TS 10th Class Maths Model Paper Set 6 with Solutions

The strategic use of TS 10th Class Maths Model Papers Set 6 can significantly enhance a student’s problem-solving skills.

TS SSC Maths Model Paper Set 6 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.
If A = {x : x is a factor of 24}, then find n(A).
Solution:
A = (1, 2, 3, 4, 6, 8, 12, 24} ; n(A) = 8
24 = 1 × 24
= 2 × 12 = 3 × 8 = 4 × 6

Question 2.
Find the roots of the Quadratic equation x² + 2x – 3 = 0.
Solution:
x² + 2x – 3 = 0
x² + 3x – x – 3 = 0
x(x + 3) – 1 (x + 3) = 0
(x + 3) (x – 1) = 0
x = – 3 (or) x = 1
Roots of Quadratic Equation are – 3, 1.

Question 3.
For what value of ‘t’ the following pair of linear equations has a no solution ?
2x – ty = 5 and 3x + 2y = 11.
Solution:
Pair of linear equations has no solution
If \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\) ⇒ \(\frac{2}{3}\) = \(\frac{-t}{2}\) ≠ \(\frac{-5}{-11}\)
– 3t = 4 ⇒ t = – \(\frac{4}{3}\)

Question 4.
In a right triangle ABC, right angled at ‘C’ in which AB =13 cm, BC = 5cm, determine the value of cos² B + sin² A.
Solution:
We have, Cos B = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac{5}{13}\)
Sin A = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac{5}{13}\)

Cos² B + Sin² A
= \(\frac{25}{169}\) +\(\frac{25}{169}\) = \(\frac{25+25}{169}\) = \(\frac{50}{169}\)

Question 5.
A point P is 25 cm from the centre O of the circle. The length of the tangent drawn from P to the circle is 24 cm. Find the radius of the circle.
Solution:
From right angled Δ AOP

OP² = OA² + AP²
(25)² = OA² + (24)²
625 = OA² + 576
OA² = 625 – 576 = 49 = 7²
OA = 7 cm
∴ The radius of the circle is 7 cm.

Question 6.
In a hemispherical bowl of 2.1 cm radius ice-cream is there. Find the volume of the bowl.
Solution:
Radius (r) = 2.1 cm
Volume (V) = \(\frac{1}{2}\)πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × (2.1)³
= 19.404 cm³

Section – II (6 × 3 = 18 Marks)

Note :

1. Answer ALL the following questions.
2. Each question carries 3 marks.

Question 7.
If x² + y² = 7xy, then prove that log(\(\frac{x+y}{3}\)) = \(\frac{1}{2}\)
Given x² + y² = 7xy, now add (2xy) on both sides. We get
x² + y² + 2xy = 7xy + 2xy = 9xy
Considering square root on both sides we get
\(\sqrt{x^2+y^2+2 x y}\) = \(\sqrt{9 x y}\)
x + y = 3\(\sqrt{x y}\)
⇒ \(\frac{x+y}{3}\) = \(\sqrt{x y}=(x y)^{\frac{1}{2}}\) considering logarithm on both sides, we get
log(\(\frac{x+y}{3}\)) = log\((x y)^{\frac{1}{2}}\) = \(\frac{1}{2}\) log(xy)
log(\(\frac{x+y}{3}\)) = \(\frac{1}{2}\) (log x + log y) (∵ log mn = log m + log n)

Question 8.
“If we multiply or divide both sides of a linear equation by a non-zero number, then the roots of that linear equation will remain the same.”
Is it true ? If so, justify with an example.
Solution:
Given statement : “If we multiply or divide both sides of a linear equation by a non-zero number, then the roots of that linear equation will remain the same”.
Consider a linear equation
x + y = 5 ………….. (1)
One of the solutions for this equation (1) is (2, 3)
Now multiply equation (1) by ‘3’.
3 (x + y) = 3 × 5
3x + 3y = 15 ………….. (2)
Now substitute the point (2, 3) in the equation (2).
∴ 3(2)+ 3 (3) = 15
16 + 9= 15 ⇒ 15 = 15
∴ (2, 3) is also a solution for the equation (2).
So the given statement is true.

Question 9.
Measures of sides of a triangle are in Arithmetic Progression. Its perimeter is 30cm, and the difference between the longest and shortest side is 4cm; then find the measures of the sides.
Solution:
Let the 3 sides of given triangle = a – d, a, a + d
Then its perimeter
= a – d + a + a + d = 30 cm.
3a = 30 cm ⇒ a = \(\frac{30}{3}\) = 10 cm.
The larger side = a + d
The shorter side = a – d
The difference between the above two
= (a + d) – (a – d) = 4 cm.
a + d – a + d = 4 cm.
2d = 40 ⇒ d = \(\frac{4}{2}\) = 2 cm.
a = 10 cm, d = 2 cm
So the sides a – d = 10 – 2 = 8 cm
and a + d=10 + 2 = 12 cm.
So 8, 10, 12 cm are the sides of the triangle.

Question 10.
A bag contains balls which are numbered from 1 to 50. A ball is drawn at random from the bag, the probability that it bears a two digit number multiple of 7.
Solution:
Number of possible outcomes = 50 Number of required outcomes = 6 {14,21,28,35,42,49}
∴ Probability of getting two digit number which is a multiple of 7
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of total outcomes }}\)
= \(\frac{6}{50}\) = \(\frac{3}{25}\)

Question 11.
If cot θ = \(\frac{7}{8}\) then,
Evaluate :
i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
ii) \(\frac{1+\cos \theta}{\sin \theta}\)
Solution:
i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
= \(\frac{1-\sin ^2 \theta}{1-\cos ^2 \theta}=\frac{\cos ^2 \theta}{\sin ^2 \theta}\) = cot² θ
= (\(\frac{7}{8}\))² = \(\frac{49}{64}\) ………… (1)
(∵ sin² θ + cos² θ = 1)

ii) \(\frac{1+\cos \theta}{\sin \theta}\) = \(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}\) = cosec θ + cot θ
Now given cot θ = \(\frac{7}{8}\)) then we have
(1 + cot² θ) = cosec² θ
⇒ 1 + (\(\frac{7}{8}\))²
= 1 + \(\frac{49}{64}\) + \(\frac{64+49}{64}\) = \(\frac{113}{64}\)
∴ cosec θ = \(\sqrt{\frac{113}{64}}\)
∴ \(\frac{1+\cos \theta}{\sin \theta}\) = cosec θ + cot θ
= \(\frac{\sqrt{113}}{8}\) + \(\frac{7}{8}\) = \(\frac{7+\sqrt{113}}{8}\)

Question 12.
A right circular cylinder has radius 3.5 cm and height 14 cm. Find curved surface area.
Solution:
radius (r) = 3.5cm; height (h) = 14cm Curved surface area of the right circular cylinder
= 2πrh = 2 × \(\frac{22}{7}\) × 3.5 × 14
= 44 × 7 = 308 cm²

Section – III (6 × 5 = 30 Marks)

Note :

1. Answer all the following questions.
2. In this section, every question has internal choice. Answer any one alternative.
3. Each question carries 5 marks.

Question 13.
A) If two dice are thrown at the same time, find the probability of getting sum of the dots cm top is prime.
Solution:
Total possible outcomes when two dice are rolled = 36
Let E be an event to get sum of the tops is prime
Total favourable outcomes : (1, 1) (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3) (5, 2), (6, 1), (5, 6), (6, 5)
Number of total favourable outcomes = 15
P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total possible outcomes }}\)
\(\frac{15}{36}\) =\(\frac{5}{12}\)

(OR)

(B) Show that \(\frac{\cos \theta}{1-\sin \theta}+\frac{1-\sin \theta}{\cos \theta}\) = 2 sec θ
Solution:

Question 14.
A) Find the ratio in which X-axis divides the line segment joining the points (2, – 3) and (5, 6). Then find the intersecting point on X-axis.
Solution:
Let X – axis divides the line segment joining points (2, – 3) and (5, 6) in the ratio m₁ : m₂
x₁ = 2 x₂ = 5
y₁ = -3 y₂ = 6
Co-ordinate of point
P = (\(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\))
= (\(\frac{m_1(5)+m_2(2)}{m_1+m_2}, \frac{m_1(6)+m_2(-3)}{m_1+m_2}\))
But this is a point on X-axis, so its y co-ordinate is zero.
\(\frac{m_1(6)+m_2(-3)}{m_1+m_2}\) = 0
6m₁ – 3m₂ = 0
6m₁ = 3m₂
\(\frac{\mathrm{m}_1}{\mathrm{~m}_2}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ Required ratio = 1 : 2

(OR)

B) DWACRA is supplied cuboidal shaped wax block with measurements 88 cm × 42 cm × 35 cm. From this how many number of cylinderical candles of 2.8 cm diametre and 8 cm of height can be prepared ?
Solution:
Shape of wax block = cuboid
Its length (l) = 88 cm ;
breath (b) = 42 cm ; height (h) = 35 cm
Then the volume of wax present in block = lbh
= 88 × 42 × 35 cm³ …………… (1)
Shape of candle = cylinder
Diameter of candle = (d) = 2.8 cm 2.8
⇒ radius = r = \(\frac{2.8}{2}\) = 1.4 cm
height (h) = 8 cm
Volume of wax required to make one candle = V = πr²h
= \(\frac{22}{7}\) × 1.4 × 1.4 × 8 cm³
∴ Total number of candles that can be
made = \(\frac{\text { Total volume of block }}{\text { Volume of each candle }}\)
= \(\frac{88 \times 42 \times 35}{\frac{22}{7} \times 1.4 \times 1.4 \times 8}\) = 2625
So 2625 candles can be made with given measurements.

Question 15.
A) Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km/hr faster than the second train. If after two hours they are 50 km. apart, find the average speed of each train.
Solution:
Let the speed of the slower train = x kmph
then speed of the faster train = x + 5 kmph.

Distance = Speed × Time
∴ Distance travelled by the first train = 2(x + 5) = 2x+ 10
Distance travelled by the second train = 2.x = 2x
By Pythagoras Theorem
(hypotenuse)² = (side)² + (side)²
⇒ (2x)² + (2x + 10)² = 50²
⇒ 4x² + (4x² + 40x + 100) = 2500
⇒ 4x² + 4x² + 40x + 100 = 2500
⇒ 8x² + 40x – 2400 = 0
= x² + 5x – 300 = 0
⇒ x² + 20x – 15x – 300 = 0
⇒ x(x + 20) – 15(x + 20) = 0
⇒ (x + 20) (x – 15) = 0
∴ x – 15 = 0 (or) x + 20 = 0
⇒ x = 15 (or) -20
But x can’t be negative.
∴ Speed of the slower train x = 15kmph.
Speed of the faster train
x + 5 = 15 + 5 = 20 kmph.

(OR)

B) In a right angle triangle, the hypotennse is 10 cm more than the shortest side. If third side is 6 cm less than the hypotenuse, find the sides of the right angle triangle.
Solution:
Let the shortest side of right angled triangle be x cm,
Hypotenuse = (x + 10) cm (given)
Third side = (x + 10 – 6) cm
= (x + 4) cm (given)
AC = (x + 10) cm, AB = x cm,
BC = (x + 4) cm
In a right angled triangle ABC

AC² = AB² + BC²
(x + 10)² = x² + (x + 4)²
x² + 20x + 100 = x² + x² + 8x + 16
⇒ x² – 12x – 84 = 0

⇒ x = 6 + 2\(\sqrt{30}\) cm (∵ x < 0)
x + 4 = 10 + 2\(\sqrt{30}\) cm
x + 10 = 16 + 2\(\sqrt{30}\) cm

Question 16.
A) Show that √5 – √3 is an irrational number.
Solution:
Suppose √5 – √3 is not an irrational number. √5 – √3 is a rational number.
Let √5 – √3 = \(\frac{p}{q}\) where q ≠ 0 and p, q e ∈ Z squaring on both sides
5 + 3 – \(2 \sqrt{15}=\frac{p^2}{q^2}\)
\(\sqrt{15}=\frac{8 q^2-p^2}{2 q^2}\)
∴ p, q ∈ Z & q ≠ 0
8q² – p² & 2q² ∈ Z and also 2q² ≠ 0.
So \(\frac{8 q^2-p^2}{2 q^2}\) JL is a rational number.
but \(\sqrt{15}\) is an irrational number.
An irrational number never be equal to a rational number.
So our supposition that √5 – √3 is not an irrational number is false.
∴ √5 – √3 is an irrational number.

(OR)

(B) Prove that :
(1 + tan² θ) + ( 1 + \(\frac{1}{\tan ^2 \theta}\)) = \(\frac{1}{\sin ^2 \theta-\sin ^4 \theta}\)
Solution:

Question 17.
A) Draw the graph of the polynomial p(x) = x² – 7x + 12, then find its zeroes from the graph.
Solution:
p(x) = x² – 7x + 12


zeroes of the polynomial is 3, 4

(OR)

B) Find the solution of x + 2y = 10 and 2x + 4y = 8 graphically.
Solution:
x + 2y = 10 ………… (1)

x	0	2	4	6
y	5	4	3	2
(x, y)	(0, 5)	(2, 4)	(4, 3)	(6, 2)

2x + 4y = 8 ……….. (2)

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

The lines are parallel
∴ No solution for the given pair of equations.

Question 18.
A) Construct a triangle of sides 4.2 cm, 5.1 cm and 6 cm. Then construct a triangle similar to it, whose sides are 2/3 of corresponding sides of the first triangle.
Solution:

Construction Steps :

1. Draw a triangle ABC, with sides AB = 4.2 cm, BC = 5.1 cm, CA = 6 cm.
2. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
3. Locate 3 points B₁, B₂, B₃ on BX so that BB₁ = B₁B₂ = B₂B₃.
4. Join B₃, C and draw a line through B₂ parallel to B₃C intersecting BC at C’.
5. Draw a line through C’ parallel to CA intersect AB at A’.
6. ΔA’BC’ is required triangle.

(OR)

B) Draw a circle of radius 6 cm and construct two tangents to the circle so that angle between the tangents is 60°.
Solution:
To begin let us consider a circle with centre ‘O’ and radius 6 cm.
Let PA and PB are two tangents draw from a point ‘P’ outside the circle and the angle between them is 60°.
In this ∠APB = 60°. Join OP.

As we know, OP is the bisector of ∠APB, then
∠OAP = ∠OPB = \(\frac{60^{\circ}}{2}\) = 30° (∵ ΔOAP≅ΔOBP)
Now in ΔOAP
sin 30° = \(\frac{\text { Opposite side }}{\text { Hypotenuse }}=\frac{\mathrm{OA}}{\mathrm{OP}}\)
\(\frac{1}{2}=\frac{6}{\mathrm{OP}}\) (From trigonometric ratio)
OP = 12 cm
Now we can draw a circle of radius 6 cm with centre ‘O’. We then mark a point at a distance of 12 cm from the centre of the circle. Join OP and complete the construction.

Hence PA and PB are the required pair of tangents to the given circle.

Part – B (20 Marks)

Note :

1. Answer all the questions.
2. Each question carries 1 mark.
3. Answers are to be written in the 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. (Marks: 20 × 1 = 20)

Question 1.
If both roots are common to the Quadratic equations
x² – 4 = 0 and x² + px – 4 = 0, then p =
A) 2
B) 0
C) 4
D) 1
Answer:
B) 0

Question 2.
The distance of the point p(x, y) from Y-axis is
A) |x|
B) |y|
C) |x + y|
D) |x – y|
Answer:
A) |x|

Question 3.
The exponential form of log_a\(\sqrt{x^4}\) = y is
A) a^y = x⁴
B) y^a = 4
C) a^y = x²
D) x^y = a²
Answer:
C) a^y = x²

Question 4.
If A ⊂ B, then A – B =
A) A
B) B
C) B – A
D) Φ
Answer:
D) Φ

Question 5.
If (a, b), (b, c) and (c, a) art the vertices of a triangle and the centroid of triangle is origin, then a³ + b³ + c³ =
A) abc
B) a + b + c
C) 3abc
D) 0
Answer:
C) 3abc

Question 6.
If \(\frac{5}{x-1}+\frac{1}{y-2}\) = 2, \(\frac{6}{x-1}+\frac{-3}{y-2}\) = 1 then x = …………..
A) 4
B) 7
C) -1
D) 3
Answer:
A) 4

Question 7.
The last (unit place) digit of 6²⁰¹⁹ in its standard form
A) 6
B) 4
C) 9
D) 19
Answer:
A) 6

Question 8.
If α, β are the zeroes of the polynomial x² + 5x + k and α – β = 3, then the value of k.
A) 6
B) 9
C) 5
D) 4
Answer:
D) 4

Question 9.
The 25^th term of -300, -290, -280, is ………
A) -60
B) -80
C) 60
D) 80
Answer:
A) -60

Question 10.
If A {x : x² – 16 = 0, x ∈ R} and
B – {x : x² – 5x + 6 = 0, x∈ R}, then A∪B is a ………..
A) Singleton set
B) Infinite set
C) Null set
D) Finite set
Answer:
D) Finite set

Question 11.
What change will be observed in the angle of elevation as we move away from the object?
A) increase
B) decrease
C) can’t be determined
D) no change
Answer:
A) increase

Question 12.
Let x₁, x₂, x₃, x₄,……. x_n be the n observations and \(\overline{\mathbf{x}}\) be the mean of n observations, then \(\sum_{i=1}^n\left(x_1-\bar{x}\right)\) = ……..
A) 0
B) n\(\overline{\mathbf{x}}\)
C) \(\frac{\bar{x}}{n}\)
D) \(\frac{2 \bar{x}}{n}\)
Answer:
A) 0

Question 13.
A 20 m long ladder is placed on a pole of 10 m height Shaking ‘α’ angle with the ground, then α =
A) 60°
B) 45°
C) 30°
D) 0°
Answer:
C) 30°

Question 14.
Tangents PA and PB inclined at an angle 60° as shown in the figure, the ratio of lengths of OA, OP and AP is

A) 1 : 2 : 3
B) 3 : 2 : 1
C) √3 : 2 : 1
D) 1 : 2 : √3
Answer:
D) 1 : 2 : √3 (Add Score)

Question 15.
ABC is a right angle triangle and ∠C = 90°, Let BC = a, CA = b, AB = c and p be the length of the perpendicular from C on AB, then ………………
A) \(\frac{1}{\mathrm{p}^2}=\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\)
B) \(\frac{1}{\mathrm{p}^2}=\frac{1}{\mathrm{~b}^2}-\frac{1}{\mathrm{a}^2}\)
C) \(\frac{1}{\mathrm{p}^2}=\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}\)
D) \(\frac{2}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\)
Answer:
C) \(\frac{1}{\mathrm{p}^2}=\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}\)

Question 16.
Let r, h had l be the radius, height and slant height of a cone respectively, then express l in terms of r and h is ……………
A) \(\sqrt{h^2-r^2}\)
B) \(\sqrt{\mathrm{r}^2+\mathrm{h}^2}\)
C) \(\sqrt{\mathrm{r}^2-\mathrm{h}^2}\)
D) \(\sqrt{4 r^2+h^2}\)
Answer:
B) \(\sqrt{\mathrm{r}^2+\mathrm{h}^2}\)

Question 17.
Number of circles passing through 3 collinear points in a plane is
A) 1
B) 0
C) 9
D) 12
Answer:
B) 0

Question 18.
Choose the correct figure for which sin A = \(\frac{5}{13}\)

Answer:

Question 19.
A letter is choosen from the word “BAHUBALI”, the probability that it was not a vowel is
A) \(\frac{1}{2}\)
B) \(\frac{3}{2}\)
C) \(\frac{4}{3}\)
D) \(\frac{3}{4}\)
Answer:
A) \(\frac{1}{2}\)

Question 20.
Which of the following statement is true ?
A) All acute angle triangles are similar.
B) All obtuse angle triangles are similar.
C) All right angle triangles are similar.
D) All isosceles right triangles are similar.
Answer:
D) All isosceles right triangles are similar.