Kerala Syllabus Class 10 Maths Model Question Paper Set 3

Practicing Maths Question Paper Class 10 Kerala Syllabus Set 3 English Medium helps identify strengths and weaknesses in a subject.

Class 10 Maths Kerala Syllabus Model Question Paper Set 3

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.

Question 1.
13th term of an arithmetic sequence is 20. What is the sum of 1st and 25th term? (1 mark)
(a) 30
(b) 50
(c) 40
(d) 20
Answer:
(c) 40

Question 2.
Consider the following statements.
Statement (A): The area of a rectangle with sides (x+5) and (x-3) is a second-degree polynomial.
Statement (B): A second-degree polynomial can be written as the product of two first-degree polynomials.
(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:
(d) Both statements are true. Statement B is not the reason for Statement A. (1 mark)

Statement B is true.
The area of a rectangle with sides (x + 5) and (x – 3) = (x + 5) (x – 3)
= x² +5x – 3x – 15
= x² + 2x -15
a second degree polynomial Statement A is true.

Question 3.
(A) The shadow of a vertical tower on level ground increases by 10 m when the altitude of the Sun changes from the angle of elevation 45° to 30°.
(a) Draw a rough diagram
Answer:

(b) Calculate the height of the tower.
Answer:
tan 45 = \(\frac{A B}{A C}\)
⇒ 1 = \(\frac{h}{x}\)
⇒ h = x
tan 30 = \(\frac{1}{\sqrt{3}}=\frac{h}{x+10}\)
⇒ x + 10 = √3h
⇒ h = 13.65 metre

OR

(B) Sum of a number and its square is 0 . If x is the number then
(a) Write the equation.
(b) What are the numbers? (3 mark)
Answer:
(a) x² + x = 0
(b) x(x + 1) = 0
⇒ x = 0, x = – 1

Question 4.
In the figure O is the centre of the circle, PA is the tangent and OP = 24 cm

(a) What are the angles of triangle ?
(b) What is the radius of the circle?
(c) Find the length of tangent (4 mark)
Answer:
(a) ∠POA = 180 – 120 = 60°
∠PAO = 90°
∠OPA = 30°

(b) 12 cm

(c) 12√3 cm

Question 5.
(A) A die in which the numbers 1 to 6 are written on the faces is thrown.
(a) What is the probability of falling an even numbered face?
(b) What is the probability of getting an odd numbered face?
(c) What is the probability of getting a prime numbered face?
Answer:
(a) Probability of falling even face = \(\frac{3}{6}=\frac{1}{2}\)
(b) Probability of falling odd face = \(\frac{3}{6}=\frac{1}{2}\)
(c) Probability of falling prime numbered face = \(\frac{3}{6}=\frac{1}{2}\)

OR

(B) A(3, 2) B(9, 2) and C(5, 7) are the vertices of a triangle.
(a) What is the length of the side parallel to x axis.
(b) What is the altitude to that side?
(c) Calculate the area of triangle ABC (4 mark)
Answer:
(a) |9 – 3| = 6
(b) |7 – 2| = 5
(c) \(\frac{1}{2}\) × 6 × 5 = 15 sq. units

Question 6.
If each side of a square is decreased by 2 metres, its area become 49 square metres. What is the side of the original square? (5 mark)
Answer:
Let, the side of the original square be x.
(x – 2)² = 49
⇒ x² – 4x + 4 = 49
⇒ x² – 4x – 45 = 0
a = 1, b = -4, c = -45

x is the length of the side of square, so it cannot be negative.
∴ x = 9

Section – B

Question 7.
In the figure AB = AC = 12 cm. If AP = 4 cm then what is the length of BC? (1 mark)

(a) 14 cm
(b) 18 cm
(c) 10 cm
(d) 16 cm
Answer:
(d) 16 cm
BP = CQ = 8, BR = CR = 8
BC = 16 cm

Question 8.
AB and CD intersect at P. O is the centre of the circle. CD is the diameter of the circle. OP = 2, PA = 9, PB = 5, then what is its radius?

(a) 7
(b) 8
(c) 4
(d) 3 (1 mark)
Answer:
(a) 7
(r + 2)(r – 2) = 9 × 5
r² – 22 = 45,
r² = 49
r = 7

Question 9.
From a wooden square prism, a square pyramid of same base area and height is carved out. Base area 400 sq.cm, height 24 cm.
(a) What is the volume?
(b) What is the slant height?
(c) Find the surface area of the pyramid. (3 mark)
Answer:
(a) \(\frac{1}{3}\) × 400 × 24 = 3200 cm³
(b) Slant height = \(\sqrt{10^2+24^2}=\sqrt{676}\) = 26 cm
(c) Surface area = 400 + 4 × \(\frac{1}{2}\) × 20 × 26 = 1440cm

Question 10.
If the endpoints of the diameter are A (-1, 0) and B (1,0):
(a) Write the coordinates of the centre of the circle.
Answer:
Centre of the circle is the midpoint of AB.
Centre = \(\left(\frac{-1+1}{2}, \frac{0+0}{2}\right)\) = (0, 0)

(b) What is the radius? (3 mark)
Answer:
AB = \(\sqrt{(1-(-1))^2+(0-0)^2}=\sqrt{(2)^2}\) = 2
Radius r = \(\frac{A B}{2}=\frac{2}{2}\) = 1

Question 11.
Draw a square with an area of 24 square centimeters. Draw a rectangle with an area of 24 square centimeters. (4 mark)
Answer:

• Consider a rectangle with length 6 cm and height 4 cm.
• Add the height of the rectangle to the length of the base. The new length is 6+4 = 10 cm.
• Now draw a semicircle with the bottom line as diameter, extend the right side of the square and meet it with the semicircle.
• This line is the side of the square.
• BSTF is the required square.

Question 12.
(A) Look at the sequence given below. It is the sequence of equilateral triangles of side 2, 4, 6………….

(a) Write the sequence of the length of altitudes?
(b) Find the altitude of 10th triangle
(c) Write the area of its tenth triangle?
(d) Write the algebraic form of the sequence of altitudes.
Answer:
(a) √3, 2√3, 3√3……………..
(b) 10√3
(c) 100√3
(d) x_n = n√3

OR

(B) Two digit numbers are written in small paper pieces and placed in a box.
(a) How many paper slips are there in the box?
(b) If one is taken from the box, what is probability of getting a number with digits same?
(c) If one is taken from the box, what is probability of getting a number in which the product of the digits a prime number.
(d) What is the probability of getting a prime number? (4 mark)
Answer:
(a) 10,11,12…99 are the numbers. There are 90 numbers

(b) Numbers with same digits are 11,22,33,44,55, 66,77,88,99
Total number of these numbers is 9
Probability = \(\frac{9}{90}=\frac{1}{10}\)

(c) In the two digit numbers with product of the digits a prime, one digit is 1 and other digit is one of the numbers 2,3,5,7
Numbers are 12, 13, 15, 17, 21, 31, 51, 71
Probability = \(\frac{8}{90}=\frac{4}{45}\)

(d) There are 25 prime numbers below 100.4 of them are one digit primes and the rest of the 21 numbers are two digit primes.
Probability is = \(\frac{21}{90}=\frac{7}{30}\)

Question 13.
Draw coordinate axes and mark the point A(-2,-2).
(a) Move 4 unit parallel to y axis in the positive direction and mark the coordinates of B.
(b) Move 6 unit to the right from B parallel to axis and mark C with the coordinates.
(c) Move 4 unit parallel to y axis up and mark D with its coordinates.
(d) Find the distance AD. (5 mark)
Answer:
Draw coordinate axes and mark the points
(a) B(-2, 2)
(b) C(4, 2)
(c) D(4, 6)
(d) Distance between A(-2, -2) and is
AD = \(\sqrt{(-2-4)^2+(-2-6)^2}=\sqrt{36+64}\) = 10

Question 14.
(A) Sum of the first 3 terms of an arithmetic sequence is 15. Sum of the first 4 terms is 28
(a) What is the 4th term?
(b) What is the sum of first 7 terms of the sequence?
(c) What is the third term ?
(d) Express the sequence algebraically
Answer:
(a) x₄ =28-15 = 13
(b) x₄ x 7 = 91
(c) 2d = 13 – 5 = 8, d = 4, x₃ =13 – 4 = 9
(d) x_n = 4n – 3

OR

(B) Integers denoted by x from -4 to 4 are written in small paper pieces and placed in a box. One is taken from the box at random.
(a) How many papers are there in the box ?
(b) What is the chance of getting an integer |x| < 3
(c) What is the probability of getting |x| = 1 (5 mark)
Answer:
(a) 9
(b) \(\frac{5}{9}\)
(c) \(\frac{2}{9}\)

Section – C

Question 15.
The radii of two hemispheres are in the ratio 1: 2. What is the ratio of their surface areas ?
(a) 1 :2
(b) 1 :8
(c) 1 :4
(d) 1:16 (1 mark)
Answer:
(c) 1 :4

Ratio of radii is 1 :2
Surface area of hemisphere = 3πr²
In surface area radius is in second degree.
So, ratio of their surface area = 1² : 2² = 1:4

Question 16.
Distance from 10th term to 15th term is 11. What is the distance from 10th term to 20th term?
(a) 33
(b) 10
(c) 22
(d) 20 (1 mark)
Answer:
(c) 22

Question 17.
Consider the arithmetic sequence 1, 4, 7, 10
(a) Write the nth term of the sequence?
(b) Find the expression for the sum of first n terms of this sequence.
(c) Calculate the sum of first 20 terms of this sequence. (3 mark)
Answer:
(a) x_n = 3n – 2

(b) Sum = (x₁ + x_n) × \(\frac{n}{2}\)
= (1 + 3n – 2) × \(\frac{n}{2}\)
= \(\frac{3 n^2}{2}-\frac{n}{2}\)

(c) \(\frac{3 \times 20^2}{2}-\frac{20}{2}\) = 590

Question 18.
Draw a circle of radius 3 cm. Draw a triangle of angles 45°, 65°, 70° and vertices on this circle. (4 mark)
Answer:
Draw a circle of radius 3 cm.

Mark the centre as O.
If 45° is the angle on the circle.
Then angle on the centre = 45° × 2 = 90°
If 65° is the angle on the circle.
Then, angle on the centre = 65 × 2 = 130°
Draw the radius OA.
Draw the radius OB which makes an angle 90° with OA.
Draw the radius OC which makes an angle 130 with OB.
Join AB,BC and AC to get the required triangle with ∠ACB = 45°, ∠CBA = 70° and ∠BAC = 65°

Question 19.
(A) A tall building and a short building are standing on a level ground. The angle of elevation of the top of the short building from the foot of the tall building is 30°. The angle of elevation of the top of the tall building from the foot of the short building is 60°. The tall building has height 50 m.
(a) Draw a rough diagram.
Answer:

(b) What is the distance between the buildings.
Answer:
tan 60° = \(\frac{50}{x}\), √3 = \(\frac{50}{x}\)
x = \(\frac{50}{\sqrt{3}}=\frac{50}{1.73}\) = 28.9 m
The distance between the buildings =28.9 m

(c) Calculate the height of the short building.
Answer:
tan 30 = \(\frac{h}{x}\)
\(\frac{1}{\sqrt{3}}=\frac{h}{28.9}\)
3h = 28.9
h = \(\frac{28.9}{1.73}\) = 16.7 m

OR

(B) A two digit number is 4 times the sum of the digits. Also the number is 3 times the product of the digits.
(a) Form an equation by taking x,y as the digits.
Answer:
Digit in tens place x, digit in one’s place y
Number is 10x + y
10x + y = 4(x + y)
10 x + y = 3 xy

(b) Make a second degree equation using the given condition.
Answer:
10x + y = 4x + 4y
6x = 3 y, y = 2x
10x + y = 3xy
⇒ 10x + 2x = 3x × 2x
12x = 6x²
6x² – 12x = 0

(c) Find the numbers. (5 mark)
Answer:
x = 0, x = 2.
Tens place cannot be 0.
Tens place = 2, one’s place 2x = 4
Number = 24

Section – D

Question 20.
AB is the diameter of the circle. P divides the diameter of the circle into a and b as shown in the diagram. PC is the perpendicular to AB. The point P divides AB in the ratio a : b. The area of the square with side PC is :

(a) a × b
(b) a + b
(c) 7
(d) a – b (1 mark)
Answer:
(a) ab
a × b = PC²

Question 21.
Read the two statements given below.
Statement 1: The numbers 7, 10, 13, 16, 19 have median = mean.
Statement 2: The numbers form an arithmetic sequence with common difference 3.

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.
Algebraic form of an arithmetic sequence is 3n + 7
(a) What is the common difference of this sequence?
(b) What-is the difference between 10th term and 20th term of this sequence?
(c) Can 100 the difference between any two terms of this sequence? Why? (3 mark)
Answer:
(a) 3
(b) x₂₀ – x₁₀ = 10d = 10 × 3 = 30
(c) No. The difference between two terms will be a multiple of 3.

Question 23.
A) The sums of the first n terms of three arithmetic sequences are S₁, S₂ and S₃.
The first term of each is 1 and common differences are 1, 2 and 3 in the order.
(a) Write the sum of first n terms of these sequences.
(b) Prove that S₁ + S₃ = 2S₂
Answer:
(a) S1 = 1 + 2 + 3 + ………. s₁ = \(\frac{n}{2}\) (n +1)
s₂ = 1 + 3 + 5 ….. s₂ = n²
s₃ = 1 + 4 + 7 ……….
nth term is 3n – 2.
Sum of first n terms = \(\frac{3 n^2}{2}-\frac{n}{2}\)

(b) S₁ + S₃ = \(\frac{1}{2}\)
On simplifying S₁ + S₃ = 2n² = 2S₂

OR

B) Number of people get vaccinated in various ages at-a community health centre is listed below.
(a) At what age limit does the median occur? What is the age of the 49th person vaccinated in the camp if all are arranged in the ascending order of the age?
(b) Calculate the median age. (3 mark)
Answer:
(a) n =109 (odd number)
\(\) th person comes in the middle. The cumulative frequency table is 55th person comes in the middle.
This belongs to the class 40- 60.
Thus, Median comes in the age limit 40 – 60.

(b) Class width is 20.
Number of people in the median class is 40.
Width of a subdivision = \(\frac{20}{40}=\frac{1}{2}\)
The age of 49 th person is \(\frac{40+40.5}{2}\) = 40.25
Median age = Age of 55th person
= 40.25 + (55 – 49) × 0.5
= 40.25 + 6 × 0.5
= 40.25 + 3
= 43.25

Question 24.
(A) Sum of the fifth and 6th terms of an arithmetic sequence is 32.
(a) What is the sum of first term and tenth term of the sequence?
(b) If the 5th term is 11 then what is its 6th term and common difference?
(c) What is the 1st term ?
(d) Express this sequence algebraically.
Answer:
(a) 32
(b) x₆ = 32 – 11 = 21, d = 21 – 11 = 10
(c) x₁ = x₅ – 4d = -29
(d) x_n = 10n – 39

OR

(B) A bag contains 4 black beads and 3.white beads. Another bag contains 4 black beads and 5 white beads. One is taken from each bag.
(a) What is the probability of getting both white?
(b) What is the probability of getting both black?
(c) What is the probability of getting one white and one black?
(d) What is the probability of getting atleast one black? (4 mark)
Answer:
(a) First bag contains 7 beads and second bag contains 9 beads.
When one from each box are taken we get = 7 × 9 = 63 pairs.
Number of pairs with both white is, = 3 × 5 = 15
Probability of getting both white is = \(\frac{15}{63}=\frac{5}{21}\)

(b) Number of pairs with both black is. = 4 × 4= 16
Probability of getting both black is = \(\frac{16}{63}\)

(c) Probability of getting one white and one black is
= \(\frac{4 \times 5+3 \times 4}{63}=\frac{32}{63}\)

(d) Probability of getting atleast one black
= \(\frac{4 \times 5+3 \times 4+4 \times 4}{63}=\frac{48}{63}=\frac{16}{21}\)

Question 25.
In the figure O is the centre of the circle, ∠BAO = 20°, ∠BCO = 10°

(a) What is the measure of angle ABC ?
Answer:
In triangle OAB.OA = OB.
Angles opposite to the equal sides are equal.
Similarly in the case of triangle OBC also.
∠ABC = 20 + 10 = 30°

(b) What is the measure of angle AOC ?
Answer:
∠AOC = 2 × 30 = 60°

(c) What is the measure of angle ADC ?
Answer:
∠ADC = 180-30 = 150°

(d) Find the angles of triangle AOC
Answer:
Triangle AOC,OA = OC
∠OAC = ∠OCA = 180 – 60 = 60°
∆OAC is an equilateral triangle. Angles are 60° each.

(e) If the diametre of the circle is 10 cm then find the length of the chord AC (5 mark)
Answer:
OA = AC = OC = 5 cm , radius 5 cm.

Question 26.
Look at the pattern given below
This pattern is made by using the terms of the arithmetic sequence
2, 5, 8, 11, 14, 17
2
5, 8
11, 14, 17
(a) Write the next line of the pattern?
Answer:
20, 23, 26, 29

(b) Find the total number of numbers in the first 9 lines
Answer:
\(\frac{9(9+1)}{2}\) = 45

(c) Which is the first number of 10th line.
Answer:
It is 46,h term of the sequence
2, 5, 8 …… x_n = 3n – 1
x₄₆ =3 × 46 – 1 = 137

(d) Which is the last number of 10th line?
Answer:
It is 55th term. x₅₅ = 3 × 55 – 1 = 164

(e) Find the sum of numbers in 10th line. (5 mark)
Answer:
(137 + 164) × 5 = 301 × 5 = 1505