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:

- Answer all the questions under Part – A on a separate answer book.
- 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 :

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

Question 1.

Find, the values of k for which the quadratic equation 4x^{2} + 5kx + 25 = 0 has equal roots.

Answer:

Comparing the given quadratic equation

4x^{2} + 5kx + 25 = 0

with ax^{2} + bx + c = 0, we get

a = 4, b = 5k, c = 25

It is given that Q.E has equal roots

b^{2} – 4ac = 0

⇒ (5k)^{2} – 4(4) (25) = 0

⇒ 25k^{2} – 400 = 0 ⇒ 25k^{2} = 400

⇒ k^{2} = \(\frac{400}{25}\) = 16

∴ k = \(\sqrt{16}\) = ± 4

Question 2.

Find the value of log_{√2} 128.

Answer:

Question 3.

For the A.P. ; -3, -7, -11, …………. ; can we find directly a_{30} – a_{20} without actually finding a_{30} and a_{20}.

Answer:

Given A.P is -3, -7, -11, …………….

Here a = -3 and d = a_{2} – a_{1} = -7 – (-3) = -4

Now, we have the formula

a_{n} = a + (n – 1) d

∴ a_{30} – a_{20} = (a + 29d) – (a + 19d)

= a + 29d – a – 19d

= 10d

= 10 × -4 = -40

Yes, we can find directly a_{30} – a_{20} without finding a_{30} and a_{20}.

Question 4.

When a dice is rolled, find the probability of getting odd prime number.

Answer:

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 4tan^{2} 45° – cosec^{2} 30° + cos^{2} 30° = \(\frac{3}{4}\).

Answer:

LHS = 4 tan^{2}45° – cosec^{2}30° + cos^{2}30°

= 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.

Answer:

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 :

- Answer ALL the following questions.
- 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.

Answer:

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) = ax^{3} + bx^{2} + cx + d is zero, then find the product of other two zeroes of p(x). (a ≠ 0)

Answer:

p(x) = ax^{3} + bx^{2} + 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.

Answer:

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.

Answer:

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)

adding (1) and (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.

Answer:

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.

Answer:

Section – III (4 × 6 = 24 Marks)

Note :

- Answer any 4 of the following questions.
- 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.

Answer:

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.

Answer:

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

By adding we get

AS + BQ + CQ + DS

= AP + BP + CR + RD

AS + DS + BQ + CQ

= AP + BP + CR + RD

AD + BC = AB + CD.

From (1), We have

AD + AD = AB + AB

2 AD = 2 AB

AD = AB ………… (2)

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.

Answer:

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)

= \(\frac{1}{2}\)(x^{2} + 4x)sq.cm

Given the area of the triangle = 48 sq.cm.

\(\frac{1}{2}\)(x^{2} + 4x) = 48

x^{2} + 4x = 48 × 2

x^{2} + 4x = 96 ⇒ x^{2} + 4x – 96 = 0

x^{2} + 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.

Answer:

Let x and x + 2 be die two consecutive positive even integers.

x^{2} + (x + 2 )^{2} = 100

after simplifications, x^{2} + 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

Answer:

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.

Answer:

Steps of construction:

- Draw a triangle ABC with

AB = 5cm, BC = 7 cm and ∠ABC = 50° - Draw a ray \(\overrightarrow{\mathrm{AX}}\) such that ∠BAX is an acute angled
- Draw A
_{1}, A_{2}, A_{3}, A_{4}, A_{5}arcs on \(\overrightarrow{\mathrm{AX}}\) such that AA_{1}= A_{1}A_{2}= …………. = A_{4}A_{5} - Join A
_{5}and B. - Draw a parallel line to A
_{5}B through A_{4}to meet AB at B’ - Draw a parallel line to BC through B’ to meet AC at C’
- ΔAB’C’ is the required similar triangle.

Part – B (20 Marks)

Note :

- Answer all the questions,
- Each question carries 1 mark,
- Answers are to be written in Question paper only,
- 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

Answer:

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) \(\frac{1}{2}\)

Answer:

D) \(\frac{1}{2}\)

Question 3.

If p(x) = x^{2} – 4x + 5, then the value of p(1) is

A) -1

B) 0

C) 1

D) 2

Answer:

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

Answer:

B) 6

Question 5.

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

A) -2

B) 0

C) 2

D) -2, 0, 2

Answer:

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^{+})

Answer:

C) 2p (p ∈ Z^{+})

Question 7.

Which of the following is true ?

A) Φ = 0

B) Φ = {}

C) Φ = {0}

D) Φ = {1}

Answer:

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) \(\frac{10 \times 100}{2}\)

Answer:

C) \(\frac{10 \times 11}{2}\)

Question 9.

What does ‘r’ represent in the general term of GP, a_{n} = ar^{n-1}

A) Radius

B) Common ratio

C) Common difference

D) Common multiple

Answer:

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°

Answer:

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) \(\frac{2}{5}\)

Answer:

A) \(\frac{4}{9}\)

Question 12.

2 – 2 sin^{2} 60° =

A) sin 60°

B) tan 60°

C) cos 60°

D) sec 60°

Answer:

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

Answer:

D) 0.5

Question 14.

The mean of the first eight multiples of 3 is

A) 8

B) 13.5

C) 13

D) 27

Answer:

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

Answer:

C) 0 ≤ P(E) ≤ 1

Question 16.

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

A) 65°

B) 35°

C) 25°

D) 55°

Answer:

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

Answer:

D) 2

Question 18.

The volume of a cylinder is given by the formula πr^{2}h, here “h” represents

A) diameter

B) height

C) radius

D) slant height

Answer:

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

Answer:

C) 81 : 256

Question 20.

In the given figure, DE || BC, if \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{3}{2}\) and EC = 3.6 cm, then AE =

A) 4.5 cm

B) 5.6 cm

C) 5.4 cm

D) 4.6 cm

Answer:

C) 5.4 cm