TS Class 10

TS 10th Class Social Model Paper 2024, TS SSC Social Model Paper 2024, TS SSC Social Question Paper 2024, TS SSC Social Previous Question Papers Pdf.

TS 10th Class Social Model Papers 2023-2024 – TS SSC Social Previous Question Papers PDF

• TS 10th Class Social Model Paper Set 1
• TS 10th Class Social Model Paper Set 2
• TS 10th Class Social Model Paper Set 3
• TS 10th Class Social Model Paper Set 4
• TS 10th Class Social Model Paper Set 5
• TS 10th Class Social Model Paper Set 6
• TS 10th Class Social Model Paper Set 7
• TS 10th Class Social Model Paper Set 8
• TS 10th Class Social Model Paper Set 9
• TS 10th Class Social Question Paper April 2023

TS 10th Class Social Model Papers, TS SSC Social Model Papers, TS SSC Social Question Papers, TS SSC Social Previous Question Papers.

TS 10th Class Model Papers

The strategic use of TS 10th Class Maths Model Papers Set 3 can significantly enhance a student’s problem-solving skills.

TS SSC Maths Model Paper Set 3 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 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.
Check whether A = {x : x² = 25 and 6x = 15} is an empty set or not ? Justify your answer.
Solution:
x² = 25 . ∴ x = \(\sqrt{25}\) = ± 5
6x = 15 ∴ x = \(\frac{15}{6}\) = \(\frac{5}{2}\)
Any value of x is not satisfying the above two equations. So, A is empty set.

Question 2.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let the first number = x
∴ Second number = 27 – x
Product of two numbers = 182
x(27 – x) = 182
27x – x² = 182
⇒ x² – 27x + 182 = 0
⇒ x² – 13x – 14x + 182 = 0
⇒ x(x- 13) – 14(x – 13) = 0
⇒ (x – 13) (x – 14) = 0
⇒ x – 13 = 0 (or) x – 14 = 0
⇒ x = 13 (or) x = 14
∴ If first number =13, then second number = 27 – 13 = 14
If first number =14, then second number = 27 – 14 = 13
∴ Required numbers = 13 and 14.

Question 3.
Verify that the points (1, 5), (2, 3) and (-2, -1) are collinear are not.
Solution:
Let A(1, 5), B(2, 3), C(-2, -1)
Area of ΔABC
= \(\frac{1}{2}\)|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
= \(\frac{1}{2}\)|1[3 – (- 1)] + 2[- 1 – 5] + (- 2) [5 – 3]|

= \(\frac{1}{2}\) × 12 = 6 sq.units
Area of ΔABC ≠ 0
∴ Given points (1, 5), (2, 3) and (-2,-1) are not collinear.

Question 4.
A person from the top of a building of height 25 m has observed another building top and bottom at an angle of elevation 45° and at an angle of depression 60° respectively. Draw a diagram for this data.
Solution:

Question 5.
Find the area of required cloth to cover the heap of grain in conical shape, of whose diameter is 8m and slant height of 3m.
Solution:
Diameter of conical shape (heap of grain) (d) = 8m
Radius (r) = \(\frac{8}{2}\) = 4 m
Slant height l = 3m
L.S.A of the conical shape (heap) = πrl
= \(\frac{22}{7}\) × 4 × 3
= \(\frac{264}{7}\) = 37.71 m²
∴ Area of required cloth to cover the heap of grain = 37.71 m²

Question 6.
ΔABC ~ ΔDEF and their areas are respectively 64 cm² and 121 cm².
If EF = 15.4 cm., then find BC.
Solution:
\(\frac{{ar}(\triangle \mathrm{ABC})}{{ar}(\triangle \mathrm{DEF})}\) = \(\left(\frac{B C}{E F}\right)^2\)
\(\frac{64}{121}\) = \(\left(\frac{B C}{15.4}\right)^2\)
= \(\frac{8}{11}\) = \(\frac{B C}{15.4}\)
⇒ BC = \(\frac{8 \times 15.4}{11}\) = 11.2 cm

Section – II (6 × 3 = 18 Marks)

Note :

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

Question 7.
Check whether the given pair of linear equations represent intersecting, parallel or coincident lines. Find the
solution if the equations are consistent.
i) 3x + 2y = 5 ; 2x – 3y = 7
ii) 2x – 3y = 5 ; 4x -6y = 15
Solution:
Given pair of linear equations
i) 3x + 2y = 5 …………….. (1)
2x – 3y = 7 …………….. (2)
3x + 2y – 5 = 0
2x – 3y – 7 = 0
a₁ = 3, b₁ = 2, c₁ = -5
a₂ = 2, b₂ = -3, c₂ -7
\(\frac{a_1}{a_2}\) = \(\frac{3}{2}\), \(\frac{b_1}{b_2}\) = \(\frac{2}{-3}\)
\(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\)
∴ Given pair of linear equations rep-resent intersecting lines.
∴ Given equations are consistent.

ii) 2x – 3y = 5 ………….. (3)
4x – 6y = 15 …………… (4)
2x – 3y – 5 = 0
4x – 6y – 15 = 0
a₁ = 2, b₁ = -3, c₁ = -5
a₂ = 4, ₂ = -6, c₂ = -15
\(\frac{a_1}{a_2}\) = \(\frac{2}{4}\) = \(\frac{2}{4}\),
\(\frac{b_1}{b_2}\) = \(\frac{-3}{-6}\) = \(\frac{1}{2}\)
\(\frac{c_1}{c_2}\) = \(\frac{-5}{-15}\) = \(\frac{1}{3}\)
\(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)
∴ Given pair of linear equations represent parallel lines.
∴ Given equations are inconsistent.

Question 8.
The 10^th term of an AP is 52 and 16^th term is 82, then find the 32^nd term.
Solution:
Given : An AP where
10^th term = 52
16^th term = 82

⇒ -6d = -30
⇒ d = \(\frac{-30}{-6}\) = 5
Substituting d = 5 in the equation (1) we get
a + 9d = 52
a + 9 × 5 = 52
⇒ a + 45 = 52 ⇒ a = 52 – 45 = 7
Now 32^nd term = a + 31 d
= 7 + 31 × 5.
= 7 + 155 = 162

Question 9.
Find the zeroes of the quadratic polynomial x² + 5x + 6 and verify the relationship between the zeroes and coefficients.
Solution:
p(x) = x² + 5x + 6
If p(x) = 0, then x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
x(x + 2) + 3(x + 2) = 0
(x + 2) (x + 3) = 0
x + 2 = 0 (or) x + 3 = 0
x = -2 (or) x = -3
∴ Zeroes of p(x) = -2, -3
Let α = -2 and β = -3
Sum of the zeroes α + β = (- 2) + (- 3)
= -5

Question 10.
A toy is in the form of a cone mounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Find the surface area of the toy. (Take π = 3.14)
Solution:
Diameter of the base of the cone d = 6 cm.
∴ Radius of the base of the cone.
r = \(\frac{d}{2}\) = \(\frac{1}{2}\) = 3 cm.

Height of the cone = h = 4 cm
Slant height of the cone
l = \(\sqrt{r^2+h^2}\)
= \(\sqrt{(3)^2+(4)^2}\) = \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5 cm
∴ C.S.A. of cone = πrl
= \(\frac{22}{7}\) × 3 × 5 = \(\frac{330}{7}\) cm²
Radius of the hemisphere
\(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm.
C.S.A. of the hemisphere = 2πr²
= 2 × \(\frac{22}{7}\) × 3 × 3 = \(\frac{396}{7}\)
Hence the surface area of the toy = C.S.A. of cone + C.S.A. of hemisphere
= \(\frac{330}{7}\) +\(\frac{396}{7}\)
= \(\frac{726}{7}\) = 103.71 cm²

Question 11.
If sec β + tan β = p, then express the value of sin β in terms of ‘p’.
Solution:
Given sec β + tan β = p ………….. (1)
sec² β – tan² β = 1
(sec β + tan β) (sec β – tan β) = 1
p(sec β – tan β) = 1
∴ sec β – tan β = \(\frac{1}{\mathrm{p}}\) ………….. (2)


Hence proved.

Question 12.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
i) a king of red colour
ii) a face card
iii) a red face card
iv) the jack of hearts.
Solution:
Out of 52 cards one card can be drawn in 52 ways.
∴ The total number of outcomes n(S) = 52

i) Let E be the event ‘The card is a king of red colour”. ,
The no.of favourable outcomes n(E) = 2
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

ii) Let F be a event “The card is a face card”.
n(F) = 12
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{12}{52}\) = \(\frac{3}{13}\)

iii) Let G be a event The card is.a red face card”.
n(G) = 6
P(G) = \(\frac{\mathrm{n}(\mathrm{G})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{6}{52}\) = \(\frac{3}{26}\)

iv) Let H be a event “The card is the jack of hearts”.
n(H) = 1
P(H) = \(\frac{\mathrm{n}(\mathrm{H})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{1}{52}\)

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) Two men on either side of a temple of 30 meter height observe its top at the angles of elevation 30° and 60° respectively. Find the distance between the two men.
Solution:
Height of the temple BD = 30 meters
Angle of elevation of one person ∠DAB = 30°
Angle of elevation of another person ∠BCD = 60°
Let the distance between the first person and the temple, AD = x and distance between the second person and the temple, CD = d

From ΔBAD
tan 30° = \(\frac{\mathrm{BD}}{\mathrm{AB}}\)
\(\frac{1}{\sqrt{3}}=\frac{30}{x}\) ⇒ x = 30√3 …………. (1)
From ΔBCD
tan 60° = \(\frac{B D}{d}\)
√3 = \(\frac{30}{\mathrm{~d}}\)
d = \(\frac{30}{\sqrt{3}}\) ………….. (2)
From (1) and (2) distance between the persons
= BC + BA = x + d
= 30√3 + \(\frac{30}{\sqrt{3}}=\frac{30 \times 4}{\sqrt{3}}=\frac{120}{\sqrt{3}}\) = 40√3m
The required distance is 40√3 m.

(OR)

B) If cosec A = √2, then find the value of \(\frac{2 \sin ^2 A+3 \cot ^2 A}{4\left(\tan ^2 A-\cos ^2 A\right)}\).
Solution:
Given cosec A = √2

Question 14.
A) Prove that √3 + √5 is an irrational number.
Solution:
Let us suppose that -√3 + √5 is rational.

Since a, b are integers, \(\frac{a^2+2 b^2}{2 a b}\) is rational and so √5 is rational.
This contradicts the fact √5 is irrational.
Hence √3 + √5 is irrational.

(OR)

B) A right circular cylinder has base radius 14 cm and Height 21 on. Find its
i) Area of the base (area of each end)
ii) Curved surface area
iii) Total surface area iv) Volume
Solution:
Given that base radius of
cylinder r = 14 cm
Height h = 21 cm
i) Area of base = πr²
= π(14)²

Question 15.
A) The sum of the third and seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Solution:
In an AP
3^rd term = a₃ = a + 2d
7^th term = a₇ = a + 6d
From given a₃ + a₇ = 6 and a₃ + a₇ = 8
Now take a₃ + a₇ = 6
a + 2d + a + 6d = 6
2a + 8d = 6
a + 4d = 3
a = 3 – 4d …………… (1)
Now take a₃ × a₇ = 8
(a + 2d) (a + 6d) = 8 …………… (2)
Substitute a = 3 – 4d in eqn. (2)
(3 – 4d + 2d) (3 – 4d + 6d) = 8
(3 – 2d) (3 + 2d) = 8
(3)² – (2d)² = 8
9 – 4d² = 8
9 – 8 = 4d²
4d² = 1 => d² = \(\frac{1}{4}\)
d = \(\frac{1}{2}\)
From (I) a = 3 – 4d
= 3 – 4 × \(\frac{1}{2}\) = 3 – 2 = 1
a = 1
Sum of 16 terms
= S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
S₁₆ = \(\frac{16}{2}\)[2 × 1 + (16 – 1) \(\frac{1}{2}\)]
= 8[\(\frac{2}{1}\) +\(\frac{15}{2}\)] = 8[latex]\frac{4+15}{2}[/latex]
= 4 × 19 = 76

(OR)

B) If the median of 60 observations, given below is 28.5 find the values of x and y.

Solution:

Class Interval	Frequency	c.f
0 – 10	5	5
10 – 20	x	5 + x
20 – 30	20	25 + x
30 – 40	15	40 + x
40 – 50	y	40 + x + y
50 – 60	5	45 + x + y

Median = l + (\(\frac{\left(\frac{n}{2}-c . f\right)}{f}\)) × h
It is given that Σf = n = 60
So, 45 + x + y = 60
x + y = 60 – 45 = 15
x + y= 15 ………….. (1)
The median is 28.5 which lies between 20 and 30.
∴ Median class = 20 – 30
Lower boundary of the median class ‘l’ = 20
\(\frac{N}{2}\) = \(\frac{60}{2}\) = 30
cf – cumulative frequency = 5 + x
h = 10
Median =l + (\(\frac{\left(\frac{n}{2}-c . f\right)}{f}\)) × h
⇒ 28.5 = 20 + \(\frac{30-5-x}{20}\) × 10
28.5 = 20 + \(\frac{25-x}{2}\)
\(\frac{25-x}{2}\) = 28.5 – 20 = 8.5
25 – x = 2 × 8.5
x = 25 – 17 = 8
also from (1); x + y = 15 8 + y = 15 ⇒ y = 7
∴ x = 8; y = 7.

Question 16.
A) From a pack of 52 playing cards, Jacks, Queens, Kings and Aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is (i) a black queen, (ii) a red card.
Solution:
Probability of black queen = \(\frac{1}{26}\)
Probability of red card = \(\frac{1}{2}\)
n(S) = 52
Probability of black queen n(a) = 2
p(a) = \(\frac{\mathrm{n}(\mathrm{a})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{2}{52}\) = \(\frac{1}{26}\)
Probability of a red card n(b) = 26
p(b) = \(\frac{\mathrm{n}(\mathrm{b})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{26}{52}\) = \(\frac{1}{2}\)

(OR)

B) If A = (x : x is a prime less than 20} and B = {x : x is a whole number less than 10},
then verify n(A∪B) = n(A) + n(B) – n(A∩B).
Solution:
Given
A = {x : x is a prime less than 20}
A = {2,3, .5, 7, 11,13, 17,19} n(A) = 8
B = {x: x is a whole number less than 10} = {0,1,2, 3,4, 5, 6, 7, 8, 9} n(B) = 10
n(A∪B) = n(A) + n(B) – n(A∩B) ………. (1)
A∩B = {2, 3, 5, 7, 11, 13, 17, 19} ∩ {0, 1,2, 3,4, 5, 6, 7, 8, 9}
= {2, 3, 5, 7}
n(A∩B) = 4
From (1) n (A∪B) = 8 + 10 – 4 = 14.

Question 17.
A) Draw the graph of p(x) = x² – 6x + 9 and find zeroes. Verify the zeroes of the polynomial.
Solution:
Given p(x) = x² – 6x + 9
Let y = x² – 6x + 9

We observe that the graph touches the X – axis at (3, 0).
So, the zeroes of the given polynomial are equal.
∴ The zeroes of p(x) are 3, 3.
Justification:
Given p(x) = x² – 6x + 9 = 0
⇒ x² – 3x – 3x + 9 = 0
⇒ x(x – 3) – 3(x – 3) = 0
⇒ (x- 3) (x – 3) = 0
x – 3 = 0 (or) x – 3 – 0
x = 3 (or) x = 3
∴ Zeroes of p(x) = 3,3

(OR)

B) Solve the following pair of linear equations by graph method.
2x + y = 6 and 2x – y + 2 = 0.
Solution:
2x + y = 6

x	0	1	2	3
y	6	4	2	0
(x, Y)	(0, 6)	(1, 4)	(2, 2)	(3, 0)

2x – y + 2 =0

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

point of intersection is (1, 4)
∴ x = 1; y = 4

Question 18.
A) Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle 60°.
Solution:
Given that, ∠APB = 60°
Join OP.

OP bisects ∠APB
∴ ∠OPA = ∠OPB = 30°

In ΔOAP sin 30° = \(\frac{\mathrm{OA}}{\mathrm{OP}}\) ⇒ \(\frac{1}{2}\) = \(\frac{5}{\mathrm{OP}}\)
∴ OP = 10 cm.
Now,

1. Draw a circle of radius 5 cm with centre ’O’.
2. Mark a point P at a distance of 10 cm from the centre of the circle.
3. Join OP and draw the perpendicular bisector of OP. Let it meet OP at M.
4. Taking M as centre and MO = MP as radius draw a circle. Which cuts the previous circle at A, B. Join PA and PB. Which are the required tangents.

(OR)

Construct a triangle PQR, in which PQ = 4 cm, QR = 6 cm and ∠PQR = 70°. Construct triangle such that each side of the new triangle is \(\frac{3}{4}\) of the triangle of PQR.
Solution:
Steps of construction :

• Draw a triangle PQR with given measurements.

• Draw \(\overrightarrow{\mathrm{PY}}\) such that ∠QPY is an acute angle.
• Locate points A₁, A₂, A₃, A₄ on \(\overrightarrow{\mathrm{PY}}\) such that PA₁ = A₁ A₂ = A₂ A₃ = A₃ A₄.
• Join A₄ and Q.
• Draw a parallel line to A₄ Q through A₃ to meet PQ at Q¹.
• Draw a parallel line to QR through Q¹ to meet PR at R¹.
• ΔPQ¹ R¹ is required similar triangle.

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.
Which of the following is an irrational number ?
A) \(\frac{2}{3}\)
B) \(\sqrt{\left(\frac{16}{25}\right)}\)
C) √8
D) \(\sqrt{0.04}\)
Answer:
C) √8

Question 2.
The product of zeroes of the cubic polynomial 2x³ – 5x² – 14x + 8 is
A) -4
B) 4
C) -7
D) 25
Answer:
A) -4

Question 3.
A pair of linear equations which satisfies dependent system
A) 2x + y- 5 = 0; 3x – 2y – 4 = 0
B) 3x + 4y = 2 ; 6x + 8y = 4
C) x + 2y = 3 ; 2x + 4y = 5
D) x + 2y – 30 =t 0 ; 3x + 6y + 60 = 0
Answer:
B) 3x + 4y = 2 ; 6x + 8y = 4

Question 4.
The n^th term of AP is t_n = a + (n – 1)d, where ‘d’ represents
A) First term
B) Common difference
C) Common ratio
D) Radius
Answer:
B) Common difference

Question 5.
The number of two digit numbers which are divisible by 3
A) 30
B) 20
C) 29
D) 31
Answer:
A) 30

Question 6.
If A = {x: x is a letter in the word MATHEMATICS} and
B = {x: x is a letter in the word MISTEACH} then
A) A⊆ B
B) B ⊆ A
C) A = B
D) B = A
Answer:
B) B ⊆ A

Question 7.
The coordinates of the centre of the circle if the ends of the diameter are (2, -5) and (-2,9)
A) (0, 0)
B) (2, -2)
C) (-5, 9)
D) (0, 2)
Answer:
D) (0, 2)

Question 8.
The Discriminant of the Quadratic equation x² + x + 1 = 0 is
A) 2
B) -3
C) 3
D) -2
Answer:
B) -3

Question 9.
The lines l and m represents.

A) Inconsistent equation
B) Consistent equations
C) Dependent lines
D) Parallel lines
Answer:
B) Consistent equations

Question 10.
The slope of a ladder making an angle 30° with the floor is
A) 1
B) \(\frac{1}{\sqrt{3}}\)
C) √3
D) \(\frac{1}{2}\)
Answer:
B) \(\frac{1}{\sqrt{3}}\)

Question 11.
The distance between the points (cos α, 0), (0, sin α) is
A) 1
B) -1
C) 0
D) -2
Answer:
A) 1

Question 12.
The Arithmetic mean of 30 students is 42. Among them two got zero marks, then Arithmetic mean of remaining students.
A) 40
B) 42
C) 45
D) 28
Answer:
C) 45

Question 13.
The probability of getting king or queen card from the deck of cards
A) \(\frac{1}{52}\)
B) \(\frac{2}{13}\)
C) \(\frac{3}{26}\)
D) \(\frac{5}{52}\)
Answer:
B) \(\frac{2}{13}\)

Question 14.
Which of the following statement Is incorrect ?
A) The ratio of surface areas of cylinder and cone is 1 : 1
B) The ratio Surface Areas of sphere and hemisphere is 2 : 1
C) The ratio Total Surface Area of sphere and hemisphere is 2 : 1
D) The ratio of volumes of cylinder and cone is 3 : 1
Answer:
A) The ratio of surface areas of cylinder and cone is 1 : 1 & C) The ratio Total Surface Area of sphere and hemisphere is 2 : 1

Question 15.
The value of sin 30°+ cos60°is
A) 1
B) 0
C) -1
D) 2
Answer:
A) 1

Question 16.
At a particular time, if the angle of elevation of the sun is 45°, then the length of the shadow of a 5 m high tree is ……………… .
A) 5√3 m
B) 10 m
C) 5 m
D) \(\frac{5}{\sqrt{3}}\)m
Answer:
C) 5 m

Question 17.
When a dice is rolled, the probability of getting a composite number is [ ]
A) \(\frac{1}{4}\)
B) \(\frac{1}{2}\)
C) \(\frac{1}{3}\)
D) \(\frac{1}{6}\)
Answer:
C) \(\frac{1}{3}\)

Question 18.
The sum of first 100 natural numbers
A) 55
B) 505
C) 5050
D) 5500
Answer:
C) 5050

Question 19.
The sides PQ and PR of right angle triangle PQR are such that PQ = 5 cm, PR = 13 cm. If ∠Q = 90° then QR = ?
A) 11.2 cm
B) 9.6 cm
C) 12 cm
D) 10.2 cm
Answer:
C) 12 cm

Question 20.
If a quadrilateral ABCD is drawii to circumscribe a circle, then AB + CD is equal to !
A) AC + BD
B) AD + BC
C) AB + AD
D) AC + BD + BC
Answer:
B) AD + BC

The strategic use of TS 10th Class Maths Model Papers Set 2 can significantly enhance a student’s problem-solving skills.

TS SSC Maths Model Paper Set 2 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.
Expand log a³b²c⁵.
Solution:
log a³b²c⁵ = log a³ + log b² + log c⁵
= 3 log a + 2 log b + 5 log c

Question 2.
If p(x) = x² + 3x + 4, then find the values of p(0) and p(1).
Solution:
p(x) = x² + 3x + 4
p(0) = (0)² + 3(0) + 4
= 0 + 0 + 4 = 4
∴ p(0) = 4
p(1) = (1)² + 3(1) + 4
= 1 + 3 + 4 = 8
∴ p(1) = 8

Question 3.
Find the 10^th term of the arithmetic progression 3, 5, 7, ……………
Solution:
3, 5, 7 ……………. A.P
a = 3, d = 2, n = 10
a_n = a + (n – 1)d
a₁₀ = 3 + (10 – 1)2
= 3 + (9 × 2)
= 3 + 18 = 21

Question 4.
Express tan θ in terms of sin θ.
Solution:
tan θ = \(\frac{\sin \theta}{\cos \theta}\)
sin² θ + cos² θ = 1
cos² θ = 1 – sin²θ
cos θ = \(\sqrt{1-\sin ^2 \theta}\)
∴ tan θ = \(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\).

Question 5.
If a dice is rolled once, then find the probability of getting an odd number.
Solution:
Total outcomes = {1, 2, 3, 4, 5, 6)
Number of Total outcomes = 6
Favourable outcomes = {1, 3, 5}
Number of favourable outcomes = 3
P (an odd number)
= \(\frac{\text { Number of favourable outcomes }}{\text { Number of Total outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 6.
“The top of a tower is observed at an angle of elevation 45° and the foot of ‘ the tower is at a distance of 30 metres from the observer”. Draw a suitable diagram for this data.
Solution:

C – Point of observation
AB – Tower
A – Top of the tower

Section – II (6 × 3 = 18 Marks)

Note :

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

Question 7.
Solve : 2x + y = 5 and 5x + 3y = 11.
Solution:
2x + y = 5 ……………. (1)
5x + 3y = 11 …………… (2)

Substituting the value of x in equation (1)
⇒ 2(4) + y = 5
⇒ 8 + y = 5 ⇒ y = 5 – 8 ∴ y = -3
∴ x = 4, y = – 3 are the solution of the equations.

Question 8.
5, 8, 11, 14, ………. is an arithmetic progression. Find the sum of first 20 terms of it.
Solution:
5, 8, 11, 14, ……………. A.P
a = 5, d = 8 – 5 = 3, n = 20
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
S₂₀ = \(\frac{20}{2}\)[(2 × 5) + (20 – 1)3]
= 10[10 + 57]
= 10 × 67 = 670

Question 9.
Write a Quadratic equation, whose roots are 3 + √5 and 3 – √5 .
Solution:
Roots are 3 + √5 and 3 – √5 .
Sum of the roots = (3 + √5) + (3 – √5) = 6
Product of roots = (3 + √5) (3 – √5)
= 9 – 5 = 4
∴ Required Quadratic equation = x² – x (sum of the roots) + Product of roots = 0
∴ x² – 6x + 4 = 0

Question 10.
A box contains 20 cards which are numbered from 1 to 20. If one card is selected at random from the box, find the probability that it bears (i) a prime number, (ii) an even number.
Solution:
Total possibilities
= {1, 2, 3, 4, 5, 6, 7, 8, 9, ……………,20}
Number of Total possibilities = 20

i) Prime numbers
= {2, 3, 5, 7, 11, 13, 17, 19}
Number of favourable outcomes = 8
P(Prime Number)
= \(\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}\)
= \(\frac{8}{20}\) = \(\frac{2}{5}\)
∴ P(Prime Number) = \(\frac{2}{5}\)

ii) Even numbers
= {2,4, 6, 8, 10,12, 14, 16, 18,20}
Number of favourable outcomes = 10
P (Even Number)
\(\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}\)
= \(\frac{10}{20}\) = \(\frac{1}{2}\)
∴ P(Prime Number) = \(\frac{1}{2}\)

Question 11.
If two persons standing on either side of a tower of height 100 metres observes the top of it with angles of elevation of 60° and 45° respectively, then find the distance between the two persons.
[Note : Consider the two persons and the tower are on the same line.]
Solution:
Tower height = 100 m
Angles of elevation = 60° and 45°

In ΔABP tan 60° = \(\frac{\mathrm{AB}}{\mathrm{PB}}\)
\(\frac{\sqrt{3}}{1}=\frac{100}{x}\)
√3 = x = 100 ⇒ x = \(\frac{100}{\sqrt{3}}\) m

In ΔABQ tan 45° = \(\frac{\mathrm{AB}}{\mathrm{BQ}}\)
\(\frac{1}{1}=\frac{100}{y}\) ⇒ y = 100 m
∴ The distance between Two persons = x + y
\(\frac{100}{\sqrt{3}}\) + 100 = \(\frac{100(\sqrt{3}+1)}{\sqrt{3}}\) m

Question 12.
In a trapezium ABCD, AB || DC. If diagonals intersect each other at point ‘O’, then show that \(\frac{\mathrm{AO}}{\mathrm{BO}}\) = \(\frac{\mathrm{CO}}{\mathrm{DO}}\)
Solution:
Given: In trapezium oABCD, AB // CD.
Diagonals AC, BD intersect at O.
R.T.P : \(\frac{\mathrm{AO}}{\mathrm{BO}}\) = \(\frac{\mathrm{CO}}{\mathrm{DO}}\)

Construction : Draw a line EF passing through the point ‘O’ and parallel to CD and AB.
Proof: In ΔACD, EO//CD
∴ \(\frac{\mathrm{AO}}{\mathrm{CO}}\) = \(\frac{\mathrm{AE}}{\mathrm{DE}}\)
[∵ line drawn parallel to one side of a triangle divides other two sides in the same ratio by Basic proportionality theorem]
In ΔABD, EO //AB
Hence, \(\frac{\mathrm{DE}}{\mathrm{AE}}\) = \(\frac{\mathrm{DO}}{\mathrm{BO}}\)
[∵ Basic proportionality theorem]
\(\frac{\mathrm{BO}}{\mathrm{DO}}\) = \(\frac{\mathrm{AE}}{\mathrm{ED}}\) ………… (2)
From (1) and (2)
\(\frac{\mathrm{AO}}{\mathrm{CO}}\) = \(\frac{\mathrm{BO}}{\mathrm{DO}}\) [∵ Alternendo]

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 sec θ + tan θ = P, then prove that sin θ = \(\frac{\mathbf{P}^2-1}{\mathbf{P}^2+1}\)
Solution:
If sec θ + tan θ = P
Then sec θ – tan θ = \(\frac{1}{\mathrm{P}}\)
[∵ sec² θ – tan² θ = 1]
by adding (+) ⇒ 2 sec θ = P + \(\frac{1}{\mathrm{P}}\)
sec θ = \(\frac{\mathrm{P}^2+1}{2 \mathrm{P}}\) ……………… (1)

(OR)

B) Find the median for the following data.

Solution:

Median = l + (\(\frac{\frac{N}{2}-c f}{f}\)) × h
l = 20, \(\frac{N}{2}\) = \(\frac{44}{2}\) = 22, cf = 16
f = 12, h = 10
Median = 20 + (\(\frac{22-16}{12}\)) × 10
20 + (\(\frac{1}{12}\) × 10)
= 20 + (\(\frac{60}{12}\)) = 20 + 5 = 25
∴ Median =25

Question 14.
A) Prove that √5 + √7 Is an irrational number.
Solution:
Suppose √5 + √7 is not an irrational number.
√5 + √7 = \(\frac{p}{q}\) ; p, q ∈ Z, q ≠ 0
Squaring on both sides
(√5 + √7)² = (\(\frac{p}{q}\))²
⇒ 5 + 7 + 2\(\sqrt{35}\) = \(\frac{p^2}{q^2}\)
⇒ 2\(\sqrt{35}\) = \(\frac{p^2}{q^2}\) – 12
⇒ \(\sqrt{35}\) = \(\frac{p^2-12 q^2}{2 q^2}\)
[p² – 12q², 2q²∈ Z and 2q² ≠ 0]
∴ \(\frac{p^2-12 q^2}{2 q^2}\) is a rational number.
But \(\sqrt{35}\) is an irrational number.
An irrational number never be-comes equal to a rational number.
So our supposition that √5 + √7 is not an irrational number is false.
∴ √5 + √7 is an irrational number.

(OR)

B) The angle of elevation of the top of a hill from the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50 m high. Find the height of the hill.
Solution:
Given height of the tower = AB = 50 m
Let height of hill be CD = h m
and distance between their feet be AC = x m
∠ACB = 30°, ∠CAD = 60°
From right angled Δ ABC, tan 30° = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)

\(\frac{1}{3}\) = \(\frac{50}{x}\) ⇒ x = 50√3 m
From right angled Δ ACD, tan 60° = \(\frac{\mathrm{CD}}{\mathrm{AC}}\)
√3 = \(\frac{h}{x}\) ⇒ h = x√3 m
h = 50√3 .√3 (∵ x = 50√3m)
h = 50 × 3
⇒ h = 150 m.

Question 15.
A) Show that the distance of the points (5, 12), (7, 24) and (35, 12) from the origin are arranged in ascending order forms an arithmetic progression. Find the common difference of the progression.
Solution:
The distance of the point (x, y) from the origin = \(\sqrt{x^2+y^2}\)
The distance of the point (5, 12) from the origin = \(\sqrt{(5)^2+(12)^2}\) = \(\sqrt{25+144}\) = \(\sqrt{169}\) = 13 units;
The distance of the point (7, 24) from the origin = \(\sqrt{(7)^2+(24)^2}\) = \(\sqrt{49+576}\) = \(\sqrt{625}\) = 25 units.
The distance of the point (35, 12) from the origin = \(\sqrt{(35)^2+(12)^2}\) = \(\sqrt{1225+144}\) = \(\sqrt{1369}\) = 37 units.,
The ascending order of the distances is 13, 25, 37
a₁ a₂ a₃
a₂ – a₁ = 25 – 13 = 12
a₃ – a₂ = 37 – 25 = 12
a₂ – a₁ = a₃ – a₂
a₁, a₂, a₃ are in A.R
Common difference (d) = 12

(OR)

B)
The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If it’s total surface area is 1628 square centimeters (cm²), then find the volume of the cylinder
(use π = \(\frac{22}{7}\))
Solution:

In the right circular cylinder
Let base radius = r
Height = h
r + h = 37 cm
Total surface area = 1628 sq.cm
Volume = ?
Total surface area of the cylinder = 2πr (r + h)
2πr(r + h) = 1628
2 × \(\frac{22}{7}\) × r × 37 = 1628
r = \(\frac{1628 \times 7}{2 \times 22 \times 37}\) = 7 cm
7 + h = 37 ⇒ h = 37 – 7 = 30 cm
∴ Volume = πr²h
= \(\frac{22}{7}\) × 7 × 7 × 30 = 4620 cm³

Question 16.
A) From the given Venn diagram, write the sets A∪B, A∩B, A – B and B – A.

Solution:
A = {1, 2, 3, 4, 5}, B = {2, 4, 6, 8}
A∪B = {1, 2, 3, 4, 5, 6, 8}
A∩B = (2, 4}
A- B = {1, 3, 5}
B – A = {6, 8}

(OR)

B) Sum of the areas of two squares is 850 m². If the difference of their perimeters is 40 m, find the sides of the two squares.
Solution:
Let the side of first square be ‘a’
the side of second square be ‘b’
Area of first square = ‘a²‘ m² ,
Area of second square = ‘b²‘ m²
Sum of the area of two squares = 850 m²
a² + b² = 850 ………… (1)
Perimeter of first square = ‘4a’m
Perimeter of second square = ‘4b’m
Difference of their perimeters = 40 m
4a – 4b = 40
a – b = 10 …………. (2)
a = b + 10
substitute this in eqn (1)
(b + 10)² + b² = 850
b² + 20b + 100 + b² = 850
2b² + 20b – 750 = 0
b² + 10b – 375 = 0
b² + 25b – 15b – 375 = 0
b(b + 25) – 15(b + 25) = 0
(b – 15) (b + 25) = 0
b = 15, b = – 25
from (2) a – 15 = 10 ⇒ a = 25
∴ The sides of two squares are 25m, 15m.

Question 17.
A) Draw the graph of the polynomial p(x) x² + 2x – 3 and find the zeroes of the polynomial from the graph.
Solution:
y = p(x) = x² + 2x – 3


Zeros of the polynomial are – 3, 1

(OR)

B) Solve the equations graphically 3x + 4y = 10 and 4x – 3y = 5.
Solution:
3x + 4y = 10 ……………… (1)

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

4x – 3y = 5. ……………… (2)

x	2	-1	5
y	1	-3	5
(x, y)	(2, 1)	(-1, -3)	(5, 5)

Intersecting point is (2, 1)
∴ x = 2, y = 1

Question 18.
A) Draw a circle of radius 4 cm. from a point 9 cm away from it’s centre, construct a pair of tangents to the circle.
Solution:

Steps of Construction :

1. Draw a circle of radius 4 cm.
2. Locate the point ‘P’ at a distance of 9 cm from the centre of the circle ‘O’ and join O, P.
3. Draw a perpendicular bisector for the line segment OP
4. Name the point of intersection of OP and its peipendicular bisector as M.
5. Consider M as centre and MO as the radius draw a circle.
6. Name the points of intersection of the two circles as A and B.
7. Draw PA and PB.
8. PA and PB are required tangents.

(OR)

B) Draw less than Ogive for the following frequency distribution. Find the median from obtained curve.

Solution:


n = 103 ⇒ \(\frac{n}{2}\) = \(\frac{103}{2}\) = 51.5
Form graph, median = 100

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 A⊂B, then A∩B is …………….
A) A
B) B
C) μ
D) Φ
Answer:
A) A

Question 2.
The coefficient of x³ in the polynomial 2x⁴ – 5x³ + 6x² + 5 is ……………..
A) -5
B) 5
C) 6
D) 2
Answer:
A) -5

Question 3.
If the slope of the line joining the points (2, 5) and (x, 3) is 2, then the value of ‘x’ is ……………
A) 0
B) 1
C) -1
D) 2
Answer:
B) 1

Question 4.
The product of prime factors of 108 is ……………
A) 2³ × 3²
B) 2² × 3²
C) 2² × 3³
D) 2³ × 3³
Answer:
C) 2² × 3³

Question 5.
The number of solutions of the pair of linear equations
3x + 2y = 6 and 6x + 4y = 18 is ……………..
A) 0
B) 1
C) 2
D) infinite
Answer:
A) 0

Question 6.
“The total cost of 2 pens and 3 books is Rs. 110”.
Linear equation representing this data is ……………..
A) x + y = 110
B) 5x = 110
C) x² + y³ = 110
D) 2x + 3y = 110
Answer:
D) 2x + 3y = 110

Question 7.
If the n^th term of an arithmetic progression is 4n – 2, then its 10^th term is ……………..
A) 38
B) 28
C) 42
D) 24
Answer:
A) 38

Question 8.
If one root of the Quadratic equation x² – kx + 36 = 0 is 4, then the value of ‘k’ is ……………
A) 12
B) 17
C) 18
D) 13
Answer:
D) 13

Question 9.
The nature of roots of the Quadratic equation x² + 6x + 9 = 0 is …………………
A) Real and distinct.
B) Real and equal.
C) No real roots.
D) One is positive and the other is negative.
Answer:
B) Real and equal.

Question 10.
The sum of first 10 natural numbers is …………….
A) 45
B) 65
C) 55
D) 35
Answer:
C) 55

Question 11.
From the given Ogive curve, the value of the median of the data is ……………..

A) 20
B) 25
C) 15
D) 30
Answer:
D) 30

Question 12.
In the formula of volume of right circular cylinder V = πr²h, the letter ‘r’ represents ……………..
A) Diameter
B) Height
C) Volume
D) Radius
Answer:
D) Radius

Question 13.
If E and \(\overline{\mathbf{E}}\) are complementary events in a random experiment and P(\(\overline{\mathbf{E}}\)) = 0.3, the value of P(E) is ……………….
A) 0.3
B) 0.7
C) 1
D) 0
Answer:
B) 0.7

Question 14.
If one letter is selected randomly from the letters of the word “COVID”, then the probability of getting a vowel is ……………..
A) \(\frac{4}{5}\)
B) \(\frac{3}{5}\)
C) \(\frac{2}{5}\)
D) \(\frac{1}{5}\)
Answer:
C) \(\frac{2}{5}\)

Question 15.
ΔABC ~ ΔDEF, if ∠A = 45° and ∠E = 75°, then ∠C is ……………….
A) 90°
B) 120°
C) 30°
D) 60°
Answer:
D) 60°

Question 16.
ΔABC is a right triangle, right angled at B. If AB = 9 cm, BC = 12 cm, then AC is …………….
A) 13 cm
B) 14 cm
C) 15 cm
D) 16 cm
Answer:
C) 15 cm

Question 17.
In the given figure OA and OB are radii. PA and PB are tangents to the circle at points A and B. If ∠AOB = 130°, then ∠APB = ?

A) 40°
B) 50°
C) 60°
D) 70°
Answer:
B) 50°

Question 18.
If sin θ = \(\frac{3}{5}\), then the value of cos θ is (θ is acute angle) ……………..
A) \(\frac{1}{5}\)
B) \(\frac{5}{3}\)
C) \(\frac{4}{5}\)
D) \(\frac{2}{5}\)
Answer:
C) \(\frac{4}{5}\)

Question 19.
The maximum number of tangents can be drawn from an external point to a circle is ……………..
A) 1
B) 2
C) 3
D) 4
Answer:
B) 2

Question 20.
If θ is acute angle, then sin θ × sec θ = ……………….
A) tan θ
B) cot θ
C) 1
D) cosec θ
Answer:
A) tan θ

The strategic use of TS 10th Class Maths Model Papers Set 1 can significantly enhance a student’s problem-solving skills.

TS SSC Maths Model Paper Set 1 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.
Ramu says, “If log₁₀ x = 0, value of x = 0″. Do you agree with him ? Give reason.
Solution:
log₁₀x = 0 [∵ log_aN = x ⇒ a^x = N]
10⁰ = x
x = 1 & x ≠ 0
∴ I don’t agree with Ramu.

Question 2.
-3, 0 and 2 are the zeroes of the polynomial p(x) = x³ + (a – 1)x² + bx + c, find a and c.
Solution:
Given solution compare with
ax³ + bx² + cx + d
Given equation p(x) = x³ + (a – 1) x² + bx + c.
Given roots -3, 0, 2
Sum of the roots ⇒ α + β + γ = \(\frac{-b}{a}\)
-3 + 0 + 2 = \(\frac{-(a-1)}{1}\)
– 1 = – a + 1 ⇒ – 1 – 1 = -a
∴ a = 2
product of the roots αβγ = \(\frac{-d}{a}\)
(-3)(0)(2) = \(\frac{-c}{1}\) ⇒ c = 0
∴ a = 2 and c = 0

Question 3.
\(\frac{1}{4}\), \(\frac{1}{16}\), \(\frac{1}{64}\), \(\frac{1}{256}\), …………….
are in G.P. justify
Solution:
To justify \(\frac{1}{4}\), \(\frac{1}{16}\), \(\frac{1}{64}\), \(\frac{1}{256}\), ……………. is a G.P
We need to show the ratio of any two successive terms is equal.
Now the common ratio = r₁ = \(\frac{1}{16}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\)
r₂ = \(\frac{1}{64}\) + \(\frac{1}{16}\) + \(\frac{1}{4}\) ⇒ r₁ = r₂
Hence it is a G.E

Question 4.
What can you say about
cot 0° = \(\frac{1}{\tan 0^{\circ}}\). Is it defined ? Why ?
Solution:
tan 0° = 0
cot 0° = \(\frac{1}{\tan 0^{\circ}}\) = \(\frac{1}{0}\) = undefined,
Reason:
Division by ‘0’ is not allowed, hence \(\frac{1}{0}\) is indeterminate.

Question 5.
Write two examples for equally likely events.
Answer:
Example 1 : Tossing a coin
Head and tails have equal chances.

Example 2 : Rolling a dice.
All faces have equal chances.

Question 6.
Draw a line segment of length 7.3 cm and divide in the ratio 3 : 4
Answer:

Section – II (6 × 3 = 18 Marks)

Note :

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

Question 7.
If A = {x : x is a factor of 12} and
B = {x : x is a factor of 6}, then find A∪B and A∩B.
Solution:
Given A = {x : x is a factor of 12}
= {1, 2, 3, 4, 6, 12}
B = {x : x is a factor of 6}
= {1, 2, 3, 6}
A∪B = {1, 2, 3, 4, 6, 12} ∪ {1, 2, 3, 6}
= {1, 2, 3, 4, 6, 12} ;
A∩B = {1, 2, 3, 4, 6, 12} ∩ {1, 2, 3, 6} = {1, 2, 3, 6}

Question 8.
For what value of’m1 in the following, mx + 4y = 10 and 9x + 12y = 30 system of equations will have no solution ? Why ?
Solution:
Given equations have no solutions. They have no solution mean they are parallel.
Given equations compare with
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel

If \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) here given
a₁ = m, b₁ = 4, c₁ = -10
a₂ = 9, b₂ = 12, c₂ = -30
\(\frac{\mathrm{a}_1}{\mathrm{a}_2}=\frac{\mathrm{b}_1}{\mathrm{~b}_2}\)
⇒ \(\frac{\mathrm{m}}{9}=\frac{4}{12}\)⇒ m = 3
∴ If m = 3 then the above system will have no solution.

Question 9.
In a flower garden, there are 23 plants in the first row, 21 plants in the sec¬ond row, 19 plants in the third row and so on. If there are 10 rows in that flower garden, then find the total num- her of plants in the last row with the help of the formula
t_n = a + (n – 1)d.
Solution:
Number of plants in the 1^st row = 23
Number of plants in the 2^nd row = 21
Number of plants in the 3^rd row = 19 and so on.
The progression is 23, 21, 19, ………….. .
Here common difference is same so the series is an A.P
Here a = 23, d = 21 – 23 = -2. n = 10
n^thterm of an A.P in t_n = a + (n – 1)d
t₁₀ = 23 + (10 – 1) (-2)
= 23 + 9(-2)
= 23 – 18 = 5 .
Number of plants in the last row is 5.

Question 10.
A box contains 4 red balls, 5 green balls and P white halls. If the probability of randomly picked ball from the box to be a red ball is \(\frac{1}{3}\), then find the number of white balls.
Solution:
P(E) = \(\frac{\text { No. of outcomes }}{\text { Sample Space }}\)
\(\frac{1}{3}\) = \(\frac{4}{4+5+P}\)
4 + 5 + P= 12
⇒ P = 3
∴ Number of white balls = 3

Question 11.
In the given figure AB, AC and PQ are tangents to a circle and AB = 6 cm. Find the perimeter of ΔAPQ.

Solution:

We know that tangents drawn from an external point to a circle are equal in length.
∴ AB = AC
AP + PB = AQ + QC
AP + PX = AQ + QX (∵ PX = BP and QX = QC)
AP + PX = AQ + QX = 6 cm
Now, Perimeter of
ΔAPQ = AP + PQ + AQ
= (AP + PX) + (QX + AQ)
= 6 + 6
= 12 cm
∴ Perimeter of ΔAPQ = 12 cm

Question 12.
If the ratio of areas of two equilateral triangles is 25 : 36, then find the ratio of heights of the triangles.
Solution:
For similar triangles
Ratio of area = Ratio of sides

∴ h₁ : h₂ = 5 : 6
∴ The ratio of heights of the triangles = 5 : 6

Section – III (6 × 5 = 30 Marks)

Question 13.
A) If the mean of the following frequency distribution is 50, then find the value of k.

Solution:

Mean = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
50 = \(\frac{4080+70 \mathrm{k}}{88+\mathrm{k}}\)
50(88 + k) =4080 + 70k
4400 + 50k = 4080 + 70k
4400 – 4080 = 70k – 50k
320 = 20k
\(\frac{320}{20}\) = k ⇒ k = 16

(OR)

B) Find the area of the segment shaded in the figure in which PQ =12 cm, PR = 5 cm and QR is the diameter of the circle with centre ‘O’. (Take π = \(\frac{22}{7}[latex] )

Solution:
To find the area of the segment shaded in the given figure.
Here ‘PQ’ = 12 cm; ‘PR’ = 5 cm; ‘QR’ is diameter
Now PQOR is a semicircle
then angle in a semicircle is 90°
then ∠QPR = 90°
∴ ΔPQR is a right angled triangle
∴ Area of ΔPQR = [latex]\frac{1}{2}\)bh
= \(\frac{1}{2}\) × PQ × PR
= \(\frac{1}{2}\) × 12 × 5 =30 cm² ………….. (1)
Now the area of shaded part = area of semicircle – area of ΔPQR
= \(\frac{1}{2}\) πr² – 30 cm² ……………. (2)
In ΔPQR, QR² = PQ² + PR² (from Pythagoras theorem)
QR² = 12² + 5²
= 144 + 25
= 169 = 13²
∴ QR = 13 then
Radius of the circle (r) = QO = \(\frac{\mathrm{QR}}{2}\) = \(\frac{13}{2}\) = 6.5 cm
then area of semicircle
\(\frac{1}{2}\) πr²
= \(\frac{1}{2}\) × \(\frac{22}{7}\) × \(\frac{13}{2}\) × \(\frac{13}{2}\) \(\frac{1}{2}\)
= 66.39 cm² – (3)
Now putting the values of (1) and (3) in (2) we get
Area of shaded part = (66.39 – 30) = 36.39 cm².

Question 14.
A) Use division algorithm to show that the square of any positive integer is of the form 5m or 5m + 1 or 5m + 4, where ‘m’ is a whole number.
Solution:
a = bq + r, 0 ≤ r < b
b = 5 so r = 0, 1, 2, 3, 4
Then ‘a’ can be of the forms
5q + 0, 5q + 1, 5q + 2, 5q + 3, 5q + 4

Case (i) When a = 5q
a² = (5q)² = 5 (5q²) = 5m
where m = 5q² ∈ W.

Case (ii) When a = 5q + 1
a² = (5q + 1)²
= 25q² + 10q + 1
= 5 (5q² + 2q) + 1
= 5m + 1 where m = 5q² + 2q ∈ W Similarly,

Case (iii) a² = (5q + 2)² = 5m + 4

Case (iv) a² = (5q + 3)² = 5m + 4

Case (v) a² = (5q + 4)²= 5m + 1

So the square of any positive integer is of the form 5m or 5m + 1 or 5m + 4 where n e W.

(OR)

B) Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point between them on the road, the angles of elevation of top of the poles are 60° and 30°. Find the height of poles.
Solution:
As shown in the figure

AD = width of road = 80 cm.
AB, DE are two poles
AB = DE (∵ they have equal heights)
‘C’ is a point on road.
∠ACB = 30°, ∠DCE = 60°
Then in ΔACB
tan C = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ tan 30 = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{A B}{A C}\) ⇒ AC = AB√3 ……… (1)

In ΔCDE
tan C = \(\frac{\mathrm{DE}}{\mathrm{CD}}\) ⇒ tan 60 = \(\frac{\mathrm{DE}}{\mathrm{CD}}\)
√3 = \(\frac{\mathrm{DE}}{\mathrm{CD}}\)
⇒ CD = \(\frac{\mathrm{DE}}{\sqrt{3}}\) ………… (2)
but AC + CD = AD
AB√3 + \(\frac{\mathrm{DE}}{\sqrt{3}}\) = 80
But DE = AB
⇒ AB√3 + \(\frac{\mathrm{AB}}{\sqrt{3}}\) = 80
⇒ \(\frac{3 \mathrm{AB}+\mathrm{AB}}{\sqrt{3}}\) = 80
⇒ 4AB = 80√3
⇒ AB = \(\frac{80 \sqrt{3}}{4}\) = 20√3
So height of the pole = 20√3 m.

Question 15.
A) On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4, respectively. Findg(x).
Solution:
Given, p(x) = x³ – 3x² + x + 2
q(x) = x – 2 and
r(x) = – 2x + 4
By division algorithm, we know that Dividend = Divisor × Quotient + Remainder
p(x) = q(x) × g(x) + r(x)
Therefore,
x³ – 3x² + x + 2
= (x – 2) x g(x) + (- 2x + 4)
⇒ x³ – 3x² + x + 2 + 2x – 4
= (x – 2) × g(x)
⇒ g(x) = \(\frac{x^3-3 x^2+3 x-2}{x-2}\)
On dividing x³ – 3x² + 3x – 2 by x – 2, we get

First term of g(x) = \(\frac{x^3}{x}\) = x²
Second term of g(x) = \(\frac{-x^2}{x}\) = -x
Third term of g(x) = \(\frac{x}{x}\) = 1
Hence, g(x) = x² – x +1.

(OR)

B) From a pack of 52 playing cards, Jacks, Queens, Kings and Aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is (i) a black queen, (ii) a red card. :
Solution:
Probability of black queen = \(\frac{1}{26}\)
Probability of red card = \(\frac{1}{2}\)
n(S) = 52
i) Probability of black queen n(a) = 2
p(a) = \(\frac{\mathrm{n}(\mathrm{a})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

ii) Probability of a red card n(b) = 26
P(b) = \(\frac{\mathrm{n}(\mathrm{b})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{26}{52}\) = \(\frac{1}{2}\)

Question 16.
A) ABCD is a trapezium in which AB||DC and its diagonals intersect each other at point ‘O’. Show that
\(\frac{A O}{B O}=\frac{C O}{D O}\).
Solution:
Given: In trapezium ABCD, AB// CD.
Diagonals AC, BD intersect at 0.
R.T.P : \(\frac{A O}{B O}=\frac{C O}{D O}\)

Construction: Draw a line EF passing through the point ‘O’ and parallel to CD and AB.
Proof: In ΔACD.E0//CD
∴ \(\frac{A O}{C O}=\frac{A E}{D E}\) ……….. (1)
[∵ line drawn parallel to one side of a triangle divides other two sides in the same ratio by Basic proportionality theorem] ,
In ΔABD, EO//AB
Hence, \(\frac{D E}{A E}=\frac{D O}{B O}\)
[∵ Basic proportionality theorem]
\(\frac{B O}{D O}=\frac{A E}{E D}\) ………… (2) [∵ Invertendo]
From (1) and (2)
\(\frac{A O}{C O}=\frac{B O}{D O}\)
\(\frac{A O}{B O}=\frac{C O}{D O}\) [∵ Alternendo]

(OR)

B) Find the area of the triangle formed by the points (2, 3), (-1, 3)and (2, -1) using Heron’s formula.
Solution:
To find the area of the triangle formed by (2, 3) (- 1, 3) and (2, – 1) using Heron’s formula.
Let the co-ordinates of A = (2, 3) ; B ='(- 1, 3); C = (2, – 1) then the sides of ΔABC are represented by as follows

AB = c, BC = a, CA = b
then the formula of the triangle usingHeron’s formula
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
where s = \(\frac{a+b+c}{2}\)
Now, we find the sides of ΔABC, using the formula \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
∴ CB = a = distance between the points, (2, – 1) and (- 1, 3)
= \(\sqrt{(2-(-1))^2+(-1-3)^2}\)
= \(\sqrt{3^2+(-4)^2}\)
= \(\sqrt{3^2+4^2}\)
= \(\sqrt{9+16}\)
= \(\sqrt{25}\)
⇒ CB = a = 5 …….. (1)

Question 17.
A) Draw the graph of p(x) = x² – 2x – 8 and find the zeroes/of the polynomial from it.
Solution:
p(x) = x² – 2x – 8


Zeroes of the given polynomial are -2, 4.

(OR)

B) Show that the following pair of equations are consistent and solve them graphically,
x + 3y = 6,
2x – 3y = 12
Solution:
x + 3y = 6

x	0	6	3
y	2	0	1
(x, y)	(0, 2)	(6, 0)	(3, 1)

2x – 3y = 12

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

Point of intersection is (6, 0)
Since the two lines are intersecting at one point, we can say that they are Consistent.
∴ x = 6, y = 0

Question 18.
A)The following table shows that ages of the patients admitted in a hospital during a year.

Draw a less than Ogive curve for the above data.
Solution:

OR

B) 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 4/5 of the corresponding sides of first triangle.
Solution:

Steps of Construction:

1. Draw a triangle ABC with AB = 5 cm, BC = 7 cm and ∠ABC = 50°.
2. Draw a ray \(\overrightarrow{\mathrm{AX}}\) such that ∠BAX is an acute angle.
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 A₅ B 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 required similar triangle.

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 die following questions in the brackets provided against them. (Marks: 20 × 1 = 20)

Question 1.
If a, b, care in A.P., then b =
A) \(\frac{a+c}{2}\)
B) a + c
C) \(\sqrt{\mathrm{ac}}\)
D) ac
Answer:
A) \(\frac{a+c}{2}\)

Question 2.
If the number of subsets of a given set is 32, then the number of elements in the set will be
A) 2
B) 4
C) 5
D) 3
Answer:
C) 5

Question 3.
The distance of (3, 4) from origin is
A) 3
B) 4
C) 5
D) 7
Answer:
C) 5

Question 4.
The sum of the roots of 6x² = 1 is
A) 0
B) \(\frac{1}{6}\)
C) –\(\frac{1}{6}\)
D) 6
Answer:
A) 0

Question 5.
If the polynomial p(x) = x⁴ – 2x³ + x² – 1 is divided by (x +1), then the degree of quotient polynomial.
A) 1
B) 3
C) 4
D) 2
Answer:
B) 3

Question 6.
The sum of a number and reciprocal is \(\frac{17}{4}\), then the number is
A) 3
B) 4
C) 5
D) 17
Answer:
B) 4

Question 7.
If log₁₀2 = 0.3010, then log₁₀32 is
A) 5.3010
B) 2.3010
C) 015050
D) 0.3010
Answer:
C) 015050

Question 8.
The point (-2, -2) is in the quadrant.
A) Q₁
B) Q₂
C) Q₃
D) Q₄
Answer:
C) Q₃

Question 9.
In a G.P., the 5^th term is 32 and 7^th term is 128, then the common ratio of G.P.
A) 2
B) 5
C) 7
D) 3
Answer:
A) 2

Question 10.
Solution to \(\frac{a^2}{x}-\frac{b^2}{y}\) = 0; \(\frac{\mathbf{a}^2 \mathbf{b}}{x}+\frac{b^2 a}{y}\) = a + b, x ≠ 0, y ≠ 0 is ……………….
A) (-a², -b²)
B) (a, b²)
C) (a, -b)
D) (a², b²)
Answer:
D) (a², b²)

Question 11.
The probability of sure event is
A) 0
B) \(\frac{1}{2}\)
C) 1
D) Undefined
Answer:
C) 1

Question 12.
\(\sqrt{1+\sin A} \cdot \sqrt{1-\sin A}\) =
A) sin A
B) 1 – sin²A
C) cos A
D) 1
Answer:
C) cos A

Question 13.
Side of a cube and diameter of a sphere are equal, then the ratio of their volume will be
A) 4 : π
B) 6 : π
C) 3 : π
D) 2 : π
Answer:
B) 6 : π

Question 14.
A dice is thrown once. The probability of getting a prime number is [ ]
A) \(\frac{1}{3}\)
B) \(\frac{1}{2}\)
C) \(\frac{2}{3}\)
D) \(\frac{1}{6}\)
Answer:
B) \(\frac{1}{2}\)

Question 15.
Volumes of two spheres are in the ratio of 8 : 27, the ratio of their surface areas is ……………….
A) 2 : 3
B) 4 : 3
C) 2 : 9
D) 4 : 9
Answer:
D) 4 : 9

Question 16.
Mean of certain number of observations is \(\). If each observation is divided by m(m ≠ 0) and then increased by n, then the mean of new observation is
A) \(\frac{\bar{x}}{n}\) + m
B) \(\overline{\mathbf{x}}+\frac{\mathbf{n}}{\mathrm{m}}\)
C) \(\overline{\mathbf{x}}+\frac{\mathbf{m}}{\mathrm{n}}\)
D) \(\frac{\bar{x}}{m}\) + n
Answer:
D) \(\frac{\bar{x}}{m}\) + n

Question 17.

A) \(\frac{1}{3}\)
B) \(\frac{3}{4}\)
C) \(\frac{1}{4}\)
D) \(\frac{2}{3}\)
Answer:
A) \(\frac{1}{3}\)

Question 18.
A ladder 15m long just reaches the top of vertical wall. If the ladder makes an angle of 60° with the wall. Then the height of the wall is
A) 15√3 m
B) \(\frac{15 \sqrt{3}}{2}\) m
C) 7.5 m
D) 15 m
Answer:
B) \(\frac{15 \sqrt{3}}{2}\) m

Question 19.
If sec θ + tan θ = x, then sec θ =
A) \(\frac{x^2+1}{x}\)
B) \(\frac{x^2+1}{2 x}\)
C) \(\frac{x^2-1}{2 x}\)
D) \(\frac{x^2-1}{x}\)
Answer:
B) \(\frac{x^2+1}{2 x}\)

Question 20.
At point ’P’ on a circle, PQ is a tangent and ‘O’ is the centre of the circle. If ΔOPQ is an isosceles triangle, then ∠OQP is equal to
A) 90°
B) 30°
C) 45°
D) 60°
Answer:
C) 45°

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