Kerala SSLC Maths Question Paper March 2021 with Answers English Medium

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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