Kerala SSLC Maths Board Model Paper March 2023 with Answers English Medium

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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