Kerala Syllabus Class 10 Maths Model Question Paper Set 4

Practicing Maths Question Paper Class 10 Kerala Syllabus Set 4 English Medium helps identify strengths and weaknesses in a subject.

Class 10 Maths Kerala Syllabus Model Question Paper Set 4

Time: 2½ Hours
Total Score: 80 Marks

Instructions:

• Use the first 15 minutes to read the questions and think about the answers
• There are 26 questions, split into four parts A, B, C, D
• Answer all questions; but in questions of the type A or B, you need answer only one of those
• You can answer the questions in any order, writing the correct question number
• Answers must be explained, whenever necessary.
• No need to simplify irrationals like √2, √3, etc using approximations unless you are asked to do so.

Section – A

Question 1.
Sum of the first n terms of an arithmetic sequence is n2 + n. What is its 13th term?
(a) 31
(b) 26
(c) 20
(d) 50 (1 mark)
Answer:
(b) 26

Question 2.
Read the two statements given below.
Statement 1: If area and perimeter of a triangle are numerically equal, that is both are same number then its inradius is 2 unit
Statement 2: If A is the area, 5 is half of the perimeter then in radius r = \(\frac{A}{s}\)

Choose the correct answer from those given below.
(a) Statement 1 is true and Statement 2 is false.
(b) Statement 1 is false and Statement 2 is true.
(c) Both the statements are true, and Statement 2 is the correct reason for Statement 1.
(d) Both the statements are true, but Statement 2 is not the correct reason for Statement 1. (1 mark)
Answer:
(c) Both the statements are true, and Statement 2 is the correct reason for Statement 1.

Question 3.
A) In the figure O is the centre of the circle. If angle ADC = 140°, angle AEC = 60° then

(a) What is the measeure of ∠APC and ∠AQC
Answer:
∠APC = 180 – 140 = 40°, ∠AQC = 40°

(b) What is the measure of angle AOC?
Answer:
∠AOC = 2 × 40 = 80°

(c) Find the angles of the quadrilateral PEQB
Answer:
In the quadrilateral ∠AEQ = ∠AEC = 60°,
∠EPB = 180 – 40 = 140°,
∠EOB = 140°
∠PBQ = 360 – (140 + 140 + 60) = 20.
Angles are 140°, 60°, 140°, 20°

OR

B) ABC is an isosceles triangle with AB = AC, ∠ABC = 50°.

(a) Name two cyclic quadrilaterals in this picture.
Answer:
Quadrilateral ABEC and quadrilateral DBES are cyclic.

(b) What is the measure of angle D?
Answer:
∠ABC = ∠ACB = 50°
∴ ∠A = 180-100 = 80°
∴ ∠D = 80°

(c) What is the measure of ∠BEC? (3 mark)
Answer:
∠BEC = 180 – 80 = 100°

Question 4.
In the figure a circle touches the sides of a triangle. If AP = 1, BQ = 2 and CR = 3 then

(a) What is the perimeter of the triangle?
Answer:
AR = 1, CQ = 3, BP = 2
Perimeter of triangle ABC = 12 cm

(b) What is the area of the triangle? (4 mark)
Answer:
The perpendicular sides of the right angled triangle ABC is 3 and 4 cm.
Area = \(\frac{1}{2}\) × 3 × 4
= 6 sq.cm

Question 5.
A) Triangle PQR is drawn by joining the midpoints of the sides of triangle ABC.

(a) How many equal triangles are there in the figure?
Answer:
4 triangles.
∆PQR, ∆APQ, ∆PCR, ∆QRB are equal triangles

(b) A fine dot is placed into the figure. What is the probability of falling the dot in triangle PQR?
Answer:
\(\frac{1}{4}\)

(c) How many parallelograms are there in the picture?
Answer:
3 Parallelograms.
PQRC, PQBR, PRQA are equal

(d) A fine dot is placed into the figure. What is the probability of falling the dot in the parallelogram PQRC ?
Answer:
To fall the dot in PQRC it is necessary to fall either in triangle PCR or triangle PQR
Probability is \(\frac{2}{4}=\frac{1}{2}\)

OR

B) Sum of the first n odd numbers is k
(a) What is the sum of first n even numbers?
Answer:
k + √k

(b) What is the sum of first n natural numbers ?
Answer:
\(\frac{k+\sqrt{k}}{2}\)

(c) Find \(\frac{1+2+3+4+\cdots+15}{16+17+18+\cdots+30}\) (4 marks)
Answer:
\(\frac{8}{23}\)

Question 6.
Angles of a pentagon are in arithmetic sequence when arranged in the ascending order. First term is 42°
(a) What is its middle term?
Answer:
Angle sum of a pentagon (n – 2) × 180
= 3 × 180
= 540

(b) What is the difference between two adjacent terms?
Answer:
Difference between third term and first term is two times common difference
2x common difference = 108 – 42 = 66,
Common difference = 33

(c) Find the largest term of the pentagon (5 mark)
Answer:
Fifth term = 108 + 2 × 33 = 108 + 66 = 174
Angles are 42°, 75°, 108°, 141°, 174°

Section – B

Question 7.
Solutions of the equation (x – 1)² = 100 are
(a) 9,-10
(b) 10,-9
(c) 10,-10
(d) 11,-9 (1 mark)
Answer:
(d) 11,-9

Question 8.
O is the centre of the circle and ∠ACB = 30°. What is ∠AOB ?

(a) 30°
(b) 15°
(c) 60°
(d) 90° (1 mark)
Answer:
(c) 60°

Question 9.
Consider the arithmetic sequence 5, 9, 13, 17, 21…
(a) Write the algebraic form of this sequence.
Answer:
x_n = dn + (f – d)
= 4n + (5 -4)
= 4n + 1

(b) What is the position of the term in the sequence whose square is 625?
Answer:
(4n + 1)² = 625,
4n + 1 = 7625 = 25,
4n = 24, n = 6

(c) Is 36 a term of this sequence. How can you realize it ?
Answer:
All terms are odd numbers. The even number 36cannot be a term of this sequence

(d) What is the position of 49 in this sequence ? (3 mark)
Answer:
4n + 1 = 49. 4n = 48, n = 12.
12th term is 49

Question 10.
A rectangular plot of land has larger side 8 meter more than twice the smaller side. Area of the land is 504 sq.meter
(a) If the smaller side is x then what is the other side?
Answer:
Smaller side is x
Other side is 2x + 8

(b) Find the length of the sides.
Answer:
x(2x + 8) = 504,
2x² + 8x = 504,
x² + 4x = 25²
x² + 4x + 4 = 256m,
(x + 2)² = 162,
x + 2 =16, x = 14
Sides are 14 meter and 36meter

(c) How much money is needed to make wall along the sides at the rate of 200 rupees per meter? (3 mark)
Answer:
Perimeter = 2(14 + 36) = 2 × 50 =100 m
Amount = perimeter × 200
Expense = 100 × 200 = 20,000 rupees

Question 11.
From a point on the ground 40 metre away from the foot of the tower sees the top of the tower at an angle of evation 30° and sees the top of the water tank on the top of the tower at an angle of elevation 45°
(a) Draw a rough diagram.
Answer:

(b) Find the height of the tower.
Answer:
In triangle ABD we have tan 45° = \(\frac{BD}{AB}\)
1 = \(\frac{h+h_1}{40}\), h + h₁ = 40
tan 30° = \(\frac{B C}{A B}, \frac{1}{\sqrt{3}}=\frac{h}{40}\)
h = \(\frac{40}{\sqrt{3}}\) = 23.1 m
Height of the tower 23.1 m

(c) Find the height of the water tank. (4 mark)
Answer:
23.1 + h₁ =40, h₁ = 40 – 23.1 = 16.9meter.

Question 12.
A) nth term of an arithmetic sequence is 2n + 1
(a) Write the sequence in the numerical form
Answer:
3, 5, 7….

(b) What is its 100th term?
Answer:
(2 × 100 ) + 1 = 201

(c) Calculate the sum of first 100 terms
Answer:
2(1 + 2 + 3 + …. + 100) + 100 = 10200

(d) Find the sum of first 100 terms of the arithmetic sequence 4, 6, 8 …………..
Answer:
The terms of 4, 6, 8,… are one more than the terms of 2n + 1
Sum is 10200 + 100 = 10300

OR

B) Two dice numbered 1 to 6 are thrown at together.
(a) Write the outcomes as pairs
Answer:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

(b) What is the probability of the occurence of equal numbers?
Answer:
\(\frac{6}{36}=\frac{1}{6}\)

(c) What is the probability of the occurence of perfect squares ?
Answer:
(1, 1), (1, 4), (4, 1), (4, 4).
Probability = \(\frac{4}{36}\)

(d) What is the probability of the occurence of multiple of 2 in one die and multiple of 3 in other die ? (4 mark)
Answer:
(2,3), (4,3), (6. 3), (2, 6), (4. 6), (6, 6), (3, 2). (3,4), (3,6). (6, 2), (6, 4)
Probability = \(\frac{11}{36}\)

Question 13.
In the figure AB is a common tangent to the circle. The line PC joins the common point of the circle to the point P. it is also a tangent to the circles.

(a) Name theiines of equal length shown in the figure.
Answer:
PA = PC, PB = PC

(b) If ∠PAC = x and ∠PBC = y then what is ∠ACB ?
Answer:
∠ACB = x + y

(c) Prove that ΔABC is a right triangle. (5 mark)
Answer:
In triangle ABC,
2x + 2y = 180,
x + y = 90
Triangle ABC is a right triangle.

Question 14.
A) y axis is the tangent to the circle at the origin of coordinates. (5,0) is the center of the circle.

(a) Write the coordinates of A.
Answer:
(10, 0)

(b) If CD is the diameter perpendicular to x axis then write the coordinates of C and D.
Answer:
C(5, 5), O(5, 5)

(c) P is a point on this circle such that OP = 8. Find the length AP.
Answer:
Triangle OPA is a right triangle. AP = 6

OR

B) In triangle ABC, AB – AC . P and Q are the mid points of the side AB and AC.
(a) Draw a rough diagram and join the points P and Q.
Answer:

(b) Prove that BPQC is a cyclic quadrilateral.
Answer:
Since AB = AC, ∠B = ∠C
Line joining the mid points of two sides of a triangle is parallel to the third side. PQ is parallel to BC.
In PBCQ. ∠B + ∠P = 180° (co interior angles)
Since ∠C = ∠B, ∠C + ∠P =180°
PQCB is a cyclic quadrilateral

(c) I f ∠A in triangle ABC is 20°, find the angles of the trapezium BPQC (5 mark)
Answer:
∠A = 20°
∠B = ∠C = \(\frac{180-20}{2}\) = 80°
∠B + ∠P = 180°, ∠P = 100°, ∠O = 100°
Angles are ∠P = 100°, ∠Q = 100°,
∠B = 80°, ∠C = 80°

Section – C

Question 15.
Which of the following is the center of the circle
(x – 1)² + y² = 1
(a) (1, 0)
(b) (1, 1)
(c) (1, 2)
(d) (1, -1) (1 mark)
Answer:
(a) (1, 0)

Question 16.
The sum and product of the solutions of a second degree polynomial are-10 and \(\frac{5}{2}\) respectively. Then the polynomial is:
(a) 2x² – 20x + 10
(b) 2x² – x + 5
(c) 2x² – 20x + 5
(d) x² – 20x + 5 (1 mark)
Answer:
(c) 2x² – 20x + 5

Sum of the Solutions is -10 and its product is \(\frac{5}{2}\)
a + b = -10, ab = \(\frac{5}{2}\)
Polynomial = x² + (a + b)x + ab
= x² – 10x + \(\frac{5}{2}\)
= 2x² – 20x + 5

Question 17.
ABCDE is a regular pentagon. AD and BD are its diagonals. The tangents intersects at a point P.

(a) What is the measure of ∠E and ∠C?
Answer:
∠E = ∠C = \(\frac{540}{5}\) = 108°

(b) Find ∠ADE, ∠BDC?
Answer:
∠ADE = 36°
∠BDC = 36°

(c) What is ∠PAB, ∠PBA?
Answer:
∠ADB = 108 – 72 = 36°
∠PAB = ∠PBA = 36°

(d) What is the angle measure of APB? (3 mark)
Answer:
∠P = 180 – 72 = 108°

Question 18.
From a wooden cube of side 12 cm, a sphere of maximum size is carved out.
(a) What is the radius of the sphere?
Answer:
Diameter of the sphere = side of the cube = 12 cm
So, radius = 6 cm

(b) What is the surface area?
Answer:
Surface area = 4πr² = 4 × π × 6² = 144π cm²

(c) Find the volume. (4 mark)
Answer:
Volume = \(\frac{4}{3}\) π × 6³ = 288π cm³

Question 19.
A) Angles of a pentagon are in an arithmetic sequence if the angle measures are arranged in an order.
Whatever be the common difference of the sequence the middle term remains same.
(a) Find the middle term of the sequence.
Answer:
Angle sum of a pentagon is (5 – 2) × 180 = 540°
Since it has 5 angles which are in an arithmetic 540
sequence x₃ = \(\frac{540}{5}\) = 108.
The middle term is 108°

(b) If the smallest angle is 42 then what is the difference between two adjacent angles ?
Answer:
x₁ = 42, x₃ = 108
3d = 108 – 42 = 66, d = 33

(c) Find the measure of all angles.
Answer:
Terms are 42°, 75°, 108°, 141°, 174°

OR

B) A bag contains 3 black balls and 2 white balls. An-other bag contains 3 black balls and 4 white balls. One is taken from each bag without looking into the bag.
(a) What is the probability of getting both balls white ?
Answer:
Total number of Outcome pairs is 5 × 7 = 35
Probability of getting both white = \(\frac{2 \times 4}{35}=\frac{8}{35}\)

(b) What is the probability of getting both balls black?
Answer:
\(\frac{3 \times 3}{35}=\frac{9}{35}\)

(c) What is the probability of getting balls of different colours? (5 mark)
Answer:
\(\frac{3 \times 4+2 \times 3}{35}=\frac{18}{35}\)

Section – D

Question 20.
In the diagram, AB is the diameter of the semicircle, PQ = \(\sqrt{14}\) RS = \(\sqrt{18}\) are perpendicular to AB. What is the length of the line AB

(a) 10
(b) 9
(c) 12
(d) 15 (1 mark)
Answer:
(b) 9

\(\sqrt{18}=\sqrt{6 \times 3}\)
\(\sqrt{14}=\sqrt{7 \times 2}\)
AB = 9

Question 21.
Read the two statement given below.
Statement (A): x² + 20x + 96 is the polynomial with the sum of solutions 8 and their product 12.
Statement (B): If a, b are the solutions of a polynomial, then the polynomial is x² +(a + b)x + ab
(a) Statement A is true, Statement B is false.
(b) Statement B is true. Statement A is false.
(c) Both statements are true. Statement B is the reason for Statement A.
(d) Both statements are true. Statement B is not the reason for Statement A. (1 mark)
Answer:
(b) Statement B is true. Statement A is false.
Statement B is true.

The sum of solutions 8 and their product 12.
a + b = 8. ab = 12
Polynomial = x² + (a + b)x + ab
= x² + 8x + 12
∴ Statement A is false.

Question 22.
When 16 added to the sum of a number of terms of the arithmetic sequence 9, 11, 13.., we get 256. How many terms are added? (3 mark)
Answer:
When 16 added to the sum of a number of terms of the arithmetic sequence 9, 11, 13,.., we get 256.
That is, sum of terms of the arithmetic sequence = 256 – 16 = 240
First term (f) = 9
Common difference (d) = 2
x_n = dn + (f – d)
= 2n + (9 – 2)
= 2n + 1
If algebraic form of an arithmetic sequence is an + b, then the sum of first n terms of the sequence
= a\(\frac{n}{2}\)(n + 1) + nb
= 2\(\frac{1}{2}\)(n + 1) + n × 7
= n(n + 1) + 7n
= n² + n + 7n
= n² + 8n
= n² + 8n = 240
= n² + 8n – 240 = 0
a = 1, b = 8, c = -240

OR

n = \(\frac{-8-32}{2}\) = -20
n is the number of terms. so cannot be negative.
Thus, n =12
12 terms are added.

Question 23.
A) In traingle ABC,AB = AC AD is the perpendicular from A to BC. This perpendicular distance from A to BC is 2 cm more than BC.
Area of the triangle is 60 sq.cm

(a) If BC = x then what is the length AD?
Answer:
BC = x, AD = x + 2

(b) Form an equation connecting the lengths BC; AD and area of the triangle.
Answer:
\(\frac{1}{2}\) x x(x + 2) = 60, x² + 2x = 120

(c) Find the length of BC.
Answer:
x² + 2x + 1 = 121, (x + 1)² = 121, x = 10 BC = 10

OR

B) The vertices of ABCD are on a circle with diameter AB. ∠BAC = 20°, AD = CD

a) What is the measure of ∠B?
b) What is the measure of ∠D?
c) Find other two angles of ABCD. (3 mark)
Answer:
(a) 180 – (20 + 90) = 180-110 = 70°
(b) ∠O = 180-70 = 110°
(c) In triangle ADC, ∠A = 35°, ∠C = 35°
In the quadrilateral
∠A = 55°, ∠B = 10°, ∠C = 90 + 35 = 125°,
∠D = 110°

Question 24.
A) Two circles intersect at B and E as in the figure. The points A – B – C are along a line. Also the points D – E – F are also on a line

(a) Prove that AD is parallel to CF
Answer:

Draw BE. ABED is a cyclic quadrilateral.
If ∠DAB = x then ∠BED = 180 – x, ∠BEF = 180 – (180 – x)=
BEFC is cyclic. ∠C = 180 – A.
In quadrilateral ADFC,
∠A + ∠C = x + 180 – x = 180°
Co interior angle sum is 180°.
AD is parallel to CF

(b) If AC = DF suggest a suitable name to the quadrilateral ADFC
Answer:
ADFC is a trapezium.
Since AC = DF is an isosceles trapezium.

(c) Prove that ADFC is a cyclic quadrilateral.
Answer:
Angles at the ends of parallel sides of an isosceles trapezium are equal.
Since ∠A = ∠D and ∠A + ∠C = 180° then ∠D + ∠C = 180°
ADFC is a cyclic quadrilateral.

OR

B) The distance between two buildings is 100 metre.The height of one building is double the height of other building.The top of the buildings can be seen at the angle of elevations 60° and 300 from a point in between the buildings.
(a) Draw a rough diagram
Answer:

(b) What is the distance from the foot of the tall tower to the point of observation.
Answer:
Let AC = h and BD = 2h

200 – 2x = 3x, 5x = 200, x = 40 m
The point of observation is at the distance 40 m from tall building.

(c) Claculate the height of the buildings. (4 mark)
Answer:
h = \(\frac{\sqrt{3} x}{2}=\frac{\sqrt{3} \times 40}{2}\) = 20√3 m
Height of the small building is 20√3 m
and height of the tall building is 40√3 m

Question 25.
The line x + y= 2 intersect x axis at A and y axis at B
(a) Write the coordinates of A and B.
Answer:
At A, y = 0, x + 0 = 2 x = 2.
The point A (2, 0)

At B, x = 0, 0 + y = 2
The point B(0, 2)

(b) Find the coordinates of the circumcentre of the triangle AOB? (5 mark)
Answer:
Circumcentre is the midpoint of the hypotenuse AB.
Midpoint of A (2, 0) and B (0, 2) is:
\(\left(\frac{2+0}{2}, \frac{0+2}{2}\right)\) = (1, 1)
Therefore the circumcentre is (1, 1)

Question 26.
The table below shows, children of a class sorted according to their scores in an examination.
table-1
(a) If the children are arranged in the ascending order of their scores, then what will be the assumed score of the 14th child?
Answer:
The cumulative frequency table is,

Scores	Number of Students
0-10	5
10-20	8
20-30	10
30-40	13
40-50	9
Total	45

The total number is 45, which is odd
\(\frac{45+1}{2}=\frac{46}{2}\) = 23

So, the 23rd child has the median score.
The 14th child is in the class 20 – 30.
The class width = 30 – 20 =10.
Number of students belonging to this class = 10
So, the width of the subdivision = \(\frac{10}{10}\) = 1
Score of 14th students = \(\frac{20+21}{2}\) = 20.5

(b) Compute the median score. (5 mark)
Answer:
Median score = Score of 23rd student
= 20.5 + (23 – 14) × 1
= 20.5 + 9
= 29.5
Only one temperature below 26 is 25.