AP 10th Class Maths Model Paper Set 9 with Solutions

Regularly solving AP 10th Class Maths Model Papers Set 9 contributes to the development of problem-solving skills.

AP SSC Maths Model Paper Set 9 with Solutions

Time : 3 Hours, 15 Minutes
Max. Marks : 100

Instructions:

1. In the duration of 3 hours 15 minutes, 15 minutes of time is allotted to read the question paper.
2. All the answers shall be written in the answer booklet only.
3. Question paper consists of 4 Sections and 33 questions.
4. Internal choice is available in Section – IV only.
5. Answers shall be written neatly and legibly.

Section – I
(12 × 1 = 12 M)

Note :

1. Answer all the questions in one word or a phrase.
2. Each question carries 1 mark.

Question 1.
If HCF (26, 91) is 13, then find LCM of (26, 91).
Solution:

Question 2.
Write an example for trinomial having degree 6.
Solution:
ax⁶ + bx² + c (any trinomial with degree 6).
3 terms with different order less than 6, one of the terms must be with order 6.

Question 3.
If P(E) the probability of an event “E”, then
A) P(E) ≥ 1
B) P(E) ≤ 0
C) 0 ≤ P(E) ≤ 1
D) P(E) ≤ 1.
Solution:
C [0 ≤ P(E) ≤ 1]

Question 4.
Define tangent to a circle.
Solution:
The line which passes through only one point of a circle is called tangent of a circle. (Or) If a line touches the circle at one point then the line is called tangent to the circle.

Question 5.
An observer 1.5 m tall is 28.5 m away from a tower of height 30 m. Find the angle of elevation of the top of tower from her eyes.
Solution:
45°

Question 6.
Match the following:

Choose the correct answer:
A) i → r, ii → p, iii → q
B) i → r, ii → q, iii → p
C) i → p, ii → q, iii → r
D) i → q, ii → r, iii → p
Solution:
D (q, r, p)

Question 7.
Draw the rough figure of the toy which is in the form of a cone mounted on a hemisphere of same radius.
Solution:

Question 8.
Statement I: Any two circles are similar.
Statement II: Any two equilateral triangles are similar.
Choose the correct option from the following.
A) Statement I is true and statement II is false
B) Statement I is false and statement II is true
C) Both statements are true
D) Both statements are false
Solution:
C) Both statements are true

Question 9.
Assertion : The pair of linear equations 2x + 3y + 6 = 0 and 4x + 6y + 7 = 0 have no
Reason : \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)
Choose the correct option from the following.
A) Both Assertion and Reason are true, Reason is supporting the Assertion
B) Both Assertion and Reason are true, but Reason is not supporting the Assertion
C) Assertion is false, but Reason is true
D) Assertion is true, but the Reason is false
Solution:
A) Both Assertion and Reason are true, Reason is supporting the Assertion

Question 10.
The quadratic polynomial whose sum of zeroes is -3 and product of zeroes is 2 is
A) x² + 3x + 2
B) x² – 3x + 2
C) x² – 3x – 2
D) x² + 3x – 2
Solution:
A) x² + 3x + 2

Question 11.
If one root of the quadratic equation x² – 7x + 12 = 0 is 4, then find the other root.
Solution:
3

Question 12.
What is the common difference in an A.P. if the first term is 6 and the nth term is 6n ?
A) 4
B) 5
C) 6
D) 0
Solution:
C) 6

Section – II
(8 × 2 = 16 M)

Note:

1. Answer all the questions.
2. Each question carries 2 marks.

Question 13.
Express (cosecθ – cotθ)² in terms of cosθ.
Solution:

Question 14.
Find the distance between (a cosθ, 0) and (0, a sinθ)
Solution:
Distance between (x₁, y₁) and (x₂, y₂) is \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
Distance between (a cos θ, 0) and (0, a sin θ) is

Question 15.
Find the zeroes of the quadratic polynomial x² + 7x + 10.
Solution:
x² + 7x + 10 = x² + 2x + 5x + 10
= x(x + 2) + 5 (x + 2)
= (x + 2) (x + 5)
To find zeroes
x² + 7x + 10 = 0
(x + 2) (x + 5) = 0
(x + 2) = 0 (OR) (x + 5) = 0
x = – 2 (OR) x = – 5
Zeroes of the Quadratic polynomial x² – 7x + 10 are -2 and – 5.

Question 16.
2 cubes each of volume 64 cm³ are joined end to end. Find the total surface area of the resulting cuboid.
Solution:
Volume of cube (a³) = 64 cm³
a = 4 cm
side of a cube = 4cm
As the cubes are joined end to end the resultant solid will be a cuboid.
Length of resulting cuboid (l) = 8 cm.
Its breadth (b) = 4 cm
Its height (h) = 4 cm
Total surface area of a cuboid = 2 (lh + bh + lb)
= 2(8 × 4 + 4 × 4 + 8 × 4)
= 2(32 + 16 + 32)
= 2(80)
= 160 cm².

Question 17.
Check whether the following is a quadratic equation or not ?
(2x – 1) (x – 3) = (x + 5) (x -1).
Solution:
(2x -1) (x – 3) = (x + 5) (x -1)
2x² – 7x + 3 = x² + 4x – 5
2x² – x² – 7x – 4x + 3 + 5 = 0
x² – 11x + 8 = 0
This is of the form ax² + bx + c = 0
So, given equation is a quadratic equation.

Question 18.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre 0 at a point Q so that OQ = 12 cm. Find the length of tangent PQ.
Solution:
Length of the tangent (PQ)
= \(\sqrt{d^2-r^2}\)
= \(\sqrt{12^2-5^2}\)
= \(\sqrt{144^2-25}\)
= \(\sqrt{119}\) cm.

Question 19.
Define similar triangles.
Solution:
Two triangles are similar, if
i) Their corresponding angles are equal (OR)
ii) Their corresponding sides are in the same ration (or proportion) (OR)
Triangle with same shape and may not same size are called similar triangles.

Question 20.
A player sitting on the top of a tower of height 20 m observes the angle of depression of a ball lying on the ground as 60°. Draw a rough diagram for this situation.
Solution:

Section – II

Note:

1. Answer all the questions.
2. Each question carries 4 marks.

Question 21.
Two dice, one blue and one grey are thrown at the same time. What is the probability that the sum of the two numbers appearing on their tops is ?
i) 6
ii) 12
iii) 9
iv) 13
Solution:
P(E) \(=\frac{\text { No.of favourable out comes to E }}{\text { Total no.of possible out comes }}\)
Total possible out comes = 36
i) Number of favourable out comes for sum 6 = 5
P(sum 6) = \(\frac{5}{36}\)
ii) Number of favourable out comes for sum 12 1
P (sum 12) = \(\frac{1}{36}\)
iii) Number of favourable out comes for sum 9 = 4
P(sum 9) = \(\frac{4}{36}\) = \(\frac{1}{9}\)
iv) Number of favourable out comes for sum 13 = 0
P(sum 13) = \(\frac{0}{36}\) = 0

Question 22.
Write the formula to find the median of a grouped data and explain the terms involved in it.
Solution:
Median = l + \(\left[\frac{\frac{n}{2}-c f}{f}\right]\) × h
Here l = lower limit (boundary) of the median class
n = sum of frequencies
cf = cumulative frequency of the class preceding the median class
f = frequency of the median class
h = class size

Question 23.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.
Solution:
Volume of wood in the entire stand
= Volume of cuboid – 4 (volume of cone)

Question 24.
Find the value of k for the quadratic equation kx (x – 2) + 6 = 0, so that it has two equal real roots.
Solution:
kx (x – 2) + 6 = 0
kx² – 2kx + 6 = 0
As the quadratic equation has equal roots b² – 4ac = 0
Here a = k, b = -2k, c = 6
b² – 4ac = (-2k)² – 4(k) (6) = O
4k² – 24k = 0
4k (k – 6) = 0
4k = 0 or k – 6 = 0
k = 0 or k = 6
But k = 0 is not possible solution
∴ k = 6

Question 25.
Prove that \(\sqrt{\frac{1-\sin \mathrm{A}}{1+\sin \mathrm{A}}}\) = sec A – tan A.
Solution:

Question 26.
How many two digit numbers are divisible by 3 ?
Solution:
The two digit numbers that are divisible by 3 are 12, 15, 18, ………. 99
This is AP because common difference is same
First term a = 12,
common difference (d) = 15 – 12 = 3
Let n^th term a_n = 99
a + (n – 1) d = a_n
12 + (n – 1) 3 = 99
(n – 1) 3 = 99 – 12
(n – 1) = \(\frac{87}{3}\)
n – 1 = 29
n = 29 + 1 = 30
∴ There are 30 two digit numbers are divisible by 3.

Question 27.
Prove that the lengths of tangents drawn from external point to a circle are equal.
Solution:
Given: PQ, PR are tangents drawn from the external point p to the circle with centre ‘O’.
To prove : QP = PR
proof : In ∆OQP and ∆ORP

OQ = OR (radii of the same circle)
∠OQP = ∠ORP = 90° (tangent ⊥ radius)
OP = OP (common side)
∴ ΔOQP ≅ ΔORP (by RHS congruence criteria)
PQ = PR (CPCT)
The lengths of the tangents drawn from an external point to a circle are equal.

Question 28.
Observe the following graph and answer the following questions.

a) Name the shape of graph.
b) How many zeroes of the polynomial are there in the graph ?
c) What are the zeroes of the polynomial in the graph ?
d) Write the sum of zeroes.
Solution:
a) Parabola
b) Two
c) -1, 4
d) (-1) + 4 = 3

Section – IV

Note:

1. Answer all the questions.
2. Each question carries 8 marks.
3. There is an internal choice for each question.

Question 29.
a) Prove that \(\sqrt{7}\) is irrational.
(OR)
b) The diagonals of a quadrilateral ABCD intersect each other at the point 0 such that \(\frac{\mathrm{A} O}{\mathrm{~B} O}\) = \(\frac{\mathrm{C} O}{\mathrm{~D} O}\). Show that ABCD is a trapezium.
Solution:
a) Let us assume \(\sqrt{7}\) is a rational number
Then \(\sqrt{7}\) = \(\frac{\mathrm{a}}{\mathrm{~b}}\). (a, b are coprimes)
squaring on both sides
\((\sqrt{7})^2\) = \(\frac{\mathrm{a}^2}{\mathrm{~b}^2}\)
7b² = a²
⇒ a² is divisible by 7, a is also divisible by 7
Let a = 7c
by substituting we get
7b² = 49 c²
⇒ b² = 7c²
⇒ b² is divisible by 7, b is also divisible by 7
Therefore, a and b have atleast 7 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that \(\sqrt{7}\) is rational.
So we conclude that \(\sqrt{7}\) is irrational.

(OR)

b) Given : ABCD is a quadrilateral
\(\frac{\mathrm{A} O}{\mathrm{~B} O}\) = \(\frac{\mathrm{C} O}{\mathrm{~D} O}\) ⇒ \(\frac{\mathrm{A} O}{\mathrm{~C} O}\) = \(\frac{\mathrm{B} O}{\mathrm{~D} O}\)

RTP : ABCD is a trapezium
Construction: Draw OP || AB
Proof: (i) in ∆ADB
OP || AB (by construction)
⇒ \(\frac{\mathrm{DP}}{\mathrm{PA}}\) = \(\frac{\mathrm{DO}}{\mathrm{BO}}\) (By BPT theorem)

Question 30.
a) Find the co-ordinates of the points of trisection of the line segment joining (-3, -5) and (-6, -8).
(OR)
b) A cow is tied to a peg at one corner of a square shaped grass field of side 20 m by means of 6 m long rope. Find:
i) the area of that part of the field in which the cow can graze.
ii) the increase in the grazing area if the rope was 11.5 m long instead of 6 m.
Solution:
a)

(OR)

b) Area of field cow
can graze by means of
6 m long rope = Area of sector AQP

Area of field cow can graze by means 11.5m long rope = Area of sector ASR
= \(\frac{90}{360}\) × πr²
= \(\frac{1}{4}\) × \(\frac{22}{7}\) (11.5) (11.5)
= 103.91 m²
Increased Area = Area of sector ASR — Area of sector AQP
= 103.91 – 28.28
= 75.63 m²

Question 31.
a) One card is drawn from a well shuffled deck of 52 cards. Calculate the probability that the card will
i) be an ace
ii) not be an ace
iii) a face card and
iv) a spade.
(OR)
b) A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m and inclined at an angle of 60° to the ground. What should be the length of the slide in each case ?
Solution:
a) Event (E₁) : getting an ace
n(E₁) = 4
n(S) = 52
P(E₁) = \(\frac{\mathrm{n}\left(\mathrm{E}_1\right)}{\mathrm{n}(\mathrm{~S})}\) = \(\frac{4}{52}\) = \(\frac{1}{13}\)
Event (E₂): getting a card not to be an ace
n(E₂) = 48
n(S) = 52
P(E₂) = \(\frac{\mathrm{n}\left(\mathrm{E}_2\right)}{\mathrm{n}(\mathrm{~S})}\) = \(\frac{48}{52}\) = \(\frac{12}{13}\)
Event (E₃) : getting a face card
n(E₃) = 12
n(S) = 52
P(E₃) = \(\frac{\mathrm{n}\left(\mathrm{E}_3\right)}{\mathrm{n}(\mathrm{~S})}\) = \(\frac{12}{52}\) = \(\frac{3}{13}\)
Event (E₄): getting a spade
n(E₄) = 13
n(S) = 52
P(E₄) = \(\frac{\mathrm{n}\left(\mathrm{E}_4\right)}{\mathrm{n}(\mathrm{~S})}\) = \(\frac{13}{52}\) = \(\frac{1}{4}\)

(OR)

b) Case (i) :
From the ∆ABC
sin 30° = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
\(\frac{1}{2}\) = \(\frac{1.5}{\mathrm{AC}}\)
AC = 3m
Length of slide = 3m

Case (ii):
From ∆PQR
sin 60° = \(\frac{\mathrm{PQ}}{\mathrm{PR}}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{3}{\mathrm{PR}}\)
PR = \(\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\) = \(2 \sqrt{3}\)m
Length of slide = 2\(\sqrt{3}\)m

Question 32.
a) Find the mean for the following data (Using step deviation method).

(OR)
b) Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.
Solution:
a)

For finding x_i
For finding u_i
For finding f_iu_i
For finding Σ f_i, Σ f_iu_i
Mean = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right] \mathrm{h}\)
= 25 + \(\left[\frac{14}{100}\right] 10\)
= 25 + 1.4 = 26.4

(OR)

b) Given a₂ = 14
a₃ = 18
d = a₃ – a₂ = 18 – 14 = 4
a + d = 14
a + 4 = 14
a = 14 – 4 = 10
Now, S_n = \(\frac{\mathrm{n}}{2}\)[2a + (n-1)d]
S_n = \(\frac{51}{2}\)[2a + (n-1)d]
= \(\frac{51}{2}\)[2(10) + 50(4)]
= \(\frac{51}{2}\)[220] = 51(110) = 5610

Question 33.
a) Solve the following pair of linear equations graphically:
2x + y- 6 = 0
4x – 2y – 4 = 0
(OR)
b) Form the pair of linear equations in the following situation and find their solution graphically.
Lata bought two pencils and three chocolates for ₹ 9 and Suma bought one pencil and two chocolates for ₹ 5. Find the price of one pencil and that of one chocolate.
Solution:
a) (i) 2x + y – 6 = 0
y = -2x + 6

x = 0 y = -2(0) + (6)
x = 1 y = -2(1) + 6 = -2 + 6 = 4
x = 2 y = -2(2) + 6 = -4 + 6 = 4

(ii) 4x – 2y – 4 = 0
-2y = -4x + 4
2y = 4x – 4
y = 2x – 2

x = 0 y = 2(0) – 2 = -2
x = 1 y = 2(1) – 2 = 0
x = 2 y = 2(2) – 2 = 2

Solution:
x = 2
y = 2

(OR)

b) For scale:
Let cost of pencil = ₹ x
cost of chacolate = ₹ y
By the problem 2x + 3y = 9 → (1)
x + 2y = 5 → (2)
2x + 3y = 9
y = \(\frac{9-2 x}{3}\)

x + 2y = 5
y = \(\frac{5-x}{2}\)

Solution:
x = 3 Cost of pencil = ₹ 3
y = 1 Cost of chacolate = ₹ 1