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Students can read Kerala SSLC Maths Board Model Paper March 2022 with Answers English Medium and Kerala SSLC Maths Previous Year Question Papers with Answers helps you to score more marks in your examinations.

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

Students can read Kerala SSLC Maths Board Model Paper March 2023 with Answers English Medium and Kerala SSLC Maths Previous Year Question Papers with Answers helps you to score more marks in your examinations.

Kerala Syllabus Class 10 Maths Board Model Paper March 2023 with Answers English Medium

Time: 2½ hrs
Score: 80

Instructions:

• Read each question carefully before answering.
• Give explanations wherever necessary.
• The first 15 minutes is cool-off time. You may use this time to read the questions and plan your answers.
• No need to simplify irrationals like √2, √3, π, etc. using approximations unless you are asked to do so.

Answer any three questions from 1 to 4. Each question carries 2 scores. (3 × 2 = 6)

Question 1.
Write the next two terms of the arithmetic sequence 5, 12, 19,…
Answer:
d = 12 – 5 = 7
The next two terms are 26, 33

Question 2.
Natural numbers from 1 to 10 are written on paper slips and put in a box. If one slip is taken from the box, without looking, then what is the probability of the number on the slip being a multiple of 3?
Answer:
\(\frac{3}{10}\)

Question 3.
ABCD is a rectangle, ∠CAB = 30°, AC = 10 centimeters.

(a) Find the length of BC.
(b) Find the length of AB.
Answer:
(a) 5 cm
(b) 5√3 cm

Question 4.
Find the median of the first 9 even numbers.
Answer:
Since the number of observations is 9 and these are the even numbers in the order from the beginning then the 5th even number is the median. It is 10.

Answer any four questions from 5 to 10. Each question carries 3 scores. (4 × 3 = 12)

Question 5.
The sum of the first 5 terms of an arithmetic sequence is 145.
(a) Find the third term.
(b) If the common difference of this sequence is 4, write the terms.
Answer:
(a) x₃ = \(\frac{145}{5}\) = 29

(b) x₁ = 29 – 2d
= 29 – 8
= 21
Terms are 21, 25, 29,…….

Question 6.
Draw a circle of radius 3 centimeters. Draw a triangle of angles 50°, 60°, 70°, and vertices on this circle.
Answer:
Steps of Construction:

• Draw the circle.
• Divide the angle around the center as twice the given angles by drawing the radii.
• Join the ends of the radii. Now we get the triangle.

Question 7.
Draw the x and y axes and mark the points A(3, 0), B(4, 1), C(2, -3).
Answer:

Question 8.
A square and two rectangles of the same height are kept together as in the picture. The width of the rectangles is 2 centimeters. The total area of the picture is 96 square centimeters.

(a) Taking the side of the square as x centimeters, write an equation representing the given details.
(b) Find the length of one side of the square.
Answer:
(a) (x + 4) × x = 96
⇒ x² + 4x = 96
⇒ x² + 4x + 4 = 100
⇒ (x + 2)² = 100

(b) x + 2 = √100 = 10
⇒ x = 8 cm

Question 9.
In triangle ABC, P(5, 0), Q(6, 1), R(3, 1) are the mid-points of sides BC, CA, and AB respectively.

(a) What is the most suitable name for the quadrilateral BPQR?
(b) Find the coordinates of B and C.
Answer:
(a) Parallelogram

(b) B(3 + 5 – 6, 1 + 0 – 1) = B(2, 0)
Join PR. The quadrilateral PCQR is a parallelogram.
C(5 + 6 – 3, 0 + 1 – 1) = C(8, 0)

Question 10.
Draw a circle of radius 3 centimeters. Mark a point 7 centimeters away from the center. Draw tangents from this point to the circle.
Answer:
Steps of Construction:

• Draw the circle, and mark the center O.
• Mark a point P at a distance 7 cm away from the center of the circle. Draw line OP.
• Draw a circle with OP as the diameter. This circle cut the first circle at A and B.
• Draw lines PA and PB.

Answer any eight questions from 11 to 21. Each question carries 4 scores. (8 × 4 = 32)

Question 11.
A sequence is written by adding 3 to the multiples of 4.
(a) Write the algebraic form of the sequence.
(b) Find the tenth term of the sequence.
(c) Is 100 a term of this sequence? Why?
Answer:
(a) 1 × 4 + 3, 2 × 4 + 3, 3 × 4 + 3,……
x_n = 4n + 3

(b) x₁₀ = 4 × 10 + 3 = 43

(c) All terms of this sequence are 3 more than a multiple of 4.
100 is not a term. It is a multiple of 4, not 3 more than a multiple of 4.

Question 12.
(a) In the picture, AB is the diameter of the semicircle. PC is perpendicular to AB. AP = 5 centimetres and PB = 3 centimetres. Find the length of the PC.

(b) Draw a square of area 15 square centimeters.
Answer:
(a) PA × PB = PC²
⇒ 5 × 3 = PC²
⇒ PC = √15 cm

(b) Construction:

Question 13.
The perimeter of a rectangle is 80 centimeters. Its area is 384 square centimeters. Find the length and breadth of the rectangle.
Answer:
2(length + breadth) = 80
⇒ length + breadth = 40
If one side is x then the other side is 40 – x
x(40 – x) = 384
⇒ 40x – x² = 384
⇒ x² – 40x = -384
⇒ x² – 40x + 400 = -384 + 400
⇒ (x – 20)² = 16
⇒ x – 20 = 4, -4
If x – 20 = 4
⇒ x = 24
One side is 24 cm, other side is 40 – 24 = 16 cm

Question 14.
In class 10A, there are 25 boys and 20 girls. In 10B, there are 26 boys and 24 girls. One student is to be selected from each class.
(a) What is the probability of both being girls?
(b) What is the probability of both being boys?
(c) What is the probability of one boy and one girl?
Answer:
There are 45 students in 10A and 50 students in 10B
The total number of pairs is 45 × 50 = 2250
(a) Number of pairs in which both are girls = 20 × 24 = 480
Probability is \(\frac{480}{2250}\)

(b) Number of pairs in which both are girls = 25 × 26 = 650
Probability is \(\frac{650}{2250}\)

(c) Number of pairs in which one boy and one girl = \(\frac{25 \times 24+20 \times 26}{2250}=\frac{1120}{2250}\)

Question 15.
ABCD is a parallelogram. BC = 15 centimetres; ∠B = 45°, AB = 10√2 centimetres. AM is perpendicular to BC.

(a) Find the length of AM and BM.
(b) What is the length of MC?
(c) Calculate the length of diagonal AC.
Answer:
(a) ΔAMB is a 45° – 45° – 90° triangle.
AM = BM = 10 cm

(b) MC = 15 – 10 = 5 cm

(c) AC² = AM² + MC² = 125
AC = √125 = 5√5 cm

Question 16.
Consider the polynomial P(x) = x² – 11x + 21
(a) Find P(2).
(b) Find P(x) – P(2).
(c) Write P(x) – P(2) as the product of two first degree polynomials.
Answer:
(a) P(2) = 2² – 11 × 2 + 21
= 4 – 22 + 21
= 3

(b) P(x) – P(2) = x² – 11x + 21 – 3 = x² – 11x + 18

(c) P(x) – P(2) = (x – 2)(x – 9)

Question 17.
The sides of rectangle ABCD are parallel to the axes. The coordinates of A and C are (3, 1) and (7, 4) respectively.

(a) Find the coordinates of B and D.
(b) Find the length of the diagonal of the rectangle.
Answer:
(a) B(7, 1), D(3, 4)

(b) AC = \(\sqrt{(7-3)^2+(4-1)^2}=\sqrt{4^2+3^2}\) = 5

Question 18.
A(1, 3), B(2, 5), C(3, 7) and D are points on a line such that AB = BC = CD.

(a) Find the coordinates of D.
(b) Find the slope of the line.
(c) Find the equation of this line.
Answer:
(a) D(4, 9)

(b) Slope = \(\frac{y_2-y_1}{x_{2^{\prime}}-x_1}=\frac{7-5}{3-2}\) = 2

(c) Let (x, y) be a point on the line.
\(\frac{y-3}{x-1}\) = 2
⇒ y – 3 = 2(x – 1)
⇒ y – 3 = 2x – 2
⇒ y = 2x + 1
or
The equation of a line is the general linear relation between the coordinates of points on the line.
Here 3 = 2 × 1 + 1
5 = 2 × 2 + 1
7 = 2 × 3 + 1
y = 2x + 1 is the equation of the line.

Question 19.
In the figure, O is the center of the circle. AB is a chord of the circle and BT is a tangent ∠ABT = 70°.

Find the measures of the angles given below.
(a) ∠OBT
(b) ∠OBA
(c) ∠AOB
(d) ∠APB
Answer:
(a) ∠OBT = 90° (Tangent is perpendicular to the radius)
(b) ∠OBA = 90° – 70° = 20°
(c) ∠AOB = 140°
(d) ∠APB = 70°

Question 20.
A square pyramid is made using a cardboard piece in the shape as shown in the figure. The side of the square is 10 centimeters. The equal sides of the triangles are 13 centimeters.

(a) Find the slant height of the square pyramid.
(b) Calculate the surface area of the square pyramid.
Answer:
(a) Half of the base edge \(\frac{a}{2}\) slant height l and lateral edge of a square pyramid make a right triangle.
13² = 5² + l²
⇒ l² = 169 – 25 = 144
⇒ l = 12 cm

(b) Surface area of the square pyramid = Base area + lateral surface area.
Surface Area = 10² + 2 × 10 × 12
= 100 + 240
= 340 sq. cm

Question 21.
Two sectors are cut out from a circle. The central angle of the larger sector is double that of the smaller. Each sector is rolled up to make a cone.

(a) The slant height of the cone made from the smaller sector is 10 centimeters. What is the slant height of the other cone?
(b) Write the ratio of the radii of the two cones.
(c) Find the ratio of the base areas of the cones.
(d) Find the ratio of their curved surface areas.
Answer:
(a) The radius of the sector becomes the slant height of the cone.
Since both sectors have the same radius then the slant height of the cones is equal. It is 10 cm.

(b) Since the ratio of central angles of the sectors is 1 : 2, the base radii of the cones are also 1 : 2 as per the relation
lx = 360r
r = \(\frac{l}{360}\) × x

(c) It is the ratio of the square of radii.
It is 1 : 4 (The area of a circle is proportional to the square of the radius)

(d) Same as the ratio of radii.
The ratio is 1 : 2.

Answer any six questions from 22 to 29. Each question carries 5 scores. (6 × 5 = 30)

Question 22.
The sum of the 8th and 19th terms of an arithmetic sequence is 125.
(a) What is the sum of the 7th and 20th terms?
(b) If the 6th term is 40, then find the 21st term.
(c) Find the sum of the first 26 terms.
Answer:
(a) In an arithmetic sequence having a definite number of terms sum of the terms equidistant from both ends is equal.
So x₇ + x₂₀ = x₈ + x₁₉ = 125

(b) x₆ + x₂₁ = 125
⇒ x₂₁ = 125 – 40 = 85

(c) 125 × 13 = 1625

Question 23.
In quadrilateral ABCD, ∠A = 95°, ∠B = 100°, ∠C = 90°.

(a) Find the measure of ∠D.
(b) If we draw a circle with BD as diameter, then check whether the vertices A and C are outside, on, or inside the circle.
(c) If a circle is drawn through points, A, B, and C, where would point D concerning that circle?
Answer:
(a) ∠D = 75°

(b) A is inside the circle.
C is On the circle.

(c) Outside the circle.

Question 24.
A body standing at the edge of a canal sees the top of a tree at the other edge at an elevation of 60°. Stepping 10 meters back, he sees the tree at an elevation of 30°. The boy is 1.5 meters tall.
(a) Draw a rough figure.
(b) Calculate the width of the canal and the height of the tree.
Answer:
(a) See diagram

(b) DE = y, FG = x
Triangle DFG is a 30°-60°-90° triangle.
x = √3y
Triangle BFG is a 30°-60°- 90° triangle.
⇒ y + 10 = √3x
⇒ y + 10 = √3 × √3 y
⇒ y + 10 = 3y
⇒ 2y =10
⇒ y = 5m
⇒ x = 5√m
The height of the boy is 5√3 + 1.5 m

Question 25.
Draw a triangle ABC in which AB = 7 centimeters, BC = 6 centimeters, AC = 5 centimeters. Draw its incircle. Measure and write the radius of the incircle.
Answer:

• Draw a triangle with the given measurements.
• Draw the bisectors of angles. Bisectors meet at a point O. It is the center of the incircle.
• Draw a circle with a center at O and a radius at the perpendicular distance from O to the sides.

Question 26.
(a) A solid metallic sphere of radius 6 centimeters is cut into two halves. What is the surface area of each hemisphere?
(b) One of these hemispheres is melted and recast to make a cone of the same radius. Find the height of the cone.
Answer:
(a) Radius = 6 cm
Total surface area = 3πr² = 108π

(b) \(\frac{2}{3} \pi \times 6^3=\frac{1}{3} \pi 6^2 \times h\)
⇒ h = 12 cm

Question 27.
Consider the circle with a center at the origin and a radius of 10 units.
(a) Find the coordinates of the points where this circle cuts the x and y axes.
(b) Check whether P(6, 8) is a point on this circle.
(c) Write the equation of this circle.
Answer:
(a) (10, 0), (0, 10), (-10, 0), (0, -10)

(b) Yes.
Radius = \(\sqrt{6^2+8^2}\) = 10

(c) x² + y² = 10²
x² + y² = 100

Question 28.
The table below shows the students of a maths club sorted according to their heights.

Height (centimeter)	Number of Students
120 – 130	2
130 – 140	7
140 – 150	10
150 – 160	5
160 – 170	1
Total	25

(a) When the heights are written in ascending order, the height of which student is taken as the median height?
(b) Find the median height.
Answer:
See the table

Height	Number of Children
Below 130	2
Below 140	9
Below 150	19
Below 160	24
Upto 170	25

(a) Since total number of students is 25, an odd number 13th height comes in the middle. The 13th height is the median height.
(b) The 13th height comes in the class 140 – 150. This is the median class.
It is assumed that the distribution of heights in the median class is in arithmetic sequence.
By this assumption, 10 cm height is divided equally among 10 students.
Each one’s share is 1.
Height of 10th student is 140 + \(\frac{1}{2}\) = 140.5 cm
In the arithmetic sequence f = 140.5, d = 1,
4th term is 13th height.
It is 140.5 + 3 × 1 = 143.5 cm.
Median = 143.5

Question 29.
Let’s find natural numbers which can be written as the sum of consecutive natural numbers.
3 = 1 + 2
5 = 2 + 3
6 = 1 + 2 + 3
7 = 3 + 4
9 = 4 + 5
10 = 1 + 2 + 3 + 4
11 = 5 + 6
12 = 3 + 4 + 5

• All odd numbers other than 1, can be written as the sum of two consecutive natural numbers.
• Even numbers, which are powers of 2 (2, 4, 8, 16 …) cannot be written as the sum of consecutive natural numbers.
• The even numbers that are not powers of 2 can be written as the sum of three or more consecutive natural numbers.

(a) Write 13 as the sum of consecutive natural numbers.
(b) Write 14 as the sum of consecutive natural numbers.
(c) Write 101 as the sum of consecutive natural numbers.
(d) Find the numbers between 20 and 100 that cannot be written as the sum of consecutive natural numbers.
Answer:
(a) 6 + 7
(b) 2 + 3 + 4 + 5
(c) 50 + 51
(d) 2⁵ = 32, 2⁶ = 64

Students can read Kerala SSLC Maths Board Model Paper March 2024 with Answers English Medium and Kerala SSLC Maths Previous Year Question Papers with Answers helps you to score more marks in your examinations.

Kerala Syllabus Class 10 Maths Board Model Paper March 2024 with Answers English Medium

Time: 2½ hours
Score: 80

Instructions:

• Read each question carefully before answering.
• Give explanations wherever necessary.
• The first 15 minutes is cool-off time. You may use this time to read the questions and plan your answers.
• There is no need to simplify irrationals like √2, √3, π, etc. using approximations unless you are asked to do so.

Answer any three questions from 1 to 4. Each question carries 2 scores. (3 × 2 = 6)

Question 1.
Consider the arithmetic sequence 1, 11, 21,…….
(a) What is its common difference?
(b) Find the 10th term of this sequence.
Answer:
(a) Common difference = 11 – 1 = 10.

(b) a₁₀ = a₁ + (10 – 1) × 10
= 1 + 9 × 10
= 1 + 90
= 91

Question 2.
In the figure, O is the circle’s center, and ∠AQB = 110°.

(a) What is the measure of ∠APB?
(b) What is the measure of ∠AOB?
Answer:
(a) ∠AQB + ∠APB = 180°.
∠APB = 180° – ∠AQB
= 180° – 110°
= 70°

(b) Minor ∠AOB = 2 × 70° = 140°.
Major ∠AOB = 2 × 110° = 220°.

Question 3.
The marks of 8 students in a Maths test are given in ascending order as below.
20, 20, 24, 32, x, 40, 45, 48
If the median mark is 34, then find the value of x.
Answer:
8 marks are there. So, the median is the average of \(\left(\frac{8}{2}\right)^{t h}\) and \(\left(\frac{8}{2}+1\right)^{t h}\) marks.
The median is the average of the 4th and 5th marks.
4th mark = 32.
5th mark = x.
It is given that median = 34
So, \(\frac{32+x}{2}\) = 34
⇒ 32 + x = 68
⇒ x = 68 – 32 = 36

Question 4.
The midpoints of the sides of a square are joined to form another square. If a dot is put inside the large square find the probability that it is within the shaded portion.

Answer:

If AB = 2a, then the area of square ABCD = 4a².
If AB = 2a, then SQ = 2a.
SQ is a diagonal of square PQRS.
So, the area of the square PQRS = \(\frac{(2 a)^2}{2}=\frac{4 a^2}{2}\) = 2a²
Thus, the area of the shaded portion = area of the square ABCD – area of the square PQRS
= 4a² – 2a²
= 2a²
Therefore, the probability that the dot will fill within the shaded portion = \(\frac{\text { Area of the shaded portion }}{\text { Area of square } A B C D}\)
= \(\frac{2 a^2}{4 a^2}=\frac{2}{4}=\frac{1}{2}\)

Answer any four questions from 5 to 10. Each question carries 3 scores. (4 × 3 = 12)

Question 5.
The algebraic expression of an Arithmetic sequence is 3n – 2.
(a) Find the first term of the sequence.
(b) Find the sum of the first 50 terms.
Answer:
(a) To get the first term of a sequence,
we have to put n = 1 in the general term.
So, the first term = 3 × 1 – 2 = 1.

(b) 50th term = 3 × 50 – 2 = 148.
The sum of first n terms = \(\frac{n}{2}\)(a₁ + a_n)
The sum of the first 50 terms = \(\frac{50}{2}\)(1 + 148)
= 25 × 149
= 3725

Question 6.
Draw a triangle of circumradius 3 centimeters and two of its angles 55° and 62½°.
Answer:
Using a compass, draw a circle of radius 3cm. Name the center of the circle as O.
If 55° is the angle on the circle, then the angle on the center is 55° × 2 = 110°.
If 62.5° is the angle on the circle, then the angle on the center is 62.5° × 2 = 125°.
Draw the radius OA.
Draw the radius OB, which makes an angle of 110° with OA.
Draw the radius OC, which makes an angle of 125° with OB:
Join AB, BC, and AC to get the required triangle.

Question 7.
One side of a rectangle is 12 centimeters longer than the other side and its area is 864 square centimetres.
(a) Form a second-degree equation by taking the smaller side as ‘x’.
(b) Calculate the lengths of the sides of the rectangle.
Answer:
(a) The smaller side = x
The larger side = 12 + x
Area = 864 cm²
⇒ x(12 + x) = 864
⇒ x² + 12x = 864

x = 24 or x = -36
But x represents length. So, it cannot be negative.
Therefore, The smaller side = x = 24.
The larger side = 12 + x = 12 + 24 = 36.

Question 8.
A parallelogram is drawn with lengths of adjacent sides 10 centimeters, 6 centimeters and the angle between them is 60°.

(a) Find the distance between the top and bottom sides of the parallelogram.
(b) Calculate the area of the parallelogram.
Answer:
(a) Consider the following picture.

We have to find the length of the DP.
ΔADP is a 30° – 60° – 90° triangle.
So, if AP = x, then AD = 2x and DP = x√3
Here, AD = 2x = 6 cm.
So, x = 3 cm
Therefore, DP = x√3 = 3√3 cm.

(b) Area of the parallelogram = AB × DP
= 10 × 3√3
= 30√3 cm²

Question 9.
Two vertices of an equilateral triangle are (0, 0) and (10, 0).

(a) Find the length of one side of this triangle.
(b) Find the height of the triangle.
(c) Find the coordinates of the third vertex.
Answer:
(a) (0, 0) and (10, 0) are the coordinates of the endpoints of a side.
So, its length = |0 – 10| = |-10| = 10 units.

(b) We have to find the length of BP.

As ΔOAB is an equilateral triangle, P is the midpoint of AB.
So, OP = 5 units.
ΔOPB is a 30° – 60° – 90° triangle.
So, if OP = x, then OB = 2x and BP = x√3
Here, OP = x = 5 units.
Therefore, BP = x√3 = 5√3 units.

(c) B = (5, 5√3)

Question 10.
A circle with a center at the origin passes through the point (4, 3).
(a) What is the radius of the circle?
(b) Write the coordinates of the points where this circle cut the axis.
Answer:
(a) From the question, it is clear that (0, 0) and (4, 3) are the endpoints of the radius.
Thus, the radius = \(\sqrt{\left(4^2+3^2\right)}\)
= \(\sqrt{(16+9)}\)
= \(\sqrt{25}\)
= 5 units

(b) Let (0, y) be the point where the circle cuts the y-axis.
The distance from the origin to this point is equal to the radius, that is, 5 units.
Thus, \(\sqrt{\left(0-0^2+0-y^2\right)}\) = 5
⇒ \(\sqrt{\left(0^2+y^2\right)}\) = 5
⇒ y = ±5
Therefore, the points where the circle cuts the axis are (0, 5) and (0, -5).

Answer any eight questions from 11 to 21. Each question carries 4 scores. (8 × 4 = 32)

Question 11.
The 3rd term of an arithmetic sequence is 16 and its 21st term is 124.
(a) Find the common difference of the sequence.
(b) Find the first term of the sequence.
(c) What is the position of280 in this sequence?
Answer:
(a) 3rd term = 16
That is a₁ + (3 – 1)d = 16
a₁ + 2d = 16 …………. (1)
21st term = 124
That is a₁ + (21 – 1) d = 124
a₁ + 20d = 124 ………….(2)
(2) – (1) gives,
a₁ – a₁ + 20d – 2d = 124 – 16
18d = 108
d = 6

(b) Using (1) we have a₁ + 2 × 6 = 16
That is, a₁ + 12 = 16
a₁ = 16 – 12
a₁ = 4
Therefore, the first term is 4.

(c) a₁ + (n – 1)d = a_n
4 + (n – 1)6 = 280
4 + 6n – 6 = 280
6n = 282
n = 47
Thus, 280 is in the 47th position of the sequence.

Question 12.
One box contains 10 paper slips numbered 1 to 10 and another box contains 20 paper slips numbered 1 to 20. One slip is taken from each box.
(a) In how many different ways can we choose a pair of slips?
(b) What is the probability of both numbers being the same?
(c) What is the probability of getting one even number and one odd number?
Answer:
(a) The number of ways in which we can select a number from the first box = 10.
The number of ways in which we can select a number from the second box = 20.
Therefore, the number of ways in which we can select a pair of slips = 10 × 20 = 200.

(b) The number of ways in which we can select a pair of slips = 10 × 20 = 200.
The number of ways in which both the numbers are the same = 10.
Thus, the probability that both the numbers are same = \(\frac{10}{200}=\frac{1}{20}\)

(c) The number of ways in which we can select an event from the 1st box and odd from the 2nd box = 5 × 10 = 50.
The number of ways in which we can select an odd from the 1st box and even from the 2nd box = 5 × 10 = 50.
Therefore, the probability of getting one even number and one odd number = \(\frac{50+50}{200}=\frac{100}{200}=\frac{1}{2}\)

Question 13.
10 added to the product of a natural number and the number 7 more than that is 304.
(a) If the first number is x, what will be the next number?
(b) Form a second-degree equation and find the two numbers.
Answer:
(a) x + 7

(b) x(x + 7) + 10 = 304
⇒ x² + 7x = 304 – 10
⇒ x² + 7x = 294
⇒ x² + 7x – 294 = 0


It is given that x is a natural number. So, it cannot be negative.
Therefore, x = 14 and x + 1 = 14 + 7 = 21.
So, the two numbers are 14, 21.

Question 14.
A ladder leans against a wall with its foot 3 meters away from the wall and makes an angle of 60° with the floor.

(a) Find the length of the ladder.
(b) The foot of the ladder is pulled to make an angle of 30° with the floor. How high will be its top from the ground?
Answer:
Consider the following picture.

(a) We have to find the length of BE.
∆BCE is a 30°-60°-90° right triangle.
So, if BC = x then BE = 2x.
Here, BC = x = 3 meters.
So, BE = 2x = 2 × 3 = 6 metres.

(b) We have to find the length of DC.
∆ACD is a 30°-60°-90° right triangle.
So, if DC = y then AD = 2y.
Here, DD = 2y = 6 metres.
So, CD = y = \(\frac{6}{2}\) = 3 metres.

Question 15.
(a) Find the distance between the points (-1, 2) and (5, 10).
(b) Prove that the line joining these points passes through the point (11, 18).
Answer:
(a) Distance = \(\sqrt{(-1-5)^2+(2-10)^2}\)
= \(\sqrt{(-6)^2+(-8)^2}\)
= 10 units.

(b) Slope of the line passing through (-1, 2) and (5, 10)
Slope = \(\frac{y_2-y_1}{x_2-x_1}=\frac{10-2}{5-(-1)}=\frac{8}{6}=\frac{4}{3}\)
The slope of the line passing through (5, 10) and (11, 18)
Slope = \(\frac{18-10}{11-5}=\frac{8}{6}=\frac{4}{3}\)
The slope is the same in both cases.
So, (11, 18) is on that line.

Question 16.
Draw a circle of radius 3 centimeters. Mark a point 7.5 centimeters away from the centre and draw the pair of tangents to the circle from this point.
Answer:

Question 17.
The incircle of a triangle touches the sides at P, Q, and R. The perimeter of the triangle is 24 centimeters and the length of AB is 7 centimeters.

(a) Prove that AP + BQ + CR = 12 centimetres.
(b) Find the length of QC.
Answer:
(a) In the figure, AP = AR, BP = BQ, CR = CQ.
It is given that perimeter = 24 cm.
That is, AB + BC + CA = 24 cm.
⇒ AP + BP + BQ + CQ + CR + AR = 24 cm
⇒ AP + BQ + BQ + CR + CR + AP = 24 cm
⇒ 2AP + 2BQ + 2CR = 24 cm
⇒ 2(AP + BQ + CR) = 24 cm
Therefore, AP + BQ + CR = 12 cm

(b) We have AP + BQ + CR = 12 cm
That is, AP + BQ + CQ = 12 cm
So, CQ = 12 – AP – BQ
⇒ CQ = 12 – AP – BP
⇒ CQ = 12 – (AP + BP)
⇒ CQ = 12 – 7 = 5 cm

Question 18.
A cone of radius 12 centimeters is to be made by folding a sector cut from a circle of radius 20 centimeters.
(a) What should be the central angle of the sector?
(b) Calculate the curved surface area of the cone.
Answer:
(a) Radius of the cone = r = 12 cm.
The radius of the circle = l = 20 cm.
Let x be the central angle of the sector.
We have, \(\frac{x}{360^{\circ}}=\frac{12}{20}\)
So, x = 216°

(b) Curved surface area of the cone = πrl
= π × 12 × 20
= 240π cm²

Question 19.
A line is drawn by joining the points (2, 3) and (5, 9).
(a) Find the slope of the line.
(b) Find the equation of the line.
(c) Check whether (1, 5) is a point on this line.
Answer:
(a) (2, 3) and (5, 9) are the two points on the line.
Therefore, slope = \(\frac{9-3}{5-2}=\frac{6}{3}\) = 2

(b) Equation of the line is y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\)(x – x₁)
So, y – 3 = 2(x – 2)
y – 3 = 2x – 4
y – 2x = 3 – 4
y – 2x = -1

(c) The equation of the line is y – 2x = -1.
If (1, 5) is a point on this line, then it has to satisfy the equation of the line.
5 – 2 × 1 = 5 – 2 = 3.
(1, 5) does not satisfy the equation of the line.
Therefore, (1, 5) is not a point on the line.

Question 20.
Consider the polynomial P(x) = 2x² – 7x + 9
(a) Find the value P(2).
(b) Find the solutions of the equation P(x) – P(2) = 0.
Answer:
(a) P(2) = 2 × 2² – 7 × 2 + 9
= 2 × 4 – 14 + 9
= 8 – 14 + 9
= 3

(b) P(x) – P(2) = 0
2x² – 7x + 9 – 3 = 0
2x² – 7x + 6 = 0

Question 21.
A solid metal hemisphere of radius 10 centimeters is melted and recast into small solid spheres of radius 1 centimeter each. How many such spheres can be made?
Answer:
The radius of the hemisphere = R = 10 cm.
The radius of the sphere = r = 1 cm.
Number of the spheres = \(\frac{\text { Volume of the hemisphere }}{\text { Volume of one sphere }}\)

Thus, 500 such spheres can be formed.

Answer any six questions from 22 to 29. Each question carries 5 scores. (6 × 5 = 30)

Question 22.
The first term of an arithmetic sequence is 5 and the common difference is 4.
(a) What is the algebraic expression for this sequence?
(b) What is the algebraic expression for the sum of the first n terms of this sequence?
(c) Find the sum of the first 20 terms of this sequence.
Answer:
From the question we have, x1 and d = 4.
(a) Algebraic expression of an AP is x_n = x₁ +(n – 1)d
So here, x_n = 5 + (n – 1)4
= 5 + 4n – 4
= 4n + 1
Therefore, the algebraic expression of the given AP is x_n = 4n + 1.

(b) The algebraic expression for the sum or the first n terms of an AP = \(\frac{n}{2}\)(2x₁ + (n – 1)d)
So here, the algebraic expression for the sum of the first n terms is S_n = \(\frac{n}{2}\)(2 × 5 + (n – 1)4)
= \(\frac{n}{2}\)(10 + 4n – 4)
= \(\frac{n}{2}\)(6 + 4n)
= n(3 + 2n)
= 3n + 2n²

(c) S₂₀ = 3 × 20 + 2 × 20²
= 60 + 2 × 400
= 60 + 800
= 860

Question 23.
A circle passes through the origin, (-3, 0) and (0, 4).

(a) Find the length of the diameter of the circle.
(b) What are the coordinates of the center?
(c) Write the equation of the circle.
Answer:
It is given that (0, 0), (-3, 0), (0, 4) are points on the circle.
(a) Let (x, y) be the centre of the circle and r be the radius of the circle.
So, the distance from (x, y) to any of these points will be equal to the radius of the circle.

Squaring on both sides we will get,
⇒ x² + y² = (x + 3)² + y²
⇒ x² = (x + 3)²
⇒ x² = x² + 6x + 9
⇒ 6x + 9 = 0
⇒ 6x = -9
⇒ x = \(\frac{-9}{6}=\frac{-3}{2}\)
Also, we can write,
\(\sqrt{x^2+y^2}=\sqrt{x^2+(y-4)^2}\)
Squaring on both sides we will get,
⇒ x² + y² = x² + (y – 4)²
⇒ y² = (y – 4)²
⇒ y² = y² – 8y + 16
⇒ -8y + 16 = 0
⇒ y = 2
Thus the coordinates of the centre of the circle
(x, y) = (\(\frac{-3}{2}\), 2)
So, the radius = \(\sqrt{x^2+y^2}=\sqrt{\left(\frac{-3}{2}\right)^2+2^2}\)
= \(\sqrt{\frac{9}{4}+4}\)
= \(\frac{5}{2}\)
Therefore, the diameter = 2 × \(\frac{5}{2}\) = 5 units.

(b) The coordinates of the centre of the circle = (\(\frac{-3}{2}\), 2)

(c) The equation of the circle with center (h, k) and radius r is (x – h)² + (y – k)² = r²
So, the equation of the given circle
\(\left(x-\frac{-3}{2}\right)^2+(y-2)^2=\left(\frac{5}{2}\right)^2\)
\(\left(x+\frac{3}{2}\right)^2+(y-2)^2=\frac{25}{4}\)

Question 24.
Draw a triangle of sides 4 centimeters, and 5 centimeters, and the angle between them is 70°. Draw the incircle of the triangle and measure its inradius.
Answer:

Inradius = 1.3 cm.

Question 25.
A toy in the shape of a cone attached to a hemisphere. Its common radius is 3 centimetres and the total height is 17 centimetres.

(a) What is the height of the cone?
(b) Find the volume of the toy.
Answer:
From the picture we have,
The radius of the cone = 3 cm
The radius of the hemisphere = 3 cm
Total height = radius of the hemisphere + height of the cone = 17 cm
(a) height of the cone = 17 – radius of the hemisphere
= 17 – 3
= 14 cm

(b) Volume of the toy = volume of the cone + volume of the hemisphere
= \(\frac{1}{3} \times \pi \times 3^2 \times 14+\frac{2}{3} \times \pi \times 3^3\)
= \(\frac{1}{3}\) × π × 3² (14 + 2 × 3)
= 3π (14 + 6)
= 3π × 20
= 60π cm³

Question 26.
The table shows the number of workers in a company sorted according to their daily wages.

Daily Wages (Rs.)	Number of Workers
800 – 900	5
900 – 1000	7
1000 – 1100	6
1100 – 1200	10
1200 – 1300	15
1300 – 1400	2

(a) If the daily wages are arranged in ascending order, what will be the assumed wage of the 19th worker?
(b) Find the median wage.
Answer:
The cumulative frequency table is

Daily Wages	Number of Workers
Below 900	5
Below 1000	12
Below 1100	18
Below 1200	28
Below 1300	43
Below 1400	45

(a) The 19th worker comes to the class 1100 – 1200.
So, the class width is 1200 – 1100 = 100.
This class includes 10 workers.
So, \(\frac{100}{10}\) is the length of a subclass.
Thus, the first subclass is 1100 – 1110.
Its midpoint is \(\frac{1110+1100}{2}\) = 1105 rupees
Therefore, the assumed age of the 19th worker = 1105.

(b) Total number of workers = n = 45.
45 is an odd number.
\(\frac{45+1}{2}=\frac{46}{2}\) = 23
So, the wage of the 23rd worker is the median wage.
The 23rd worker comes to the class 1100 – 1200.
Median wage = 1105 + (23 – 19) × 10
= 1105 + 4 × 10
= 1105 + 40
= 1145 rupees

Question 27.
A boy 1.5 meters tall, standing at the top of a building 8.5 meters high, sees the top of a tower at an elevation of 40° and the bottom of the tower at a depression of 50°.
(a) Draw a rough figure using the given details.
(b) How far is the building from the tower?
(c) Find the height of the tower.
(tan 40° = 0.84, tan 50° = 1.2)
Answer:
(a)

(b) We have to find the length of AF.
∠FCE = 50°. So, ∠CFA= 50°. (AF and CE are parallel).
Consider ∆CAF, tan 50° = \(\frac{A C}{A F}\)
So, AF = \(\frac{A C}{\tan 50^{\circ}}\)
= \(\frac{10}{1.2}\)
= 8.33 metre.

(c) We have to find the length of DF.
DF = DE + EF = DE + 10.
Consider ∆CED,
tan 40° = \(\frac{D E}{C E}\)
So, DE = CE × tan 40°
= 8.33 × 0.84
= 6.99
Therefore, DF = DE + 10
= 6.99 + 10
= 16.99 metre.

Question 28.
Two circles meet at point C. AB and CD are common tangents to the circles.

(a) Prove that D is the midpoint of AB.
(b) Find the measure of ∠ACB.
Answer:
It is given that AB and CD are common tangents to the circle.
(a) From the picture it is clear that D is a point outside both circles.
So, DA = DC and DB = DC.
Thus, DA = DB.
Therefore, D is the midpoint of AB.

(b) If ∠ADC = x then ∠BDC = 180° – x
So, ∠BCD = \(\frac{x}{2}\) (as DB = DC)
and ∠ACD = \(\frac{180^{\circ}-x}{2}\) (as DA = DC)
∠ACB = ∠BCD + ∠ACD
= \(\frac{x}{2}+\frac{180^{\circ}-x}{2}\)
= 90°

Question 29.
See the pattern given below.
1 + 2 + 1 = 4
1 + 2 + 3 + 2 + 1 = 9
1 + 2 + 3 + 4 + 3 + 2 + 1 = 16
1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 25
_________________________________
_________________________________
(a) Write the 5th line of the pattern.
(b) Find the sum of the line
1 + 2 + 3 + ……. + 13 + 14 + 15 + 14 + 13 + …….. + 2 + 1
(c) Find the middle number of the line that gives the sum 400.
(d) Find the value of n if
1 + 2 + 3 + …… + (3n – 2) + (3n – 1) + (3n – 2) + ……. + 2 + 1 = 2500
Answer:
(a) 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 36

(b) 15² = 225

(c) The middle number of the line that gives the sum 400 = \(\sqrt{400}\) = 20.

(d) (3n – 1)² = 2500
So, (3n – 1) = \(\sqrt{2500}\) = 50
⇒ (3n – 1) = 50
⇒ 3n = 50 + 1 = 51
⇒ n = 17

Students can read Kerala SSLC Maths Question Paper March 2021 with Answers English Medium and Kerala SSLC Maths Previous Year Question Papers with Answers helps you to score more marks in your examinations.

Kerala Syllabus Class 10 Maths Question Paper March 2021 with Answers English Medium

Time: 2½ hours
Total Score: 80

Instructions:

• 20 minutes is given as tool-off lime.
• Use cool-off time to read the questions and plan your answers.
• Attempt the questions according to the instructions.
• Keep in mind, the score and time while answering the questions.
• The maximum score for questions from 1 to 45 will be 80.
• There is no need to simplify irrationals like √2, √3, π, etc., using approximations unless you are asked to do so.

For questions from 1 to 5, choose the correct answer from the brackets. Each carries 1 score. (5 × 1 = 5)

Question 1.
Arithmetic sequence with common difference 2 is: (1)
[7, 10, 13,…; 7, 5, 3,…; 7, 9, 11,…; 2, 5, 8]
Answer:
7, 9, 11,………..

Question 2.
Which is always a cyclic quadrilateral? (1)
[Parallelogram; Square; Trapezium; Rhombus]
Answer:
Square

Question 3.
Which among the following is a point on the x-axis? (1)
[(2, 0); (0, 2); (1, 1); (3, 4)]
Answer:
(2, 0)

Question 4.
The measure of the smallest angle of a right-angled triangle is 30°. The length of its smallest side is 6 centimeters. What is the length of its largest side? (1)
(6, 3, 18, 12)

Answer:
12

Question 5.
What is the slope of the line passing through the points (2, 5) and (3, 7)? (1)
(2, 3, 4, 5)
Answer:
2

Questions from 6 to 10 carry 2 scores each. (5 × 2 = 10)

Question 6.
Write the first term and common difference of the arithmetic sequence 3n + 2. (2)
Answer:
f = 5, d = 3

Question 7.
In the figure, AB is the diameter of the circle. C is a point on the circle. One of the angles ∠ACB and ∠ADB is twice the other. (2)

Write the measures of the angles ∠ACB and ∠ADB.
Answer:
∠ACB = 90°, ∠ADB = 45°

Question 8.
One is asked to say a natural number less than 10.
(a) What is the probability of it being an odd number? (1)
(b) What is the probability that it will not be an even number? (1)
Answer:
Odd numbers are 1, 3, 5, 7, 9.
There are 5 in number.
(a) \(\frac{5}{9}\)
(b) \(\frac{5}{9}\)

Question 9.
In the figure, AB and CD are the diameters of the circle. The coordinates of B are (3, 0). Write the coordinates of O and C. (2)

Answer:
O(0, 0), C(0, 3)

Question 10.
Write x² – 1 as the product of two first-degree polynomials. (2)
Answer:
x² – 1 = (x – 1)(x + 1)

Questions from 11 to 20 carry 3 scores each. (10 × 3 = 30)

Question 11.
(a) What is the tenth term of the arithmetic sequence a + 1, a + 2, a + 3, ….? (1)
(b) What is its common difference? (1)
(c) Write the algebraic form of the above sequence. (1)
Answer:
(a) x₁₀ = a + 10
(b) d = 1
(c) x_n = a + n

Question 12.
Draw a triangle of circumradius 3 centimeters and two of the angles 40° and 50°. (3)
Answer:

• Draw a circle of radius 3 cm, center O.
• Mark a point A on the circle.
• Draw the radius OA.
• Mark a point B on the circle such that ∠AOB = 2 × 40 = 80°.
• Mark a point C on the circle such that ∠BOC = 2 × 50 = 100°.
• Complete ΔABC.

Question 13.
(a) Write the sequence of even numbers. (1)
(b) One added to the product of two consecutive even numbers gives 289. Form a second-degree equation to solve this problem. (2)
Answer:
(a) 2, 4, 6, 8,…..
(b) Let x and x + 2 be the even numbers.
x(x + 2) + 1 = 289
⇒ x² + 2x = 288

Question 14.
In the figure chords AB and CD intersect at P. AB = 10 centimeters, PB = 4 centimeters, and PC = 3 centimeters.

(a) What is the length of PA? (1)
(b) Find the length of PD. (2)
Answer:
(a) PA = AB – PB
= 10 – 4
= 6 cm
(b) PA × PB = PC × PD
⇒ 6 × 4 = 3 × PD
⇒ PD = 8 cm

Question 15.
P is at a distance of 13 centimeters from the center of a circle of radius 5 centimeters.
(a) How many tangents can be drawn from the point P to the circle? (1)
(b) Find the lengths of the tangents. (2)
Answer:
(a) Two tangents can be drawn.
(b) Length of tangent is \(\sqrt{13^2-5^2}\) = 12 cm

Question 16.
ABCD is a square, coordinates of A are (1, -5). Diagonals of the square intersect at P(1, 0). Write the coordinates of B, C, and D. (3)

Answer:
PA = 5, Therefore PB = 5
B(6, 0), C(1, 5), D(-4, 0)

Question 17.
In the figure ∠B = 90°, AB = 7 centimetres, BC = 24 centimetres, AC = 25 centimetres.

(a) sin A = \(\frac{24}{K}\), what number is K? (1)
(b) Write cos C and sin C. (2)
Answer:
(a) k = 25
(b) cos C = \(\frac{24}{25}\) , sin C = \(\frac{7}{25}\)

Question 18.
A Sector with a central angle of 120° and a radius of 12 centimeters is rolled up into a cone.
(a) What is the slant height of the cone? (1)
(b) Find the radius of the cone. (2)
Answer:
(a) l = 12 cm
(b) \(\frac{x}{360}=\frac{\mathrm{r}}{l}\)
⇒ l × x = 36 × r
⇒ 12 × 120 = 360 × r
⇒ r = 4 cm

Question 19.
(a) In the figure OA is the radius of the circle. PQis the tangent through A. What is the measure of ∠OAP? (1)

(b) Draw a circle of radius 3 centimeters and mark a point A On it. Drawn the tangent through A. (2)
Answer:
(a) ∠OAP = 90°
(b) Draw a circle of radius 3 cm and center O.
Draw the radius OA.
Draw perpendicular to OA at A.
This perpendicular line is tangent to the circle at A.

Question 20.
ABCD is a rectangle. P is the mid-point of CD. If we put a dot in the figure without looking into it:

(a) What is the probability that it would be inside triangle APB? (1)
(b) What is the probability that it would be inside triangle ADP? (2)
Answer:
Draw a line parallel to AD through P and cut the side AB at Q.
Now we get four triangles of an equal area.
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{4}\)

Questions from 21 to 30 carry 4 scores each. (10 × 4 = 40)

Question 21.
(a) Write the 20th term of the arithmetic sequence 5, 10, 15,………. (1)
(b) Find the sum of the first 20 terms of the arithmetic sequence 5, 10, 15,…… (2)
(c) What is the sum of the first 20 terms of the arithmetic sequence 4, 9, 14,…..? (1)
Answer:
(a) x_n = 5n
x₂₀ = 5 × 20 = 100
(b) Sum = (x₁ + x₂₀) × \(\frac{20}{2}\)
= (5 + 100) × 10
= 1050
(c) All terms of the sequence 4, 9, 14, 19,….. are 1 less than the terms of 5, 10, 15, 20,……
Taking first 20 terms.
The required sum is 1050 – 20 = 1030.

Question 22.

In figure C, D, E, and G are points on the circle. ∠D = 70°. For the angles given in column I choose suitable measures from column II. (4)

Column I	Column II
∠ECG	120°
∠EBG	60°
∠EAG	110°
	180°

Answer:
∠ECG = 110°
∠EBG = 120°
∠EAG = 60°

Question 23.
Fill up the empty cells of the given square such that the numbers in each row, each column, and both diagonals form arithmetic sequences. (4)

3		13
7

Answer:
3, A, 13 are in an arithmetic sequence.
A = \(\frac{3+13}{2}\) = 8
3, B, 7 are in an arithmetic sequence.
B = \(\frac{3+7}{2}\) = 5
7, C, 13 are in an arithmetic sequence, C = 10
5, 10, D are in an arithmetic sequence,
D = 15, E = 12, F = 17

3	A	13
B	C	D
7	E	F

Question 24.
In the figure ∠B = 90°. BC = 1 centimetre, sin A = \(\frac{1}{2}\).

(a) What is the length of AC? (1)
(b) Find the length of AB. (1)
(c) What is the measure of ∠A? (1)
(d) sin 60° = _______________ (1)
Answer:
(a) AC = 2 cm
(b) AB = √3
(c) ∠A = 30°
(d) sin 60° = \(\frac{\sqrt{3}}{2}\)

Question 25.
Draw a circle of radius 3 centimeters. Mark a point P outside the circle at a distance of 7 centimeters from the center. Draw tangents from P to the circle. Measure the length of the tangents. (4)
Answer:

• Draw a circle of radius 3 cm and center O.
• Mark a point P at a distance of 7 cm from the center of the circle. Draw the line OP, and find its center C.
• With C as the center and CP as the radius, draw a circle that intersects the first circle at A and B.
• Draw the lines PA and PB which are the tangents to the circle.

Question 26.
Scores of 10 students are given below:
11, 32, 33, 35, 39, 41, 45, 47, 48, 49
(a) Find the mean score. (2)
(b) Find the median score. (2)
Answer:
(a) Mean = \(\frac{11+32+33+35+39+41+45+47+48+49}{10}\)
= \(\frac{380}{10}\)
= 38

(b) The date is already in the ascending order.
5th and 6th come in the middle.
Median is \(\frac{39+41}{2}\) = 40

Question 27.
Draw the x and y axes. Mark the point (2, 3). Draw a circle with the origin as a center and pass through the point (2, 3). (4)
Answer:
The drawing is given below.

Question 28.
(a) The perimeter of a rectangle is 40 centimeters. The length of its smaller side is 7 centimeters. What is the length of its larger side? (1)
(b) Find the sides of a rectangle with a perimeter of 40 centimeters and an area of 96 square centimeters. (3)
Answer:
(a) 20 – 7 = 13 cm
b) Sides are 10 + x, 10 – x
⇒ (10 + x) (10 – x) = 96
⇒ 10² – x² = 96
⇒ x² = 4
⇒ x = 2
Sides are 10 + 2 = 12 cm, 10 – 2 = 8 cm.

Question 29.
One is asked to say a two-digit number,
(a) What is the probability of both digits being the same? (2)
(b) What is the probability of the first digit being twice the second? (2)
Answer:
(a) Total number of two-digit numbers is 90.
The probability of getting both digits same is \(\frac{9}{90}\).

(b) Favourable outcomes are 21, 42, 63, 84.
The probability of getting the first digit being twice the second is \(\frac{4}{90}\).

Question 30.
(a) P(x) = x² – 5x + 9, find P(2) and P(3). (2)
(b) Write P(x) – P(2) as the product of two first degree polynomials. (2)
Answer:
(a) p(2) = 2² – 5 × 2 + 9 = 3
p(3) = 3² – 5 × 3 + 9
= 9 – 15 + 9
= 3
(b) p(x) – p(1) = x² – 5x + 9 – 3
= x² – 5x + 6
= (x – 3)(x – 2)

Questions from 31 to 45 carry 5 scores each. (15 × 5 = 75)

Question 31.

(a) Write the fifth line of the pattern. (1)
(b) How many numbers are there in the tenth line? (1)
(c) How many numbers are there in the first ten lines altogether? (2)
(d) What is the first number in the eleventh line? (1)
Answer:
(a) 11, 12, 13, 14, 15
(b) 10
(c) 1 + 2 + 3 + 4 +……+ 10 = \(\frac{10(10+1)}{2}\) = 55
(d) 55 + 1 = 56

Question 32.
(a) In the figure area of the rectangle ABCD is 8 square centimetres and BC = BP. What is the area of the shaded square? (1)

(b) Draw a rectangle of an area of 8 square centimeters. Draw a square having the same area as the rectangle. (4)
Answer:
(a) 8 sq. cm
(b) Construction steps are given below:

• Draw a rectangle of side AB = 5 cm and 3 cm.
• It is marked as ABCD as in the given picture.
• Produce AB to P such that BP = BC.
• With AP as the diameter draw a semicircle.
• Produce BC to intersect the semicircle at E.
• Complete the square with side BE.

Question 33.
A man standing at the edge of a river sees the top of a tree at an elevation of 60°. Stepping 20 metres back he sees it at an elevation of 30°. Draw a rough figure and find the width of the river. (5)
Answer:
(a) The picture is given below.

(b) In the figure,

∠ACD = 20°, ∠D = ∠CAD = 30°, CD = AC = 20 m
Width of the river BC = 10 meters.

Question 34.
The sides of a rectangle are parallel to the axes. One pair of its opposite vertices are A(2, 4) and C(6, 12).
(a) Write the coordinates of the other two vertices. (2)
(b) Write the coordinates of the mid-point of AC. (2)
(c) x coordinate of a point on AC is ‘a’. What is its y coordinate? (1)
Answer:
(a) Coordinates of B is (6, 4)
Coordiante of D is (2, 12)
(b) \(\left(\frac{2+6}{2}, \frac{4+12}{2}\right)\) = (4, 8)
(c) 2a

Question 35.
In figure AB, BC and AC touch the circle at the points Z, X, and Y. ∠ZXY = 60° and ∠XZY = 50°. Find the measures of ∠A, ∠B, and ∠C. (5)

Answer:
(a) ∠AZY = ∠ZXY = 60°
∠A = 180° – (60° + 60°) = 60°

(b) ∠CXY = ∠XZY = 50°
∠C = 180° – (50° + 50°) = 80°

(c) ∠B = 180° – (60° + 80°) = 40°

Question 36.
(a) Radius of a solid metal cone is 5 centimeters, its slant height is 13 centimeters. Find its height. (2)
(b) Find the volume of the cone. (1)
(c) It is melted and recast into small cones of radius 1 centimeter and height one centimeter. How many cones will we get? (2)
Answer:
(a) h = \(\sqrt{13^2-5^2}\) = 12 cm.
(b) Volume = \(\frac{1}{3}\) × π × 5² × 12 = 100π cubic cm.
(c) Volume of small cone = \(\frac{1}{3}\) × π × 1² = \(\frac{\pi}{3}\)
Number of cones = 100π ÷ \(\frac{\pi}{3}\) = 300 cubic cm.

Question 37.
A circle is drawn with (1, 1) as center. (4, 5) is a point on the circle.

(a) Find the radius of the circle. (1)
(b) Write the equation of the circle. (2)
(c) The x coordinate of a point on the circle is 6. What is the y coordinate of that point? (2)
Answer:
(a) r = \(\sqrt{(4-1)^2+(5-1)^2}\)
= \(\sqrt{3^2+4^2}\)
= 5

(b) (x – 1)² + (y – 1)² = 5²

(c) When x = 6, (6 – 1)² + (y – 1)² = 25
⇒ (y – 1)² = 0
⇒ y = 1

Question 38.
The diametres of the two spheres are in the ratio 1 : 2.
(a) What is the ratio of their radii? (1)
(b) Find the ratio of their surface areas. (3)
(c) If the surface area of the first sphere is 10π square centimeters. What is the surface area of the second sphere? (1)
Answer:
(a) 1 : 2
(b) r₁ = 1r, r₂ = 2r
Ratio of surface area 4πr² : 4π(2r)² = 1 : 4
(c) The surface area of the second sphere is 4 × 10π = 40π

Question 39.
(a) What is the remainder on dividing the terms of the arithmetic sequence 100, 109, 118,….. by 9? (1)
(b) Write the sequence of three-digit numbers, which are multiple of 9. (2)
(c) What is the position of 999 in the arithmetic sequence of three-digit numbers which are multiples of 9? (2)
Answer:
(a) 1
(b) 108, 117, 126,……
(c) The algebraic form of this sequence is 9n + 99.
9n + 99 = 999
⇒ 9n = 900
⇒ n = 100
There are 100 terms in this sequence.

Question 40.
In the figure AB = AC = 4 centimetres, ∠A = 120°.

(a) ∠B = _________ (1)
(b) Find the perpendicular distance from A to BC. (2)
(c) Find the area of the triangle. (2)
Answer:
(a) 30°
(b) 2 cm
(c) BC = 4√3 cm
Area of the triangle = \(\frac{1}{2}\) × 4√3 × 2 = 4√3 sq.cm

Question 41.
(a) In the figure, a circle with center O touches the sides of the triangle ABC at the points P, Q, and R. If ∠B = 50°, what is ∠POR? (1)

(b) Draw a circle of radius 2.5 centimeters. Draw a triangle of angles 50°, 60°, and 70° with all its sides touching the circle. (4)
Answer:
(a) BPOR is cyclic ∠POR = 180° – 50° = 130°.
(b) Draw a circle of radius 2.5 cm.
Divide the angle around the centre at the measures 180° – 50° = 130°, 180° – 60° = 120° and 180° – 40° = 140°.
Draw tangents at the ends of the radii to the circle.
The tangents enclose the required triangle.

Question 42.
In the figure, O is the center of the circle. A, B, C, and D are points on the circle. ∠AOB = 80°.

(a) Write the measures of ∠ACB, ∠ADB, and ∠ADP. (3)
(b) Find ∠CQD + ∠P. (2)
Answer:
(a) ∠ACB = 40°
∠ADB = 40°
∠ADP = 180° – 40° = 140°

(b) ∠CQD + ∠P + 2 × 140° = 360°
∠CQD + ∠P = 360° – 280° = 80°

Question 43.
A box is to be made by cutting off small squares from each corner of a square of thick paper and bending upwards. The height of the box is to be 10 centimeters and volume 1 liter.
(a) What should be the length of a side of the square cut-off? (1)
(b) What should be the length of a side of the square, thick paper sheet? (4)
Answer:
(a) 10 cm

(b) 1 litre = 1000 cubic cm3
⇒ (x – 20)² × 10 = 1000
⇒ (x – 20)² = 100
⇒ x – 20 = 10
⇒ x = 30 cm

Question 44.
The table below shows children of a class sorted according to their scores in an examination.

Scores	Number of Children
0 – 10	5
10 – 20	8
20 – 30	10
30 – 40	13
40 – 50	9
Total	45

(a) If the children are arranged in the ascending order of their scores, then what will be the assumed score of the 14th child? (1)
(b) Compute the median score. (4)
Answer:
The cumulative frequency table is given below.

Scores	Number of Children
Below 10	5
Below 20	13
Below 30	23
Below 40	36
Upto 50	45

n = 45, odd number.
23rd score is considered as the middle scorp or median.
It is in between the 20th and 30th scores.
This class contains 10 scores and 10 children.
When 10 scores are divided equally among 10 children each one’s share is 1.
Score of 14th student is assumed to be 20 + \(\frac{1}{2}\) = 20.5
We assume that the marks in the median class (20 – 30) are in an arithmetic sequence of f = 20.5 and common difference 1.
Its 10th term is the mark of the 23rd student.
x₁₀ = f + 9d
= 20.5 + 9 × 1
= 29.5

Question 45.
Read the following passage. Understand the mathematical concept in it and answer the questions that follow.
The circle passing through all the three vertices of a triangle is its circumcircle. Like this, the circle touching all three sides of a triangle is its incircle. The point of intersection of the angle bisectors is the incentre.

The distance from the center of the circle to the touching point is the radius.
The area of triangle ABC is the sum of the areas of the triangles OBC, OAC, and OAB.
If the radius of the incircle is taken as r and the sides of the triangle as a, b, and c.
The area of triangle ABC = \(\frac{1}{2} a r+\frac{1}{2} b r+\frac{1}{2} c r\)
= \(\frac{1}{2}\) × r(a + b + c)
= \(\mathrm{r} \frac{(a+b+c)}{2}\)
= r × s
Here s = \(\frac{a+b+c}{2}\) (half of perimeter)
(a) A circle touching all three sides of a triangle is:
[Circumcircle, Incircle, Semicircle, Ellipse]
(b) A circle passing through all three vertices of a triangle is:
[Circumcircle, Incircle, Semicircle, Ellipse]
(c) If the radius of the incircle is taken as r and the half of the perimeter as s then the area of the triangle is:
[r + s, \(\frac{r}{s}\), r × s, r² × s]
(d) The perimeter of a triangle is 20 centimeters and the radius of its incircle is 2 centimeters. What is the area of the triangle, (in square centimeters)
(40, 20, 10, 5)
(e) The Area of a triangle is 24 square centimeters and its perimeter is 24 centimeters. The radius of the incircle is _____________ centimeters.
(1, 2, 1.5, 2.5)
Answer:
(a) Incircle
(b) Circumcircle
(c) A = r × s
(d) 20

Students can read Kerala SSLC Maths Question Paper March 2022 with Answers English Medium and Kerala SSLC Maths Previous Year Question Papers with Answers helps you to score more marks in your examinations.

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)

Students can read Kerala SSLC Maths Question Paper March 2023 with Answers English Medium and Kerala SSLC Maths Previous Year Question Papers with Answers helps you to score more marks in your examinations.

Kerala Syllabus Class 10 Maths Question Paper March 2023 with Answers English Medium

Time: 2½ Hours
Total Score: 80

Instructions:

• Read each question carefully before answering.
• Give explanations wherever necessary.
• The first 15 minutes is cool-off time. You may use this time to read the questions and plan your answers.
• There is no need to simplify irrationals like √2, √3, π, etc., using approximations unless you are asked to do so.

Answer any three questions from 1 to 4. Each question carries 2 scores. (3 × 2 = 6)

Question 1.
7, 13, 19,…. is an arithmetic sequence.
(a) What is its common difference?
(b) Find its 11th term.
Answer:
(a) d = 6

(b) x_n = 7 + 10 × 6 = 67

Question 2.
The weights of 11 players of a football team are given in kilograms:
55, 65, 56, 70, 62, 54, 64, 58, 68, 65, 60
Find the median of the weights of players.
Answer:
54, 55, 56, 58, 60, 62, 64, 65, 65, 68, 70
Median weight is the weight of the 6th player.
∴ The median is 62.

Question 3.
O is the center of the circle. A dot is put inside the circle without looking at it.

(a) What is the probability that the dot to be within the unshaded part?
(b) What is the probability that the dot to be within the shaded part?
Answer:
120° is \(\frac{1}{3}\) of 360°
(a) \(\frac{1}{3}\)
(b) \(\frac{3}{2}\)

Question 4.
AB is a chord of a circle of radius 3 centimeters. Chord AB makes a right angle at the center. What is the length of AB?

Answer:
∆OAB is a 45° – 45° – 90° triangle.
∴ AB = 3√2 cm.

Answer any four questions from 5 to 10. Each question carries 3 scores. (4 × 3 = 12)

Question 5.
A(3, 9), and C(8, 12) are the coordinates of two opposite vertices of a rectangle whose sides are parallel to the coordinate axes.

(a) Find the coordinates of the other two vertices of the rectangle.
(b) Find the lengths of the sides of the rectangle.
Answer:
(a) B(8, 9), D(3, 12)
(b) AB = |8 – 3| = 5
BC = |12 – 9| = 9

Question 6.
Draw a circle of radius 4 centimeters. Draw a triangle whose vertices are on this circle and two of the angles 40° and 60°.
Answer:

1. Draw a circle of radius 4 cm.
2. Divide the angle around the center as 80° and 120° by drawing the radii.
3. Join the ends of radii. It makes the triangle.

Question 7.
Find the lengths of the sides of the rectangle whose perimeter is 80 centimeters and area is 351 square centimeters.
Answer:
2 × (length + breadth) = 80
⇒ length + breadth = 40
Sides are 20 – x, 20 + x
(20 – x)(20 + x) = 351
⇒ 20² – x² = 351
⇒ 400 – 351 = x²
⇒ x² = 49
⇒ x = 7
Sides are 13 cm, 27 cm.

Question 8.
(4, 5) and (8, 11) are coordinates of two points on a line.
(a) Find the slope of the line.
(b) Find the equation of the line.
Answer:
(a) Slope = \(\frac{11-5}{8-4}=\frac{3}{2}\)
(b) Let(x, y) be a point on the line.
\(\frac{y-5}{x-4}=\frac{3}{2}\)
⇒ 2(y – 5) = 3(x – 4)
⇒ 2y – 3x = -2
⇒ 3x – 2y = 2
This is the equation of a line.

Question 9.
6th term of an arithmetic sequence is 46. Its common difference is 8.
(a) What is its 16th term?
(b) Find its 21st term.
Answer:
(a) x₁₆ = x₆ + 10d
= 46 + 10 × 8
= 126
(b) x₂₁ = x₁₆ + 5d
= 126 + 40
= 166

Question 10.
The sides of a right triangle are 9 centimeters, 12 centimeters, and 15 centimeters.

(a) Find the area of the triangle.
(b) Calculate the radius of the triangle.
Answer:
(a) Area = \(\frac{1}{2}\) × 9 × 12
= 9 × 6
= 54 cm
(b) A = rs.
where A is the area, r is the radius of the incircle, and s is the semiperimeter.
s = \(\frac{12+15+9}{2}\) = 18
r = \(\frac{A}{8}=\frac{54}{18}\) = 3 cm

Answer any eight questions from 11 to 21. Each question carries 4 scores. (8 × 4 = 32)

Question 11.
P(x) = x² – 4x + 4
(a) What is P(1)?
(b) Write a first degree factor of P(x) – P(1).
(c) Write the polynomial P(x) – P(1) as the product of two first degree polynomials.
Answer:
(a) p(1) = 12 – 4 × 1 + 4 = 1
(b) x – 1 is a factor of p(x) – p(1)
(c) p(x) – p(1) = x² – 4x + 4 – 1
= x² – 4x + 3
= (x – 3)(x – 1)

Question 12.
A cone is made by rolling up a semicircle of radius 20 centimeters.
(a) What is the slant height of the cone?
(b) Find the radius of the cone.
(c) Calculate the curved surface area of the cone.
Answer:
(a) l = 20 cm
(b) lx = 360r
⇒ 20 × 180 = 360 × r
⇒ r = 10 cm
(c) Curved surface area = πrl = 200π

Question 13.
Draw a circle of radius 2.5 centimeters. Mark a point 6.5 centimeters away from the center. Draw the tangents to the circle from this point. Measure and write the lengths of the tangents.
Answer:

1. Draw a circle of radius 2.4 cm, and mark the center as O.
2. Draw a circle with OP as the diameter.
3. This circle cuts the first circle at A and B.
4. Draw PA and PB tangents to the circle from P, the outer point itemize.

Question 14.
The sum of the first 7 terms of an arithmetic sequence is 140. The sum of the first 11 terms of the same arithmetic sequence is 440.
(a) What is the 4th term of this arithmetic sequence?
(b) Find its 6th term.
(c) What is the common difference?
(d) Find the first term of this sequence.
Answer:
(a) x₄ = \(\frac{140}{7}\) = 20
(b) x₆ = \(\frac{440}{11}\) = 40
(c) x₆ – x₄ = 2d
⇒ 20 = 2d
⇒ d = 10
(d) x₁ = x₄ – 3d
= 20 – 30
= -10

Question 15.
A box contains 4 slips numbered 1, 2, 3, 4 and another contains 5 slips numbered 1, 2, 3, 4, 5. One slip is taken from each box without looking at it.
(a) In how many different ways we can choose the slips?
(b) What is the probability of both numbers being odd?
(c) What is the probability of both numbers being the same?
Answer:
(a) 2 × 5 = 10
(b) 2 × 3 = 6
Probability of getting odd is \(\frac{6}{20}=\frac{3}{10}\)
(c) \(\frac{4}{20}=\frac{2}{10}\)

Question 16.
In a right triangle, one of the perpendicular sides is 2 centimeters more than that of the other. The area of the triangle is 24 square centimeters. Find the lengths of the perpendicular sides of the right triangle.
Answer:
The sides are x and x + 2
\(\frac{1}{2}\) × x × (x + 2) = 24
⇒ x² + 2x + 1 = 49
⇒ (x + 1)² = 49
⇒ x + 1 = 7
⇒ x = 6
The sides are 6 cm and 8 cm itemized.

Question 17.
Draw the co-ordinate axes and mark the points A(0, 0), B(4, 4), C(8, 0) and D(4, -4).
(a) Write the suitable name of the quadrilateral ABCD.
(b) Find the length of the diagonal BD.
Answer:
Draw coordinate axes and mark the points.
(a) Square
(b) Diagonal 8

Question 18.
Diagonals AC and BD of the cyclic quadrilateral ABCD cuts at P.

PA = 12 centimetres; PC = 2 centimetres; BD = 11 centimetres.
(a) If PB = x, then write PD in terms of x.
(b) Find the lengths of PB and PD.
Answer:
(a) PD = 11 – x
(b) PA × PC = PB × PD
⇒ 12 × 2 = x × (11 – x)
⇒ 24 = 11x – x²
⇒ x² – 11x + 24 = 0
PB = 3 cm, PD = 8 cm

Question 19.
BC is a chord of the circle centered at O. BC = 10 centimeters, ∠A = 60°. Find the radius of the circle.

Answer:
Draw a rough diagram.
Mark the center O.
Draw diameter through B as BD.
Join DC.
Triangle BCD is a right triangle.
CD = \(\frac{10}{\sqrt{3}}\)
Diameter BD = \(\frac{20}{\sqrt{3}}\)
Radius is \(\frac{10}{\sqrt{3}}\)

Question 20.
In the figure, the coordinates of 3 vertices of the parallelogram ABCD are given.

(a) Write the coordinates of C.
(b) Calculate the length of the diagonal AC.
(c) Find the coordinates of the point of intersection of the diagonals.
Answer:
(a) C(10 + 12 – 7, 11 + 7 – 5) = C(15, 13)
(b) AC = \(\sqrt{(15-7)^2+(13-5)^2}\)
= \(\sqrt{64+64}\)
= \(\sqrt{128}\)
= 8\(\sqrt{2}\)
(c) \(\left(\frac{7+15}{2}, \frac{5+13}{2}\right)\) = (11, 9)

Question 21.
A square pyramid is made by cutting out a paper as in the figure. The side of the square is 40 centimeters. The height of the triangle is 25 centimeters.

(a) What is the slant height of the square pyramid?
(b) Find the height of the pyramid.
(c) Calculate the volume of the pyramid.
Answer:
(a) 25
(b) h = \(\sqrt{25^2-20^2}\) = 15
(c) Volume = \(\frac{1}{3}\) × 40² × 15 = 8000 cm³

Answer any six questions from 22 to 29. Each question carries 5 scopes. (6 × 5 = 30)

Question 22.
The daily wages of 99 workers in a factory are 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 based on their daily wages, at what position does the median wage fall?
(b) What is the median class?
(c) Find the median of the wages.
Answer:

Daily Wages	Number of Workers
Below 600	8
Below 700	21
Below 800	41
Below 900	66
Below 1000	85
Below 1100	99

(a) n = 99 (odd number)
The 50th wage comes in the middle.
50th wage is median.

(b) 800 – 900 is the median class.

(c) When 100 rupees is divided equally among 25 workers then each one’s wage is 4.
Wages in the median class are assumed in an arithmetic sequence with first term 800 + 2 = 802 and common difference 4.
802 is considered as the wage of the 42nd worker as per the table.
9th term of this arithmetic sequence will be median.
∴ Median = 802 + 8 × 4 = 802 + 32 = 834

Question 23.
Draw a rectangle of an area of 24 square centimeters. Draw a square of area equal to the area of this rectangle.
Answer:

1. Draw a rectangle of area 24 sq. cm. We can draw this rectangle using sides 8 cm and 3 cm or any other.
2. Let it be ABCD with AB on the base side and BC on the width. Procure AB to such that BC = BE.
3. With AE as diameter draw a semicircle.
4. Produce BC to the semicircle and mark the point on the semicircle as F. BF² = 8 × 3
5. Draw a square with side BF. The area of this square will be 24.

Question 24.
In the figure, (0, 6) and (8, 0) are coordinates of the points A and B. A circle of diameter AB is to be drawn.

(a) Find the coordinates of the center of the circle.
(b) Find the radius of the circle.
(c) What is the equation of the circle?
Answer:
(a) AOB is a right triangle.
The circumcentre will be the midpoint of the hypotenuse. It is (4, 3).
(b) The radius of the circle is 5.
(c) (x – 4)² + (y – 3)² = 5²
⇒ x² + y² – 8x – 6y = 0

Question 25.
PA and PB arc two tangents to the circle centered at O.

∠ACB = 105°. Find the angles given below.
(a) ∠ADB = ______________________
(b) ∠AOB = ______________________
(c) ∠APB = ______________________
(d) ∠ABP = ______________________
(e) ∠ABO = ______________________
Answer:
(a) ∠ADB = 180° – 105° = 75°
(b) ∠AOB = 2 × 75° = 150°
(c) ∠APB = 180° – 150° = 30°
(d) Draw AB.
∠ABP = ∠BDA = 75°
(e) ∠ABO = 90° – 75° = 15°

Question 26.
There are two cylindrical wooden blocks with a diameter of 60 centimeters and a height of 60 centimeters. The largest cone is carved out from one block and the largest sphere from the other.
(a) What is the volume of the cylinder?
(b) Find the volume of the cone.
(c) Find the radius of the sphere.
(d) Calculate the volume of the sphere.
(e) Find the ratio of the volumes of the cone and the sphere.
Answer:
(a) π × 30² × 60 = 54000π cm³
(b) \(\frac{1}{3}\) × 54000π = 18000π cm³
(c) 30 cm
(d) \(\frac{4}{3}\)π × 30³ = 36000π
(e) 18000 : 36000 = 1 : 2

Question 27.
(a) Find the sum of the first 20 natural numbers.
(b) Write the algebraic expression of the arithmetic sequence 5, 9, 13,……….
(c) Find the sum of the first 20 terms of the arithmetic sequence 5, 9, 13,……….
Answer:
(a) \(\frac{20 \times 21}{2}\) = 210
(b) 4n + 1
(c) 4 × 210 + 20 = 860

Question 28.
A child sees the top of a telephone tower at an elevation of 80°. Stepping 20 meters back, he sees it at an elevation of 40°.
(a) Draw a rough figure.
(b) Calculate the height of the tower.
[sin 40° = 0.64; cos 40° = 0.77; tan 40° = 0.84; sin 80° = 0.98; cos 80° = 0.17; tan 80° = 5.7]
Answer:
(a)

(b) sin 80° = \(\frac{h}{20}\)
⇒ h = 20 × 0.98 = 19.6 meter

Question 29.
Diagonals of a quadrilateral are the lines joining its opposite vertices. What about the diagonals of a polygon?
The lines from one vertex to the adjacent two vertices are not diagonals. They are the sides of the polygon. Lines to all other vertices are diagonals.
In a quadrilateral, only one diagonal can be drawn from one vertex. If we draw from all 4 vertices, we get 4 diagonals.
But 2 among them are the same. In a pentagon, from one vertex, 2 diagonals can be drawn.
Therefore the total number of lines is 5 × 2 = 10
But 5 among them are the same.
So number of diagonals in a pentagon = \(\frac{5 \times 2}{2}\) = 5
Now complete the table given below:

Polygon	Number of Sides	Number of Diagonals from One Vertex	Total number of Diagonals
Quadrilateral	4	1	\(\frac{4 \times 1}{2}\) = 2
Pentagon	5	2	\(\frac{5 \times 2}{2}\) = 5
Hexagon	6	3	\(\frac{6 \times 3}{2}\) = 9
Heptagon	7	………	………
Decagon	10	………	………
n Sided Polygon	n	n – 3	………

Answer:
(a) Heptogon → 7 → 4 → 14
(b) Decagon → 10 → 7 → 35
(c) n sided polygon → n → n – 3 → \(\frac{\mathrm{n}(\mathrm{n}-3)}{2}\)

Students can read Kerala SSLC Maths Question Paper March 2024 with Answers English Medium and Kerala SSLC Maths Previous Year Question Papers with Answers helps you to score more marks in your examinations.

Kerala Syllabus Class 10 Maths Question Paper March 2024 with Answers English Medium

Time: 2½ Hours
Total Score: 80

Instructions:

• Read each question carefully before answering.
• Give explanations wherever necessary.
• The first 15 minutes is cool-off time. You may use this time to read the questions and plan your answers.
• There is no need to simplify irrationals like √2, √3, π, etc., using approximations unless you are asked to do so.

Answer any three questions from 1 to 4. Each question carries 2 scores. (3 × 2 = 6)

Question 1.
See the figure.

If a circle is drawn with AB as its diameter, what are the positions of points P and Q for this circle?
Answer:
Point P is inside the circle (since ∠APB > 90°).
Point Q is outside the circle (since ∠AQB < 90°).

Question 2.
The hemoglobin levels in grams per decilitres of seven students are given below:
12.9, 12.0, 12.6, 12.5, 14.1, 13.7, 13.4
Find the median hemoglobin level.
Answer:
Arrange the given data in increasing order:
12.0, 12.5, 12.6, 12.9, 13.4, 13.7, 14.1
Here there are 7 observations. So, the median is the 4 th observation.
Median = 12.9

Question 3.
The sequence of perimeters of squares of sides 1 centimeter, 2 centimeters, 3 centimeters, and so on form an arithmetic sequence.
(a) Write the sequence.
(b) What is the common difference?
Answer:
(a) Sequence of perimeters of squares of sides 1 cm, 2 cm, 3 cm, etc.. = 4, 8, 12,…

(b) Common difference = 4.

Question 4.
A rectangular portion is shaded in a square of side 5 centimeters as shown in the figure. Adot is put inside the square without looking. Find the probability of the dot to be in the shaded region.

Answer:
Total area of the square = 5² = 25 cm²
Area of the shaded rectangular region = 2 × 5 = 10 cm²
∴ Probability of dot to be in the shaded region = \(\frac{10}{25}=\frac{2}{5}\)

Answer any four questions from 5 to 10. Each question carries 3 scores. (4 × 3 = 12)

Question 5.
Draw the coordinate axes and mark the points A(0, 0), B(2, 3) and C(4, 0). What is the perpendicular distance from B to AC?
Answer:

The perpendicular distance from B to AC = 3

Question 6.
Ajay is 10 years older than Renuka. The product of their ages is 144.
(a) Taking the age of Renuka as x, what is the age of Ajay in terms of x?
(b) Find their ages.
Answer:
(a) Ajay’s age = x + 10
(b) Product of their ages = 144
⇒ x(x + 10) = 144
⇒ x² + 10x = 144
⇒ x² + 10x + 5² = 144 + 5²
⇒ (x + 5)² = 13²
⇒ x = 8
So, Renuka’s age is 8 and Ajay’s age is 18.

Question 7.
Draw a rectangle of sides 4 centimeters and 3 centimeters. Draw a square of the same area.
Answer:

Question 8.
Prove that the points (3, 5), (6, 7) and (9, 9) are on the same line.
Answer:
Let A(3, 5), B(6, 7), C(9, 9).
Slope of AB = \(\frac{7-5}{6-3}=\frac{2}{3}\)
Slope of BC = \(\frac{9-7}{9-6}=\frac{2}{3}\)
Hence, A, B, and C lie on the same line.

Question 9.
The nth term of an arithmetic sequence is 4n + 1.
(a) Write the common difference of the sequence.
(b) Write the first term of the sequence.
(c) What is the remainder obtained when the terms of this sequence are divided by 4?
Answer:
(a) nth term = 4n + 1
1st term = 4 × 1 + 1 = 5
2nd term = 4 × 2 + 1 = 9
Common difference = 9 – 5 = 4.

(b) First term = 5

(c) Terms of the sequence are 5, 9, 13,…..
When each term is divided by 4, we get the remainder 1.

Question 10.
AB, BC, and CA are tangents to the circle centered at O, touching the circle at P, Q, and R respectively, as shown in the figure.

(a) Find ∠QOR.
(b) Find the angles of triangle ABC.
Answer:
(a) ∠QOR = 360 – (110 + 100)
= 360 – 210
= 150

(b) ∠POQ + ∠B = 180
⇒ 110 + ∠B = 180
⇒ ∠B = 180 – 110 = 70
∠POR + ∠A = 180
⇒ 100 + ∠A = 180
⇒ ∠A = 180 – 100 = 80
∠ROQ + ∠C = 180
⇒ 150 + ∠C = 180
⇒ ∠C = 180 – 150 = 30

Answer any eight questions from 11 to 21. Each question carries 4 scores. (8 × 4 = 32)

Question 11.
Numbers from 1 to 50 are written on slips of paper and put in a box. Without looking, a slip is to be drawn from it.
(a) What is the probability that it is a multiple of 4?
(b) What is the probability that it is a multiple of 6?
(c) What is the probability that it is a multiple of 4 and 6?
Answer:
(a) Multiple of 4 from 1 to 50 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48
∴ Probability that it is a multiple of 4 = \(\frac{12}{50}=\frac{6}{25}\)

(b) Multiple of 6 from 1 to 50 are 6, 12, 18, 24, 30, 36, 42, 48
∴ Probability that it is a multiple of 4 = \(\frac{8}{50}=\frac{4}{25}\)

(c) Multiple of 4 and 6 from 1 to 50 are 12, 24, 36, 48
∴ Probability that it is a multiple of 4 and 6 = \(\frac{4}{50}=\frac{2}{25}\)

Question 12.
Draw a circle of radius 2.5 centimeters and mark a point 6 centimeters away from the center of the circle.
(a) How many tangents can be drawn to the circle from this point?
(b) Draw the tangents to the circle from this point.
Answer:
(a) 2

(b)

Question 13.
Consider the arithmetic sequence 8, 14, 20,…
(a) Is 25 a term of this sequence?
(b) Check if 144 is a term of this sequence.
(c) Prove that there are no perfect squares in this sequence.
Answer:
Given the arithmetic sequence 8, 14, 20,…..
Common difference (d) = 6
(a) No, 25 is not a term of the sequence.
Because each term is divided by the common difference 6 leaves the remainder 2.
But when 25 is divided by 6 leaves remainder is 1.

(b) No, 144 is not a term of the sequence. Because 144 leaves the remainder 0 on division by 6.

(c) Perfect squares are 1, 4, 9, 16, 25,…..
\(\frac{(6 n)^2}{6}\) remainder is 0
\(\frac{(6 n \pm 1)^2}{6}\) remainder is 1
\(\frac{(6 n \pm 2)^2}{6}\) remainder is 4
\(\frac{(6 n \pm 3)^2}{6}\) remainder is 3
Continue like this, the remainder cannot be 2.
Hence the proof.

Question 14.
A(2, 3), B(8, 5), and C(4, 7) are the coordinates of the vertices of triangle ABC. P is the midpoint of AB and Q is the midpoint of BC.
(a) Find the coordinates of P and Q.
(b) Find the distance between P and Q.
Answer:
A(2, 3), B(8, 5), C(4, 7)
(a) P is the midpoint of AB. So,
P = \(\left(\frac{2+8}{2}, \frac{3+5}{2}\right)\) = (5, 4)
Q is the midpoint of BC. So,
P = \(\left(\frac{8+4}{2}, \frac{5+7}{2}\right)\) = (6, 6)

(b) Distance between P and Q = \(\sqrt{(6-5)^2+(6-4)^2}\)
= \(\sqrt{1+4}\)
= \(\sqrt{5}\)

Question 15.
From a circle of radius 15 centimeters, a sector of central angle 120° is cut out and rolled up to make a cone.
(a) What is the slant height of the cone?
(b) What is the base radius of the cone?
(c) Calculate die curved surface area of the cone.
Answer:
(a) Slant height (l) = 15 cm

(b) x = 120°
\(\frac{r}{l}=\frac{x}{360}\)
\(\frac{r}{15}=\frac{120}{360}\)
r = \(\frac{15 \times 120}{360}\) = 5 cm

(c) Curved surface area of the cone = πrl
= π × 5 × 15
= 75π cm²

Question 16.
The diagonal of a rectangle is 9 centimetres and it makes an angle 49° with one side. Find the length of the sides of the rectangle.
(sin 49°= 0.75, cos 49° = 0.66)

Answer:
sin 49 = \(\frac{R Q}{9}\)
0.75 = \(\frac{R Q}{9}\)
RQ = 9 × 0.75 = 6.75 cm
cos 49 = \(\frac{P Q}{9}\)
0.66 = \(\frac{P Q}{9}\)
PQ = 9 × 0.66 = 5.94 cm

Question 17.
ABCDEF is a regular hexagon with its origin as the center. The coordinates of the point A is (4, 0).

(a) What are the coordinates of the point D?
(b) Find the length of BG.
(c) Write the coordinates of the points B and E.
Answer:
(a) Coordinate of point D = (-4, 0)

(b) Since ABCDEF is a regular hexagon, each angle is 120°.
∴ ∠OAB = 60°
Also, ∠AOB = 60°
Consider ∆BOG,
∠GOB = 60°
∠BGO = 90°
∴ ∠GBO = 30°
So, the sides are in the ratio 1 : √3 : 2
∴ BG = 2√3

(c) B(2, 2√3)
E(-2, -2√3)

Question 18.
The square of a number is equal to 12 added to the number. Find the number.
Answer:
Let x be the number.
Given, x² = 12 + x
⇒ x² – x – 12 = 0
⇒ (x – 4)(x + 3) = 0
⇒ x = 4, -3

Question 19.
Consider the polynomial p(x) = x² – 5x + 6
(a) Write p(x) as the product of two first-degree polynomials.
(b) Find the solutions of the equation p(x) = 0.
Answer:
(a) Let p(x) = x² – 5x + 6 = (x – a)(x – b)
x2 – 5x + 6 = x² – (a + b)x + ab
⇒ a + b = 5 and ab = 6
⇒ a = 2, b = 3
∴ x² – 5x + 6 = (x – 2)(x – 3)

(b) p(x) = 0
⇒ x² – 5x + 6 = 0
⇒ (x – 2) (x – 3) = 0
⇒ x = 2, 3

Question 20.
The diameters of the two hemispheres are in the ratio 5 : 3.
(a) Write the ratio of their radii.
(b) Find the ratio of their surface areas.
(c) If the surface area of the first hemisphere is 100 square centimeters, what is the surface area of the other?
Answer:
(a) Ratio of their radii = 5 : 3

(b) Surface area of hemisphere = 3πr²
\(3 \pi r_1^2: 3 \pi r_2^2\)
= 3π × 5² : 3π × 3²
= 25 : 9

(c) 25 : 9 = 100 : x
⇒ x = \(\frac{9 \times 100}{25}\) = 36 sq.cm

Question 21.
The central angle of the AXB is 110° and the central angle of the arc CYD is 80°. Find the angles of triangle APD.

Answer:
Since the central angle of AXB is 110°, ∠ADB = 55°
Since the central angle of CYD is 80°, ∠DAC = 40°
∴ ∠APD = 180° – (55° + 40°)
= 180° – 95°
= 85°

Answer any six questions from 22 to 29. Each question carries 5 scores. (6 × 5 = 30)

Question 22.
Draw a triangle of sides 5 centimeters, 6 centimeters, and 7 centimeters. Draw the incircle of the triangle. Measure the radius of the incircle.
Answer:

Radius = 1.6 cm

Question 23.
The ages of the workers of an organization are arranged as follows:

Age	Number of Workers
20 – 30	9
30 – 40	10
40 – 50	8
50 – 60	5
60 – 70	1

(a) If the workers are arranged in order of their wages, the age of which worker is taken as the median age?
(b) Find the median age.
Answer:

Age	Number	Age	Number
20 – 30	9	Below 30	9
30 – 40	10	Below 40	19
40 – 50	8	Below 50	27
50 – 60	5	Below 60	32
60 – 70	1	Below 70	33
Total	33

(a) \(\frac{33+1}{2}\) = 17
So, the age of the 17th worker is taken as the median age.

(b) d = \(\frac{40-30}{10}\) = 1
Age of 10th worker = \(\frac{30+31}{2}\) = 30.5
Age of 17th worker = 30.5 + 7 × 1 = 37.5
So, the median age = 37.5

Question 24.
From a point on the ground at a distance of 100 meters away from a tower, the top of the tower is seen at an angle of elevation 45°. From the top of the tower, a car is seen on the opposite side of the tower at an angle of depression 25°.
(a) Draw a rough figure showing the details given in the question.
(b) Find the height of the tower.
(c) What is the distance of the car from the tower?
(sin 65° = 0.91, cos 65° = 0.42, tan 65° = 2.14)
Answer:
(a)

(b) In ∆ABC, angles are 45°, 45°, 90°
So sides are in the ratio 1 : 1 : √2
So, the height of the tower AC = 100 m

(c) tan 65 = \(\frac{C D}{100}\)
CD = 100 × tan 65
= 100 × 2.14
= 214 m

Question 25.
The third term of an arithmetic sequence is 26 and its eighth term is 61.
(a) Find the common difference of the sequence.
(b) What is its first term?
(c) Write the algebraic form of the sequence.
(d) Find the sum of the first 15 terms of the sequence.
Answer:
(a) 3rd term = 26
8th term = 61
Common difference (d) = \(\frac{\text { Term difference }}{\text { Position difference }}\)
= \(\frac{61-26}{8-3}\)
= 7

(b) First term = 3rd term – 2d
= 26 – 2 × 7
= 26 – 14
= 12

(c) Algebraic form = dn + (f – d)
= 7n + (12 – 7)
= 7n + 5

(d) Sum of first 15 terms = \(\frac{n}{2}\)[2f + (n – 1)d]
= \(\frac{15}{2}\) [2 × 12 + (15 – 1)7]
= \(\frac{15}{2}\) [24 + 98]
= 915

Question 26.
A vessel (without lid) in the shape of a square pyramid, made from a metallic sheet with a base perimeter of 80 centimeters and a slant height of 26 centimeters.
(a) How many square centimeters of the metallic sheet was needed to make the vessel?
(b) Calculate the height of the vessel.
(c) What is the capacity of the vessel in liters?
Answer:
(a) Base perimeter = 80 cm
⇒ 4a = 80
⇒ a = 20 cm
Slant height, l = 26 cm
∴ Lateral surface area = 2al
= 2 × 20 × 26
= 1040 cm²

(b) Height h = \(\sqrt{l^2-\left(\frac{a}{2}\right)^2}\)
= \(\sqrt{26^2-10^2}\)
= \(\sqrt{676-100}\)
= \(\sqrt{576}\)
= 24 cm

(c) Volume of the vessel = \(\frac{1}{3}\)a²h
= \(\frac{1}{3}\) × 20 × 20 × 24
= 3200 cm³
= \(\frac{3200}{1000}\) l
= 3.2 l

Question 27.
C and D are points on a semicircle with AB as diameter. ∠BDC = 125°. The CD is parallel to AB.

Find the measures of:
(a) ∠BAC
(b) ∠ACB
(c) ∠ACD
(d) ∠ABD
Answer:
(a) ∠BAC = 180° – 125° = 55°

(b) ∠ACB = 90°

(c) In ΔACB, ∠BAC = 55°
∠ACB = 90°
∴ ∠ABC = 180° – (55° + 90°) = 35°
AB parallel to CD.
∴ ∠BCD = 35°
∴ ∠ACD = ∠ACB + ∠BCD
= 90° + 35°
= 125°

(d) In ΔBCD,
∠BCD = 35°
∠CDB = 125°
∴ ∠CBD = 180° – (35° + 125°) = 20°
∴ ∠ABD = ∠ABC + ∠CBD
= 35° + 20°
= 55°

Question 28.
The equation of a line is 2x – y – 2 = 0.
(a) Check whether the point (3, 4) is on this line.
(b) Find the coordinates of the points where this line cuts the x and y axes.
Answer:
(a) Equation of the line is 2x – y – 2 = 0 …………(1)
Consider the point (3, 4).
Substitute x = 3, y = 4 in (1),
2 × 3 – 4 – 2 = 6 – 4 – 2 = 0
So, (3, 4) is a point on the line 2x – y – 2 = 0.

(b) If the line cuts the x-axis its y-coordinate is 0.
2x – y – 2 = 0
⇒ 2x – 0 – 2 = 0
⇒ 2x = 2
⇒ x = 1
So, the coordinates of the point where the line cuts the x-axis is (1, 0).
If the line cuts the y-axis its x-coordinate is 0.
∴ 2x – y – 2 = 0
⇒ 2 × 0 – y – 2 = 0
⇒ -y = 2
⇒ y = -2
So, the coordinates of the point where the line cuts the y-axis is (0, -2).

Question 29.
Consider the sequence: 2, 6, 18, 54,……
First term = 2
Second term = 2 × 3 = 6
Third term = 6 × 3 = 18
Fourth term = 18 × 3 = 54 and so on.
Sequences starting with a non-zero number, and each succeeding term got by multiplying the preceding term by a fixed number except zero, are called geometric sequences. The fixed number multiplied is the common ratio of the sequence. Thus, in the geometric sequence 2, 6, 18, 54,…. the first term is 2 and the common ratio is 3.
(a) The first term of a geometric sequence is 3 and the common ratio is 2. Find its second and third terms.
(b) Which of the following is a geometric sequence?
(i) 2, 4, 6, 8,………
(ii) 2, 4, 8, 16,…….
(iii) 1, 4, 9, 16,…….
(c) What is the common ratio of the geometric sequence 5, 20, 80, 320,…..
(d) Write the next term of the geometric sequence 3, 9, 27,……
Answer:
(a) Second term = 3 × 2 = 6
Third term = 6 × 2 = 12

(b) (ii) 2, 4, 8, 16,….. is a geometric sequence with a common ratio of 2.

(c) Common ratio = \(\frac{20}{5}\) = 4.

(d) Next term = 27 × 3 = 81.