Kerala Syllabus Class 10 Maths Model Question Paper Set 5

Practicing Maths Question Paper Class 10 Kerala Syllabus Set 5 English Medium helps identify strengths and weaknesses in a subject.

Class 10 Maths Kerala Syllabus Model Question Paper Set 5

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.
O is the center of a circle. If ∠OAB = 40° and C is a point on the circle then ∠ACB is. (1 mark)

(a) 40°
(b) 50°
(c) 80°
(d) 100°
Answer:
(b) 50°

Question 2.
Read the two statements given below.
Statement 1: The coordinates of the midpoint of a line joining (-3, 6) and (1, -2) are (-1, 2).
Statement 2: A midpoint divides a line segment joining two points in the ratio 1:1.

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) Radius of the big circle is equal to the diameter of small circle.

A) fine dot is placed into the figure without looking The probability of falling the dot in the unshaded part?
OR
B) The factors of 6 are written in different paper pieces and kept in a box. One is taken from the box without looking.
(a) What is the probability of getting an even number?
(b) What is the probability of getting a prime number? (3 mark)
Answer:
(A) \(\frac{3}{4}\)
OR
(B) (a) \(\frac{2}{4}=\frac{1}{2}\)
(b) \(\frac{2}{4}=\frac{1}{2}\)

Question 4.
In the figure PA and QB are parallel tangents. Another line PQ touch the circle at R and cut the parallel tangents.

(a) Draw a rough diagram and join OA,OR and OB.
(b) Name the equal triangles in the figure.
(c) Find the measure of ∠POQ. (4 mark)
Answer:
(a) Draw the diagram yourself
(b) PA = PR, OA = OR and OP is common.
Triangle PAO and triangle PRO are equal. Similarly AQRO and QBO are equal.
(c) If ∠POA = ∠POR = x and ∠QOR = ∠QOB = y then 2x + 2y= 180
x + y = 90
∠POQ = 90°

Question 5.
A) The product of first term and third term of an arithmetic sequence with common difference 1 is 143. Find the first three terms of the sequence.
Answer:
If first term be x, then third term = x + 2
x(x + 2) = 143
⇒ x² + 2x = 143
⇒ x² + 2x -143 = 0
a = 1, b = 2, c = -143

If first term = 11, then first three terms of the sequence 11,12, 13
If first term x = -13 then first three terms of the sequence -13, -12, -11

OR

B) In the diagram, a line passing through the center of a circle divides a chord into two. If so find the radius of the circle? (4 mark)

Answer:
If we extend the line OP that intersects the two ends of the circle, we get another chord CD.

AB and CD are two chords that meet at the point P.
Therefore, PA × PB = PC × PD.
That is,
4 × 6 = (r + 5) × (r – 5)
24 = r² – 52
24 = r² – 25
24 + 25 = r²
49 = r²
r = ±7
r is the radius of the circle, it will always a positive number
⇒ r = 7 cm

Question 6.
The table below shows the students of a maths club sorted according to their heights. (5 Marks)

Height (centimeter)	Number of Students
120-130	2
130-140	7
140-150	10
150-160	5
160-170	1
Total	25

(a) When the heights are written in ascending order, the height of which student is taken as the median height?
Answer:

Height (centimeter)	Number of Students
Below 130	2
Below 140	9
Below 150	19
Below 160	24
Below 170	25

Here total number of students = 25, which is an odd number.
So, there is only one middle value \(\frac{25+1}{2}=\frac{26}{2}\) = 13
So, the height of the 13th student is the median height.

(b) Find the median height.
Answer:
Median class is 140 – 150
Width of median class = 150 – 140 = 10
Number of students in this median class = 10
So, the width of the subdivisions = \(\frac{10}{10}\) = 1
Height of 10th student = \(\frac{140+141}{2}\) = 140.5
Height of the 13th student
= 140.5 + (13 – 10) × 1
= 140.5 + 3
= 143.5

Section – B

Question 7.
The vertices of a square are on a circle. BP is the tangent to the circle at P. What is the measure of ∠PBC. (1 mark)

(a) 50°
(b) 40°
(c) 45°
(d) 30° (1 mark)
Answer:
(c) 45°

Question 8.
The algebraic form of the sequence \(\frac{1}{7}, \frac{3}{7}, \frac{5}{7}\)…………. (1 mark)
(a) \(\frac{n}{7}\)
(b) \(\frac{2 n+1}{7}\)
(c) \(\frac{2 n-1}{7}\)
(d) None of these
Answer:
(c) \(\frac{2 n-1}{7}\)

Question 9.
Sum of the areas of two squares is 130. Side of one square is 2 more than the side of the other square.
(a) If the side of the small square is x then what is the side of the big square?
Answer:
Side of the big square is x + 2

(b) Form a second degree equation using the condition. (3 mark)
Answer:
x² +(x + 2)² = 130
x² + x² + 4x + 4 = 130
2x² + 4x + 4 -130 = 0
2x² + 4x -126 = 0
x² + 2x – 63 = 0

Question 10.
Chords AB and CD intersect at point P. AB = 13 cm, CD = 15 cm, PD = 3 cm.

(a) What is the length of PC?
(b) If PB = x, find PA?
(c) What is the length of PB? (3 mark)
Answer:
(a) CD= PC + PD
15 = PC + 3
PC = 15 – 3 = 12

(b) AB = PA + PB
13 = PA + x
PA = 13 – x

(c) PC x PD = PA × PB
12 × 3 = (13- ) × x
36 = 13x – x²
x² – 13x + 36 = 0
⇒ x = 4 or x = 9
Therefore, PB = 4 cm

Question 11.
Consider the line 2x + 3y = 6
(a) What are the points at which the line cut the coordinate axes?
(b) Write the slope of this line. (4 mark)
Answer:
(a) When the line cut x axis y = 0
2x + 3 × 0 = 6
2x = 6 x = 3
The point on x axis is (3, 0)
When the line cut y axis x = 0.
2 × 0 + 3y = 6
3y = 6
y = 2
The point is (0, 2)

(b) Slope = \(\frac{0-2}{3-0}=\frac{-2}{3}\)

Question 12.
A) The diagonal of the rectangle ABCD is 12 cm ∠BAC = 30°

(a) What is the length of the side AB ?
(b) What is the length of the side BC ?
(c) Calculate the area of the rectangle
Answer:
(a) 6√3 cm
(b) 6 cm
(c) 36√3 cm²

OR

B) Draw the coordinates axes and mark the points on the plane.
(a) A(1, 0), B(6, 0), C(8, 3) and D(3, 3)
(b) Write the most suitable name to ABCD
(c) Calculate the area of ABCD (4 mark)
Answer:
(a) Mark the points using co – ordinate axes.
(b) Parallelogram
(c) 5 × 3 = 15 sq. units

Question 13.
In the figure, chord AB and CD are extended to meet at P. PB = 14 centimetres, AB = 5 centimetres, CD = 15 centimetres. Find the length of PC.

Answer:
Let the length of PC be x,
PD = x + 15
PB = 14, AB = 5
⇒ PA = 14 – 5 = 9
PA × PB = PC × PD
⇒ 9 × 14 = x(x + 15)
⇒ x² + 15x = 126
⇒ x² + 15x – 126 = 0
⇒ a = 1, b = 15, c = 126

Since, x is the length of PC and hence not negative. So, x = 6
∴ the length of PC = 6 centimetres.

Question 14.
A) In the figure 0 is the centre of the circle, then

(a) What kind of triangle is OAC ?
(b) What is the measure of angle ABC ?
(c) What is the measure of angle ADC ?
(d) If the radius of the circle is 6 cm then what is the length of the chord AC
Answer:
(a) OA = OC, ∠OAC = ∠OCA = 45°,
∠AOC = 90°
ΔOAC is an isosceles right triangle

(b) ∠ABC = \(\frac{1}{2}\)AOC = 45°

(c) ∠ADC = 180 – 45 = 135°

(d) AC = \(\sqrt{6^2+6^2}\) = 6√2 cm

OR

B) ABCD is a cyclic quadrilateral. AB is the diametre of the circle, AD = CD and ∠ADC = 130°

(a) What is the measure of ∠ACB ?
(b) What is the measure of ∠ABC ?
(c) Find ∠DCB
(d) What is the measure of ∠BAD ? (5 mark)
Answer:
(a) ∠ACB = 90° (Angle in the semicircle)
(b) ∠ABC = 180-130 = 50°
(c) Since CD = AD, the angles opposite to the equal sides of triangle ADC, are equal.
∠DCA = 25°
∠DCB = 90 + 25 = 115°
(d) ∠BAD = 180-115 = 65°

Section – C

Question 15.
First 10 terms of the arithmetic sequence 3n+l are written in small paper pieces separately and kept in a box.One is taken from the box without looking. What is the probability of getting odd number. (1 mark)
(a) \(\frac{1}{2}\)
(b) \(\frac{3}{5}\)
(c) \(\frac{5}{11}\)
(d) \(\frac{3}{8}\)
Answer:
(a) \(\frac{1}{2}\)

Question 16.
Length of the diagonal of a rectangle is 12 cm and it makes 30° with the longest side. What is the breadth of the rectangle?
(a) 7
(b) 6
(c) 5
(d) 4 (1 mark)
Answer:
(b) 6

Question 17.
The sum of the first 5 terms of an arithmetic sequence is 65 and the sum of the first 9 terms is 189.
(a) What is the 3rd term of the sequence ?
(b) What is the 5th term of the sequence ?
(c) What is the common difference of the sequence? (3 mark)
Answer:
Third term = \(\frac{65}{5}\) = 13
Fifth term = \(\frac{189}{9}\) = 21
2 x common difference = 21 – 13 = 8,
Common difference is 4

Question 18.
The top of a 30 high building can be seen from a point at some distance from the foot of the building is at an angle of elevation 30°. When the point of observation is some distance closer to the building the angle become 60°.
(a) Draw a rough diagram
Answer:

(b) What is the distance from the foot of the building to the second point of observation.
Answer:
In ∆ABP, tan60° = \(\frac{30}{y}\), √3 = \(\frac{30}{y}\)
y = \(\frac{30}{\sqrt{3}}\) = 10√3 m

The second point is at the distance
10 × 1.732 = 17.32 m away from the building.

(c) What is the distance between two points from which the top of the building is observed
Answer:
tan 30 = \(\frac{30}{x+y}\)
\(\frac{1}{\sqrt{3}}=\frac{30}{x+y}\)
30√3 = x + y = x + 17.3
x = 30√3 – 17.3
= 34.66 m
The distance between the points is 34.66 m.

(d) What is the distance from the foot of the tower to the first point of observation. (4 mark)
Answer:
Distance is x + y = 30√3 = 51.96 m

Question 19.
A) A child sees the top of a telephone tower at an elevation of 80°. Stepping 20 metres back, he sees it at an elevation of 40°.
(a) Draw a rough figure.
Answer:

(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:
sin 80° = \(=\frac{B C}{C D}=\frac{B C}{20}\)
0.98 = \(\frac{B C}{20}\)
⇒ BC = 0.98 × 20 = 19.6 CM

OR

B) In a class of 60 students, 30 are boys. In another class of 50 students, 20 are girls. If one student is to be selected from each class,
(a) How many pairs are possible?
(b) What is the probability of one boy and one girl?
(c) What is the probability of both being boys?
(d) What is the probability of at least one girl? (5 mark)
Answer:
a) Total number of pairs = 60 × 50 = 3000

b) Probability of one boy and one girl
= \(\frac{3000-(900+600)}{3000}\)
= \(\frac{3000-1500}{3000}=\frac{1500}{3000}=\frac{1}{2}\)

c) Probability of both being boys
= \(\frac{30 \times 30}{3000}=\frac{900}{3000}\)
= \(\frac{9}{30}=\frac{3}{10}\)

d) Probability of atleast one girl
= \(\frac{3000-900}{3000}=\frac{2100}{3000}\)
= \(\frac{21}{30}=\frac{7}{10}\)

Section – D

Question 20.
What is the altitude of an equilateral triangle of side 6 cm?
(a) 3√3 cm
(b) 2√3 cm
(C) 5√3 cm
(d) √2 (1 mark)
Answer:
(a) 3√3 cm

Question 21.
Please read the explanations below.
In a circle, two chords AB and CD meet at a point P inside the circle. AP = 4 cm PB = 6 cm, CP = 3 cm, PD = ?
Statement (A): PD = 8 cm
Statement (B): If two chords intersect inside a circle, then
AP × PB = CP × PD

(a) Statement A is correct, Statement B is incorrect
(b) Statement B is correct, Statement A is incorrect
(c) Both are correct. Statement A is the reason for Statement B
(d) Both are correct. Statement A is not the reason for Statement B (1 mark)
Answer:
(c) Both are correct. Statement A is the reason for Statement B

Question 22.
AD and BC are the common tangents to two circles. The centres of the circles are M and N respectively. ∠APB = 40°

(a) Which lines shown in the figure are of equal length?
(b) Prove that AD = BC
(c) Calculate ∠AMB and ∠CND (3 mark)
Answer:
(a) PA = PB, PD = PC

(b) By adding the equalities,
PA + PD = PB + PC
AD = BC

(c) ∠AMB = 180 – 40 = 140°
∠CND = 180 – 40 = 140°

Question 23.
A) Hypotenuse of a right angled triangle is 1 less than twice its small side. Third side is 1 more than its small side.
(a) If the small side is x what is the length of other two sides.
(b) Form an equation connecting the length of the sides.
(c) Calculate the length of the sides of the triangle.
Answer:
(a) Hypotenuse = 2x – 1, Third side= x + 1,

(b) (2x – 1)² = x² + (x + 1)²
4x – 4x +1 = x + x + 2x + 1
2x² – 6x = 0

(c) x = 3.
Sides are:
Hypotenuse, 2x – 1 = 6 – 1 = 5 cm
Other two sides are 3 cm, 4 cm

OR

B) The mid points of the two sides and one vertex of a square are joined in such a way as to get a triangle which is coloured in the picture.

Answer:
(a) Unshaded part t = (\(\frac{1}{2}\) × a × \(\frac{a}{2}\)) × 2 + \(\frac{1}{2} \times \frac{a}{2} \times \frac{a}{2}\)
= \(\frac{a^2}{2}+\frac{a^2}{8}=\frac{5 a^2}{8}\)

(b) Area of shaded part = a² – \(\frac{5 a^2}{8}=\frac{3 a^2}{8}\)

(c) Probability = \(\frac{3 a^2}{8}\) + a² = \(\frac{3}{8}\)

Question 24.
A) Consider the arithmetic sequence 4, 7, 10….
(a) What is its 13th term?
(b) What is the sum of its first 25 terms?
(c) Find the sum of first 25 terms of the arithmetic sequence 8, 14, 20
Answer:
Consider the arithmetic sequence 4,7,10
(a) 13th term = first term + 12 times common difference
13th term =4 + 12 × 3 = 40

(b) Sum of first 25 terms = 13th term × 25
= 40 × 25 = 1000

(c) Each term is two times the terms of the first sequence.
Sum = 2 × 1000 = 2000

OR

B) Consider the arithmetic sequence -1,3,7…
(a) What is the common difference?
(b) Write the nth term of this sequence.
(c) Is 95 a term of the sequence?
(d) Calculate the sum of the terms upto 95. (4 mark)
Answer:
(a) d = 4
(b) 4n – 5
(c) 4n – 5 = 95 ⇒ 4n = 100, n = 25
(d) Sum = 4(1 + 2 + 3 + …………. + 25) – 5 × 25
= 4 × 325 – 125
= 1175

Question 25.
∆ABC is an equilateral triangle. A(2, 2), B(6, 2)
(a) Find the length of the side.
(b) Calculate the altitude to the side
(c) Write the coordinates of C
(d) Calculate the area of the triangle (5 mark)
Answer:
(a) AB = |6-2| = 4
(b) 2√3
(c) Draw CP perpendicular to AB. Triangle CPA.is a 30° – 60° – 90° triangle, P(4, 0) Coordinates of C are (4, 2 + 2√3 )
(d) \(\frac{1}{2}\) × 4 × 2√3
= 4√3 sq. units

Question 26.
In the figure BD = CD, ∠DBC = 25°

(a) What is the measure of ∠BDC ?
(b) What is the measure of ∠BAC ?
(c) What is the measure of ∠EBC ? (5 mark)
Answer:
(a) In triangle BDC, BD = CD.
Angle opposite to these sides are equal.
∠BCD = 25°
∠BOC = 180 – (25 + 25) = 130°

(b) ∠BAC = 180 – 130 = 50°

(c) ∠EBC = ∠BAC = 50°
∠EBC = 180 – (90 + 50)
= 180 – 140
= 40°