Kerala SSLC Maths Question Paper March 2023 with Answers English Medium

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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 → \(\frac{\mathrm{n}(\mathrm{n}-3)}{2}\)