the figure, from a solid cube of side 7 cm, a cylinder of radius 2.1 cm and height 7 cm is scooped out. Find the total surface area of the remaining solid.
OR
Atent is in the shape of a right circular cylinder up to a height of 3 m and then a right circular cone, with a maximum height of 13.5 m above the ground. Calculate the cost of painting the inner side of the tent at the rate of ₹ 2 per square metre, if the radius of the base is 14 m.
33. Agirl on a ship standing on a wooden platform, which is 50 m above water level, observes the angle of elevation of the top of a hill as 30∘ and the angle of depression of the base of the hill as 60∘. Calculate the distance of the hill from the platform and the height of the hill.
OR
From the top of a tower, the angles of depression of two objects on the same side of the tower are found to be �� and ��(�� >��). If the distance between the objects is ' p ' metres,
3
5
Show that the height ' h ' of the tower is given by ℎ = ��tan ��tan �� tan �� tan �� .Also determine the height of the tower, if p=50m,�� =60∘ ,�� =30∘ 5
34. Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio. 5
35. Solve the pair of linear equations 3�� 2 5�� 3 = 2 and �� 3 + �� 2 = 13 6 by substitution method. 5
Section E
Section E consists of 3 case study-based questions of 4 marks each.
36. Read the following text carefully and answer the questions that follow:
Aball is thrown in the air so that �� seconds after it is thrown, its height ℎ metre above its starting point is given by the polynomial ℎ =25�� 5��2 .
Observe the graph of the polynomial and answer the following questions:
i.Write zeroes of the given polynomial.
ii.Find the maximum height achieved by ball.
iii.a.After throwing upward, how much time did the ball take to reach to the height of 30 m? OR
iii. b. Find the two different values of t when the height of the ball was 20 m.
37. Read the following text carefully and answer the questions that follow:
Lokesh, a production manager in Mumbai, hires a taxi every day to go to his office. The taxi charges in Mumbai consists of a fixed charges together with the charges for the distance covered. His office is at a distance of 10 km from his home. For a distance of 10 km to his office, Lokesh paid ₹ 105. While coming back home, he took another route. He covered a distance of 15 km and the charges paid by him were ₹ 155.
i.What are the fixed charges?
ii.What are the charges per km?
iii.If fixed charges are ₹ 20 and charges per km are ₹ 10, then how much Lokesh have to pay for travelling a distance of 10 km?
OR
Find the total amount paid by Lokesh for travelling 10 km from home to office and 25 km from office to home. [Fixed charges and charges per km are as in (i) & (ii).]
38. Read the following text carefully and answer the questions that follow:
Essel World is one of India's largest amusement parks that offers a diverse range of thrilling rides, water attractions and entertainment options for visitors of all ages. The park is known for its iconic "Water Kingdom" section, making it a popular destination for family outings and fun-filled adventure. The ticket charges for the park are ₹ 150 per child and ₹ 250 per adult.
On a day, the cashier of the park found that 300 tickets were sold and an amount of ₹ 55,000 was collected. Based on the above, answer the following questions:
i.If the number of children visited be x and the number of adults visited be y, then write the given situation algebraically.
ii.a. How many children visited the amusement park that day? OR
ii. b. How many adults visited the amusement park that day?
iii.How much amount will be collected if 250 children and 100 adults visit the amusement park?
Marking Scheme
SECTIONA
∴ Total number of elementary events =8
Proabibilty of the event = Number of favourble outcomes
Total number of possible outcomes
i.Since the bag contains 3 red balls, therefore, one red ball can be drawn in 3 ways. ∴
ii.Since the bag contains 5 black balls along with 3 red balls, therefore one black (not red) ball can be drawn in 5 ways.
AB=CD, BC=DA (opposite sides are equal) and AC≠BD (Diagonals are unequal)
Hence, the quadrilateral ABCD is a parallelogram. 1
OR
Given (7, -1) and (-3, -4)
So ��1 =7,��1 = 1
��2 = 3,��2 = 4
Using section formula
�� = ��1��2+��2��1 ��1+��2
�� = 2( 3)+3(7) 2+3 = 6+21 5
�� = 15 5 =3 1
So coordinates of intersection point (3, 11 5 ).
25. The general term of anAP is given by ���� =��+(�� 1)��
Substituting in (i), we get d=4
26. Let 3 √5 be a rational number.
( 3�� �� ) and �� are integers, so ( 3�� �� �� ) is a rational number, but √5 is an irrational number. This contradiction arises because of our wrong assumption. So (3 √5) is an irrational number.
27. Consider,
Let O be the common centre of the two circles andAB be the chord of the larger circle which touches the smaller circle at C .
Then, AB=46cm.
Join OAand OC
Then, OA=��cm and OC=7cm.
Now, OC⊥AB and OC bisectsAB. [as perpendicular to a chord through the chord bisects it]
In right, △������, we have,
OA2 =OC2 +AC2 [by Pythagoras' theorem]
⇒OA=√����2 +����2
=√����2 +(1 2����)2 [∵�� is the midpoint of ����]
⇒�� =√72 +232 cm
⇒�� =√578cm =17√2
⇒�� =17√2cm 2
OR
Given a circle with centre O and an external point T and two tangents TP and TQ to the circle, where P,Q are the points of contact.
To Prove: ∠PTQ=2∠OPQ
Proof: Let ∠PTQ=��
Since TP, TQ are tangents drawn from point T to the circle.
TP=TQ
∴ TPQ is an isoscles triangle
∴∠TPQ=∠TQP= 1 2(180∘ ��)=90∘ �� 2
Since, TP is a tangent to the circle at point of contact P
∴∠OPT=90∘
∠OPQ=∠OPT ∠TPQ=
Thus, ∠PTQ=2∠OPQ Hence proved.
29. Let us denote the incomes of the two-person by ₹ 9�� and ₹ 7�� and their expenditures by ₹ 4�� and ₹ 3�� respectively.
Then the equations formed in the situation is given by:
9�� 4�� =2000… (1) and 7�� 3�� =2000… (2)
Step 1: Multiply Equation (1) by 3 and Equation (2) by 4 to make the coefficients of �� equal. Then, we get the equations:
27�� 12�� =6000... (3)
28�� 12�� =8000... (4)
Step 2: Subtract Equation (3) from Equation (4) to eliminate ��, because the coefficients of �� are the same. So, we get (28�� 27��) (12�� 12��)=8000 6000 i.e., �� =2000 1
Step 3: Substituting this value of �� in (1), we get 9(2000) 4�� =2000 i.e., �� =4000
So, the solution of the equations is �� =2000,��=4000.
Therefore, the monthly incomes of the persons are ₹18,000 and ₹14,000 respectively. 1
The given pair of linear equations 2��+��=4 and 2�� �� =4 Table for line 2��+�� =4
And table for line 2�� ��=4
So, the Graphical representation of both lines is as above.
Here, both lines and ��-axis form a △������
Hence, the vertices of a △ABC are A(0,4),B(2,0) and C(0, 4) whereAand C are obtained by putting �� =0 in the given equations and �� is obtained by solving them together.
∴ Required area of △ABC=2×Area of △AOB
△ABC=2×(1 2 ×4×2)=8 sq. units.
Hence, the required area of the triangle is 8 sq units. 1
30. The greatest number of cartons is the HCF of 144 and 90
Now the prime factorisation of 144 and 90 are
144=16×9=24 ×32
90=2×3×3×5=2×32 ×5 2
HCF (144,90)=2×32 =18
∴ The greatest number of cartons each stack would have =18. 1
31. Section formula
If point ��(��,��) divides ���� in the ratio ��:��, then:
��(��,��)=(����2+����1 ��+�� , ����2+����1 ��+�� ) where ��(��1,��1),��(��2,��2).
Here, ��(��,10),��(3, 8),��( 4,6).
Use y-coordinate to find ratio
6= ��( 8)+��(10) ��+��
Multiply through:
6(��+��)= 8��+10��
6��+6�� = 8��+10��
6��+8�� =10�� 6��
14�� =4�� ⇒ �� �� = 2 7
So, ratio ����:���� =2:7.
Use ��-coordinate to find ��
4= ��(3)+��(��) ��+��
Substitute ��:��=2:7.
So ��=2,�� =7
4= 2(3)+7�� 2+7
4= 6+7�� 9
Multiply both sides:
36=6+7��
7k= 42 k= 6
32. We have;
ACube:
Cube's side,�� =7����
ACylinder:
Cylinder's Radius, �� =2.1����
Cylinder's Height, ℎ =7����
SECTION D
∵Acylinder is scooped out from a cube,
∴ TSAof the resulting solid: = TSAof whole Cube + CSAof the scooped out Cylinder 2× (Area of upper circle or Area of lower circle)
=6��2 +2����ℎ 2×(����2)
=6×(7)2 +2×
=6×49+(44 7 ×147) (44
=294+924 2772
=294+64.68 =358.68cm2
Hence, the total surface area of the remaining solid is 35868cm2
OR
Height of the cylinder =3m.
Total height of the tent above the ground =13.5m
Height of the cone =(13.5 3)m=10.5m
Radius of the cylinder = Radius of cone = 14 m
Curved surface area of the cylinder =2����ℎ =(2× 22 7 ×14×3)m2 =264m2
∴ Curved surface area of the cone =������ =(22 7 ×14×17.5)m2 =770m2
Let S be the total area which is to be painted. Then,
S = Curved surface area of the cylinder + Curved surface area of the cone
⇒S=(264+770)m2 =1034m2 1
Hence, Cost of painting =S× Rate =₹(1034×2)=₹2068
Here, CD=CE+ED=h+50
Now, In △ABD
60∘ = ���� ����
= 50 ��
In △CEA tan 30∘ = ���� ���� 1
Let us suppose thatAand B are the objects �� m apart, DC is the tower of height ℎ
Suppose BC=�� m
Substituting the value of �� in (i), we get
Dividing equation (i) by equation (ii),
⇒ area(△������)
area(△������) = 1 2 ×����×���� 1 2 ×����×����
⇒ area(△������)
area(△������) = ���� ���� (iii)
Similarly, In △������,
Area of △������ = 1 2 ×����×���� (iv)
In △������,
Area of △������ = 1 2 ×����×���� (v)
Dividing equation (iv) by equation (v),
⇒ area(△������)
area(△������) = 1 2×����×���� 1 2×����×����
⇒ area(△������)
area(△������) = ���� ���� (vi)
△������ and △������ lie on the same base DE and between two parallel lines DE and BC .
∴Area(△������)=Area(△������)
From equation (iii),
⇒ area(△������)
area(△������) = ���� ���� ---(vii)
From equation (vi) and equation (vii), ���� ���� = ���� ����
∴ If a line is drawn parallel to one side of a triangle to intersect the other two sides in two points, then the other two sides are divided in the same ratio. 1 35. 3�� 2 5�� 3 = 2;�� 3 + �� 2 = 13 6
The given system of linear equation is 3�� 2 5�� 3 = 2 …. (i) �� 3 + �� 2 = 13 6 ……. (ii)
From (i) ⇒ 9�� 10�� = 12 … (iii)
From (ii) ⇒2��+3�� =13 (iv)
From equation (iii)
9�� 10�� = 12
9�� =10�� 12
�� = 10�� 12 9
Substituting the value of �� in equation (iv), we get 2(10�� 12 9 )+3�� =13
20�� 24+27�� =117
47�� =117+24
�� = 141 47 �� =3
Substituting the value of �� in equation (iv), we get
2��+3×3=13
2��+9=13
2�� =13 9
�� = 4 2 =2
Therefore, the solution is �� =2,�� =3
SECTION E
36. i.Zeroes of the polynomial are 0 and 5 1 ii.Maximum height achieved by ball =25× 5 2 5×(5 2)2 = 125 4 or 3125m
iii.a. 5t2 +25t=30
⇒t2 5t+6=0
⇒(t 2)(t 3)=0
t≠3,t=2
iii. b. 5t2 +25t =20
⇒t2 5t+4=0
⇒(t 4)(t 1)=0 ⇒t=4,1
37. i.Let the fixed charge be ₹ �� and per kilometre charge be ₹ ��.
∴��+10�� =105… (i)
��+15�� =155…...(ii)
From (i) and (ii)
5�� =50
y= 50 5 =10
From equation (i)
��+100=105
�� =105 100=5
Fixed charges = ₹ 5
ii.Let the fixed charge be ₹ �� and per kilometre charge be ₹ ��
∴��+10�� =105 (i)
��+15�� =155 ...(ii)
From (i) and (ii)
5�� =50 y= 50 5 =10
iii.Let the fixed charge be ₹ �� and per kilometre charge be ₹ ��
Amount to be paid for travelling a distance of 10 km is ��+10�� If �� =20 and �� =10
38. i. ��+�� =300...(i)
150��+250�� =55000 ...(ii)
ii.a. Solving equation (i) and (ii)
Substitute (i) in (ii):
150��+250(300 ��)=55,000
150��+75,000 250�� =55,000
100��+75,000=55,000
100�� = 20,000 �� =200
b. Solving equation (i)
TIME: 3 Hrs
General Instructions:
MATHEMATICS STANDARD
SAMPLE QUESTION PAPER - 5
CLASS – X (2025-26)
Read the following instructions carefully and follow them:
1. This question paper contains 38 questions.All Questions are compulsory
2. This question paper is divided into 5 SectionsA, B, C, D, and E.
MAX. MARKS: 80
3. In SectionA, Question numbers 1-18 are multiple-choice questions (MCQs) and questions number 19 and 20 areAssertion-Reason based questions of 1 mark each.
4. In Section B, Questions numbers 21-25 are very short answer (VSA) type questions, carrying 02 marks each.
5. In Section C, Questions numbers 26-31 are short answer (SA) type questions, carrying 03 marks each.
6. In Section D, Questions numbers 32-35 are long answer (LA) type questions, carrying 05 marks each.
7. In Section E, Questions numbers 36-38 are case study-based questions carrying 4 marks each with subparts of the values of 1, 1, and 2 marks respectively.
8. Draw neat and clean figures wherever required. Take �� = 22 7 wherever required if not stated.
9. Use of calculators is not allowed.
SectionA
SectionAconsists of 20 questions of 1 mark each.
1. In a cyclic quadrilateral ABCD, it is being given that ∠A=(x+y+10)∘ ∠B=(y+ 20)∘ ,∠C=(x+y 30)∘ and ∠D=(x+y)∘. Then, ∠B= ?
2. The discriminant of the quadratic equation ��2 4��+3=0 is:
2
3. What is the common difference of anAP in which ��18 ��14 =32 ? A) -8 B) 4 C) -4
8
4. In Figure, a circle touches the side DF of △EDF at H and touches ED and EF produced at K and M respectively. If EK=9cm, then the perimeter of △EDF is
5. In the given figure, DE‖BC. Find the value of ��.
√3
√5
6. For what value of ��, the product of zeroes of the polynomial ����2 4�� 7 is 2?
7. The 5th term of anAP is 20 and the sum of its 7th and 11th terms is 64. The common difference is
8. Thequadraticequation2��2 3��+2=0 has:
Real and Distinct roots
No Real roots
Real roots
Real and Equal roots
9. If ��(��) denotes the probability of an event ��, then
The median and mode respectively of a frequency distribution are 26 and 29. Then its mean is:
11. The HCF of two numbers is 27 and their LCM is 162. If one of the numbers is 54, what is the other number:
12. Find the modal class for the following distribution:
19. Assertion (A): The constant difference between any two terms of anAP is commonly known as common difference.
Reason (R): The common difference of 2, 4, 6, 8 thisA.P. is 2.
A) BothAand R are true and R is the correct explanation ofA.
B) BothAand R are true but R is not the correct explanation ofA.
C)Ais true but R is false.
D)Ais false but R is true.
20. Assertion (A): A sphere of radius 7 cm is mounted on the solid cone of radius 6 cm and height 8 cm. The volume of the combined solid is 1737.97cm3
Reason (R): Volume of sphere is 4 3 ����3 .
A) Both �� and �� are true and �� is the correct explanation ofA.
B) BothAand R are true but R is not the correct explanation ofA.
C)Ais true but R is false.
Section B
Section B consists of 5 questions of 2 marks each.
21. Find the LCM of the following polynomials: (��+1)2 and (2��2 +3��+1). 2
22. a. If �� and �� are the zeros of the quadratic polynomial ��(��)=
26.
b. The graph of the polynomial ��(��)=����2 +����+�� is as shown below (Fig.). Write the signs of ' �� ' and ��2 4����.
Section C consists of 6 questions of 3 marks each.
a. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun's altitude is 30∘, than when it is 60∘. Find the height of the tower.
b.Akite is flying, attached to a thread which is 165 m long. The thread makes an angle of 30∘ with the ground. Find the height of the kite from the ground, assuming that there is no slack in the thread.
28. Amomento is made as shown in the figure. Its base PBCR is silver plated from the front side. Find the area which is silver plated. (Use ��
30. Below figure shows the cross-section of railway tunnel. The radius OAof the circular part is 2 m. If ∠AOB=90∘, calculate i.the height of the tunnel
ii.the perimeter of the cross-section
iii.the area of the cross-section 3
31. If �� and �� are zeroes of the quadratic polynomial 4��2 +4��+1, then form a quadratic polynomial whose zeroes are 2�� and 2��. 3
Section D
Section D consists of 4 questions of 5 marks each
32. a.Arectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find the length and breadth of the park.
OR
b. If the roots of the quadratic equation (�� ��)(�� ��)+(�� ��)(�� ��)+(�� ��)(�� ��)=0 are equal. Then show that �� =�� =c
33. a. From the top of a building 15 m high, the angle of elevation of the top of a tower is found to be 30∘. Form the bottom of the same building, the angle of elevation of the top ofthetoweris foundto be 45∘.Determinetheheight ofthetowerandthedistancebetween the tower and the building.
OR
b.Aparachutist is descending vertically and makes angles of elevation of 45∘ and 60∘ at two observing points 100 m apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of the point where he falls on the ground from the just observation point.
34. 250 apples of a box were weighed and the distribution of masses of the apples is given in the following table: Mass (in g) 80 - 100 100 - 120 120 - 140 140 - 160 160 - 180 Apples 20 60 70 �� 60
i.Find the value of �� and the mean mass of the apples. ii.Find the modal mass of the apples. 5
35. An iron pillar has some part in the form of a right circular cylinder and remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high, and the conical part is 36 cm high. Find the weight of the pillar if one cubic cm of iron weighs 7.8 grams.
Section E
Section E consists of 3 case study-based questions of 4 marks each.
36. Read the following text carefully and answer the questions that follow:
Mary and John are very excited because they are going to go on a dive to see a sunken ship. The dive is quite shallow which is unusual because most sunken ship dives are found at depths that are too deep for two junior divers. However, this one is at 40 feet, so the two divers can go to see it.
They have the following map to chart their course. John wants to figure out exactly how far the boat will be from the sunken ship. Use the information in this lesson to help John figure out the following.
i.What are the coordinates of the boat and the sunken ship respectively?
ii.How much distance will Mary and John swim through the water from the boat to the sunken ship?
iii. a. If each square represents 160 cubic feet of water, how many cubic feet of water will Mary and John swim through from the boat to the sunken ship? OR
b. If the distance between the points (��, 1) and (3,2) is 5, then what is the value of �� ?
37. Read the following text carefully and answer the questions that follow:
Statue of a Pineapple: The Big Pineapple is a heritage-listed tourist attraction at Nambour Connection Road, Woombye, Sunshine Coast Region, Queensland,Australia. It was designed by Peddle Thorp and Harvey, Paul Luff, and Gary Smallcombe andAssociates. It is also known as Sunshine Plantation. It was added to the Queensland Heritage Register on
6 March 2009.
Kavita last year visited Nambour and wanted to find the height of a statue of a pineapple. She measured the pineapple's shadow and her own shadow. Her height is 156 cm and casts a shadow of 39 cm. The length of shadow of pineapple is 4 m.
i.What is the height of the pineapple?
ii.What is the height Kavita in metres?
iii. a. Write the type of triangles used to solve this problem. OR
b. Which similarity criterion of triangle is used?
38. Road SafetyAwareness Campaign:
To reduce road accidents, the traffic department conducted a Road SafetyAwareness Campaign in a city.
The campaign focused on different age groups, educating them about traffic rules, the importance of helmets, seat belts, and safe driving practices. The following table represents the number of participants in different age groups:
i.What percentage of total participants belong to the 30–40 years age group?
ii.Find the class mark for the age group that has 25 participants.
iii.Calculate the mode of the given data.
Calculate the mode of the given data
(OR)
Marking Scheme
SECTIONA
SECTION B
21. Step 1: Factor the polynomials completely (��+1)2 =(��+1)(��+1)
Factor the quadratic 2��2 +3��+1 :
Find two numbers whose product is 2×1=2 and sum is 3. These are 2 and 1.
Rewrite:
Step 2: List the factors
Step 3: Find LCM by taking the highest powers of all factors
• ��+1 appears as squared in (��+1)2
• 2��+1 appears only once in the second polynomial
Since the parabola cuts the ��-axis at two points, this means that the polynomial will have two real solutions.
Hence ��2 4���� >0
Hence a>0 and ��2 4���� >0
=
Multiplying and dividing by (
⇒ our assumption was wrong
⇒5+2√3 is an irrational number.
=20, Hence from (i) we get
b. LetAbe the position of the kite. Let O be the position of the observer and OAbe the thread.
30. Let OC be perpendicular toAB.
Radius OA=OB=2m
∠AOB=90∘
In ∠AOB, by pythagoras theorem
AB2 =OA2 +OB2
⇒AB2 =22 +22
⇒AB2 =8
⇒AB=√8=2√2m
Area of △
Now, the height from the baseAB to the top of the tunnel is from the midpoint ofAB vertically upwards to the circular arc.
The highest point is a vertical line through O up to the arc. This is just the radius (2 m). ButAB is not at the same level as O; it's below O.
The vertical distance from O down toAB is the length from O to the midpoint ofAB, which is ���� 2 vertically downward. But since angle at O is 90°, OAB forms a triangle below O.
However, the height of the triangle from O to AB is: Let ℎ be the perpendicular from O toAB. In a right triangle, the altitude from the right-angle vertex to the hypotenuse is
ℎ = ����×���� ���� . So,
ℎ = 2×2
2√2 =√2 m
since the highest point is at the top of the circle, and the base isAB (which is below O by h), the total height from AB to the top is:
Height = radius + ℎ=2+√2 m
ii.perimeter of cross- section =AB+ arc length of major arcAB
33.b. Let PC be the height ℎ of the parachutist and makes an angle of elevations between 45∘ and 60∘ respectively at two observing points 100 m apart from each other
Let AB=100
Let distance (BC)=��m
In △������ tan 60∘ = ���� ����
⇒√3= ℎ �� ⇒�� = ℎ √3 (1)
In △������
tan 45∘ = ���� ����
⇒1= ℎ 100+��
⇒ℎ=100+��
⇒ℎ=100+ ℎ √3 [From (1)]
⇒√3ℎ =100√3+ℎ [Multiply by √3 ]
⇒√3ℎ ℎ =100√3
⇒ℎ(√3 1)=100√3
⇒ℎ = 100√3 √3 1 × √3+1 √3+1 ----(2)
⇒ℎ= 100(3+√3) 3 1
⇒ℎ=50(3+1.732)
⇒ℎ=50×4732=2366m 1
Put value of ℎ from equation (2) in equation (1)
∴ Height of parachutist =2366m
of first point from Where he falls =136.6m
i.Given, total number of apples =250
Modal class is (120-140), Since it consists highest frequency
∴�� =120
ℎ = class size = 20
��1 = frequency of modal class =70
��0 = frequency of class preceding the modal class =60
��2 = frequency of class succeeding the modal class =40
On putting these values in (i), we get
Modal mass or mode
=120+( 70 60 2×70 60 40)×20
=120+ 10 40 ×20
=120+ 10 2
=120+5
= 125
35. Let us suppose that ��1 cm and ��2 cm denote the radii of the base of the cylinder and cone respectively. Then, ��1 =��2 =8cm
Let us suppose that ℎ1 and ℎ2 cm be the heights of the cylinder and the cone respectively. Then,
∴ Total volume of the iron = Volume of the cylinder + Volume of the cone
weight of the pillar = Volume × Weight per cm3
(22×64×36)×78gms
3953664gms=3953664kg
7 or -1
37. Kavita's height =156cm
Kavita's shadow =39cm
Pineapple's shadow =4m=400cm
(i)Height of the pineapple
Height of Pineapple
Shadow of Pineapple = Height of Kavita Shadow of Kavita
iii.a. Type of triangles used
The triangles formed (height vs shadow) are right-angled triangles (one angle = 90° at the ground, another angle common due to sunlight).
iii.b. ByAAsimilarity, the triangles are similar right triangles. 2
38. i. Identify the values and apply the percentage formula
Number of participants in the 30-40 years age group =36
Total number of participants =15+28+36+25+16=120
Percentage =( Participants in 30 40 years group Total participants )×100
Percentage =(36 120)×100
Percentage =30%
30% of the total participants belong to the 30 40 years age group.
ii.Find the class mark.
The frequency 25 corresponds to the age group 40-50 years.
Class Mark = Lower Limit + Upper Limit 2
Class Mark == 40+50 2 = 90 2 =45
The class mark for the frequency 25 is 45.
iii. a. From the given data:
�� =30 (Lower boundary of the modal class)
ℎ =10 (Class width)
��1 =36 (Frequency of modal class)
��0 =28 (Frequency of class before modal class)
��2 =25 (Frequency of class after modal class)
Find the mode.
Mode =��+( ��1 ��0 2��1 ��0 ��2)×ℎ
Mode =30+( 36 28 2(36) 28 25)×10≈34.2
Thus, the approximate mode is 34.2 years.
iii. b. From the given data:
�� =20 (Lower boundary of the modal class)
ℎ =10 (Class width)
��1 =36 (Frequency of modal class)
��0 =15 (Frequency of class before modal class)
��2 =28 (Frequency of class after modal class)
Find the mode.
Mode =��+( ��1 ��0 2��1 ��0 ��2)×ℎ
Mode =20+( 36 15 2(36) 15 28)×10≈2724
Thus, the approximate mode is 27.24 years.
TIME: 3 Hrs
General Instructions:
MATHEMATICS STANDARD
SAMPLE QUESTION PAPER - 6
CLASS – X (2025-26)
Read the following instructions carefully and follow them:
1. This question paper contains 38 questions.All Questions are compulsory
2. This question paper is divided into 5 SectionsA, B, C, D, and E.
MAX. MARKS: 80
3. In SectionA, Question numbers 1-18 are multiple-choice questions (MCQs) and questions number 19 and 20 areAssertion-Reason based questions of 1 mark each.
4. In Section B, Questions numbers 21-25 are very short answer (VSA) type questions, carrying 02 marks each.
5. In Section C, Questions numbers 26-31 are short answer (SA) type questions, carrying 03 marks each.
6. In Section D, Questions numbers 32-35 are long answer (LA) type questions, carrying 05 marks each.
7. In Section E, Questions numbers 36-38 are case study-based questions carrying 4 marks each with subparts of the values of 1, 1, and 2 marks respectively.
8. Draw neat and clean figures wherever required. Take �� = 22 7 wherever required if not stated.
9. Use of calculators is not allowed.
SectionA
SectionAconsists of 20 questions of 1 mark each.
1. Find the area of the segment if the area of the sector is 44 m2 and the part of a triangle in the sector is 12 m2 .
2. Apart of monthly expenses of a family on milk is fixed which is ₹ 700 and remaining varies with quantity of milk taken extra at the rate of ₹ 25 per litre. Taking quantity of milk required extra as �� litres and total expenditure on milk as ₹ ��, write a linear equation from the above information. A) 25��+��=700 B) 20��+10�� =300 C) 20��+�� =500 D) ��+25�� =900 1
3. The value of k for which the system of equations kx+2y=5 and 3x+4y=1 have no solution, is
4. The height of a tower is 20 m. The length of its shadow made on the level ground when the Sun's altitude is 60∘, is:
5. The HCF of 95 and 152, is
6. If the two zeroes of a quadratic polynomial are ±√5, then the quadratic polynomial is:
��2 +5
7. The common difference of anA.P. in which ��20 ��15 =20, is
If �� tan
sin
, then �� is equal to
9. Number of zeroes of the polynomial ��(��) shown in the Figure, are:
10. In the given figure, △ABC∼△DCB, then AB
11. In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT=30∘, then AB:AT is
12. If the points (6,1),(8,2),(9,4) and (��,3), taken in order are the vertices of a parallelogram, then the value of ' �� ' is
13. In the given figure, △ABC∼△QPR. If AC=6cm,BC=5cm,QR=3cm and PR= x; then the value of x is:
14. Adice is thrown once. The probability of getting an odd number is
15. sec4 A sec2 A is equal to
16. If �� and �� are the zeroes of the polynomial 2��
17. The median group in the following frequency distribution is:
18. The value of a so that the point (3,A) lies on the line represented by 2�� 3�� =5 is A) 1 3 B) 1 3 C) -1 D) 1
19. Assertion (A): PointAis on the y -axis at a distance of 4 units from the origin. If the coordinates of the point B are ( 3,0 ), then the length ofAB is 5 units.
Reason (R): Distance between points A(��1,��1) and B(��2,��2) is
2 ��1)2 +(��2 ��1)2
A)Both A and R are true and R is the correct explanation ofA.
B)Both A and R are true but R is not the correct explanation ofA.
C)Ais true but R is false.
D)Ais false but R is true. 1
20. Assertion (A): The arithmetic mean of the following given frequency distribution table is 13.81.
Reason (R): �� =
A)Both A and R are true and R is the correct explanation ofA.
B)Both A and R are true but R is not the correct explanation ofA.
C)Ais true but R is false.
D)Ais false but R is true.
Section B
Section B consists of 5 questions of 2 marks each.
21. One card is drawn from a well-shuffled deck of 52 playing cards. Find the probability of getting the following:
i.a king of red colour
ii.a red face card
OR
From a well-shuffled deck of 52 playing cards, all black queens and red kings are removed. One card is selected at random from the remaining cards. Find the probability that the selected card is: i.an ace. 2
ii.a jack of red colour.
iii.a king of spade.
22. Find a point on the ��-axis which is equidistant from the points ��(6,5) and ��( 4,3)
23. Prove that 11+3√2 is an irrational number, given that √2 is an irrational number. 2
24. 5 books and 7 pens together cost Rs. 79 whereas 7 books and 5 pens together cost Rs.77.
25. The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2:3. Determine the fraction.
Section C
Section C consists of 6 questions of 3 marks each.
26. Aplane left 30 minutes later than the schedule time and in order to reach its destination 1500 km away in time it has to increase its speed by 250km/hr from its usual speed.
Aman sold a chair and a table together for ₹1520 thereby making a profit of 25% on the chair and 10% on table. By selling them together for ₹1535 he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each.
31. Find the missing frequencies in the following frequency distribution table, if N=100 and median is 32.
Find the mean marks per student, using assumed-mean method:
Section D
Section D consists of 4 questions of 5 marks each
32. The mode of the following frequency distribution is 55. Find the missing frequencies �� and ��.
Following is the distribution of I.Q. of 100 students. Find the median I.Q.
33. In anA.P., the nth term is
and the
34. If the price of a book is reduced by ₹ 5, a person can buy 4 more books for ₹ 600. Find the original price of the book. OR
The difference of squares of two numbers is 204. The square of the smaller number is 4 less than 10 times the larger number. Find the two numbers.
35. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30∘, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60∘. Find the further time taken by the car to reach the foot of the tower from this point.
OR
An aeroplane when flying at a height of 3000 metres from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are 60∘ and 45∘ respectively. Find the vertical distance between the aeroplanes at that instant. [Take √3=1.73 ]
Section E
Section E consists of 3 case study-based questions of 4 marks each.
36. Read the following text carefully and answer the questions that follow:
Swimmer in Distress:Alifeguard located 20 metre from the water spots a swimmer in distress. The swimmer is 30 metre from shore and 100 metre east of the lifeguard. Suppose the lifeguard runs and then swims to the swimmer in a direct line, as shown in the figure.
i.How far east from his original position will he enter the water? (Hint: Find the value of
x in the sketch.)
ii.Which similarity criterion of triangle is used?
iii.What is the distance of swimmer from the shore?
37. Read the following text carefully and answer the questions that follow:
For the inauguration of 'Earth Day' week in a school, badges were given to volunteers. Organisers purchased these badges from an NGO, who made these badges in the form of a circle inscribed in a square of side 8 cm.
i.What is the area of square �������� ?
ii.What is the length of diagonalAC of squareABCD?
iii.Find the area of sector OPRQO.
Find the area of remaining part of squareABCD when area of circle is excluded.
O is the centre of the circle and ∠AOB=90∘ :
38. Read the following text carefully and answer the questions that follow: Saving money is a good habit and it should be inculcated in children from the beginning. Mrs. Pushpa brought a piggy bank for her childAkshar. He puts one five-rupee coin of his savings in the piggy bank on the first day. He increases his savings by one five-rupee coin daily.
i.If the piggy bank can hold 190 coins of five rupees in all, find the number of days he can contribute to put the five-rupee coins into it.
ii.Find the total money he saved.
iii.How much moneyAkshar saves in 10 days?
Marking Scheme
21. Total number of playing cards in a well-shuffled deck =52 i.Total number of kings of red colour =2 (hearts (∨), diamonds (∙) )
ii.Total number of red face cards = 6 (Kings, queens and jacks are called face cards)
Probability of getting a red face card = Favourabe
i. ��( an ace )= 4 48 or 1 12
ii. P( jack of red colour )= 2 48 or 1 24
iii. P( king of spade )= 1 48
22. We have to find a point on the ��-axis which is equidistant from the points ��(6,5) and ��( 4,3)
We know that a point on y-axis is of the form (0,��). So, let the required point be P(0,��)
Then,
Solve equation and find the value of ��
So, the required point is (0,9).
23. Let us assume that 11+3√2 be a rational number.
⇒11+3√2= �� ��, where �� and �� are integers, ��≠0
⇒√2= �� 11�� 3��
RHS is a rational number but LHS is irrational.
∴ Our assumption was wrong. Hence, 11+3√2 is an irrational number.
24. Let the cost of 1 book be Rs.�� and that of 1 pen be Rs.��.
Then, according to the question,
5��+7�� =79… (1) and 7��+5�� =77...(2)
Let us draw the graphs of the equation (1) and (2) be finding two solutions. These two solutions of the equations (1) and (2) are given below in table 1 and table 2 respectively.
For equation (1) 5��+7�� =79
Table 1 of solutions
For equation (2) 7��+5�� =77
⇒5��=77 7�� ⇒��= 77 7�� 5
Table 2 of solutions
We plot the points A(6,7) and B( 8,13) on a graph paper and join these points to form the line AB representing the equation (1) as shown in the figure.Also, we plot the points C(1,14) and D( 4,21) on the same graph paper and join these points to form the line CD representing the equation (2) as shown in the same figure.
In the figure, we observe that the two lines intersect at the points A(6,7). So, �� =6 and ��=7 is the required solution of the pair of linear equation formed, i.e., the cost of 1 book is Rs. 6 and of 1 pen is Rs.7.
Therefore the cost of 1 book and 2 pens =6+2×7= Rs. 20.
25. Let the numerator and denominator of fraction be �� and �� respectively.
Then, the fraction is �� �� .
As per first condition
The sum of the numerator and denominator of a fraction is 4 more than twice the numerator.
��+��=2��+4
⇒ ��+�� =4 (i)
According to the second condition,
If the numerator and denominator are increased by 3, they are in the ratio 2:3.
��+3
��+3 = 2 3
⇒3��+9=2��+6 ----(ii)
⇒3�� 2�� = 3
Multiply (i) by -2, we get 2��+2�� =8 (iii)
Adding (ii) and (iii), we get and 3�� 2�� = 3+8
⇒�� =5
Substituting �� =5 in (i), we get 5 ��=4
�� =9
Hence, the required fraction is 5 9 1
SECTION C
26 Let usual speed =��km/hr
New speed =(��+250)km/hr
Total distance = 1500 km
Time taken by usual speed = 1500 �� hr
Time taken by new speed = 1500 ��+250 hr 1
According to question,
1500
�� 1500 ��+250 = 1 2
⇒ 1500��+1500×250 1500�� ��2 +250�� = 1 2
⇒��2 +250�� =750000
⇒��2 +250�� 750000=0
⇒��2 +1000�� 750�� 750000=0
⇒��(��+1000) 750(��+1000)=0
⇒�� =750 or �� = 1000
Therefore, usual speed is 750km/hr, 1000 is neglected.
∵ Time taken by the car to travel a distance PQ=6 seconds.
∴ Time taken by the car to travel a distance BQ, i.e. 1 2PQ= 1 2 ×6=3 seconds.
Hence, the further time taken by the car to reach the foot of the tower is 3 seconds. OR
Let C and D is the position of two aeroplanes. The height of the aeroplane which is at point D is 3000 m and it passes another aeroplane vertically which is at point C.
Let BC=��m. It is also given that the angles of elevation of two planes from the pointAon the ground is 45∘ and 60∘ respectively.
In right triangle ������, we have
In right triangleABD , we have
Comparing (i) and (ii), we get
Hence, vertical distance between the aeroplane =CD=BD BC
Hence, the required distance between the two aeroplanes is =1267.8m
ii.Total money she saved =5+10+15+20+⋯=5+10+15+20+⋯ upto 19 terms = 19 2 [2×5+(19 1)5] = 19 2 [100]= 1900 2 =950 and total money she shaved = ₹ 950 1
iii.Money saved in 10 days
���� = �� 2[2��+(�� 1)��]
⇒��10 = 10 2 [2×5+(10 1)×5]
⇒��10 =5[10+45}]
⇒S10 =275
Money saved in 10 days = ₹ 275 2 OR
Number of coins in piggy bank on 15th day
���� = �� 2[2��+(�� 1)��]
⇒��15 = 15 2 [2×1+(15 1)×1]
⇒��15 = 15 2 [2+14]
⇒S15 =120
So, there are 120 coins on 15th day.
MATHEMATICS STANDARD
SAMPLE QUESTION PAPER - 7
CLASS – X (2025-26)
TIME: 3 hours
General Instructions:
1. This question paper contains 38 questions.All Questions are compulsory
2. This question paper is divided into 5 SectionsA, B, C, D, and E.
MAX. MARKS: 80
3. In SectionA, Question numbers 1-18 are multiple-choice questions (MCQs) and questions number 19 and 20 areAssertion-Reason based questions of 1 mark each.
4. In Section B, Questions numbers 21-25 are very short answer (VSA) type questions, carrying 02 marks each.
5. In Section C, Questions numbers 26-31 are short answer (SA) type questions, carrying 03 marks each.
6. In Section D, Questions numbers 32-35 are long answer (LA) type questions, carrying 05 marks each.
7. In Section E, Questions numbers 36-38 are case study-based questions carrying 4 marks each with subparts of the values of 1, 1, and 2 marks respectively.
8. Draw neat and clean figures wherever required. Take �� = 22 7 wherever required if not stated.
9. Use of calculators is not allowed.
A
Questions no. 1 to 18 are multiple choice questions (MCQs) of 1 mark each. 1. The exponent of 3 in the prime factorization of 864 is:
1 2. If the polynomial 3��3 4��2 17�� �� is exactly divisible by �� 3, then the value of ‘��’is
3. For the pair of linear equations:
3�� + 4�� = 10
6�� + 8�� = 20
What is the nature of their graph?
A)Intersecting lines B)Coincident lines C)Parallel lines D)None of the above 1
Section
4. The value of �� for which the system of equations 3�� ���� = 7 and 6�� + 10�� = 3 is inconsistent, is
5. The discriminant of the equation (2�� + ��)�� = ��2 +2���� is ______
6.
If the angles of a right-angled triangle are inA.P. then the angles of that triangle will be A) 45°,45°,90 B) 30°,60°,90° C) 40°,50°,90°
20°,70°,90°
7. In an isosceles triangleABC, ifAC = BC andAB2 = 2AC2 then the measure of ∠C is
8. If the points (��,��), (1,2) and (7,0) are collinear, then the relation between ‘ �� ’ and ‘ �� ’ is given by
and
, then the value of (��+��) is
11. The _____________ is the line drawn from the eye of an observer to the point in the object viewed by the observer A)Horizontal lineB)Line sightC)None of theseD)Vertical line
12. In the given figure, if QP = 4.5 cm, then the measure of QR is equal to
13. The part of the circular region enclosed by a chord and the corresponding arc of a circle is called
Test tube available in a chemistry lab is a combination of A) a hemisphere and a cylinderB)a hemisphere and a cone C) a cone and a cylinder D)None of these
15. Arectangular piece of paper is 44 cm long and 18 cm wide. If a cylinder is formed by rolling the paper along its length, then the radius of the base of the cylinder is
17. If the mean of ��,��,�� is ��, then what is the mean of �� and ��
18. Acard is drawn at random from a well shuffled deck of 52 cards. The probability that it will be a spade or a king is
Assertion-Reason Type Questions
In Question 19 and 20, anAssertion (A) statement is followed by a statement of Reason (R). Select the correct option out of the following :
(A)BothAssertion (A) and Reason (R) are true and Reason (R) is the correct explanation ofAssertion (A).
(B)BothAssertion (A) and Reason (R) are true but Reason (R) is not the correct explanation ofAssertion (A).
(C)Assertion (A) is true but Reason (R) is false.
(D)Assertion (A) is false but Reason (R) is true.
19. Assertion (A): The perimeter of ABC, with AB = 2 cm, ∠�� =90°, and BC = 3 cm is a rational number. Reason (R): The sum of the squares of two rational numbers is
20. Assertion (A): Point P (0,2) is the point of intersection of ��-axis with 3��+2�� =4 Reason (R): The distance of point P(0,2) from ��-axis is 2 units.
of
of
Find the value of ′��′ if the points (7, –2), (5, 1), (3, ��) are collinear.
24. a.Acone and a sphere have equal radii and equal volume.What is the ratio of the diameter of the sphere to the height of cone?
b. The largest sphere is to be curved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere.
25. It is known that a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non-defective bulb?
Section C
Section C consists of 6 questions of 3 marks each.
26. Atwo-digit number is such that the sum of the digits is 10, and the difference between the digits is 4. Find the number.
27. If –4 is a root of the quadratic equation ��2 + ���� 4 = 0 and the quadratic equation ��2 + ���� + �� = 0 has equal roots. Find the value of ��.
28. a. Determine anA.P. whose third term is 9 and when fifth term is subtracted from 8th term, we get 6
b. If the sum of Rs 1890 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 50 less than its preceding prize. Then find the value of each of the prizes.
29. In the given figureA, B and C are points on OP, OQ and OR respectively such that AB||PQ andAC||PR. Prove that BC||QR.
30. Prove the identity 2(sin6 ��+cos6 ��) 3(sin4 ��+cos4 ��)+1=0
InABC right angles at B, if tanA= 1 √3 find value of: ��������.�������� + ��������.��������
31. In figure, O is the centre of a circular arc andAOB is a straight line. Find the perimeter and the area of the shaded region correct to one decimal place. (Take �� = 3.142)
Section D
Section D consists of 4 questions of 5 marks each
32. A motorboat, whose speed is 15 km/hr in still water, goes 30 km downstream and comes back in a total time of 4 hr 30 minutes, find the speed of the stream.
33. ABCD is a trapezium in whichAB || DC and P and Q are points on AD and BC, respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, findAD.
34.
35.
a. If the points ��(��, 11) , ��(5,��) , ��(2,15) and ��(1,1) are the vertices of a parallelogram ��������, find the values of �� and ��
b. PointAis on ��-axis, point B is on ��-axis and the point �� lies on line segment ����, such that ��(4,5) and ����:����=5:3. Find the coordinates of point �� and ��.
a.An iron pillar has some part in the form of a right circular cylinder and remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high, and the conical part is 36 cm high. Find the weight of the pillar if one cubic cm of iron weighs 7.8 grams.
b.Asolid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and total surface area of the solid (Use�� = 22 7 )
Section E
Section E consists of 3 case-study based questions of 4 marks each.
36. There is fire incident in the house. The house door is locked so, the fireman is trying to enter the house from the window. He places the ladder against the wall such that its top reaches the window as shown in the figure.
i.If window is 6 m above the ground and angle made by the foot of ladder to the ground is 30°, then find the length of the ladder?
ii.If the height of the window is 8 m above the ground and angle of elevation is observed to be 45°, then find the horizontal distance between the foot of ladder and wall?
iii. a. If the fireman gets a 9 m long ladder and window is at 6 m height, then how far should the ladder be placed?
b. If the fireman gets a 27 m long ladder and window is at 12 m height, then how far should the ladder be placed?
37 Avantika joins four cubical open boxes of edge 20 cm each to make a pot for planting saplings of pudina in her kitchen garden. The saplings are cylindrical in shape with diameter 14.2 cm and height 11 cm.
i.IfAvantika wants to paint the outer surface of the pot, then how much area she needs to paint?
ii.What is the volume of the pot formed?
iii. a. IfAvantika decorates the four walls of the pot with coloured square paper of side 10 cm each, then how many pieces of papers would be required?
b. IfAvantika decorates the four walls of the pot with coloured square paper of side 20 cm each, then how many pieces of papers would be required?
38. 100 M RACE:
Astopwatch was used to find the time that it took a group of students to run 100 m.
i.What will be the upper limit of the modal class?
ii. a. How many students finished the race within 1 minute? OR
b. How many students finished the race within 40 seconds?
iii.Estimate the mean time taken by a student to finish the race.
Product of the zeroes:
���� =(√3+√5)(√5 √3)
Using the identity (��+��)(�� ��)=��2 ��2: =(√5)2 (√3)2 =5 3=2
Form the quadratic polynomial.
Aquadratic polynomial with roots �� and �� is given by: ��2 ( sum of roots )��+( product of roots )=0
Substituting values: ��2 (2√5)��+2=0
Thus, the required quadratic polynomial is:
2 2√5��+2
22. Using the formula for the general term of anA.P
�� =��+(�� 1)�� Given: ��25 ��20 =45
Substituting the formula: [��+24��] [��+19��]=45 Simplifies to:
23. Formation of equation (7, 2),(5,1),(3,��) Area of the triangle
= 1 2[8 2��]=4 ��
To find ��
If the points are collinear, then area of the triangle =0 ⇒4 �� =0 ⇒�� =4
24. a. Find relation between radius and height of the cone.
Let the radius of both sphere & cone be r
Let the height of the cone be ℎ.
Volume of sphere = 4 3 ����3 and volume of cone = 1 3 ����2ℎ
Out of 600 electric bulbs one bulb can be chosen in 600 ways.
Total number of elementary events =600
There are 588(=600 12) non-defective bulbs out of which one bulb can be chosen in 588 ways.
Favourable number of elementary events =588
Find the required probability.
Hence, P(Getting a non-defective bulb)=
Section C consists of 6 questions of 3 marks each.
26. Form equations from given conditions
Let the �� be the tens digit of two-digit number and �� be the ones digit of two-digit number
Sum of digit is 10
So, ��+��=10 …(1)
Difference between digits is 4
�� =4 ……(2)
Adding Equation (1) and Equation (2) to get �� and �� (��+��)+(�� ��)=10+4 2�� =14 �� =7
Substituting �� =7 into Equation (1) to find ��
27. Find �� using the first equation
Since 4 is a root of ��2 +���� 4=0, it must satisfy the equation: ( 4)2 +��( 4) 4=0
��=3
Use the condition for equal roots in the second
For the equation ��2 +����+�� =0 to have equal roots, the discriminant (Δ) must be
zero:
Δ=��2 4�� =0
Substituting �� =3 :
32 4�� =0
9 4�� =0
4�� =9 �� = 9 4
28. a. Using the formula for the general term
The general term of anA.P. is given by:
���� =��+(�� 1)��
Given that the third term is 9:
��+2�� =9
Given that the difference between the eighth and fifth terms is 6:
(��+7��) (��+4��)=6
3�� =6
Solving for ��
From equation 2:
�� = 6 3 =2 1
Find �� and writing theA.P
Substituting �� =2 into Equation 1:
��+2(2)=9
��+4=9
�� =5
TheA.P. is 5,7,9,11,13,….
b.Defining the given information
Let the highest prize be ��, and the prizes form an arithmetic progression (A.P.) with a common difference of �� = 50
Number of prizes �� =7.
Total sum of prizes:
��7 =1890
Formula for the sum of anA.P.:
���� = �� 2[2��+(�� 1)��]
1890= 7 2[2��+6( 50)]
Solving for �� and find the sequence
Simplify the equation:
1890= 7 2[2�� 300]
1890×2=7(2�� 300)
3780=14�� 2100
14�� =5880
�� =420
So, �� =420 and �� = 50
Series will be 420, 370, 320,….,120
The prizes are Rs. 420, 370, 320, 270, 220, 170, and 120.
29. Given:
equation (i) and (ii),
30. a. Simplify LHS
b. Figure and findAC
Consider a triangleABC in which ∠B=90∘
Let BC=�� and AB=√3��
Using Pythagoras theorem
For ∠C, Base =BC, Perpendicular =AB and Hypotenuse =AC
Put all the value to solve given expression
31. Find radius of semi-circle.
AC=12cm,BC=16cm
In △ACB, by Pythagoras theorem
AB2 =AC2 +BC2
⇒����2 =122 +162
FindArea of shaded region.
∴Area of shaded region =Area of semi-circle Area of △ACB
Find perimeter of shaded region. Perimeter of shaded region
=AC+BC+ circumference of semi-circle
=12+16+��(10)
=28+3.142×10
=28+3142=5942cm
Section D
Section D consists of 4 questions of 5 marks each
32. Formation of equation:
Let the speed of the stream be ��km/h
Speed of the boat in still water =15km/h
Speed downstream =15+��km/h
Speed upstream =15 ��km/h
Distance one way =30km Total time for the journey =4hr30min= 9 2 hours
Solve of ��
Find the LCM of the denominators:
��)+30(15+��) (15+��)(15 ��) = 9 2
Expanding the numerator:
1800=2025 9��2
9��2 =225
��2 =25
�� =5
Since speed cannot be negative, we take �� =5km/h.
33. Figure and Formation of given things:
In trapeziumABCD
AB||CD (Given)
PQ||DC (Given)
And PD = 18 cm, BQ = 35 cm and QC = 15 cm
To find:AD
For finding
∴AB||CD||PQ (i)
In △������, OQ || CD [From (i)]
∴ ���� ���� = ���� ���� (ii)[By BPT]
Similarly, in △������, PO ||AB [From (i)]
∴ ���� ���� = ���� ���� (iii)[By BPT]
TofindAD
From (ii) and (iii) ����
⇒ ���� 18 = 35 15 ⇒AP= 35 15 ×18=7×6
a. Figure and Formation of given things
LetA(��, 11),B(5,��),C(2,15) andD(1,1) bethegivenpoints.Weknowthatdiagonals of parallelogram bisect each other. 2.5
Therefore, Coordinates of mid-point of AC= Coordinates of mid-point of BD (��+2 2 , 15 11 2 )=(5+1 2 , ��+1 2 )
⇒ ��+2 2 =3 and ��+1 2 =2
⇒��+2=6 and ��+1=4
⇒�� =6 2 and �� =4 1
⇒�� =4 and �� =3
Hence value of �� and �� is equal to 4 and 3 respectively. 2.5
b. Figure and given
Let coordinates ofAare (��,0) and coordinates of B are (0,��)
To find �� coordinate
Using section formula, we get
4= 5×0+3×�� 5+3
⇒32=3�� ⇒�� = 32 3
To find �� coordinate Similarly,
5= 5×��+3×0
5+3
⇒40=5��
⇒��=8
∴ Coordinate ofAare (32 3 ,0) and coordinates of B are (0,8).
35. a. Draw figure and assume all the required dimension.
Let us suppose that ��1 cm and ��2 cm denote the radii of the base of the cylinder and cone respectively. Then,
��1 =��2 =8cm
Let us suppose that ℎ1 and ℎ2 cm be the heights of the cylinder and the cone respectively.
and ℎ2 =36cm
Find the volume of the cylinder and volume of the cone.
∴ Volume of the cylinder =����1 2ℎ1 cm3
=(��×8×8×240)cm3
=(��×64×240)cm3 Now, Volume of the cone = 1 3
Find the total volume of the iron
∴ Total volume of the iron = Volume of the cylinder + Volume of the cone
=(��×64×240+ 1 3 ��×64×36)cm3
=��×64×(240+12)cm3
= 22 7 ×64×252cm3
=22×64×36cm3
Find the total weight of the pillar.
Total weight of the pilar = Volume × Weight per cm3
=(22×64×36)×78 gms
=3953664 gms
=395.3664 kg
b. Draw figure and write all the given dimension
Diameter of the cylinder =7cm
Therefore radius of the cylinder = 7 2 cm
Total height of the solid =19cm
Therefore, Height of the cylinder portion =19 7=12cm
Also, radius of hemisphere = 7 2 cm
Find volume of the solid figure.
Let V be the volume and S be the surface area of the solid. Then,
V= Volume of the cylinder + Volume of two hemispheres
⇒ V={22 7 ×(72)2 ×(12+ 4 3 × 7 2)}cm3
= 22 7 × 7 2 × 7 2 × 50 3 cm3 =64166cm3
Find the surface area of the solid figure.
S= Curved surface area of cylinder + Surface area of two hemispheres
⇒ S=(2����ℎ+2×2����2)cm2
⇒ S=2����(ℎ+2��)cm2
⇒ S=2× 22 7 × 7 2 ×(12+2× 7 2)cm2
=(2× 22 7 × 7 2 ×19)cm2
=418cm2 2
Section E
Section E consists of 3 case-study based questions of 4 marks each.
36. i.Find the length of ladder
Let ���� be the length of the ladder In right-angled △������, ⇒sin 30
ii.Find the distance between the foot of ladder and wall? 1
Let ���� be the horizontal distance between the foot of ladder and wall.
In right-angled △������. tan45°= ���� ����
⇒1= 8 AB
⇒AB=8m
iii a. Find the distance where ladder will be placed?
Let the required distance be ��
In right-angled △������ (9)2 =��2 +(6)2
⇒ 81 36=��2 ⇒ 45=��2 ⇒ �� =3√5��
iii b. Find the distance where ladder will be placed?
Let the required distance be ��
In right-angled △������ (27)2 =��2 +(12)2
⇒ 729 144=��2 ⇒ 585=��2 ⇒ �� = √585��
37. i.Find the area to be paint.
Given, edge of each cubical box (��)=20cm
∴Area to be painted = 14 ×Area of square face =14��2 =14(20)2 =5600cm2
ii.Find volume of pot.
Length of the pot (��)=20cm
Breadth of the pot (��)=20×4=80cm and height of the pot (ℎ)=20cm
Volume of pot =��×��×ℎ =20×80×20
=32000cm3
iii.Find the area of four walls.
Area of four walls =2(��+��)×ℎ
=2(20+80)×20=4000cm2
a. Find the required number of pieces of paper.
Given, side of coloured square paper =10cm
Now, area of a square paper =(10)2 =100cm2
∴ Number of pieces of paper
b. Find the required number of pieces of paper.
Given, side of coloured square paper =20cm
38. i.Identify the modal class
Modal class = 40 – 60
Upper limit = 60
Hence, the upper limit of the modal class is 60.
ii. a. Find the sum of the students who finished the race in 1 minute. Number of students who finished the race within 1 minute = 8 + 10 + 13 = 31 (OR)
b. Find the sum of the students who finished the race in 40 seconds. Number of students who finished the race within 40 seconds = 8 + 10 = 18
iii.Find ∑�� and ∑����
TIME: 3 hours
General Instructions:
MATHEMATICS STANDARD
SAMPLE QUESTION PAPER - 8
CLASS – X (2025-26)
MAX. MARKS: 80
1. This question paper contains 38 questions.All Questions are compulsory
2. This question paper is divided into 5 SectionsA, B, C, D, and E.
3. In SectionA, Question numbers 1-18 are multiple-choice questions (MCQs) and questions number 19 and 20 areAssertion-Reason based questions of 1 mark each.
4. In Section B, Questions numbers 21-25 are very short answer (VSA) type questions, carrying 02 marks each.
5. In Section C, Questions numbers 26-31 are short answer (SA) type questions, carrying 03 marks each.
6. In Section D, Questions numbers 32-35 are long answer (LA) type questions, carrying 05 marks each.
7. In Section E, Questions numbers 36-38 are case study-based questions carrying 4 marks each with subparts of the values of 1, 1, and 2 marks respectively.
8. Draw neat and clean figures wherever required. Take �� = 22 7 wherever required if not stated.
9. Use of calculators is not allowed.
SectionA
Questions no. 1 to 18 are multiple choice questions (MCQs) of 1 mark each.
1. The smallest number of 5 digits exactly divisible by 12, 15, 118 and 27 is
The number of zeroes that the polynomial ��(��)= (�� 2)2 4 can have
3. If ‘2’is the zero of both the polynomials 3��
the value of �� 2�� is
4. 5 notebooks and 7 pens together cost ₹ 85, whereas 7 notebooks and 5 pens together cost ₹ 95. Find the cost of one notebook.
5. Aquadratic equation ����2 +����+�� =0 has real and distinct roots, if
6. The first and last terms of anA.P. are 1 and 11. If their sum is 36, then the number of terms will be
The 17th term of anAP exceeds its 10th term by 7, then the common difference is
8. In the given figure XY || BC. IfAX = 3 cm, XB = 1.5 cm and BC = 6 cm, then XY is equal to
9. If the distance between the points (��, 5) and (2,7) is 13 units, then the value of ‘ �� ’ is
10. The distance between the points (��1,��1) and (��2,��2) is given by
The value of cosec
12. Apole 10 m high cast a shadow 10 m long on the ground, then the sun’s elevation is
13. In the given figure, ifAQ = 4cm, QR = 7 cm, DS = 3 cm, then �� is equal to
The angle described by the minute hand between 4:00 pm and 4:25 pm is
15. The volume of the cuboid whose length, breadth and height are 12cm, 8cm and 6cm is
16. The longest rod that can be placed inside the cube is
In the formula ��
for finding the mean of grouped frequency distribution,
18. The probability that a leap year will have 53 Sundays or 53 Mondays is
Assertion-Reason Type Questions
In Question 19 and 20, anAssertion (A) statement is followed by a statement of Reason (R). Select the correct option out of the following :
(A)BothAssertion (A) and Reason (R) are true and Reason (R) is the correct explanation ofAssertion (A).
(B)BothAssertion (A) and Reason (R) are true but Reason (R) is not the correct explanation ofAssertion (A).
(C)Assertion (A) is true but Reason (R) is false.
(D)Assertion (A) is false but Reason (R) is true.
19. Assertion (A): If the Mean and the Median of a distribution are 169 and 170 respectively, then its Mode is 172.
Reason(R): The relation between Mean, Median and Mode is Mode =3 Median 2 Mean.
20. Assertion (A): The probability of randomly drawing a card with an even number from a box containing cards numbered 1 to 100 is 1 2 .
Reason (��):P(event)
Section B
Section B consists of 5 questions of 2 marks each.
21. Three years ago, Rashmi was thrice as old as Nazma.Ten years later, Rashmi will be twice as old as Nazma. How old are Rashmi and Nazma now
22. a. If tanA= cot B, prove that ��+�� =90°.
In the given figure, find the perimeter ofABC, ifAP = 10 cm.
24. The area enclosed between the concentric circles is 770 cm2. If the radius of the outer circle is 21 cm, find the radius of the inner circle.
25. Find the mean of the following distribution.
Section C consists of 6 questions of 3 marks each.
26. a. Find the values of �� for which the given equation has real and equal roots:
b. Sum of the areas of two squares is 400 cm2. If the difference of their perimeters is 16 cm, find the sides of the two squares.
27. In a quadrilateral ABCD, P, Q, R, S are the mid-points of the sides AB, BC, CD and DA respectively. Prove that PQRS is a parallelogram.
28. Show that four points (0, 1),(6,7),( 2,3) and (8,3) are the vertices of a rectangle. Also, find its area. 3
29. a.At the foot of a mountain the elevation of its summit is 45°.After ascending 1000 m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain. OR
b.Aflagstaff stands on the top of a 5 m high tower. From a point on the ground, the angle of elevation of the top of the flagstaff is 60° and from the same point, the angle of elevation of the top of the tower is 45°. Find the height of the flagstaff.
30. Compare the modal ages of two groups of students appearing for an entrance test:
31. In a game, the entry fee is Rs 5. The game consists of tossing a coin 3 times. If one or two heads show, Shweta gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose.
For tossing a coin three times, find the probability that she i.loses the entry fee.
ii.gets double entry fee.
iii.just gets her entry fee.
Section D
Section D consists of 4 questions of 5 marks each
32. The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle.
33. a. If the sum of the first �� terms of anA.P. is 4�� ��2, what is the first term? What is the sum of first two terms? What is the second term? Similarly, find the third, the tenth and the ��th terms.
34.
b. The ratio of the 11th term to 17th term of an A.P. is 3:4. Find the ratio of 5th term to 21st term of the sameA.P.Also, find the ratio of the sum of first 5 terms to that of first 21 terms.
a. In a trapezium ABCD, AB | | DC and DC = 2AB. EF | | AB, where E and F lie on BC andAD respectively such that. Diagonal DB intersects EF at G. Prove that 7EF = 11AB. OR
b. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆ PQR. 5
35. In the given figure,ABCD is a rectangle withAB = 80 cm and BC = 70 cm, ∠������ = 90° and DE = 42 cm.Asemicircle is drawn, taking BC as diameter. Find the area of the shaded region.
E
Section E consists of 3 case-study based questions of 4 marks each.
36 Two friends Soniya and Deepak have some saving in their piggy bank. They decided to count the total coins they both had.After counting they find they have fifty 1 coins, fortyeight 2 coins, thirty-six 5 coins, twenty-eight 10 coins and eight 20 coins. Now, they said toAnnu, another friend, to choose a coin randomly.
Section
i.Find the probability that the coin chosen is Rs 5 coin.
ii.Find the probability that the of at least Rs 10
iii. Find the probability that not a Rs 10 coin
Find the probability that not a Rs 20 coin.
37 In a park, four poles are standing at positionsA, B, C and D around the fountain such that the cloth joining the polesAB, BC, CD and DAtouches the fountain at P, Q, R and S respectively as shown in the figure.
i. a. If DR = 7 cm andAD = 11 cm, then findAP?
b. If DR = 8 cm andAD = 21 cm, then findAP?
ii.If O is the centre of fountain with ∠������ =60°,thenfind ∠������.
iii.IF O is the centre of the circular fountain, then ∠������=
38. In structural design, a structure is composed of triangles that are interconnecting.Atruss a series of triangle in same plane end is one of the major types of engineering structures and is especially used in the design of bridges and buildings. Trusses are designed to support loads, such as the weight of people.Atruss is exclusively made of long, straight members connected by joints at the end of each member.
i.What is the length ofAC?
ii.What is the length of BC?
iii.If sinA= sin C, what will be the length of BC? OR If ∠�� =60°, what will be the length of BC?
Marking Scheme
Section B consists of 5 questions of 2 marks each.
21. Form equations from given conditions
Let Rashmi’s present age be �� years and Nazma’s present age be �� years.
Three years ago, Rashmi was thrice as old as Nazma.
So, �� 3=3(�� 3)
⇒�� =3�� 6 ….. (1)
Ten years later, Rashmi will be twice as old as Nazma
Solving equation (1) and (2) we get
=42,��=16
Thus, Rashmi is 42 years old, and Nazma is 16 years old.
22. a. Use property of complementary angles
Given: tan A=cot B
23. Draw figure and use tangents property:
∵BC touches the circle at R.
∵ Tangents drawn from external point to the circle are equal.
∴AP=AQ,BR=BP And ���� =����
Find the perimeter of ������
∴ Perimeter of △������ =����+����+����
=����+(����+����)+����
=����+����+����+����
=����+���� =2���� =2×10=20cm
24. Draw figure and form equation according to given
Radius of outer circle (��)=21cm let radius of inner circle =��cm Given,
Area enclosed between concentric circle =770cm2
⇒����2 ����2 =770
Solve equation and find final answer
⇒����2 ����2 =770
⇒ 22 7 ×21×21 22 7 ×��2 =770
⇒ 22 7 (441 ��2)=770
441 ��2 = 770×7 22
⇒
⇒��2 =196
⇒�� =√196=14cm
∴ radius of inner circle =14cm
25. Find Σ�� and ∑����:
Section C consists of 6 questions of 3 marks each.
a. Step 1: Write the discriminant condition
= 8±4√3 2 =4±2√3 OR
b. Formulation of equation:
Let the sides of the two squares be ��cm and ��cm, where �� >��
The sum of the areas of the two squares is given as:
��2 +��2 =400 ……. (1)
The perimeter of a square is given by 4× (side length).
The difference of the perimeters is given as:
4�� 4�� =16
Dividing by 4:
�� ��=4 (2)
Solve for ��and��
From �� �� =4, express �� in terms of �� :
�� =��+4
Substituting into the area equation:
(��+4)2 +��2 =400
Expanding:
��2 +8��+16+��2 =400
2��2 +8��+16=400
2��2 +8�� 384=0
Dividing by 2:
��2 +4�� 192=0
Solve the quadratic equation by factorisation
Find two numbers whose product is 192 and sum is 4:
(��+16)(�� 12)=0
Thus,
��+16=0 or �� 12=0
�� = 16 or ��=12
Since the side length cannot be negative, we take �� =12cm
From �� =��+4 :
�� =12+4=16
The sides of the squares are 16 cm and 12 cm.
27. Figure and statement.
To Prove: PQRS is a parallelogram
Construction: JoinAC
Proving SR||AC
Proof: In △ DAC ���� ���� = ���� ���� =1
[∵�� and �� are mid-points of ���� and ����]
⇒SR||AC..........(i) [by converse of B.P.T]
Proving SR||����
In △BAC,���� ���� = ���� ���� =1[∵P and Q are midpoints ofAB and BC]
⇒SR||����......(ii) [By converse of B.P.T]
Solving (i) and (ii) to PQRS is a parallelogram
From (i) and (ii), we get
SR||PQ … (iii)
Similarly, join B to D and PS||QR
⇒∴PQRS is a parallelogram.
28. To proveADBC is a parallelogram
Let A(0 1),B(6,7),C( 2,3) and D(8,3) be the given points. Then, AD=√(8 0)2 +(3+1)2 =√64+16=4√5 BC=√(6+2)2 +(7 3)2 =√64+16=4√5 AC=√( 2 0)2 +(3+1)2 =√4+16=2√5 and BD=√(8 6)2 +(3 7)2 =√4+16=2√5
Therefore, AD=BC and AC=BD
So,ADBC is a parallelogram
To proveADBC is a rectangle
To find the area of rectangle Area
Let
be the mountain of height ℎ and C be its foot.
Find the required height
Now, ���� =���� ����
⇒���� =ℎ 500√3[ from (i) ]
In △������, we have, tan 60∘ = ���� ����
⇒√3= h BF BG [∵BG=EF and BF=EG
⇒√3= ℎ 500 ℎ 500√3 ⇒ℎ 500=√3ℎ 500×3
⇒ 500+1500=ℎ(√3 1)⇒1000=ℎ×0.73
⇒ℎ = 1000 073 =136986m
∴ the height of the mountain is 1369.86 m.
b. Draw
Find the distance of point P from tower = ��
Let the height of flagstaff =CB=ℎm
height of tower =AB=5m
Height of top of flagstaff from ground =AC=(ℎ+5)m
Let distance of point �� from tower =��
Find the height of the flagstaff
Using △������, �� ℎ+5 =cot60∘
⇒ �� ℎ+5 = 1 √3
30.
⇒�� = ℎ+5 √3 (ii)
From (i) and (ii), we get
ℎ+5 √3 =5
⇒ℎ+5=5√3
∴ℎ =5√3 5
=5(1.73 1)=5×.73=3.65m
∴ the height of the flagstaff is 3.65 m approx.
Identify the modal class for each group
The modal class is the class interval with the highest frequency.
GroupA: The highest frequency is 78 (in the 18-20 age group).
Group B: The highest frequency is 89 (in the 18-20 age group).
Thus, for both groups, the modal class is 18 20 years.
Apply the mode formula
Mode =��+( ��1 ��0 (2��1 ��0 ��2))×ℎ
Where:
�� = lower boundary of the modal class
��1 = frequency of the modal class
��0 = frequency of the class before modal class
��2 = frequency of the class after modal class
ℎ = class width (assumed to be 2 years)
Compute mode for groupA:
�� =18,��1 =78,��0 =50,��2 =46,ℎ =2
Mode�� =18+( 78 50 (2×78 50 46))×2
=18+( 28 (156 96))×2
=18+(28 60 ×2)
=18+093
=1893 years
Compute mode for group B:
�� =18,��1 =89,��0 =54,��2 =40,ℎ =2
Mode�� =18+( 89 54 (2×89 54 40))×2
=18+( 35 (178 94))×2
=18+(35 84 ×2)
=18+083
=18.83 years
Compare the modal ages
Modal age for Group ��=18.93 years
Modal age for Group �� =1883 years
Both groups have a similar modal age, with GroupA's modal age (18.93 years) slightly higher than Group B's (18.83 years).
31. Find the probability when Shweta losses the entry fee.
Possible outcomes when a coin is tossed 3 times:
HHH, HHT, HTH, THH, TTH, THT, HTT, TTT
⇒ Total number of outcomes = 8
i.Shweta will lose the entry fee if she gets 'TTT'.
��(Shweta losses the entry fee)= 1 8
Find the probability when Shweta will get double the entry fee.
Possible outcomes when a coin is tossed 3 times:
HHH, HHT, HTH, THH, TTH, THT, HTT, TTT
⇒ Total number of outcomes = 8
ii.Shweta will get double the entry fee if she gets HHH, ��(Shweta will get double the entry fee)= 1 8
Find the probability when Shweta will get her entry fee.
Possible outcomes when a coin is tossed 3 times:
HHH, HHT, HTH, THH, TTH, THT, HTT, TTT
⇒ Total number of outcomes = 8
iii.Shweta will get her entry fee, if she gets HHH, HTH, THH, TTH, THT or TTT
��(Shweta will get her entry fee) = 6 8 = 3 4
Section D
Section D consists of 4 questions of 5 marks each
32. Formation of 1st equation
Let the base and altitude of the right-angled triangle be �� and �� cm, respectively
Therefore, the hypotenuse will be (��+2)cm
∴(��+2)2 =��2 +��2 …………. (1) 1.5
Formation of 2nd equation
Again, the hypotenuse exceeds twice the length of the altitude by 1 cm.
∴ℎ =(2��+1)
⇒��+2=2��+1
⇒�� =2�� 1 …………… (2)
Solve for �� and y
Putting the value of �� in (1), we get: (2�� 1+2)2 =��2 +(2�� 1)2
⇒(2��+1)2 =��2 +4��2 4��+1
⇒4��2 +4��+1=5��2 4��+1
⇒ ��2 +8�� =0
⇒��2 8��=0
⇒��(�� 8)=0
⇒�� =8 cm
∴�� =16 1=15cm
∴ℎ =16+1=17cm
Thus, the base, altitude and hypotenuse of the triangle are 15cm,8cm and 17 cm respectively. 2
33. a. Finding the first term
The sum of the first �� terms of theA.P. is given by: ���� =4�� ��2
The first term is ��1 : ��1 =4(1) (1)2 =4 1=3
First term �� =3 1
Finding the sum of the first two term
Sum of first two terms:
��2 =4(2) (2)2 =8 4=4
Sum of first two terms = 4. 1
Finding the Second, Third, Tenth, and nth terms
The ��th term is given by:
���� =���� ���� 1
Finding the second term (��2) :
��2 =��2 ��1 =4 3=1
Finding the third term (��3) :
��3 =4(3) (3)2 =12 9=3
��3 =��3 ��2 =3 4= 1
Finding the tenth term ( ��10 ):
��10 =4(10) (10)2 =40 100= 60
��10 =��10 ��9
First, find ��9 :
��9 =4(9) (9)2 =36 81= 45 Now.
��10 = 60 ( 45)= 15
Finding the general ��th term (����) : ���� =���� ���� 1
Therefore, ΔABC ~ ΔPQR [By SAS similarity criterion]
35. Find area of rectangleABCD. Length of rectangleABCD
∴Area of rectangleABCD
=��������
=80×70
=5600cm2
FindAE
In right-angled △������,
AE2 =(AD2 DE2)
=(702 422)
=(70+42)(70 42)
=112×28
AE=√4×28×28
���� =2×28
���� =56 cm
Find area of semi-circle.
∴ Area of △������
= 1 2 ��������
= 1 2 ×42×56
=1176cm2
Area of semi-circle = 1 2 ��×(70 2 )2
={1 2 × 22 7 ×35×35}cm2
=1925cm2
Find the area of shaded region.
Thus,Area of the shaded region
= Area of rectangle ABCD ( Area of △AED+ Area of semi-circle )
=5600 (1176+1925)
=5600 3101
=2499cm2
Section E
Section E consists of 3 case-study based questions of 4 marks each.
36. i.Find probability for Rs 5 coin.
Total number of coins: 50+48+36+28+8=170
Number of Rs 5 coins =36
∴ Required probability = 36 170 = 18 85
ii.Find probability for denomination at least Rs 10 coin
Total number of coins of ₹10 and ₹20 = 28 + 8 = 36
∴ Required probability = 36 170 = 18 85
iii. a. Find the probability for getting Rs 10 coin.
Number of ₹10 coins =28
Required (coin is of ₹10) = 28 170
Find the probability for not getting Rs 10 coin.
∴ Required probability =1 P( coin is of ₹10) =1 28 170 = 142 170 = 71 85
b. Find the probability for getting Rs 20 coin.
Number of ₹20 coins =8
Required (coin is of ₹20) = 8 170
Find the probability for not getting Rs 20 coin.
37. i. a. FindAP.
Using tangent property: AP =AS =AD - DS =AD - DR =11 7=4cm
b. FindAP.
Using tangent property:
ii.Find ∠������.
In quadrilateral ��������,∠������ =60∘ (Given) And ∠������ =∠������ =90
(Since, radius at the point of contact is perpendicular to tangent.)
iii.Find ∠������.
Here, OS the is radius of circle.
TIME: 3 hours
General Instructions:
MATHEMATICS STANDARD
SAMPLE QUESTION PAPER - 9
CLASS – X (2025-26)
MAX. MARKS: 80
1. This question paper contains 38 questions.All Questions are compulsory
2. This question paper is divided into 5 SectionsA, B, C, D, and E.
3. In SectionA, Question numbers 1-18 are multiple-choice questions (MCQs) and questions number 19 and 20 areAssertion-Reason based questions of 1 mark each.
4. In Section B, Questions numbers 21-25 are very short answer (VSA) type questions, carrying 02 marks each.
5. In Section C, Questions numbers 26-31 are short answer (SA) type questions, carrying 03 marks each.
6. In Section D, Questions numbers 32-35 are long answer (LA) type questions, carrying 05 marks each.
7. In Section E, Questions numbers 36-38 are case study-based questions carrying 4 marks each with subparts of the values of 1, 1, and 2 marks respectively.
8. Draw neat and clean figures wherever required. Take �� = 22 7 wherever required if not stated.
9. Use of calculators is not allowed.
SectionA
Questions no. 1 to 18 are multiple choice questions (MCQs) of 1 mark each.
1. Anumber when divided by 61 gives 27 as quotient and 32 as remainder, then the number is:
2. Aquadratic polynomial with zeroes 1 4 and – 1 is.
3. The value of k for which the pair of linear equations 5�� + 2�� 7 =
and 2�� + ���� + 1 = 0 don't have a solution, is:
4. If one root of the equation then the value 4��2 2��+(λ 4)=0 be the reciprocal of the other, then the value of λ is
5. If ��,�� and �� are in A. P., then the value of
6. The sum of first 24 terms of the list of numbers whose nth term is given by ���� =3+2�� is
7. In the given figure if Δ������ ∼Δ������ then DE is equal to
8. If one end of a diameter of a circle is (4, 6) and the centre is ( – 4, 7), then the other end is
( 12,8)
8, 12)
(8,10)
9. The point ( – 3, 5) lies in the ___________ quadrant.
(secθ+cosθ)(secθ cosθ)=
(8, 6)
11. IfA, B and C are interior angles of a triangleABC, then the value of tan(
) is
12. If the angle of depression of an object from a 75m high tower is 30°, then the distance of the object from the tower is
13. In the given figure, if OQ = 3 cm, OP = 5 m, then the length of PR is
The area of a square that can be inscribed in a circle of radius 10 cm is
15. The radius of the circle whose area is equal to the sum of the areas of the two circles of radii 24cm and 7cm is
16. Asolid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1cm and the height of the cone is equal to its radius. The volume of the solid is
17. Abox contains 3 blue balls, 2 white balls and 4 red balls. If a ball is drawn at random from the box, the probability of getting a white ball is
18. For the following distribution:
Find the sum of lower limits of median class and modal class.
19. Assertion (A): If the median of a series exceeds the mean by 3, then mode exceeds mean by 10.
Reason (R): If mode =12.4 and mean =10.5, then the median is 11.13.
A)BothAand R are true and R is the correct explanation ofA
B)BothAand R are true but R is NOT the correct explanation ofA
C)Ais true but R is false
D)Ais false and R is True
20. Assertion (A): The polynomial ��(��)=��2 +3��+3 has two real zeroes. Reason (��):Aquadratic polynomial can have at most two real zeroes.
A)BothAand R are true and R is the correct explanation ofA
B)BothAand R are true but R is NOT the correct explanation ofA
C)Ais true but R is false
D)Ais false and R is True
Section B
Section B consists of 5 questions of 2 marks each.
25. The mean and median of 100 observations are 50 and 52 respectively. The value of the largest observation is 100. It was later found that it is 110 not 100. Find the true mean and median.
Section C
an elevation of 45°. Both the boys are on opposite sides of both the kites. Find the length of the string that the second boy must have so that the two kites meet.
OR
Around balloon of radius �� subtends an angle at the eye of the observer while the angle of elevation of its centre is ��. Prove that the height of the centre of the balloon is ��sinβcosec �� 2
30. Four right circular cylindrical vessels each having diameter 21 cm and height 38 cm are full of ice cream. The ice cream is to be filled in cones of height 12 cm and diameter 7 cm having a hemispherical shape on the top. Find the total number of such cones which can be filled with ice cream.
31. Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another. What is the probability that both will visit the shop on: i.the same day?
ii.different days?
iii.consecutive days?
OR
Abox contains cards bearing numbers 6 to 70. If one cards is drawn at random from the box, find the probability that it bears.
i.a one-digit number.
ii.a number divisible by 5.
iii.an odd number less than 30.
Section D
Section D consists of 4 questions of 5 marks each
32. Prove that √�� + √�� is irrational, where ��,�� are primes
Prove √3 is irrational by contradiction method. 5
33. When a polynomial ��(��) is divided by ��2 5, the quotient is ��2 2�� 3 and remainder is zero. Find the polynomial and all its zeroes.
If are the zeroes of polynomial ��(��) = 3��2 + 2�� + 1, find the polynomial whose zeroes are 1 �� 1+�� and 1 ��
35. InABC,AD is a median. Prove that
Section E
Section E consists of 3 case-study based questions of 4 marks each.
36 Aseminar is being conducted by an educational organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.
i.Find the sum of the powers of each prime factor of 108?
ii.Find the product of HCF and LCM of 60, 84 and 108.
iii.What is the minimum number of rooms required during the event?
Find the minimum number of rooms required when the number of participants in Hindi, English and Mathematics are 72, 96, and 132 respectively.
37. Your younger sister wants to buy an electric car and plans to take loan from a bank for her electric car. She repays her total loan of 327600 by paying every month starting with the first instalment of Rs 1200 and it increases the instalment by Rs 200 every month.
i. Find the difference of the amount in 4th and 6th instalment paid by younger sister ii. In how many instalments, she clear her total bank loan?
iii. Find the sum of the first seven instalments.
Find the sum of the first five instalments
38. In a classroom, 4 friends Ravi, Vinod, Raghav and Vithal are seated at the points A(2,3),B(7,8),C(10,5) and D(5,0) respectively.
Based on the above, answer the following questions:
i. Find the distance between Ravi and Raghav.
ii. Find the distance between Vinod and Vithal.
iii. Show thatABCD is a rectangle.
Marking Scheme
654=302×2+50
Divide 302 by 50 :
302=50×6+2
Divide 50 by 2 :
50=2×25+0
So,
GCD(302,654)=2
22. Factorise the given quadratic equation.
3��2 √6�� √6��+2=0
√3��(√3�� √2) √2(√3�� √2)=0
√2)(√3�� √2)=0
Solve for ��:
Check for real roots using the discriminant
The discriminant (Δ) of a quadratic equation ����2 +����+�� =0 is given by:
Δ=��2 4����
For the given equation:
�� =1,�� = 6,�� =6
Calculate the discriminant:
Δ=( 6)2 4(1)(6)
Δ=36 24=12
Since Δ>0, the equation has real and distinct roots.
Solve the quadratic equation using the quadratic formula
The quadratic formula is:
Substituting values:
�� = ( 6)±√12 2(1) �� = 6±2√3 2
�� =3±√3
Thus, the two roots are: �� =3+√3 or �� =3 √3
23. Find �� coordinate
The given vertices of triangle are (3, 5),( 7,4) and (10, 2).
Let (��,��) be the coordinates of the centroid. Then
To find �� coordinate
∴ The coordinates of the centroid are (2, 1)
24. Draw figure and formula Pythagoras theorem use tangent property.
∴ PQ is the tangent, and OP is the radius through the point of contact. [The tangent at any point of a circle is perpendicular to the radius through the point of contact] Pythagoras theorem
Find����: By Pythagoras theorem in right △������, OQ2 =OP2 +PQ2
⇒ (12)2 =(5)2 +PQ2
⇒ 144=25+PQ2
⇒ ����2 =144 25
⇒ PQ2 =119
⇒ ���� =√119cm
Hence, the length PQ is √119 cm.
25. Given data:
Mean of 100 observations = 50
Median of 100 observations = 52
Largest observation recorded = 100
Correct value of the largest observation = 110
Find the true mean:
Since mean of 100 observations is 50.
⇒ 50= Σ����
���� =5000
Find the true median:
Since only the largest value is being corrected, which is at the end of the data, it does not affect the middle values.
The median remains unchanged at 52.
Section C consists of 6 questions of 3 marks each.
26. Eliminating one variable
Multiply the first equation by �� and the second equation by ��, then subtract:
Subtracting the two equations:
27. Expressing the given condition using the general formula
The general term of anA.P. is given by:
Given that the 9th term is 0:
Finding the 19th and 29th terms
The 19th term:
��19 =��+(19 1)��
��19 =��+18��
Substituting �� = 8�� :
��19 = 8��+18�� =10��
The 29th term:
��29 =��+(29 1)��
��29 =��+28��
Substituting �� = 8�� : ��29 = 8��+28�� =20��
Proving the required condition
To prove that the 29th term is double the 19th term: ��29 =2��19
20�� =2(10��) 20�� =20��
Hence, proved.
Figure and given statement.
∠1+∠2=∠1+∠3
⇒∠2=∠3
To prove ����2 =����⋅����
In △������ and △������, we have ∠2=∠3 (proved) and ∠������ =∠������ =90∘
29. Draw the figure
Find DF
Let �� be the position of the first boy and ���� be the building on the roof of which second boy is standing. Let the required length of the string be �� m.
In, △������ we have, sin 45∘ = ���� �� ⇒ 1 √2 = ���� �� ⇒���� = �� √2
In △������, we have sin 30∘ = ���� 100 ⇒ 1 2 = ����+20 100 [∵���� =����+����]
⇒���� =50 20=30
Find the final answer
∴30= �� √2 From (i) we have
⇒�� =30√2=30×1.41=42.32m
∴ to meet the kites the second boy must have 42.32 m long string.
Draw figure
Find OP
Let O be the centre of the balloon of radius �� and P the eye of the observer.
Let PA,PB be tangents from P to the balloon. Then, ∠������ =��.
∴∠������ =∠������ = �� 2
Let OL be perpendicular from �� on the horizontal PX. We are given that the angle of the elevation of the centre of the balloon is �� i.e, ∠������ =��.
In △������, we have
Final answer
Hence, the height of the centre of the balloon is ��sin ��cosec �� 2 1 30. Find the volume of four cylindrical vessels 1
In △������, we have sin �� = ���� ���� ⇒����=����sin �� =��cosec �� 2 sin �� [Using equation (i)]
According to question it is given that
Diameter of cylinder =21cm,
Therefore, radius of cylinder = 21 2 cm
Height of cylinder =38cm.
Therefore, Volume of four cylindrical vessels
=4×��(21 2)2 ×38cm3 (i)
Find the volume of one ice-cream cone
Dimensions of cone with hemispherical end:
Diameter of base of cone = diameter of hemisphere =7cm
Height of cone =12cm
Volume of one ice cream cone = volume of conical part + volume of hemispherical part = 1 3 ��(72)2 ×12+ 2 3 ��(72)3 = 1 3 ��(72)2 [12+2× 7 2] = 1 3 ��(72)2 ×19 ……..(ii) 1
Find the total number of ice cream cones.
Total number of ice cream cones
= volume of four cylindrical vessels volume of one ice cream cone = 4��(21 2)2 ×38 1 3��(72)2 ×19 [fromeq(i)and(ii)]
= 4×3×441×2
49 =216
Hence, 216 cones can be filled.
31. Find the probability when two customers can visit the shop on same day of the week.
Total number of days to visit the shop =6
Two customers can visit the shop on same days is, 6×6=36 ways
So total number of outcomes =36
i.Two customers can visit the shop on same day of the week is 6 ways i.e. (M, M), (T, T), (W, W), (Th, Th), (F, F), (S, S)
Favourable
ways =6
Find the probability when both will reach on different day.
Find the probability when both will reach on consecutive days.
iii.Two customers can visit the shop on consecutive days in 5 ways i.e. (M, T), (T, M), (T, W), (W, T), (W, Th), (Th, W) , (Th, F), (F, Th), (F, S), (S, F)
Favourable number of ways =10
��(both will reach on consecutive days)=
Find the probability for getting a one-digit number.
i.Let �� be the event of getting a one-digit number.
Number of possible outcomes =70 6+1=65
The outcomes favourable to E are 6,7,8 and 9
∴ Number of favourable outcomes =4
��(��)=��(Getting a one-digit number)= 4 65
Find the probability for getting a number divisible by 5.
ii.Let F be the event of getting a number divisible by 5.
Number of possible outcomes =65
The outcomes favourable to the event F are 10,15,20, ,65,70
∴ Number of outcomes favourable to F=13
P(F)=P(Getting a number divisible by 5)= 13 65 = 1 5
Find the probability for getting an odd number less than 30.
iii.Let G be the event of getting an odd number less than 30.
Number of possible outcomes =65
The outcomes favourable to the event G are 7,9,11,13,…..,29.
∴ Number of favourable outcomes =12
P(G)=P(Getting an odd number less than 30)= 12 65
Section D
Section D consists of 4 questions of 5 marks each 32. We prove by contradiction. Consider two cases. Case 1: �� =�� Then
Call it �� ∈��.
Square both sides;
The right-hand side is rational, so √���� would be rational. But �� and �� are distinct primes, so ���� is not a perfect square; therefore √���� is irrational. This contradiction shows our assumption was false.
Combining both cases, for any primes ��,�� the number √��+√�� is irrational.
OR
Let us assume the contrary that root 3 is rational.
Then √3 = �� �� , where ��,�� are the integers i.e., ��,�� ∈ �� and co-primes, i.e., GCD
(��,��) = 1.
= ��
��
By squaring both sides, we get,
��2 = 3��2
��2 3 = ��2 (1)
(1)shows that 3 is a factor of ��. (Since we know that by theorem, if a is a prime number and if a divides ��2, then a divides ��, where a is a positive integer)
Here 3 is the prime number that divides ��2, then 3 divides �� and thus 3 is a factor of ��.
Since 3 is a factor of p, we can write �� = 3�� (where c is a constant). Substituting �� = 3�� in (1), we get, (3��)2 3 = ��2 9��2 3 = ��2
3��2 = ��2 ��2 = ��2 3 (2)
Hence 3 is a factor of �� (from 2)
Equation (1) shows 3 is a factor of �� and Equation (2) shows that 3 is a factor of q. This is the contradiction to our assumption that �� and �� are co-primes. So, √3 is not a rational number. Therefore, √3 is irrational. 1
33. Writing the polynomial equation. 1
We are given that when the polynomial ��(��) is divided by ��2 5, the quotient is ��2 2�� 3 and the remainder is zero. This means that:
��(��)=(��2 5)×(��2 2�� 3)
Expanding the polynomial:
��(��)=(��2 5)(��2 2�� 3)
Expanding each term:
So, the required polynomial is:
=
Finding the zeroes of ��(��):
Since ��(��) is a product of two quadratic polynomials, we first solve:
��2 5=0
��2 =5⇒�� =±√5
Now, solving: ��2 2�� 3=0
Factoring: (�� 3)(��+1)=0
So, �� =3 and �� = 1
Thus, the four zeroes of ��(��) are: √5, √5,3, 1
Using Vieta’s formulas
Since �� and �� are the roots of 3��2 +2��+1=0, we use Vieta's formulas: Sum of roots:
+�� = 2 3 Product of roots:
Finding the sum of the new roots The new roots are
Using the sum formula: 1.5
Using the identity: 1 �� 1+�� + 1 �� 1+�� = (1 ��)(1+��)+(1 ��)(1+��) (1+��)(1+��)
Expanding the numerator: (1 ��)(1+��)+(1 ��)(1+��)
=1+�� �� ����+1+�� �� ���� =2 2����
Expanding the denominator: (1+��)(1+��)=1+��+��+����
Using Vieta's formulas ��+�� = 2 3 and ���� = 1 3 :
Finding the product of the new roots
Using the product formula:
Expanding:
′ = 2 2 3 =3
Forming the required polynomial
Since the new polynomial has zeroes ��′ =2 and ��′ =3, the required polynomial is:
34.
Substituting �� = 2�� �� 5 and forming the quadratic equation
Given equation: ( 2�� �� 5)2 +5( 2�� �� 5) 24=0
Substituting �� = 2�� �� 5
Let: �� = 2�� �� 5
Rewriting the equation in terms of �� :
��2 +5�� 24=0
Solve the Quadratic Equation
Factorizing:
(��+8)(�� 3)=0
Solving for �� : ��+8=0 ⇒��= 8 �� 3=0 ⇒��=3
Solve for ��
Since �� = 2�� �� 5, we substitute each value of �� :
Case 1: �� = 8 2�� �� 5 = 8
2�� = 8(�� 5)
2�� = 8��+40
2��+8�� =40
10�� =40 �� =4 1
Case 2: �� =3
2�� �� 5 =3
2�� =3(�� 5)
2�� =3�� 15
2�� 3�� = 15 �� = 15 1
�� =15
�� =4, �� =15
Since the given condition is �� ≠5, both values �� =4 and �� =15 are valid. 35. Figuregiven,andconstruction
Construction: - Draw ���� ⊥����
Given,AD is a median
To prove AE2 =AD2 DE2
In △������
By using Pythagoras theorem, we get
AD2 =AE2 +DE2
AE2 =AD2 DE2 (i)
To prove AB2 =AD2 +CD2 2CD×DE
In △������,
By using Pythagoras theorem, we get
AB2 =AE2 +BE2
⇒AB2 =AD2 DE2 +(BD DE)2[ from (i) ]
⇒AB2 =AD2 DE2 +BD2 +DE2 2BD×DE
⇒AB2 =AD2 +CD2 2CD×DE… (i) [BD=CD] 1.5
To prove AB2 +AC2 =2AD2 +2CD2
In △������,
By using Pythagoras theorem, we get:
AC2 =AE2 +EC2
⇒AC2 =AD2 DE2 +(DE+DC)2[ from (i)]
⇒AC2 =AD2 DE2 +DE2 +DC2 +2DE×DC
⇒AC2 =AD2 +DC2 +2DE×DC (ii)
Add equations (i) and (ii)
Therefore, AB2 +AC2 =2AD2 +2CD2
Section E
Section E consists of 3 case-study based questions of 4 marks each.
36. i.Find the sum of the powers of each prime factor of 108?
Prime factorisation of 108
=2×2×3×3×3
=22 ×33
∴ Required sum of the powers =2+3=5
ii.Find the product of HCF and LCM of 60, 84 and 108.
HCF(60,84,108)=22 ×3=12 and LCM(60,84,108)=3780
HCF ×LCM=12×3780 =45360 1
iii.What is the minimum number of rooms required during the event? Given.
The number of participants in Hindi =60
The number of participants in English =84 and the number of participants in Mathematics =108
.. Total number of participants =60+84+108 =252
Hence, minimum number of rooms required during event = Total number of participants Maximum number of participants that can be accommodated in each room = 252 12 =21
The number of participants in Hindi =72
The number of participants in English =96 and the number of participants in Mathematics =132
.. Total number of participants =72+96+132
=300
Hence, minimum number of rooms required during event = Total number of participants
Maximum number of participants that can be accommodated in each room
= 300 12 =25
37. i. Find the difference of the amount in 4th and 6th instalment paid by younger sister
Let �� and �� be the first term and common difference of anAP.
Then, ��4 =��+(4 1)�� [���� =��+(�� 1)��)
=��+3��
Similarly, ��6 =��+5��.
∴ Required difference =��6 ��4
=(��+5��) (��+3��)=2��
=2×200=400 1
ii. In how many instalment, she clear her total bank loan?
Here, �� =1200,�� =200 ∴The sum of first seven instalments is
Find the sum of the first seven instalments.
�� =1200,�� =200
The sum of first five instalments is
MOST EXPECTED QUESTIONS
1. Prove that 5 √3 is an irrational number.
2. Amita, Sneha, and Raghav start preparing cards for all persons of an old age home. In order to complete one card, they take 10,16 and 20 minutes respectively. If all of them started together, after what time will they start preparing a new card together?
3. If LCM(��,18)=36 and hence HCF(��,18)=2 then "��" is?
4. Iftwo positiveintegers �� and �� are expressiblein terms ofprimes as �� =��2��3 and ��=��3��,What can you say about their LCM and HCF.
5. Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial ��(��)= ����2 +����+��
6. Atwo-digit number is 4 more than 6 times the sum of its digit. If 18 is subtracted from the number, the digits are reversed. Find the number.
7. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
8. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km down-stream. Determine the speed of the stream and that of the boat in still water.
9. 6 men and 10 women can finish making pots in 8 days, while the 4 men and 6 women can finish it in 12 days. Find the time taken by one man alone and that of one woman alone to finish the work.
10. Places �� and �� are 100 km apart on a highway. One car starts from �� and another from �� at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours if they travel towards each other they meet in 1 hour. What are the speeds of the two cars?
11. 5 years hence, the age of Jacob will be three times that of his son. 5 years ago, Jacobs age was seven times that of his son. What are their present ages?
12. Find the value of �� and �� for which the system of equations represents coincident lines 2��+3��= 9,(��+��)��+(��+��+)��=3(��+��+1)
13. A train travels 480 km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. What is the speed of the train?
14. The sum of the areas of two squares is 468m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
15. Two water taps together can fill a tank in 93 8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can fill the tank.
16. In a flight of 600 km, an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200km/hr and the time of flight increased by 30 minutes. Find the original duration of the flight.
17. In a class test the sum of Shefali's marks in mathematics and English is 30 . Had she got 2 marks more in mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in 2 subjects.
18. The summer 4th and 8th term of anAP is 24 and the sum of 6th and 10th term is 44. Find the first three terms of theAP.
19. Find the middle term of theAP 213,205,197,…..37 ?
20. How many multiples of 4 lie between 10 and 250?
21. If the sum of the first �� terms of an ���� is the same as the sum of its first �� terms, show that the sum of its first (��+��) terms are zero.
22. If the sum of the first �� terms of an ���� is 4�� ��2, what is the first term (that is S1 )? What is the sum of the first two terms? What is the second term? Similarly find the 3rd, the 10th and the nth terms.
23. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
24. Prove the Basic Proportionality theorem.
25. The diagonals of a quadrilateral �������� intersect each other at the point �� such that ���� ���� = ���� ���� . Show that �������� is a trapezium.
26. In Fig. altitudes ���� and ���� of △������ intersect each other at the point ��. Show that:
(i) △AEP∼△CDP (ii) △ABD∼△CBE (iii) △AEP∼△ADB (iv) △PDC∼△BEC
27. In the figure, DE‖AC and DF‖AE. Prove that BF FE = BE EC .
28. Sides ���� and ���� and median ���� of a triangle ������ are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that △ABC∼△PQR
29. In Figure given, PQL and PRM are tangents to the circle with centre O at the points �� and ��, respectively and �� is a point on the circle such that ∠SQL=50∘ and ∠SRM=60∘. Find ∠QSR ?
30. Find the coordinates of the points of trisection (i.e., points dividing into three equal parts) of the line segment joining the points ��(2, 2) and ��( 7,4).
31. If (1,2),(4,��),(��,6) and (3,5) are the vertices of a parallelogram taken in order, find �� and ��.
32. Prove that sec ��+tan �� 1 tan �� sec ��+1 = cos �� 1 sin ��
33. Prove that (1+tan2A 1+cot2A)=(1 tan A 1 cot A)2 =tan2A
34. If sin ��+cos �� =√3, then prove that tan ��+cot �� =1.
35. The angle of elevation of an aeroplane from point �� on the ground is 60∘. After a flight of 15 seconds, the angle of elevation changes to 30∘. If the aeroplane is flying at a constant height of 1500√3m, find the speed of the plane in km/hr.
36. The angle of elevation of the top of a tower from two points distant �� and �� from its foot are complementary. Prove that the height of the tower is √st .
37. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60∘. After some time, the angle of elevation reduces to 30∘. Find the distance travelled by the balloon during the interval.
38. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ=2∠OPQ
39. Prove that the parallelogram circumscribing a circle is a rhombus.
40. AtriangleABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides ���� and ����
41. In the figure, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP=2r, show that ∠OTS=∠OST=30∘
42. Acanal is 300 cm wide and 120 cm deep. The water in the canal is flowing at a speed of 20km/h. How much area will it irrigate in 20 minutes if 8 cm of standing water is desired?
43. Two cones have their heights in the ratio 1:3 and radii in the ratio 3:1. What is the ratio of their volumes?
44. Awooden article was m ade by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10 cm , and its base is of radius 3.5 cm , find the total surface area of the article.
45. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm . Find its surface area.
46. Ahorse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long. Find:
(i)the area of that part of the field in which the horse can graze. (ii)the increase in the grazing area if the rope were 10 m long instead of 5 m . (Use �� =314 )
47. Asquare is inscribed in a circle. Calculate the ratio of the area of the circle and the square.
48. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
49. If the median of a distribution given below is 28.5 then, find the value of an �� & ��
50. Two dice are numbered 1,2,3,4,5,6 and 1,1,2,2,3,3, respectively. They are thrown, and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
SOLUTIONS
1. Solution:
Let us assume on the contrary that 5 √3 is rational. Then, there exist prime positive integers �� and �� such that
5 √3= a b
⇒ 5 a b =√3
⇒ √3 is rational [∵a,b are integers.: 5b a b is a rational number ]
This contradicts the fact that √3 is irrational.
So, our assumption is incorrect. Hence, 5 √3 is an irrational number.
2. Solution:
The three of then take 16,10 and 20 minutes to make a card.
If they started at time t=0, the next they start making the cards at the same time would be the LCM of 16,10 and 20, which is 80
Thus, at the end of the 80th minute, they will start making the cards together. How did we get the LCM as 80
16∗5=80
10∗8=80
20∗4=80
Thus, 80 is the answer.
3. Solution:
�� =L.C.M×H.C.F= First number × Second number
Hence, required number = 36×2 18 =4
4. Solution:
�� =��2��3 ,��=��3��
LCM is the product of highest powers of individual variables �� and ��, and HCF is the product of lowest powers of the individual variable p and q.
Let the side of the first square be ' �� ' �� and that of the second be ' �� ' ��
Area of the first square =��2 sq m.
Area of the second square =A2 sq m.
Their perimeters would be 4a and 4Arespectively.
Given 4A 4a=24
�� �� =6 (1)
��2 +��2 =468 (2)
From (1), A=a+6
Substituting for �� in (2), we get
(��+6)2 +��2 =468
��2 +12��+36+��2 =468
2��2 +12��+36=468
��2 +6��+18=234
��2 +6�� 216=0
��2 +18�� 12�� 216=0
��(��+18) 12(��+18)=0
(�� 12)(��+18)=0
�� =12, 18
So, the side of the first square is 12 m . and the side of the second square is 18 m.
15.
Solution:
Consider that the tap with smaller diameter fills the tank in x hours.
Then, the tap with larger diameter fills the tank in �� 10 hours.
This shows that the tap with a smaller diameter can fill 1 �� part of the tank in 1 hour. Similarly, the tap with larger diameter can fill 1 �� 10 part of the tank in 1 hour.
It is given that the tank is filled in 75 8 hours that is, the taps fill 8 75 part of the tank in 1 hour. Then,
∴�� = 240 4 =60 which is the required number of multiples.
21.
Solution:
Let theAP be denoted as ��1,��2,��3,…,����,… with a common difference ��.
Sum of first �� terms: ���� = �� 2 (2��1 +(�� 1)��)
Sum of the first �� terms:
�� = �� 2(2��1 +(�� 1)��)
Given that the 2 sums are equal. Hence, �� 2 (2��1 +(�� 1)��)= �� 2(2��1 +(�� 1)��) 2��1 ��+(��2 ��)�� =2��1��+
∴2��1 =��(1 �� ��)
Sum of first (��+��) terms is given by
��+�� = ��+�� 2 [2��1 +(��+�� 1)��]
Substituting
2��1 =��(1 �� ��),
0
Hence, proved that the sum of the first (��+��) terms is 0.
22.
Solution:
If Sum of first �� terms of anAP ���� =4�� ��2
Then, the first term will be when n=1 in the above expression
��1 =�� =4(1) (1)2 =3
Sum of first two terms,
�� =2,��2 =4(2) 22 =4
So, the second term will be ��2 =��2 ��1 =4 3=1
Similarly
�� =3,��3 =4(3) 32 =3
⇒��3 =��3 ��2 =3 4= 1
�� =9,��9 =4(9) 92 = 45
�� =10,��10 =4(10) (10)2 = 60
⇒��10 =��10 ��9 = 60 ( 45)= 15
���� =���� ��(�� 1)
⇒���� =4�� ��2 (4(�� 1) (�� 1)2)
⇒���� =5 2��
23.
Solution:
It can be observed that the number of trees planted by the students is in anAP. 1, 2, 3, 4, 5 12
First term, �� =1
Common difference, �� =2 1=1
���� = �� 2[2��+(�� 1)��]
��12 = 12 2 [2(1)+(12 1)(1)]
=6(2+11) =6(13) =78
Therefore, number of trees planted by 1 section of the classes = 78
Number of trees planted by 3 sections of the classes =3×78=234
Therefore, 234 trees will be planted by the students.
24.
Solution:
Basic Proportionality Theorem states that, if a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides of the triangle in proportion.
LetABC be the triangle.
The line �� parallel to BC intersect AB at D andAC at E .
To prove AD DB = AE EC
Join BE, CD
Draw EF⊥AB,DG⊥CA
Since EF⊥AB, EF is the height of trianglesADE and DBE
Area of △ADE= 1 2 × base × height = 1 2 AD×EF
Area of △DBE= 1 2 ×DB×EF
area of area of △ADE= 1 2 ×AD×EF 1 2 ×DB×EF = AD DB ……….(1)
Similarly,
area of area of △ADE= 1 2 ×AE×DG 1 2 ×EC×DG = AE EC ………(2)
But △DBE and △DCE are the same base DE and between the same parallel straight line BC and DE
Area of △DBE= area of △DCE ……..(3)
From 1, 2, 3, we have AD
DB = AE EC
Hence proved.
25. Solution: Given:
The diagonals of a quadrilateral �������� intersect each other at the point �� such that AO BO = CO DO
i.e., AO CO = BO DO
To Prove:ABCD is a trapezium
Construction:
Draw OE || DC such that E lies on BC.
Proof:
In △BDC, By Basic Proportionality Theorem,
BO
OD = BE EC
But, AO
CO = BO DO (Given)
∴ From (1) and (2) AO
CO = BE EC
Hence, By Converse of Basic Proportionality Theorem, OE ||AB
Now Since,AB||OE||DC
∴AB || DC
Hence,ABCD is a trapezium.
26.
Solution:
In △AEP and △CDP,
∠APE=∠CPD (Vertically opposite angle)
∠AEP=∠CDP=90∘
∴ ByAAcriterion of similarity, △AEP∼△CDP
In △������ and △������
∠ADB=∠CEB=90∘
∠B is common
∴ ByAAcriterion of similarity, △ABD∼△CBE
In △������ and △������
∠AEP=∠ADB=90∘
∠A is common
∴ ByAAcriterion of similarity, △AEP∼△ADB
△PDC and △BEC
∠PDC=∠BEC=90∘
∠C is common
∴ ByAAcriterion of similarity, △PDC∼△BEC
27.
Solution:
In △������
DE ||AC
Line drawn parallel to one side of triangle, insects the other two sides. It divides the other side in same ratio.
BE
EC = BD DA
In △AEB
DF ||AE
Line drawn parallel to one side of triangle, intersects the other sides. It divides the other sides in same ratio.
BF
FE = BD DA
From (i) & (ii)
����
���� = ���� ����
∴ Hence proved.
28.
Solution:
Given: △ABC&△PQR
���� is the median of △������
PM is the median of △PQR
AB
PQ = AC PR = AD PM →1
To prove: △ABC∼△PQR
Proof:Let us extendAD to point D such that that AD=DE and PM upto point L such that PM=ML
Join B to E,C to E,&Q to L and R to L
We know that medians is the bisector of opposite side
Hence BD=DC & AD=DE (By construction)
Hence in quadrilateralABEC , diagonalsAE and BC bisect each other at point D
∴ABEC is a parallelogram
∴AC=BE&AB=EC( opposite sides of a parallelogram are equal) →2
Similarly we can prove that PQLR is a parallelogram.
PR=QL,PQ=LR (opposite sides of a parallelogram are equal) →3
Given that
= ����
= ���� ����( from 1 )
⇒
=
���� = ���� ����( from 2 and 3)
⇒ ���� ���� = ���� ���� = 2���� 2����
⇒ ���� ���� = ���� ���� = ���� ����
(�������� =����,���� =����+���� =����+���� =2���� and ���� =����,���� =
2����)
∴△ABE∼△PQL (By SSS similarity criteria)
We know that corresponding angles of similar triangles are equal
∴∠BAE=∠QPL→4
Similarly, we can prove that
△AEC∼△PLR
We know that corresponding angles of similar triangles are equal
∠CAE=∠RPL→5
Adding 4 and 5, we get
∠BAE+∠CAE=∠QPL+∠RPL
⇒∠CAB=∠RPQ→6
In △������ and △������ ���� ���� = ���� ����( from 1 )
∠������ =∠������( from 6 )
∴△������ ∼△������ (By SAS similarity criteria)
Hence, proved
Solution:
Step-1: Write the given angles referring to given diagram.
Given ∠SRM=60∘ ,∠LQM=50∘
Step -2: Use the geometric rules and property to find the val ue of ∠QSR.
If, ∠SRM=60∘
⇒∠SRM+∠SRO=90∘
∴∠SRM=30∘
For ΔOSR,
OS=OR [radii of same circle]
∠OSR=∠SRO=30∘ [angles opposite to equal sides are equal]
As, ∠SQL=50∘
⇒∠SQL+∠SQO=90∘
∴∠SQO=40∘
For △OSQ,
OS=OQ [radii of same circle]
∠SQO=∠OSQ=40∘ [angles opposite to equal sides are equal]
∴∠QSR=∠OSR+∠OSQ =30∘ +40∘ =70∘
FinalAnswer: Hence, the value of ∠QSR is 70∘ .
30.
Solution:
Given:Aline segment joining the points A(2, 2) and B( 7,4). Let �� and �� be the points on ���� such that, AP=PQ=QB
Therefore,
�� and �� divides ���� internally in the ratio 1:2 and 2:1 respectively. As we know that if a point ( h,k ) divides a line joining the point ( x1,y1 ) and (��2,��2) in the ration
��:��, then coordinates of the point is given as(h,k)=(mx2 +nx1 m+n , my2 +ny1 m+n )
