TS 10th Class Maths Question Paper June 2023

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TS 10th Class Maths Question Paper June 2023

Time: 3 Hours
Maximum Marks: 80

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 mean of the factors of 24.
Solution:
Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24
Mean = \(\frac{\Sigma \mathrm{x}}{\dot{\mathrm{n}}}\)
= \(\frac{1+2+3+4+6+8+12+24}{8}\)
= \(\frac{60}{8}\) = 7.5
∴ Mean of the factors of 24 = 7.5

Question 2.
Express ‘tan θ’ in terms of cos θ’.
Solution:

(OR)

Question 3.
Find the value of

Solution:

Question 4.
If the pair of linear equations 6x – 4y + 10 = 0 and 3x + ky + 6 = 0 represents parallel lines graphically, then find the value of ‘k’.
Solution:
Given pair of linear equations
6x – 4y + 10 = 0 and
3x + ky + 6 = 0 represents parallel lines.

Question 5.
In ΔABC, DE is a line such that AD = 3 cm, AB = 5 cm, AE = 6 cm and AC = 10 cm. Is DE || BC?
Justify.

Solution:
In the given figure,
AD = 3 cm, AB = 5 cm
BD = AB – AD = 5 – 3 = 2
AE = 6 cm, AC = 10 cm
EC = AC – AE = 10 – 6 = 4
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{3}{2}\) and \(\frac{\mathrm{AE}}{\mathrm{EC}}\) = \(\frac{6}{4}\) = \(\frac{3}{2}\)
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)
By converse of Basic proportionality theorem DE || BC.

Question 6.
A = {x: x is a factor of 18}, B = {x: x is a factor of 36}. Is A ⊂ B? Justify?
Solution:
A = {x: x is a factor of 18}
A = {1, 2, 3, 6, 9, 18}
B = {x : x is a factor of 36}
B = {1, 2, 3, 4, 6, 9, 12, 18, 36}
Every element of get A is also an elementof B. So A ⊂ B.

Section – II (6 × 4 = 24 Marks)

Note :

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

Question 7.
If 2304 = 2^x × 3^y, then find the value of log_y x.
Solution:
2304 = 2^x × 3^y
= 2⁸ × 3² = 2^x × 3^y

Prime factorisation of composite number is unique.
∴ x = 8, y = 2
∴ log_yx = log₂8
∴ log₂ 2³ = 3log₂ 2 = 3(1) = 3
[log_ax^m = mlog_ax] 2034 = 2⁸ × 3²
If 2304 = 2^x × 3y then log_yx = 3

Question 8.
Write the formula for mode of a grouped data and explain each term of it.
Solution:
Mode = l + \(\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right)\) × h
Where l = lower boundary of the model class
h = size of the model class
f₁ = frequency of model class
f₀ = frequency of the class preceding the model class
f₂ = frequency of the class succeeding the model class

Question 9.
If the zeroes of the polynomial x³ – 9x² + 26x – 24 are α – β, α, α + β, then two find the values of α and β.
Solution:
Zeroes of the given polynomial
P(x) = x³ – 9x² + 26x – 24
are α – β, α, α + β
Sum of the zeroes = \(\frac{-\mathrm{b}}{\mathrm{a}}\)
(α – β) + α + (α + β) = \(\frac{-(-9)}{1}\)
3α = 9
∴ α = \(\frac{9}{3}\) = 3
∴ α = 3
Product of the zeroes = \(\frac{-\mathrm{d}}{\mathrm{a}}\)
(α – β)α(α + β) = \(\frac{-(-24)}{1}\)
(3 – β)3(3 + β) = 24
(3² – β²)3 = 24
9 – β² = \(\frac{24}{3}\) = 8
9 – 8 = β²
1 = β² ⇒ β = \(\sqrt{1}\) = ±1
∴ α = 3, β = ±1

Question 10.
If 6 times of 6^th term of an arithmetic progression is equal to 9 times of 9^th term of it, then show that 15^th term of that A.P. is zero.
Solution:
n^th term of an A.P = a_n = a + (n – 1)d
6^th term of an A.P = a₆ = a + 5d → (1)
9^th term of an A.P = a₉ = a + 8d → (2)
According to the problem,
6 times of 6^th term = 9 times of 9^th term
6a₆ = 9a₉
6(a + d) = 9(a + 8d)
6a + 30d = 9a + 72d
6a + 30d – 9a – 72d = 0
-3a – 42d = 0
3(a + 14d) = 0
a + 14d = 0
∴ a₁₅ = 0
∴ 15^th term of the given A.P is zero.

Question 11.
If the area of the triangle formed by joining the points A(x, y), B(3, 2) and C(-2, 4) is 10 square units, then show that 2x – 5y + 4 = 0.
Solution:
Let A(x, y), B (3, 2) and C (-2, 4) and
Area of ∆ABC = 10 sq. units.
Area of ∆ABC

Question 12.
If one card is randomly selected from a well shuffled deck of cards, then find the probability of getting –
i) a face card
ii) a jack of hearts and
iii) an ace card
Solution:
Number of a cards in a deck n(S) = 52

i) Number of face card in the deck n(F) = 12
The probability of getting face card
P(F) = \(\frac{\mathrm{n}(\mathrm{~F})}{\mathrm{n}(\mathrm{~S})}\) = \(\frac{12}{52}\) = \(\frac{3}{13}\)

ii) Number of ajack of hearts card in the deck n(J) = 1
The probability of getting a jack of hearts card
P(J) = \(\frac{\mathrm{n}(\mathrm{~J})}{\mathrm{n}(\mathrm{~S})}\) = \(\frac{1}{52}\)

iii) Number of an ace cards in a deck n(A) = 4
The probability of getting an ace card
P(A) = \(\frac{\mathrm{n}(\mathrm{~A})}{\mathrm{n}(\mathrm{~S})}\) = \(\frac{4}{52}\) = \(\frac{1}{13}\)

Section – III (4 × 6 = 24 Marks)

Note:

1. Answer any 4 questions from the given six questions.
2. Each question carries 6 marks.

Question 13.
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

Graph of given polynomial intersect X – axis at (-3, 0) and (1, 0).
∴ Zeroes of given polynomial p(x) = x² + 2x – 3 are -3, 1.

Question 14.
The numerator of a fraction is 3 less than its denominator. If 2 is added to both numerator and denominator, the sum of the new fraction formed and original fraction is \(\frac{29}{20}\), then find the original fraction.
Solution:
Let the numerator of a fraction = x and denominator = y
But the numerator is 3 less than the denominator.
∴ x = y – 3 ⇒ y = x + 3
∴ Required fraction \(\frac{x}{y}\) = \(\frac{x}{x+3}\)
If 2 added to both numerator and denominator, then formed new fraction.
⇒ \(\frac{x+2}{(x+3)+2}\) = \(\frac{x+2}{x+5}\)
According to problem, the sum of the new fraction formed and original fraction = \(\frac{29}{20}\)

Question 15.
Find the arithmetic mean of the following data.

Solution:

Question 16.
Construct a circle of radius 5 cm. Then construct a pair of tangent to the circle such that the angle between them is 60°.
Solution:

Steps ofconstruction:

1. Draw a circle with centre ‘O’ and radius 5 cm.
2. Take a point ‘P’ out side the circle such that OP = 10 cm.
3. Draw the perpendicular bisector to OP which bisects it at M.
4. Taking M as centre and PM as radius draw a another circle. It intersect previous circle at A and B.
5. Join P to A and B.
6. PA and PB are the required tangents.

Question 17.
If two boys standing on either side of their school building of height 20 m. observed the top of it with angles of elevation of 30° and 60° respectively, then find the distance between the two boys.
Solution:
AB = Height of the school building = 20 m
CD = Distance between two boys.

Question 18.
Show that the quadrilateral fòrmed by joining the points (-4, 2), (4, 4), (2, 12) and (-6, 10) taken in order is a square.
Solution:

Part – B (20 Marks)

Note:

1. All questions are to be answered.
2. Each question carries 1 mark.
3. Añswers are to be written in the Question paper only.
4. Marks will not be given for over-writing, rewriting or erased answers.

I. Write the CAPITAL LETTERS (A, B, C, D) of the correct answer in the brackets provided against each question. (20 × 1 = 20 M)

Question 1.
Slope of the line passing through the points (5, 8) and (6, 10) is ……… ( )
A) 2
B) 1/2
C) 3/4
D) 4/3
Solution:
A) 2

Question 2.
ΔABC ~ ΔPQR. If ∠A + ∠C = 110° , then ∠Q=…….. ( )
A) 110°
B) 90°
C) 70°
D) 20°
Solution:
C) 70°

Question 3.
The roster form of the set A={x : x ∈ N, -2 ≤ x ≤ 2}
A) {-2, 2}
B) {=2, -1, 0, 1, 2}
C) {-1, 0, 1}
D) {-2, -1, 1, 2}
Solution:
There is no correct answer in the given four options. Correct answer is {1, 2}.

Question 4.
The quadratic equation having the roots 2 and -2 is ….. ( )
A) x² + 4x – 4 = 0
B) x² – 2x – 4 = 0
C) x² + 2x – 4 = 0
D) x² – 4 = 0
Solution:
D) x² – 4 = 0

Question 5.
The 8^th term of the geometric progression 512, 256, 128,……… is …….. ( )
A) 64
B) 8
C) 16
D) 4
Solution:
D) 4

Question 6.
If sin A = cos B, where A and B are acute angles, then ………. ( )
A) A + B = 180°
B) AB = 90°
C) A – B = 90°
D) A = B
Solution:
B) AB = 90°

Question 7.
In a random experiment, E and \(\overline{\mathrm{E}}\) are complementary events. If P(E) = \(\frac{1}{6}\), then P(\(\overline{\mathrm{E}}\)) is ( )
A) 0
B) 1
C) 5/6
D) 4/6
Solution:
C) 5/6

Question 8.
The ratio of total surface areas of a hemisphere and a sphére with equal radii is …….. ( )
A) 3 : 4
B) 1 : 2
C) 4 : 3
D) 2 : 1
Solution:
A) 3 : 4

Question 9.
If the length and breadth of a rectangle are (x + 5) and (x + 2) respectively (x > 0), then its area is represented by …….
A) x² + 5x + 10
B) x² + 7x + 10
C) x² + 2x + 10
D) x² + 10x + 10
Solution:
B) x² + 7x + 10

Question 10.
If the quadratic equation x² + kx + 9 = 0 has equal roots, then the value of ’k’ is…. ( )
A) 3
B) -3
C) -6
D) 9
Solution:
C) -6

Question 11.
In the given figure, AP and AQ are two tangents to a circle with centre ‘O’ such that ∠POQ = 125°, then ∠PAQ is ……. ( )

A) 55°
B) 25°
C) 35°
D) 45°
Solution:
A) 55°

Question 12.
The value of sin²29° + sin²61° of is ………
A) -1
B) 2
C) 0
D) 1
Solution:
D) 1

Question 13.
The distance of the point (log₂8, log₃81) from the origin is ….. . ( )
A) 3
B) 5
C) 4
D) 2
Solution:
B) 5

Question 14.
The mode of the values of sin 90°, cos 90°, tan 60°, sec 60°, cosec 90° is ( )
A) 0
B) 2
C) 1
D) \(\sqrt{3}\)
Solution:
C) 1

Question 15.
Which of the following is not a linear equation in two variables? ( )
A) 5 + 4x = y + 3
B) x + 2y = y – x
C) x + y = 0
D) 3 – x = y² + 4
Solution:
D) 3 – x = y² + 4

Question 16.
Which of the following can’t be the probability of an event? ( )
A) 5/4
B) 0.75
C) 46%
D) 2 : 3
Solution:
A) 5/4

Question 17.
The total surface area of a right circular cylinder is 2πr(h + r). In this formula, ‘h’ represents ……. ( )
A) radius
B) diameter
C) height
D) slant height
Solution:
C) height

Question 18.
If θ=45°, then the value of \(\frac{1-\cos 2 \theta}{\sin 2 \theta}\) is …….( )
A) 2
B) 1
C) -1
D) 0
Solution:
B) 1

Question 19.
A bag contains 10 black balls. If one ball is selected randomly from the bag, then the probability of getting a white ball is …… ( )
A) 0
B) 1/2
C) 1/3
D) 1/10
Solution:
A) 0

Question 20.
If 125 = 7q + r, where q and r are quotient and remainder respectively, then the value of ‘r is ……. ( )
A) 0
B) 1
C) 3
D) 6
Solution:
D) 6