Kerala SSLC Maths Question Paper March 2022 with Answers English Medium

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Kerala Syllabus Class 10 Maths Question Paper March 2022 with Answers English Medium

Time: 2½ Hours
Total Score: 80

Instructions:

• In addition to the writing time, there is a 15-minute ‘cool-off time’. Use this time to familiarize yourself with the questions and plan your answer.
• Questions with different scores are given as distinct parts.
• Read the instructions carefully before answering the questions.
• Keep in mind, the score and time while answering the questions.
• The maximum score for questions from 1 to 24 will be 40.
• There is no need to simplify irrationality like √2, √3, π, etc, using approximations unless you are asked to do so.

Part – I
Questions from 1 to 10 carry 1 score each.

A. Answer any four questions from 1 to 6. (4 × 1 = 4)

Question 1.
What is the common difference between the arithmetic sequences 3, 7, and 11?
Answer:
d = 7 – 3 = 4

Question 2.
In the figure ∠C = 110°. Find the measure of ∠A.

Answer:
∠A = 180° – 100° = 70°

Question 3.
A box contains 7 white balls and 3 black balls. If a ball is taken from it, what is the probability of it being black?
Answer:
\(\frac{3}{10}\)

Question 4.
Find the distance between the points (0, 0) and (4, 0).
Answer:
4

Question 5.
From the circle of radius 12 centimeters, a sector of central angle 90° is cut out and made into a cone. What is the base radius of this cone?
Answer:
3 cm

Question 6.
If (x – 1) is a factor of the polynomial p(x), write P(1).
Answer:
0

B. Answer all questions from 7 to 10, Choose the correct answers from the bracket. (4 × 1 = 4)

Question 7.
What is the value of tan x if x = 30°?
\(\left(\frac{1}{2} ; \frac{1}{\sqrt{2}} ; \frac{1}{\sqrt{3}} ; \sqrt{3}\right)\)
Answer:
\(\frac{1}{\sqrt{3}}\)

Question 8.
If the perimeter of a triangle is 24 centimeters and its inradius is 2 centimeters, find its area in square centimeters.
(12; 20; 24; 26)
Answer:
24 sq. cm

Question 9.
The lateral faces of a square pyramid are equilateral triangles. If the length of one base edge is 20 centimeters, what will be the measure of its slant height in centimeters?

(10; 10√2; 10√3; 20)
Answer:
10√3

Question 10.
The equation of a line is 2x + y = 5 if the x coordinate of a point on this line is 2, what is the coordinate of this point?
(0; 1; -1; 2)
Answer:
1

Part – II
Questions from, 11 to 18 carry 2 scores each.

A. Answer any three questions from 11 to 15. (3 × 2 = 6)

Question 11.
5, 8, 11,…… is an arithmetic sequence.
(a) What is the 20th term? (1)
(b) What is the algebraic expression for this sequence? (1)
Answer:
(a) x₂₀ = f + 19d
= 5 + 19 × 3
= 62
(b) 3n + 2

Question 12.
A triangle is drawn by joining the mid-point of one side of a parallelogram and the endpoints of the opposite side. The triangle is shaded as shown in the figure.

(a) What is the area of the triangle, if the area of the parallelogram is 50 square centimetres? (1)
(b) Find the probability of a dot put without looking, into the figure. (1)
Answer:
(a) \(\frac{50}{2}\) = 25 cm²
(b) \(\frac{1}{2}\)

Question 13.
A ladder leans against a wall. The ladder makes an angle of 60° with the floor. The length of the ladder is 6 meters.

(a) What is the height of the top of the ladder from the ground? (1)
(b) How far is the foot of the ladder from the wall? (1)
Answer:
(a) 3√3 meter
(b) 3 meter

Question 14.
Write the second-degree polynomial x2 + x as the product of two first-degree polynomials. (2)
Answer:
x² + x = x(x + 1)

Question 15.
The weight of 7 pupils in a class is given (in kilograms). Find the median weight. (2)
35, 43, 38, 45, 32, 44, 42
Answer:
32, 35, 38, 42, 43, 44, 45
The median is 42.

B. Answer any two questions from 16 to 18. (2 × 2 = 4)

Question 16.
The algebraic expression for the sum of n terms of an arithmetic sequence is n2 + n.
(a) Find the first term of this arithmetic sequence. (1)
(b) Find the sum of the first 10 terms of this arithmetic sequence. (1)
Answer:
(a) First term x₁ = 12 + 1 = 2
(b) Sum of the first 10 terms = 10² + 10 = 110

Question 17.
In the figure PA = 4 centimetres, AB = 5 centimetres and PC is a tangent to the circle. Find the length of the PC. (2)

Answer:

PA × PB = PC²
⇒ 4 × 9 = PC²
⇒ PC = √36
⇒ PC = 6

Question 18.
Find the coordinates of the point that divides the line joining the points (1, 2) and (7, 5) in the ratio 2 : 1. (2)
Answer:
x = x₁ + \(\frac{p}{p+q}\) (x₂ – x₁)
= 1 + \(\frac{2}{3}\) × 6
= 5
y = y₁ + \(\frac{p}{p+q}\) (y₂ – y₁)
= 2 + \(\frac{2}{3}\) × 3
= 4

Part – III
Questions 19 to 25 carry 4 scores each.

A. Answer any three questions from 19 to 23. (3 × 4 = 12)

Question 19.
Draw a triangle of circumradius 3 centimeters and two of its angles 50° and 60°. (4)
Answer:
Draw a circle of radius 3 cm.
Divide the angle around the center as 2 × 50 and 2 × 60 by drawing radii.
Join the ends of the radius.
This makes a triangle with angles of 50° and 60°.

Question 20.
A strip of width 4 centimeters is attached to one side of a square to form a rectangle. The area of the new rectangle is 77 square centimeters.

(a) If we take the width of the new rectangle as x, what will be its length? (1)
(b) Find the measure of the side of the square by constructing an equation. (3)
Answer:
(a) x + 4
(b) x(x + 4) = 77
⇒ x² + 4x + 4 = 81
⇒ (x + 2)² = 9²
⇒ x + 2 = 9
⇒ x = 7

Question 21.
Draw a circle of radius 2.5 centimeters and mark a point 6 centimeters away from the center of the circle. Draw tangents to the circle from this point. (4)
Answer:
Draw a circle of radius 2.5 cm and center O. Mark a point P at the distance 6 cm away from the center of the circle.
Draw a circle with a diameter OP. This circle cut the first circle at A and B. Join PA and PB. These are the tangents.

Question 22.
Find the surface area of a cone having a base radius of 9 centimeters and a height of 12 centimeters. (4)
Answer:
l² = 12² + 9² = 225
⇒ l = 15
Total Surface Area = Base Area + Curved Surface Area
= π × 9² + π × 9 × 15
= 81π + 135π
= 216π cm²

Question 23.
The coordinates of three vertices of a parallelogram are given.

(a) Find the coordinates of the vertex C. (2)
(b) Find the coordinates of the midpoint of the diagonal AC. (2)
Answer:
(a) C(8 + 4 – 2, 4 + 6 – 2) = C(10, 8)
(b) Midpoint of the diagonal AC = \(\left(\frac{10+2}{2}, \frac{8+2}{2}\right)\) = (6, 5)

B. Answer any one question from 24 and 25. (1 × 4 = 4)

Question 24.
One box contains four slips numbered 1, 2, 3, and 4 and another box contains five slips numbered 5, 6, 7, 8, 9. If one slip is taken from each box.
(a) How many number pairs are possible? (1)
(b) What is the probability of both being odd? (1)
(c) What is the probability of getting the sum of the numbers 10? (2)
Answer:
(a) The number of pairs = 4 × 5 = 20
(b) Number of favourable outcomes = 2 × 3 = 6
Probability of getting both odd = \(\frac{6}{20}=\frac{3}{10}\)
(c) Favourable outcomes (1, 9), (2, 8), (3, 7), (4, 6)
Probability of getting the sum 10 is \(\frac{4}{20}=\frac{1}{5}\)

Question 25.
Two sides of a parallelogram are 20 centimeters and 10 centimeters. If the angle between them is 40°.

(a) What is the height of the parallelogram? (2)
(b) Find the area of the parallelogram. (2)
(sin 40° = 0.64; cos 40° = 0.77; tan 40° = 0.84)
Answer:

(a) sin 40° = \(\frac{h}{10}\)
h = 10 × sin 40°
= 10 × 0.64
= 6.4 cm
(b) Area = 20 × h
= 20 × 6.4
= 128 sq. cm

Part – IV
Questions from 26 to 32 carry 6 scores each.

A. Answer any three questions from 26 to 29. (3 × 6 = 18)

Question 26.
(a) In the figure AB is the diameter of the circle. Line CD is perpendicular to AB. AP = 8 centimetres and PB = 2 centimetres. Find the length of the PC. (2)

(b) Draw a rectangle of sides 5 centimeters and 3 centimeters. Draw a square of the same area. (4)
Answer:
(a) PC² = 8 × 2
PC = √16 = 4
(b) 1. Draw the rectangle ABCD. AB = 5 cm, BC = 3 cm.
2. Produce AB and mark the point E such that BC = BE.
3. Draw a semicircle of diameter AE. Produce BC, meet the semicircle at F.
4. BA × BE = BF² can be written as AB × BC = BF². AB × BC is the area of the rectangle.
5. Draw a square of side BF. The area of the rectangle is equal to the area of the square as per the relation AB × BC = BF².

Question 27.
In the figure, AC is the diameter of the circle. Given that AC = 20 centimeters. ∠BAC = 60° and ∠ACD = 45°.

(a) What is the measure of ∠ADC? (1)
(b) Find the perimeter of the quadrilateral ABCD. (5)
Answer:
(a) ∠ADC = 90°
(b) ∠ABC = 90°
AB = 10
BC = 10√3
CD = 10√2
AD = 10√2
Perimeter of ABCD is 10 + 10√3 + 10√2 + 10√2.

Question 28.
The rectangle has sides parallel to the axes. The coordinates of one pair of opposite vertices are (2, 1) and (7, 5).

(a) Find the coordinates of the other two opposite vertices. (2)
(b) Find the length and breadth of the rectangle. (2)
(c) Find the length of the diagonal AC. (2)
Answer:
(a) Co-ordinates of B(7, 1)
Co-ordinates of D: D(2, 5)
(b) Length = 5 cm, Breadth = 4 cm
(c) AC² = 5² + 4²
AC = √41 cm

Question 29.
The radius of a solid metal sphere is 6 centimeters.
(a) Find the volume of the sphere. (3)
(b) This sphere is melted and recast into a solid cone of radius 6 centimeters. Find the height of the cone. (3)
Answer:
(a) Volume of the sphere = \(\frac{4}{3}\)πr³ = 288π
(b) The volume of the melted sphere is equal to the volume of the cone so formed.
Volume of the cone = \(\frac{1}{3}\)πr²h = 12πh
∴ 288π = 12πh
⇒ h = 24 cm

B. Answer any two questions from 30 to 32. (2 × 6 = 12)

Question 30.
The product of a number and 5 more than that number gives 104.
(a) If we take the first number as ‘x’, what will be the second number? (1)
(b) Form a second-degree equation using the given details. (2)
(c) Find the number. (3)
Answer:
(a) x + 5
(b) x(x + 5) = 104
⇒ x² + 5x – 104 = 0
(c) x = \(\frac{-5 \pm \sqrt{5^2-4 \times 1 \times-104}}{2 \times 1}\) = 8, -13

Question 31.
Consider the second-degree polynomial p(x) = x² – 3x + 5.
(a) Find p(1). (1)
(b) Write one first-degree factor of the polynomial p(x) – p(1). (1)
(c) Write p(x) – p(1) as the product of two first-degree factors and find the solutions of the equation p(x) – p(1) = 0. (4)
Answer:
(a) p1) = 1³ – 3 × 1 + 5
= 1 – 3 + 5
= 3
(b) x – 1
(c) p(x) – p(1) = x² – 3x + 5 – 3 = x² – 3x + 2
P(x) – P(1) = (x – 1)(x – 2)
Solutions are 1, 2.

Question 32.
The table below shows the households of an area sorted according to consumption of electricity.

Consumption (in units)	Number of Households
100 – 120	4
120 – 140	8
140 – 160	7
160 – 180	10
180 – 200	6
200 – 220	4
220 – 240	6

(a) If the households are arranged according to the consumption of electricity, the consumption of which house is taken as median? (1)
(b) What is the consumption of the 20th household according to our assumption? (2)
(c) What is the median consumption? (3)
Answer:
Prepare cumulative frequency table.

Consumption	Number of Households
Below 120	4
Below 140	12
Below 160	19
Below 180	29
Below 200	35
Below 220	39
Below 240	45

(a) Since the total number of houses is 45, an odd number the 23rd house comes in the middle.
The consumption of the 23rd house is considered the median.
(b) From the cumulative frequency table we can say median domes between 160-180.
When 20 units between 160 and 180 are divided into 10 equal parts each one share is \(\frac{20}{10}\) = 2.
We assume that the consumption of electricity in the median class ate in an arithmetic sequence.
Consumption of 20th house is assumed as 160 + \(\frac{2}{2}\) = 160 + 1 = 161
(c) First f = 161 and common difference 2.
The median is the consumption of the 23rd house.
It is f + 3d = 161 + 3 × 2 = 167.

Part – V
Questions from 33 to 35 carry 8 scores each.

A. Answer any two questions from 33 to 35. (2 × 8 = 16)

Question 33.
6, 10, 14,….. is an arithmetic sequence.
(a) Find the sum of the first 15 terms of the arithmetic sequence. (4)
(b) What is the difference between the first term and the 16th term? (2)
(c) Find the difference between the sum of the first 15 terms and the sum of the next 15 terms. (2)
Answer:
(a) We consider the first 15 terms.
The 8th term is the middle term.
x₈ = 6 + 7 × 4 = 34
x₁ + x₁₅ = 2 × x₈ = 68
Sum = (x₁ + x₁₅) × \(\frac{15}{2}\)
= 68 × \(\frac{15}{2}\)
= 510
(b) x₁₆ – x₁ = 15d
= 15 × 4
= 60
(c) x₁₇ – x₂ = 60, x₁₈ – x₃ = 60………x₃₀ – x₁₅ = 60
The required difference is 15 × 60 = 900

Question 34.
(a) The two tangents AC and BC of the circle with center O meet at C. What is the measure of ∠OAC? If ∠AOB = 110°, find the measure of ∠ACB. (2)

(b) Draw a circle of radius 2.5 centimeters. Draw a triangle with angles 50°, 60°, 70°, and all its sides are tangents to this circle. (6)
Answer:
(a) ∠OAC = 90°
∠ACB = 180° – 110° = 70°
(b) Draw a circle of radius 2.5 cm.
Divide the angle around the center as 180° – 50° = 130° and 180° – 60° = 120° by drawing the radii.
Draw tangents to the circle at the ends of the radii.
The tangents make the triangle.

Question 35.
(a) Draw the coordinate axes and mark the points (2, 1) and (4, 3). (3)
(b) Find the slope of the line joining these points. (2)
(c) The center of a circle is (3, 2) and the coordinates of one end of its diameter are (1, 2). Find the coordinates of the other end of the diameter. (3)
Answer:
(a) Draw the figure as given below.

(b) Slope = \(\frac{3-1}{4-2}\) = 1
(c) (5, 2)