Practicing Maths Question Paper Class 10 Kerala Syllabus Set 2 English Medium helps identify strengths and weaknesses in a subject.
Maths Class 10 Kerala Syllabus Model Question Paper Set 2
Time: 2½ Hours
Total Score: 80 Marks
Instructions:
- Use the first 15 minutes to read the questions and think about the answers
- There are 26 questions, split into four parts A, B, C, D
- Answer all questions; but in questions of the type A or B, you need answer only one of those
- You can answer the questions in any order, writing the correct question number
- Answers must be explained, whenever necessary.
- No need to simplify irrationals like √2, √3, etc using approximations unless you are asked to do so.
Section – A
Question 1.
The first three digit term of the arithmetic sequence 1, 6, 11, 16… is
(a) 101
(b) 102
(c) 100
(d) 103 (1 mark)
Answer:
(a) 101
Question 2.
Consider the following statements.
Statement (A): A long wire of length a cut equally and ends are joined to make a square pyramid. The pyramid will have slant \(\frac{a}{2}\)√3.
Statement (B): The lateral faces are equilateral triangle and have height √3 times half of its side
(a) Statement A is true, Statement B is false.
(b) Statement B is true, Statement A is false.
(c) Both statements are true. Statement B is the reason for Statement A.
(d) Both statements are true. Statement B is not the reason for Statement A. (1 mark)
Answer:
(a) Statement A is true, Statement B is false.
Since square pyramid have 8 edges (4 base edges and 4 lateral edges), the wire of length a is cut into 8 equal pieces. Each edge = \(\frac{a}{8}\) . So, lateral faces are equilateral triangles. Thus, first part of statement (A) is true.
Height of the equilateral triangle
= \(\frac{\sqrt{3}}{2} \times \frac{a}{8}\) = √3 × \(\frac{1}{2} \times \frac{a}{8}\)
So, height of the equilateral triangle is √3 times half of its side.
Hence statement (B) is true.
Slant height = Height of the equilateral triangle
= \(\frac{\sqrt{3}}{2} \times \frac{a}{8}=\frac{a \sqrt{3}}{16} \neq \frac{a}{2} \sqrt{3}\)
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Question 3.
A) In the figure a, b, c are the sides opposite to A, B and C, AP is the altitude to BC

(a) Write AP in terms of c and sin B
Answer:
sin B = \(\frac{A P}{c}\)
AP = c sin B
(b) Write the expression for the area of triangle ABC
Answer:
Area = \(\frac{1}{2}\) × BC × AP
A = \(\frac{1}{2}\) × a × c sin B sq.unit
(c) If a = 12, c = 6 and ∠B = 60° then the area of triangle ABC
Answer:
Area = \(\frac{1}{2}\) × 12 × 6 sin 60
= 36√3 sq.units
OR
B) In the figure ∠B =140°, AC = 12 cm

(a) What is the measure of ∠P?
Answer:
180 – 140 = 40°
(b) What is the radius of the circle?
[sin 40 = 0.64, cos 40 = 76, tan 40 = .83] (3 mark)
Answer:
sin 40° = \(\frac{12}{AP}\)
0.64 = \(\frac{12}{AP}\), AP = 18.75, r = 9.37 cm
Question 4.
P(6, 8) is a point on a circle with center at the origin. Q is a point on the circle such that OQ makes angle 30° with x axis as in the figure

(a) What is the radius of the circle?
Answer:
\(\sqrt{6^2+8^2}\) = 10
(b) What are the points where the circle cut the axes?
Answer:
(10, 0), (0, 10), (-10, 0), (0, -10)
(c) Write the coordinates of Q (4 mark)
Answer:
Draw perpendicular from Q to x axis. It is QA Triangle QAO is a 30° – 60° – 90° right triangle. Q(-5√3, 5)
Question 5.
A) The vertices of a triangle are A (-3, 2), B(3, -4) and C (1, 5).
(a) Find the midpoint of the sides AB.
Answer:
Midpoint of AB = \(\left(\frac{-3+3}{2}, \frac{2 \pm 4}{2}\right)\) = (0, -1)
(b) Calculate the length of median to the side AB.
Answer:
Median is the line joining a vertex to the midpoint of opposite side.
Length of the median to the side AB
= \(\sqrt{(1-0)^2+(5–1)^2}=\sqrt{37}\)
(c) Find the centroid.
Answer:
The centroid divides the median in the ratio 1 :2
Let the point be G (x, y) then,
x = 1 – a, y = 5 – b

⇒ y = 5 – 4
= 1
Therefore the centroid = (\(\frac{1}{3}\), 0)
OR
B) Third term of an arithmetic sequence is 12 and its fifth term20
(a) What is the common difference?
Answer:
Difference between fifth term and third term is two times common difference.
2 × common difference = 20 – 12 = 8,
Common difference = 4
(b) What is the first term ?
Answer:
First term = third term – 2 common difference
= 12 – 8 = 4
(c) Write the sequence numerically (4 mark)
Answer:
4, 8, 12…
Question 6.
Numbers 1, 2, 3, … 25 are written in small paper pieces and put in a box.One is taken from it without looking into the box. (5mark)
(a) What is the probability of getting an even number?
Answer:
\(\frac{12}{25}\)
(b) What is the probability of getting an odd number?
Answer:
\(\frac{13}{25}\)
(c) What is the probability of getting a prime number?
Answer:
\(\frac{9}{25}\)
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Section – B
Question 7.
A bridge of length 600 meter is built across a river. It makes 45° with the stream. The width of the river is

(a) 300√2 m
(b) 200 m
(c) 300√3 m
(d) 100√2 m (1 mark)
Answer:
(c) 300√3 m
Question 8.
Choose the correct answer from the options given below.
PA, PB are the tangents from P to the circle. If ∠APB = 40° then what is the measure of ∠AOB? (1 mark)

(a) 140°
(b) 150°
(c) 120°
(d) 110°
Answer:
(a) 140°
Tangents from the outer point to the circle and radii form a cyclic quadrilateral.
Question 9.
Sum of the area of two squares is 80.The sum of its perimetres is 48.
(a) If the sides are x and y then what is x + y?
Answer:
Sides are x and y.
4x + 4y = 48,
x + y = 12
(b) Form an equation representing the sum of the areas
Answer:
Sides are x and 12 – x
x2 + (12 – x)2 = 80
x2 + 144 – 24x + x2 = 80
2x2 – 24x = -64
x2 – 12x = -32
(c) Find the side of the the squares. (3 mark)
Answer:
x2 – 12x + 36 = -32 + 36
(x – 6)2 = 4, x – 6 = 2
x = 8 cm, y = 12 – 8 = 4 cm
Question 10.
Write the polynomial x2 + 5x – 84 as the product of two first degree polynomials. (3 mark)
Answer:
x2 + 5x – 84 = (A + a)(A + b)
= x2 + (a + b)x + ab
⇒ a + b = 5 ,ab = -84
⇒ (a – b)2 = (a + b)2 – 4ab
⇒ (a – b)2 = 52 – 4 × (-84)
= 25 + 336 = 361
⇒ a – b = ±19
Take a – b = 19
a + b = 5, a – b = ±19
⇒ a = \(\frac{1}{2}\) (5 + 19) = \(\frac{1}{2}\) × 24 = 12
b = \(\frac{1}{2}\) (5 – 19) = \(\frac{1}{2}\) × (-14) = -7
Take a – b = -19, a + b = 5, a – b = -19
⇒ a = \(\frac{1}{2}\) (5 + (-19) = \(\frac{1}{2}\) × (-14) = -7
b = \(\frac{1}{2}\) (5 – (-19)) = \(\frac{1}{2}\) × (5 + 19) = 12
x2 + 5x – 84 = (x + 12)(x – 7)
Question 11.
A man standing on the bank of a river sees the top of the tree at an angle of elevation 50°. Stepping 20 m back finds the angle to sees at an angle of elevation 30°.
(a) Draw a rough diagram.
Answer:

(b) Find the width of the river.
Answer:
tan 50° = \(\frac{B C}{A B}\)
tan 30° = \(\frac{B C}{B D}\)
1.19 = \(\frac{h}{x}, \frac{1}{\sqrt{3}}=\frac{h}{x+20}\)
h = 1.19x, h = \(\frac{x+20}{\sqrt{3}}\)
1.19x = \(\frac{x+20}{\sqrt{3}}\)
1.19 × √3x = x + 20,
1.19 × 1.73x = x + 20,
x = 18.89 m
Width of the river 18.89 metre
(c) Calculate the height of the tree. (4 mark)
Answer:
When x = 18.89, h =1.19
x = 1.19 × 18.89
= 22.47 metre.
Question 12.
A) Sum of the areas of two squares is 468sq.cm.The difference between the perimetres is 24 cm.
(a) If the small side is x then what is the length of the big side ?
Answer:
Length of the big side bey, difference between the perimeter,
4y – 4x = 24,
4 (x – y) = 24, x – y = 6, y = x + 6
(b) What is the perimetre of the big square?
Answer:
4x + 24
(c) Write the length of the sides the squares in x.
Answer:
Side of the small square is A, Side of the big square is \(\frac{4 x+24}{4}\) = x + 6
(d) Form a second degree equation and find the length of the small square.
Answer:
x2 + (x + 6)2 = 468,
x2 + x2 + 12x + 36 = 468,
2x2 + 12x = 432
x2 + 6x = 216,
x2 + 6x + 9 = 225.
(x + 3)2 = 225
(x + 3) = 15, x = 12
Side of the small square is 12 cm
(e) Find the length of the big square.
Answer:
Length of the big square is 12 + 6 = 18 cm
OR
B) AB is the diametre of the circle. CD is a chord of length equal to radius of the circle.

(a) What is the measure of ∠COD?
Answer:
Draw OC, OD, OCD is an equilateral traingle.
∠COD = 60°
(b) What is the measure of ∠CBD?
Answer:
∠CBD = \(\frac{1}{2}\) × 60 = 30°
(c) What is the measure of ∠BCP?
Answer:
∠BCA = 90° (angle in the semicircle).
∴ BCP = 90°
(d) Find the measure of ∠CPD (4 mark)
Answer:
In traingle BCP,
∠CPD = ∠CPB
= 180 – (90 + 30)
= 60°
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Question 13.
The chords AB and CD intersect at P.

(a) Draw, AC and BD. Establish the triangles PAC and PBD as similar triangles, and establish the relationship between PA, PB, PC and PD.
Answer:
Consider ΔPCA and ΔPBD
∠C = ∠B, ∠A = ∠D (Angles at the end of an arc are equal)
\(\frac{P A}{P D}=\frac{P C}{P B}\)
PA × PB = PC × PD

(b) PA = 3 cm PB = 5 cm. What is the area of the square with sides PC and PD?
Answer:
3 × 5 = PC × PD
PC × PD = 15
The area of the rectangle with sides PC, PD is 15 sq.cm
(c) Draw a rectangle with sides of length 5 cm and 3 cm. Draw a rectangle with length side 6 cm and area,equal to the area of the first rectangle. (5 mark)
Answer:
- Consider a rectangle with length 5 cm and width 3 cm.
- Consider another length 6 cm.
- In the square, we can draw a line extending 3 cm from the bottom edge to the left side and 6 cm from the left edge to the bottom:
- Now draw a circle through the left, right, and bottom corners. Draw a line through the rectangle where the left side intersects the circle.
- Now, let’s draw the required square by marking the length of the rectangle we have obtained.

Here ABCD is the required rectangle.
Question 14.
A) The numbers 21, 22, 23 … 250 are written in small paper pieces and placed in a box.
(a) Write the sequence of numbers comes in the right end of these numbers ?
Answer:
2, 4, 8, 6, 2, 4, 8, 6…
(b) If one is taken from the box at random, then what is the probability of getting a number with 4 in ones place?
Answer:
Up to 248, the set of digits 2,4,8,6 repeats 12 times. One’s place of 249 is 2 and one’s place of 250 is 4 .
Probability of getting 4 in one’s place is = \(\frac{13}{250}\)
(c) What is the probability of getting a number with 8 in ones place?
Answer:
Probability of getting 8 in one’s place is = \(\frac{12}{50}\)
(d) What is the probability of getting a number with 2 in ones place?
Answer:
Probability of getting 2 in one’s place is = \(\frac{13}{50}\)
(e) What is the probability of not getting a number with 2 in ones place?
Answer:
Proability of not getting 2 in one’s place is = 1 – \(\frac{13}{50}=\frac{37}{50}\)
OR
B) Two digit numbers are written in small paper pieces and placed in a box.One is taken from the box at random
(a) How many multiples of 5 are there in the box?
Answer:
10, 11, 12, 99 are the two digit numbers.
Number of two digit numbers is 90
Multiples of five are 10, 15, 20. ..95
Number of numbers = 18
(b) What is the probability of getting a multiple of 5 ?
Answer:
Probability of getting a multiple of five = \(\frac{18}{90}\)
(c) What is the probability of not getting a multiple of 5 ? (5 mark)
Answer:
Probability of not getting a multiple of 5
= 1 – \(\frac{18}{90}=\frac{72}{90}=\frac{8}{10}\)
Section – C
Question 15.
a. a – 1, a – 2 ………. is an arithmetic sequence. What is its nth term?
(a) a + n + 1
(b) a + n – 1
(c) a – n – 1
(d) a – n + 1 (1 mark)
Answer:
(d) a – n + 1
Question 16.
In the given figure BC is the diameter of a circle and AB = AC. Then ∠ABC is

(a) 30°
(b) 45
(c) 60°
(d) 90° (1 mark)
Answer:
(b) 45
Question 17.
Consider the arithmetic sequence 6, 10, 14,…
(a) What is the common difference of this sequence?
Answer:
d = 4
(b) Find the sum of first n terms of this sequence
Answer:
xn = 4n + 2
Sum = (6 + 4n + 2) × \(\frac{n}{2}\)
= 2n2 + 4n
(c) Can sum of some terms of this sequence 1225? How do you know this?
Answer:
All terms are even numbers. Sum of even num-bers cannot be an odd number. 1225 cannot be the sum
Question 18.
ABCD is a parallelogram. A(1, 1), B(1, 1) and C(1, 4)
(a) What is the length of the side parallel to x axis ?
Answer:
AB = |7 – 1| = 6
(b) Find the length of other side
Answer:
BC = \(\frac{1}{2}\)
= 3√5
(c) Find the coordinates of the vertex D
Answer:
D(-5, 4)
(d) Calculate the area of the parallelogram.(4 mark)
Answer:
Distance between the parallel sides AB and CD is 4 – 1 = 3
Area = 6 × 3 = 18 sq. units
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Question 19.
A) A person, 1.75 metres tall, standing at the foot of a tower sees the top of a hill 40 metres away at an elevation of 60°. From the top of the tower, he sees it at an elevation of 50°. Calculate the heights of the tower and the hill.
Answer:

\(\frac{D E}{\mathrm{CE}}\) = tan 50° ⇒ DE = 40 × tan 50° = 47.67 m CE
\(\frac{D F}{\mathrm{BE}}\) = tan 60° ⇒ DF = 40 × tan 60 = 69.28 m
EF = 69.28 – 47.67
= 21.61m
Height of the tower AC = 21.61 + 1.75
= 23.36 m
Height of the hill DG = 47.67 + 23.36
= 71. 03 m
OR
B) 700 rupees is used for purchasing 7 prizes . Cost of each prize is 20 rupees more than the cost of the prize just below it.
(a) What is the cost of fourth prize?
Answer:
Fourth term = \(\frac{700}{7}\) = 100
(b) What is the cost of first prize?
Answer:
First prize is 3 × 20 more than fourth term. It is 100 + 3 × 30= 16
(c) Write costs of prizes in the descending order. (5 mark)
Answer:
160, 140, 120, 100, 80, 60, 40
Section – D
Question 20.
What is the median of the following data? 21, 6, 14, 9, 5
(a) 9
(b) 8
(c) 5
(d) 11 (1 mark)
Answer:
(a) 9
Question 21.
Read the two statements given below.
Statement 1: The midpoint of the line joining (2, 3) and (4, 7) is (3, 5).
Statement 2: The midpoint of the line joining (x1, y1) and (x2, y2) is given by \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
Choose the correct answer from those given below.
(a) Statement 1 is true and Statement 2 is false.
(b) Statement 1 is false and Statement 2 is true.
(c) Both the statements are true, and Statement 2 is the correct reason for Statement 1.
(d) Both the statements are true, but Statement 2 is not the correct reason for Statement 1. (1 mark)
Answer:
(c) Both the statements are true, and Statement 2 is the correct reason for Statement 1.
Question 22.
The vertices of a quadrilateral ABCD are on a circle. ∠A = x, ∠B = 3y, ∠C = 4x, ∠D = 2y

(a) Find x, y
(b) Find the angles of ABCD (3 mark)
Answer:
a) x + 4x = 180 ⇒ 5x = 180, x = 36
Angles are 36°, 144°
b) 2y + 3y = 180 ⇒ 5y = 180, y = 36°
Angles are 72°, 108°
Question 23.
A) Algebraic form of an arithmetic sequence is \(\frac{n+1}{3}\)
(a) What is its common difference ?
Answer:
\(\frac{1}{3}\)
(b) At what position 15 becomes a term of the sequence?
Answer:
44th term is 15
OR
B) The numbers 2,3,4 are written in small paper pieces kept in one box. Fractions \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\) are written in paper pieces and kept in another box. One is taken from each box and find the product
(a) What are the outcome pairs
Answer:

Products are 1, \(\frac{2}{3}\), \(\frac{1}{2}\), \(\frac{3}{2}\), 1, \(\frac{3}{4}\), 2, \(\frac{4}{3}\), 1
(b) What is the probability of getting an integer as the product. (3 mark)
Answer:
\(\frac{4}{9}\)
Question 24.
A) First term of an arithmetic sequence is \(\frac{1}{2}\) and common difference \(\frac{3}{4}\)
a) What is the algebra of this sequence?
Answer:
xn = dn + (f – d)
= \(\frac{3}{4}\)n + \(\left(\frac{1}{2}-\frac{3}{4}\right)\)
= \(\frac{3 n-1}{4}\)
b) At what position 50 becomes a term of the sequence
Answer:
\(\frac{3 n-1}{4}\) = 50
⇒ 3n – 1 = 200
3n = 201
n = 67
c) What is the sum of first 11 terms?
Answer:
Sum = \(\frac{3(1+2+3+\cdots+11)-11}{4}=\frac{187}{4}\)
OR
B) In the figure PA is the tangent from P to A on the circle and OA is the radius. If OP = 18 and ∠OPA = 40° then
(a) If the workers are arranged on the basis of their daily wages, at what position does the median wage fall?
(b) What is the median class?
(c) Find the median of the wages. (5 mark)
[sin 40° = 0.64, cos 40° = 0.76, tan 40° = 0.83]

(a) What is the length of tangent?
Answer:
cos 40 = \(\frac{O A}{O P}=\frac{O A}{18}\)
0.76 = \(\frac{A P}{18}\)
AP = 0.76 × 18
= 13.68 cm
(b) What is the radius of the circle? (4 mark)
Answer:
sin 40 = \(\frac{O A}{O P}=\frac{O A}{18}\)
OA = 11.52 cm
Question 25.
A cone made of wax has base radius 15 cm and slant height 25 cm.
(a) What is the height of the cone?
Answer:
Height = \(\sqrt{25^2-15^2}\) = 20 cm
(b) What is the volume of the cone?
Answer:
Volume = \(\frac{1}{3}\) × π × 152 × 20
= 1500 π cm3
(c) If it is melted to form small cones of radius 1 cm and height 4 cm, what is the volume of the small cone?
Answer:
Volume of the small cone
= \(\frac{1}{3}\) × π × 12 × 4
= \(\frac{4}{3}\) π cm3
(d) How many small cones can be made? (5 mark)
Answer:
Number of the small cone

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Question 26.
The daily wages of 99 workers in a factory is shown in the table.
| Daily wages | Number of Workers |
| 500-600 | 8 |
| 600- 700 | 13 |
| 700-800 | 20 |
| 800 – 900 | 25 |
| 900- 1000 | 19 |
| 1000- 1100 | 14 |
(a) If the workers are arranged on the basis of their daily wages, at what position does the median wage fall?
Answer:
| Daily wages | Number of Workers |
| Below 600 | 8 |
| Below 700 | 21 |
| Below 800 | 41 |
| Below 900 | 66 |
| Below 1000 | 85 |
| Below 1100 | 99 |
Here n = 99, an odd number 99 + 1 = 100 = 50
So, 50th worker wage is the median wage That is, the median wage falls at the position.
(b) What is the median class?
Answer:
800-900 is the median class
(c) Find the median of the wages. (5 mark)
Answer:
Class width = 900 – 800 = 100
Number of workers in median class = frequency of the class = 25
Length of the subdivision = \(\frac{100}{25}\) = 4
Wage of 42nd worker = \(\frac{800+804}{2}\) = 802
Wage of 50th worker = 802 + (50 – 42) × 4
= 802 + 32
= 834






































