Kerala SSLC Maths Board Model Paper March 2022 with Answers English Medium

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

Time: 2 hours
Score: 80

Instructions:

• There is a ‘Cool-off time’ of 15 minutes in addition to the writing time. Use this time to get familiar with questions and to 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.
• No need to simplify irrationality like √2, √3, π, etc, using approximations unless asked.

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.
Write the first three terms of the arithmetic sequence with first term 6 and common difference 4.
Answer:
6, 10, 14, 18,……

Question 2.

Suppose we draw a circle with AB as diameter. Among the points C, D, E which lies on the circles?
Answer:
D is on the circle.

Question 3.
A line parallel to x-axis passes through the point (2, 1). Find the co-ordinates of the point on this line intersecting with y-axis.
Answer:
(0, 1)

Question 4.

In the figure, AB is the diameter of the circle. Calculate the probability of a dot put inside the circle, without looking, to be within the non-shaded region.
Answer:
\(\frac{1}{2}\)

Question 5.
The radii of two hemispheres are in the ratio 1 : 2. What is the ratio of their surface areas? (1)
Answer:
1² : 2² = 1 : 4

Question 6.
p(x) = x² + 2x. Find the number to be subtracted from p(x) to get a polynomial for which x – 1 is a factor. (1)
Answer:
3

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

Question 7.

From the figure, which among the following is tan x?
\(\left[\frac{3}{5} ; \frac{4}{5} ; \frac{3}{4} ; \frac{4}{3}\right]\)
Answer:
\(\frac{3}{4}\)

Question 8.

In the figure chord, BA extended and the tangent at C meets at P. PA = 4 centimeters and PB = 9 centimeters. What is the area of the square with a side PC?
[6, 36, 13, √6]
Answer:
36

Question 9.
The equation of a line is y = 2x. Which of the following is not a point on this line?
[(1, 2); (5, 10); (\(\frac{1}{2}\), 1); (3, 1)]
Answer:
(3, 1)

Question 10.

The figure shows one lateral face of a square pyramid. Its sides are 5 centimeters, 5 centimeters, and 6 centimeters. What would be the slant height of a square pyramid in centimetres?
[3; 4; 5; 6]
Answer:
4 cm

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.
The sum of the first seven terms of an arithmetic sequence is 84. Find its 4th term. (2)
Answer:
x₄ = \(\frac{84}{7}\) = 12

Question 12.
A box contains 3 red balls and 6 green balls.
(a) What is the probability of getting a red ball from this box? (1)
(b) How many more red balls should be added so that the probability of getting a red ball is \(\frac{1}{2}\)?
Answer:
(a) \(\frac{3}{9}=\frac{1}{3}\)

(b) When 3 red balls are added the number of red balls becomes 6 and the total number of balls is 12.
Probability is \(\frac{6}{12}=\frac{1}{2}\).

Question 13.

In the right triangle ABC, ∠A = 40° and AC = 20 centimeters. Calculate the length of the side BC.
(sin 40° = 0.64; cos 40° = 0.76)
Answer:
sin 40° = \(\frac{B C}{20}\)
⇒ 0.64 = \(\frac{B C}{20}\)
⇒ BC = 0.64 × 20 = 12.8 cm

Question 14.
Write the polynomial x², \(-\frac{1}{4}\) as the product of two first degree polynomials.
Answer:
\(x^2-\frac{1}{4}=x^2-\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\)

Question 15.
The scores of 10 students in an examination are given below.
30, 28, 25, 32, 20, 36, 24, 33, 27, 38
Calculate the median score.
Answer:
Ascending order: 20, 24, 25, 27, 28, 30, 32, 33, 36, 38
5th and 6th score comes in the middle. These are 28 and 30.
Median = \(\frac{28+30}{2}\) = 29

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

Question 16.
The expression for the sum of the first ‘n’ terms of an arithmetic sequence is 2n² + 4n. Find the first term and common difference of this sequence. (2)
Answer:
x₁ = 2 × 1² + 4 × 1 = 6
When n = 2
The sum of the first 2 terms is 2 × 2² + 4 × 2 = 16.
x₂ = 16 – x₁ = 16 – 6 = 10
d = x₂ – x₁ = 10 – 6 = 4

Question 17.
Find the radius of the incircle of a triangle with a perimeter of 42 centimeters and an area of 84 square centimeters. (2)
Answer:
s = \(\frac{42}{2}\) = 21
r = \(\frac{A}{s}=\frac{84}{21}\) = 4 cm

Question 18.
Find the equation of the circle with the center at the origin and radius 5. (2)
Answer:
x² + y² = 5²
⇒ x² + y² = 25

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 rectangle of sides 4 centimeters and 3 centimeters. Draw a square of area equal to the area of this rectangle.
Answer:
Draw the rectangle ABCD with sides 4 cm and 3 cm.
AB = 4 cm and BC = 3 cm.
Produce AB to E such that BC = BE.
Draw a semicircle with diameter AE.
Produce BC which intersects the semicircle at F.
Complete the square with side BF. This is the square BFGE.

The area of the rectangle and square are equal according to the relation BA × BE = BF².

Question 20.
A rectangle is to be made with a perimeter of 60 meters and an area of 189 square meters. What should be the length of its sides?
Answer:
2(length + breadth) = 60
Therefore (length + breadth) = 30
If one side is x the other side will be 30 – x
x(30 – x) = 189
⇒ 30x – x² = 189
⇒ -x² + 30x = 189
⇒ x² – 30x = -189
⇒ x² – 30x + 225 = -189 + 225
⇒ (x – 15)² = 36
⇒ x – 15 = 6
⇒ x = 21
The sides are 21 cm and 9 cm.

Question 21.

(a) Tangents at points A and B on the circle meet at P. If PA = 5 cm, what is PB? (1)
(b) Draw a circle of radius 3 centimeters. Mark a point P, at a distance 7 centimeters away from the center of the circle. Then construct tangents from P to the circle. (3)
Answer:
(a) PB = 5 cm

(b) Draw a circle of radius 3 cm.
Mark the center as O and point P at a distance of 7 cm away from the center of the circle.
Draw a circle with diameter OP which cut the first circle at A and B.
Draw the lines PA and PB. Then PA and PB are the required tangents.

Question 22.

(a) What are the coordinates of the fourth vertex of the parallelogram? (2)
(b) What are the coordinates of the point of intersection of its diagonals? (2)
Answer:
(a) Since it is a parallelogram opposite sides are parallel.
The shift of the x coordinates of the points (0, 0) and (4, 2) is 4.
So the shift of x coordinates of the opposite side is also 4.
The x co-ordinates of fourth vertex is 2 + 4 = 6
Similarly y co-ordinates = 8 + 2 = 10
The unknown vertex is (6, 10).

(b) The diagonals of a parallelogram bisect each other.
The intersecting point will be the midpoint of a diagonal.
It is \(\left(\frac{2+4}{2}, \frac{8+2}{2}\right)\).
The intersecting point is (3, 5).

Question 23.
From a cube of side 6 centimeters, the largest sphere is carved out.
(a) What is the volume of the sphere? (3)
(b) This sphere is cut into two halves. What is the volume of one hemisphere? (1)
Answer:
(a) Radius of the sphere is 3 cm.
(The side of the cube becomes the diameter of the sphere)
Volume = \(\frac{4}{3}\) × π × 3³ = 36π cubic cm

(b) 18π cubic cm

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

Question 24.
Natural numbers from 1 to 10 are written on paper slips and are put in a box. Another box contains paperslips with numbers less than 10 which are multiples of 3. One slip is taken from each box.
(a) What is the probability of both being odd? (3)
(b) What is the probability of getting at least one even? (1)
Answer:
Box A: 1, 2, 3,….. 10
Box B: 3, 6, 9
The total number of output pairs is 10 × 3 = 30
(a) The number of odds in A is 5.
Number of odds in B is 2.
Probability of getting both odd is \(\frac{5 \times 2}{30}=\frac{1}{3}\).

(b) Probability of getting at least one event is 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\).

Question 25.

In the figure, the radius of the circle with center O is 7 centimeters.
∠BOC = 100°.
(a) Find ∠A. (1)
(b) Find the length of BC. (3)
(sin 50° = 0.76; cos 50° = 0.64; tan 50° = 1.19)
Answer:
Draw a rough diagram. In this diagram mark the perpendicular OP to BC.
The foot of perpendicular bisect BC. OP bisects the 100° angle given in the figure.

(a) ∠A = 50°

(b) sin 50° = \(\frac{B P}{O B}\)
⇒ 0.76 = \(\frac{B P}{7}\)
⇒ BP = 7 × 0.76 = 5.32
BC = 2 × 5.32 = 10.64 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.

In the rectangle shown above, its sides are parallel to the axes.
(a) Find the coordinates of the remaining two vertices of the rectangle. (2)
(b) Find the length of one diagonal. (2)
(c) Find the coordinates of the center of its circumcircle. (2)
Answer:
(a) B(9, 2), D(1, 8)

(b) AB = 8, BC = 6
Using Pythagoras theorem
AC² = 8² + 6² = 100
⇒ AC = √100 = 10

(c) Circumcenter is the midpoint of the diagonal.
O\(\left(\frac{1+9}{2}, \frac{2+8}{2}\right)\) = O(5, 5)

Question 27.

(a) In Figure, A, B, C, and D are points on the circle with center O. ∠AOB = 140°. Find the measures of ∠ACB and ∠ADB. (2)
(b) Draw a triangle of circum radius 3.5 centimeters and two angles 50° and 70°. (2)
Answer:
(a) ∠ACB = 70°
∠ADB = 180° – 70° = 110°

(b) Draw a circle of radius 3.5 cm.
Divide the angle around the center as 2 × 50°, 2 × 70° by drawing radii.
Angles are 100°, 140°, 120°.
Draw a triangle by joining the ends of the radii.

Question 28.

(a) In the figure ∠A = 45°, ∠B = 90°, AB = 10 centimetres. What is the length of the AC? (1)
(b) A boy sees the top of a tower at an elevation of 60°. Stepping 20 meters back, he sees it at an elevation of 30°. Find the height of the tower. (5)
Answer:
(a) 10√2 cm

(b)

In triangle BDC, if BD = x then h = √3x
In triangle ADC, 20 + x = √3h
Therefore, 20 + x = √3 × √3x
⇒ 20 + x = 3x
⇒ 2x = 20
⇒ x = 10 meter
h = √3x = 10√3 meter

Question 29.
From a circular metal sheet of radius 30 centimeters, a sector of central angle 120° is cut out and made into a cone.
(a) What are the slant height and base radius of this cone? (2)
(b) Calculate the area of the curved surface of this cone. (2)
(c) What would be the radius of the cone that can be made by rolling up the remaining sector? (2)
Answer:
(a) Slant height l = 30 cm
If x is the central angle of the sector and r is the radius of the cone then lx = 360r
30 × 120 = 360 × r
r = 10 cm

(b) Curved surface area = πrl
= π × 10 × 30
= 300π sq. cm

(c) The central angle of the remaining part is 360 – 120 = 240°
lx = 360r
r = \(\frac{30 \times 240}{360}\) = 20 cm

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

Question 30.
(a) Find 1 + 2 + 3 +……….+ 10. (2)
(b) How many consecutive natural numbers starting from 1 should be added to get 300? (4)
Answer:
(a) Sum = \(\frac{n(n+1)}{2}=\frac{10 \times 11}{2}\) = 55

(b) \(\frac{n(n+1)}{2}\) = 300
⇒ n(n + 1) = 600
⇒ n² + n = 600
⇒ n² + n – 600 = 0
⇒ n = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
⇒ n = 24

Question 31.
p(x) = x² – 5x + 6
(a) Find p(2). (2)
(b) Write p(x) as the product of two first-degree polynomials. (2)
(c) Find the solutions of the equation x² – 5x + 6 = 0. (2)
Answer:
(a) p(2) = 2² – 5 × 2 + 6 = 0

(b) Since p(2) = 0, x – 2 is a factor.
Another factor is ax + b.
(x – 2) (ax + b) = x² – 5x +6
a = 1, -2b = 6 ⇒ b = -3
Another factor is x – 3.
p(x) = (x – 2)(x – 3)

(c) x² – 5x + 6 = 0
⇒ (x – 2)(x – 3) = 0
⇒ x = 2 and x = 3

Question 32.
The daily wages of workers in a firm are given below.

Daily Wages	No. of Workers
500 – 600	5
600 – 700	7
700 – 800	10
800 – 900	8
900 – 1000	5
Total	35

(a) If workers are arranged according to their wages (lower to higher).
(i) Which position is taken as median? (1)
(ii) What will be the assumed wage of the 13th worker? (2)
(b) Calculate the median of daily wages. (3)
Answer:

Daily Wages	No. of Workers
Below 600	5
Below 700	12
Below 800	22
Below 900	30
Upto 1000	35

n = 35 number of workers is odd.
\(\frac{35+1}{2}\)th worker comes in the middle.
The wage of the 18th worker is the median wage.
(a) The wage of the 18th worker is the median wage.

(b) Median wage comes in the class 700 – 800.
When 100 rupees is divided equally among 10 workers, each share is 10.
It is assumed that the distribution of wages in the median class is in an arithmetic sequence.
Wage of 13th worker is 700 + \(\frac{10}{2}\) = 705

(c) f = 705, d = 10.
The sixth term of the arithmetic sequence will be the median.
x₆ = f + 5d
= 705 + 5 × 10
= 755
According to the assumption, x6 is the wage of the 18th worker. It is 755 rupees.
∴ The median is 755.

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.
(a) Consider the arithmetic sequence 4, 7, 10,……. what is the algebraic expression for this sequence? (2)
(b) Write the 20th term of this sequence. What is the smallest three-digit number which is a term of this sequence? (2)
(c) Find the sum of the first 20 terms of this sequence. (2)
What is the difference between the sum of the first 20 terms of this sequence and the sum of the first 20 terms of the arithmetic sequence with algebraic form 3n + 2?
Answer:
(a) x_n – dn + (f – d) = 3n + 1

(b) x₂₀ = 3 × 20 + 1 = 61
3n + 1 > 99
⇒ 3n > 98
⇒ n > = 32.6
So n = 33, x₃₃ is the first three-digit term of this sequence.
x₃₃ = 3 × 33 + 1 = 100

(c) Sum of first 20 terms is (x₁ + x₂₀) × \(\frac{20}{2}\)
= (4 + 61) × 10
= 650
3n + 2 = (3n + 1) + 1.
The sum of the first 20 terms of the arithmetic sequence with the nth term 3n + 2 is 650 + 20 = 670
So the difference is 20.

Question 34.

(a) In the figure, the incircle of triangles A, B, and C touches its sides at the points P, Q and R. O is the center of the circle. (4)
(i) Find ∠OQB.
(ii) Examine whether quadrilateral POQB is cyclic.
(iii) If ∠B = 50°, then ∠POQ = _____________
(b) Draw a triangle with a radius of the incircle 2.5 centimeters and two angles 50°, 60°. (4)
Answer:
(a) (i) ∠OQB = 90°
(ii) In POQB the sum of ∠P and ∠Q is 90° + 90° = 180°.
The opposite angle sum is 180°. The quadrilateral is cyclic.
(iii) ∠POQ = 180°

(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 radii.
Now we can see three radii.
Draw tangents to the circle at the ends of the radii.
The tangents make the required triangle.

Question 35.
(a) Draw the x and y axis. Mark the points (1, 2) and (3, 5). (3)
(b) Find the slope of the line, passing through the points (1, 2) and (3, 5). (2)
(c) The x coordinate of a point on this line is 21. What is its y coordinate? (3)
Answer:
(a) Use the given graph paper

(b) Slope = \(\frac{y_2-y_1}{x_2-x_1}=\frac{5-2}{3-1}=\frac{3}{2}\)

(c) Consider the point (1, 2) and (21, y).
Slope = \(\frac{3}{2}=\frac{y-2}{21-1}\)
⇒ \(\frac{y-2}{20}=\frac{3}{2}\)
⇒ 2(y – 2) = 3 × 20
⇒ 2y – 4 = 60
⇒ 2y = 64
⇒ y = 32