Kerala SSLC Maths Question Paper March 2024 with Answers English Medium

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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.