Kerala SSLC Maths Board Model Paper March 2024 with Answers English Medium

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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