# TS 10th Class Maths Model Paper Set 9 with Solutions

The strategic use of TS 10th Class Maths Model Papers Set 9 can significantly enhance a student’s problem-solving skills.

## TS SSC Maths Model Paper Set 9 with Solutions

Time: 3 Hours
Maximum Marks: 80

General Instructions:

1. Answer all the questions under Part – A on a separate answer book.
2. Write the answers to the questions under Part – B on the question paper itself and attach it to the answer book of Part – A.

Part – A (60 Marks)
Section – I (6 × 2 = 12 Marks)

Note :

1. Answer ALL the following questions,
2. Each question carries 2 marks.

Question 1.
Find, the values of k for which the quadratic equation 4x2 + 5kx + 25 = 0 has equal roots.
4x2 + 5kx + 25 = 0
with ax2 + bx + c = 0, we get
a = 4, b = 5k, c = 25
It is given that Q.E has equal roots
b2 – 4ac = 0
⇒ (5k)2 – 4(4) (25) = 0
⇒ 25k2 – 400 = 0 ⇒ 25k2 = 400
⇒ k2 = $$\frac{400}{25}$$ = 16
∴ k = $$\sqrt{16}$$ = ± 4

Question 2.
Find the value of log√2 128.

Question 3.
For the A.P. ; -3, -7, -11, …………. ; can we find directly a30 – a20 without actually finding a30 and a20.
Given A.P is -3, -7, -11, …………….
Here a = -3 and d = a2 – a1 = -7 – (-3) = -4
Now, we have the formula
an = a + (n – 1) d
∴ a30 – a20 = (a + 29d) – (a + 19d)
= a + 29d – a – 19d
= 10d
= 10 × -4 = -40
Yes, we can find directly a30 – a20 without finding a30 and a20.

Question 4.
When a dice is rolled, find the probability of getting odd prime number.
When a dice is rolled,
The probability of getting odd prime number
= $$\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}$$ = $$\frac{2}{6}$$ = $$\frac{1}{3}$$

Question 5.
Prove that 4tan2 45° – cosec2 30° + cos2 30° = $$\frac{3}{4}$$.
LHS = 4 tan245° – cosec230° + cos230°
= 4(1)2 – (2)2 + ($$\frac{\sqrt{3}}{2}$$)2
= 4 – 4 + $$\frac{3}{4}$$ = $$\frac{3}{4}$$ = RHS
∴ LHS = RHS

Question 6.
Find the mean of prime numbers which are less than 30.
Prime numbers that are less than 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Their mean = $$\frac{\text { Sum of observations }}{\text { number of observations }}$$
= $$\frac{2+3+5+7+11+13+17+19+23+29}{10}$$
= $$\frac{129}{10}$$ = 12.9

Section – II (6 × 4 = 24 Marks)

Note :

1. Answer ALL the following questions.
2. Each question carries 4 marks.

Question 7.
If A = {1, 2, 3, 4,} B = {2, 4, 6, 8,10}, then represent the Venn diagram of A – B.
Given A = {1, 2, 3, 4}, B = {2,4,6,8,10}
Venn – Euler diagram

A – B = {1, 3}

Question 8.
If one of the zeroes of the cubic polynomial p(x) = ax3 + bx2 + cx + d is zero, then find the product of other two zeroes of p(x). (a ≠ 0)
p(x) = ax3 + bx2 + cx + d
Let α, β, γ be the zeroes of p(x)
It is given one of the zeroes say γ = 0
we have αβ + βγ + γα = $$\frac{\mathrm{c}}{\mathrm{a}}$$
αβ + β(0) + 0(α) = $$\frac{\mathrm{c}}{\mathrm{a}}$$
αβ + 0 + 0 = $$\frac{\mathrm{c}}{\mathrm{a}}$$
αβ = $$\frac{\mathrm{c}}{\mathrm{a}}$$ [∵ γ = 0]
Thus the product of the remaining two zeroes = $$\frac{\mathrm{c}}{\mathrm{a}}$$

Question 9.
Lalitha says that HCF and LCM of the numbers 80 and 60 are 20 and 120 respectively. Do you agree with her ? Justify.
We know that
LCM × HCF = Product of two numbers
⇒ 20 × 120 = 80 × 60
⇒ 2400 = 4800 which is not true
∴ I don’t agree with Lalitha.

Question 10.
If cosec (A + B) = -1 and cot (A – B) = √3 0° < A + B ≤ 90°, A > B then find A and B.
Suppose cosec (A + B) = 1
but cosec (90°) = 1
∴ cosec (A + B) = cosec 90°
Since 0° < A + B < 90°, we get
A + B = 90° ……………. (1)
Also, cot (A – B) = √3
but cot 30° = √3
cot (A – B) = cos 30°
since A – B is also acute,
A – B = 30° ………… (2)
A + B + A – B = 90° + 30°
2A = 120°
A = $$\frac{120}{2}$$ = 60°
From (1), we get 60° + B = 90°
B = 90° – 60° – 30°
∴ A = 60°, B = 30°

Question 11.
A bag contains 7 red, 5 white and 6 black balls. A, ball is drawn from the bag at random, find the probability that the ball drawn is not black.
A bag contains 7 red, 5 white and 6 black balls. Total number of possible outcomes
The number of favourable outcomes when the drawn ball is not black = 7 + 5 = 12.
∴ The probability that the ball drawn is not black
= $$\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}$$
= $$\frac{12}{18}$$ = $$\frac{2}{3}$$

Question 12.
The angle of elevation of the top of a tower from a point on the ground which is 50 m away from the
foot of the tower is 45°. Draw the diagram for the situation.

Section – III (4 × 6 = 24 Marks)

Note :

1. Answer any 4 of the following questions.
2. Each question carries 6 marks.

Question 13.
The below distribution gives the weight of 40 students in a class. Find the median weight of the students.

 Weight (in kg) Number of students LCF 30 – 35 4 4 35 – 40 5 9 40 – 45 10 19 cf 45 – 50 8(f) 27 median 50 – 55 8 35 55 – 60 5 40(n)

Here, $$\frac{n}{2}$$ = $$\frac{40}{2}$$ = 20 ; median class = (45 – 50) ; l = 45, f = 8, cf = 19, h = 5
Median = l + $$\left[\frac{\mathrm{n}-\mathrm{cf}}{\mathrm{f}}\right]$$ × h = 45 + ($$\frac{20-19}{8}$$) × 5 = 45 + $$\frac{5}{8}$$ = 45 + 0.625 = 45.625

Question 14.
Prove that the parallelogram circumscribing a circle is a rhombus.

Given : ABCD is a parallelogram circumscribing a circle.
Proof : ABCD is a rhombus.
To prove : ABCD is a parallelogram
AB = CD and AD = BC ……… (1)
The lengths of tangents to a circle from an external point are equal.
Thus,
AS = AP; BQ = BP; CQ – CR;
DS = RD
AS + BQ + CQ + DS
= AP + BP + CR + RD
AS + DS + BQ + CQ
= AP + BP + CR + RD
AD + BC = AB + CD.
From (1), We have
From (1) and (2), we conclude that
AB = BC = CD = AD
∴ ABCD is a rhombus.
Hence proved.

Question 15.
The base of a triangle is 4 cm longer than its altitude. If tire area of die triangle is 48 sq. cm then find its base and altitude.
Let the altitude of a triangle be x cm
Its base = (x + 4) cm
Area of the triangle = $$\frac{1}{2}$$ × base × altitude
= $$\frac{1}{2}$$ × x × (x + 4)
= $$\frac{1}{2}$$(x2 + 4x)sq.cm
Given the area of the triangle = 48 sq.cm.
$$\frac{1}{2}$$(x2 + 4x) = 48
x2 + 4x = 48 × 2
x2 + 4x = 96 ⇒ x2 + 4x – 96 = 0
x2 + 12x – 8x -96
(x + 12)(x – 8) = 0
Either x + 12 = 0 or x – 8 = 0
x + 12 = 0 gives x = -12
x – 8 = 0 gives x = 8
x = -12 is not possible
∴ Altitude = x = 8cm
∴ Base= x + 4 = 8 + 4 = 12 cm
Base = 12 cm : Altitude = 8 cm

Question 16.
Sum of the squares of two consecutive positive even integers is 100; find those numbers by using quadratic equations.
Let x and x + 2 be die two consecutive positive even integers.
x2 + (x + 2 )2 = 100
after simplifications, x2 + 2x – 48 = 0
(x + 8) (x – 6) = 0
x = -8 or x = 6
∴ The two consecutive positive even integers are 6 and 8

Question 17.
Solve the following pair of linear equations by graph method. 2x + y = 6 and 2x – y + 2 = 0
Given pair of linear equations is 2x + y = 6 and 2x – y + 2 = 0
2x + y = 6

 x 0 1 2 y 6 4 2 (x, y) (0, 6) (1, 4) (2, 2)

2x – y + 2 = 0

 x 0 1 2 y 2 4 6 (x, y) (0, 2) (1, 4) (2, 6)

From the graph, the two straight lines represented by given equations intersect at the point (1, 4) ∴ The solution of the given pair of linear equations is x = 1, y = 4.

Question 18.
Construct a triangle ABC in which AB = 5 cm, BC = 7 cm and ∠ABC = 50°, then construct a triangle similar to it, whose sides are $$\frac{4}{5}$$ of the corresponding sides of first triangle.
Steps of construction:

1. Draw a triangle ABC with
AB = 5cm, BC = 7 cm and ∠ABC = 50°
2. Draw a ray $$\overrightarrow{\mathrm{AX}}$$ such that ∠BAX is an acute angled
3. Draw A1, A2, A3, A4, A5 arcs on $$\overrightarrow{\mathrm{AX}}$$ such that AA1 = A1A2 = …………. = A4A5
4. Join A5 and B.
5. Draw a parallel line to A5B through A4 to meet AB at B’
6. Draw a parallel line to BC through B’ to meet AC at C’
7. ΔAB’C’ is the required similar triangle.

Part – B (20 Marks)

Note :

2. Each question carries 1 mark,
3. Answers are to be written in Question paper only,
4. Marks will not be awarded in any case of over writing, rewriting or erased answers.

Note : Write the capital letters (A, B, C, D) showing the correct answer for the following questions in the brackets provided against them.

Question 1.
If set A and set B are disjoint sets and n(A) = 6, n(B) = 5, then n(A∪B)
A) 11
B) 6
C) 5
D) 1
A) 11

Question 2.
The decimal expansion of 0.225 in its rational form is
A) 225
B) $$\frac{225}{10^2}$$
C) $$\frac{225}{10^4}$$
D) $$\frac{1}{2}$$
D) $$\frac{1}{2}$$

Question 3.
If p(x) = x2 – 4x + 5, then the value of p(1) is
A) -1
B) 0
C) 1
D) 2
D) 2

Question 4.
If 2x + 3y = 8 and 4x + py = 16 has infinite solutions, then p = ……………
A) 8
B) 6
C) 10
D) 16
B) 6

Question 5.
From the graph, the zeroes of the polynomial are …………..

A) -2
B) 0
C) 2
D) -2, 0, 2
D) -2, 0, 2

Question 6.
Every even positive integer can be written in the form of ………………
A) 2p + 1 (p ∈ Z+)
B) 2p – 1 (p ∈ Z+)
C) 2p (p ∈ Z+)
D) 3p (p ∈ Z+)
C) 2p (p ∈ Z+)

Question 7.
Which of the following is true ?
A) Φ = 0
B) Φ = {}
C) Φ = {0}
D) Φ = {1}
B) Φ = {}

Question 8.
Sum of the first 10 natural numbers is
A) $$\frac{10 \times 9}{2}$$
B) $$\frac{10 \times 10}{2}$$
C) $$\frac{10 \times 11}{2}$$
D) $$\frac{10 \times 100}{2}$$
C) $$\frac{10 \times 11}{2}$$

Question 9.
What does ‘r’ represent in the general term of GP, an = arn-1
B) Common ratio
C) Common difference
D) Common multiple
B) Common ratio

Question 10.
If slope of a line is ‘1’, then the angle between the line and X – axis is
A) 45°
B) 30°
C) 60°
D) 90°
A) 45°

Question 11.
From a set of single digit natural numbers, if a number chosen at random, then the probability that the number chosen is a multiple of 2, is
A) $$\frac{4}{9}$$
B) $$\frac{1}{3}$$
C) $$\frac{9}{4}$$
D) $$\frac{2}{5}$$
A) $$\frac{4}{9}$$

Question 12.
2 – 2 sin2 60° =
A) sin 60°
B) tan 60°
C) cos 60°
D) sec 60°
C) cos 60°

Question 13.
If 14 is deleted from the data 12, 14, 15, 16, 17, 18, 19 and 20, then the median increases by
A) 1
B) 1.5
C) 2
D) 0.5
D) 0.5

Question 14.
The mean of the first eight multiples of 3 is
A) 8
B) 13.5
C) 13
D) 27
B) 13.5

Question 15.
If P(E) the probability of an event, then
A) P(E) ≥ 1
B) P(E) ≤ 0
C) 0 ≤ P(E) ≤ 1
D) P(E) ≤ 1
C) 0 ≤ P(E) ≤ 1

Question 16.
From the given figure, ∠XOY = 130°, then ∠XPO =

A) 65°
B) 35°
C) 25°
D) 55°
C) 25°

Question 17.
The value of cos 15° × cos 45° × 2cosec 75° is
A) $$\frac{2}{\sqrt{3}}$$
B) $$\frac{\sqrt{3}}{2}$$
C) $$\frac{1}{\sqrt{3}}$$
D) 2
D) 2

Question 18.
The volume of a cylinder is given by the formula πr2h, here “h” represents
A) diameter
B) height
D) slant height
B) height

Question 19.
ΔABC ~ ΔXYZ, AB : XY = 9 : 16,then ar (ΔABC) : ar (ΔXYZ) is
A) 3 : 4
B) 4 : 3
C) 81 : 256
D) 256 : 81
In the given figure, DE || BC, if $$\frac{\mathrm{AD}}{\mathrm{DB}}$$ = $$\frac{3}{2}$$ and EC = 3.6 cm, then AE =