AP 10th Class Maths Model Paper Set 8 with Solutions

Regularly solving AP 10th Class Maths Model Papers Set 8 contributes to the development of problem-solving skills.

AP SSC Maths Model Paper Set 8 with Solutions

Time : 3.15 Hours]
[Max. Marks : 100

Instructions:

1. In the duration of 3 hrs. 13 minutes, 15 minutes of time is allotted to read the question paper.
2. All 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 phrase.
2.  Each question carries 1 mark.

Question 1.
Find the prime factorization of 30. (Ch.No-1)
Solution:
30 = 2 × 15
= 2 × 3 × 5
∴ 2, 3, & 5 are the prime factors of 30.

Question 2.
Assertion : Sum of the zeroes of a Quadratic polynomial (Ch.No-2)
2x² – 3x – 4 is \(\frac{-3}{2}\).
Reason: Sum of the zeroes of a Quadratic polynomial
ax² + bx + c is \(\frac{\mathrm{c}}{2}\)
Now, choose the correct answer 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 true, but the Reason is false.
D) Assertion is false, but the reason is true.
Solution:
C) Assertion is true, but the Reason is false.

Question 3.
The general form of linear equation in two variables is ……… (Ch.No-3)
Solution:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

Question 4.
If n^th terms of an A.P is an = 2n – 6 then (Ch.No-5)
Match the following.

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

Question 5.
Statement – I: All similar triangles are congruent. (Ch.No-6)
Statement – II: All right angled isosceles triangles are similar.
Now, choose the correct answer.
A) Both statements are true.
B) Statement I is true and Statement II is false.
C) Statement I is false and statement II is false.
D) Both statements are false.
Solution:
C) Statement I is false and Statement II is false.

Question 6.
A person standing 20 meters away from the base of a building observes that the elevation to the top of the building is 45° then the height of the building is ……… (Ch.No-9)
Solution:
In triangle ABC,
AB is the base of a building BC is the height of a building

Question 7.
How many tangents can a circle have ? (Ch.No-10)
Solution:
Infinite.

Question 8.
Draw a rough figure of cylinder with height h cm and base radius r cm. (Ch.No-12)
Solution:

Question 9.
If p (E) = 0.05, what is the probability of ‘not E’ ? (Ch.No-14)
Solution:
Given P (E) = 0.05
we know that
P(E) + P \((\overline{\mathrm{E}})\) = 1
⇒ P \((\overline{\mathrm{E}})\) = 1 – P (E)
= 1 – 0.05
∴ P \((\overline{\mathrm{E}})\) = 0.95

Question 10.
Zero of the polynomial of ax + b is
A) \(\frac{\mathrm{b}}{\mathrm{a}}\)
B) \(\frac{\mathrm{a}}{\mathrm{b}}\)
C) \(\frac{-\mathrm{a}}{\mathrm{~b}}\)
D) \(\frac{-\mathrm{b}}{\mathrm{a}}\)
Solution:
D) \(\frac{-\mathrm{b}}{\mathrm{a}}\)

Question 11.
If 4 cot A = 3 then tan A = (Ch.No-8)
A) \(\frac{3}{5}\)
B) \(\frac{4}{5}\)
C) \(\frac{4}{3}\)
D) \(\frac{3}{4}\)
Solution:
C) \(\frac{4}{3}\)

Question 12.
If x = \(\frac{1}{\mathrm{x}}\) then the roots are (Ch.No-4)
A) 1
B) -1
C) A, B
D) None
Answer:
C) A, B

Section – II
[8 × 2 = 16 M]

Note :

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

Question 13.
Find the volume of a cylinder with radius of base 6 cm and height 7 cm. (Ch.No 12)
Solution:
Given Radius (r) = 6 cm
Height (h) = 7 cm
The volume of a cylinder = πr²h

∴ Hence, the volume of a cylinder is 793 cm³.

Question 14.
Find a Quadratic polynomial whose sum and product of the zeroes are 3 and 2 respectively. (Ch.No-2)
Solution:
Let the quadratic polynomial be ax² + bx + c, and its zeros be α and β we have sum of the zeros α + β = 3 product of the zeros αβ = 2
The quadratic polynomial is,
x² – (α + β) x + αβ
⇒ x² – 3x + 2.

Question 15.
Check whether the following are Quadratic Equations or not. (Ch.No-4)
i) (x – 2)² + 1 = 2x – 3
ii) x (x + 1) + 8 = (x + 2) (x – 2)
Solution:
i) Given equation is,
(x – 2)² + 1 = 2x – 3
⇒ x² + 4 – 4x + 1 = 2x – 3
⇒ x² – 4x + 5 – 2x + 3 = 0
⇒ x² – 6x + 8 = 0
∴ It is a quadratic equation.

ii) Given equation is,
x (x + 1) + 8 = (x + 2) (x – 2)
⇒ x² + x + 8 = x² – (2)²

⇒ x + 8 + 4 = 0
x + 12 = 0
∴ It is not a quadratic equation.

Question 16.
Give an example for (Ch.No-6)
i) Similar figures
ii) Non similar figures
Solution:
i) Any two squares

ii) Triangle and parallelogram

Question 17.
Find the coordinates of mid point of the line segment joining (cos 0, 0) and (0, sin 90°). (Ch.No-7)
Solution:

Question 18.
Express the ratios cos A and tan A in terms of sin A. (Ch.No-8)
Solution:

Question 19.
Draw a diagram for the following situation. (Ch.No-9)
A boy observed the top of an electric pole at an angle of elevation of 60° when the observation point is 8 meters away from the foot of the pole.
Solution:

Question 20.
Calculate the length of tangent from a point 15 cm away from the centre of a circle of radius 9 cm. (Ch.No 10)
Solution:
Given radius (OA) = 9 cm
Centre (OQ) = 15 cm
We know that tangent to a circle makes a right angle with radius ∠OPQ = 90°
By pythagoras we have
OQ² = OP² + PQ²
⇒ (15)² = (9)² + PQ²
⇒ PQ² = 225 – 81
⇒ PQ² = 144
⇒ PQ = \(\sqrt{144}\) = 12 cm
∴ Hence the length of a tangent = 12 cm.

Section – III
[8 × 4 = 32 M]

Note:

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

Question 21.
One card is drawn from a well – shuffled deck of 52 cards. Find the probability of getting
i) a king of black colour (Ch.No-14)
ii) a red face card.
Solution:
i) A king of Black colour
Favourable outcome = 2

ii) A red face card
Favourable out come = 6

Question 22.
Write the formula to find the mode of grouped data and explain the terms involved in it. (Ch.No-13)
Solution:
Mode = l + \(\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right]\) × h
l = lower limit of the modal class.
h = size of the class interval.
f₁ = frequency of the modal class.
f₀ = frequency of the class preceding the modal class.
f₂ = frequency of the class succeeding the modal class.

Question 23.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of n. (Ch.No 12)
Solution:
Given r = 1 cm
h = 1 cm
Now, volume of solid volume of conical part + volume of hemišpherical part
We know that,
the volume of cone = \(\frac{1}{3}\) πr²h
The volume of the hemisphere = \(\frac{2}{3}\) πr³
Volume of solid = \(\frac{1}{3}\) πr²h + \(\frac{2}{3}\) πr³

Question 24.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let the first number be x.
and the second number be 27 – x
The product of two numbers = 182
⇒ x (27 – x) = 182
⇒ 27x – x² = 182
⇒ x² – 27x + 182 = 0
⇒ x² – 13x – 14x + 182 = 0
⇒ (x – 13) – 14 (x – 13) = 0
⇒ (x – 13) (x – 14) = 0
x – 13 = 0 x – 14 = 0
∴ x = 13 x = 14
∴ x = 13 or x = 14
∴ First number = 13
second number = 27 – 13 = 14
∴ Hence, the numbers are 13 and 14.

Question 25.
Prove that

Solution:

Question 26.
Find the sum of odd numbers between 0 and 50. (Ch.No-5)
Solution:
The odd numbers between 0 and 50 are 1, 3, 5, 7, 9, ….. 49.
Here, First term a = 1
Common difference d = a₂ – a₁
= 3 – 1 = 2
last term l = 49.
We know that
l = a + (n – 1)d
⇒ 49 = 1 + (n – 1)2
⇒ 49 = 1 + 2n – 2
⇒ 2n – 1 = 49
⇒ 2n = 49 + 1
⇒ 2n = 50
⇒ n = \(\frac{50}{2}\) = 25
∴ n = 25
we know the sum of nth term is, n

Question 27.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre. (Ch.No-10)
Solution:
Let, ‘O’ is the centre of the given circle A tangent PR has been drawn touching the circle at point P.
Draw QP ⊥ RP at point P, such that point Q lies on the circle.
∠OPR = 90°
Also ∠QPR = 90°
∴ ∠OPR =∠QPR

∴ Now, the above cases is possible only when centre O lies on the line QP.
∴ Hence, perpendicular at the point of contact to the tangent to a circle passes through the center of the circle.

Question 28.
Due to heavy storm an electric wire got bent as shown in the figure. It followed a mathematical shape. Answer the following questions below. (Ch.No-2)

a) Name the shape in which the wire is bent.
b) How many zeroes are there for the polynomial (Shape of the wire)
c) The zeroes of the polynomial are
d) Sum of the zeroes of the polynomial
Solution:
a – parabola
b – 2
c – -1, 3
d – 2.

Section – IV
[5 × 8 = 40 M]

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 2 + 5\(\sqrt{3}\) is irrational.
(OR)
b) Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. (Ch.No-6)
Solution:
a) Let us assume that 2 + \(\sqrt{3}\) is a rational number.
So, 2 + 5\(\sqrt{3}\) = \(\frac{a}{b}\) (a, b are co – primes)
∴ 5\(\sqrt{3}\) = \(\frac{a}{b}\) – 2
∴ 5\(\sqrt{3}\) = \(\frac{a-2 b}{b}\)
∴ \(\sqrt{3}\) = \(\frac{a-2 b}{5 b}\)
Since, \(\frac{a-2 b}{5 b}\) ‘is rátional and \(\sqrt{3}\) is also rational. But it is given that \(\sqrt{3}\) is an irrational number. By our assumption is wrong, which is a contradiction.
∴ Hence, 2 + 5\(\sqrt{3}\) is an irrational number.

(OR)

b) Proof: We are given a triangle ABC.
Construction : If DE || BC which intersects other two sides AB at D and AC at. R.
RTP : \(\frac{A D}{D B}\) = \(\frac{A E}{E C}\)

Let us join BE and CD and then draw DM ⊥ AC and EN ⊥ AB.

Here, Δ BDE and Δ DEC are on the same base DE and between the same parallels BC and DE.
So, ar (BDE) = ar (DEC) …….. (3)
From (1), (2) & (3) we have
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{AC}}\)

Question 30.
a) Find the area of a rhombus if its vertices are (-4, -7), (-1, 2), (8, 5) and (5, -4) taken in order.
(Ch.No-7)
(OR)
b) A horse is tied to a peg at one corner of a square shaped grass field of side 15m by means of a 5 m long rope. Find (Ch.No-11)
i) the area of that part of the field in which the horse can graze.
ii) the increase in the grazing area if the rope was 10 m log instead of 5 m.
Solution:
a) Let the given points are
A = (-4, -7), B = (-1, 2), C = (8, 5) and D = (5, -4).
Its taken in an order.
Length of the diagonals AC and BD.

(OR)

b) As the horse is tied at one end of a square field, it will graze only a quarter of the field with radius 5 m.
Here, the length of rope will be the radius of a circle i.e., r = 5m

It is also known that the side of square field = 15 m.

i) Area of circle = πr²
= \(\frac{22}{7}\) × (5)²
= \(\frac{22}{7}\) × 25
= \(\frac{550}{7}\)
= 78.5 m²
Now, the area of the part of the field where the horse can graze = \(\frac{1}{4}\) (area of circle)
= \(\frac{1}{4}\)(78.5)
\(\frac{78.5}{4}\)
= 19.625 m²

ii) If the rope is increased to 10 m Area of circle = πr²
= \(\frac{22}{7}\) × (10)²
= (3.14) (100)
= 314 m²
Now, the area of the part of the field where the horse can graze
= \(\frac{1}{4}\) (area of circle)
= \(\frac{314}{4}\)
= 78.5 m²
∴ Increasing in the grazing area
= 78.5 – 19.625
= 58.875 m²
∴ Hence, the rope is increasing to 10 m grazing area is 58.875 m².

Question 31.
a) A box contains 100 discs which are numbered from 1 to 100. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number,
(ii) a perfect square,
(iii) a number divisible by 5,
(iv) a number divisible by 10. (Ch.No-14)
(OR)
b) The angles of depression of the top and bottom of an 8m tall building from the top of a multi – storeyed building are 30° and 45° respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (Ch.No-9)
Solution:
a) The total number of discs = 100
P(E) \(=\frac{\text { No.of favourable outcomes }}{\text { Total no. of outcomes }}\)
Total no. of outcomes

i) Total number of discs having two digit numbers = 90
Since 1 to 9 are single – digit numbers
So, total 2 – digit numbers are 99 – 9 = 90.
p (bearing a two digit number)

= \(\frac{9}{10}\) = 0.9

ii) Total number of perfect square numbers = 10

iii) Total number which are divisible by 5 = 20.

iv) Total numbers which are divisible by 10 = 40.

(OR)

b) PB denotes the Multi-storyed building and AD denotes the 8m tall building.

Observe PD is a transversal to the parallel lines PQ & DC.
∴ ∠QPD and ∠PDC are alternate angles and so are equal, so ∠PDC = 30° & ∠PAB = 45°.
In right ∆ PDC, we have
Tan θ = \(\frac{\mathrm{DC}}{\mathrm{PC}}\)
Tan 30° = \(\frac{\mathrm{DC}}{\mathrm{PC}}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{DC}}{\mathrm{PC}}\)
∴ PC = \(\sqrt{3}\)DC.
In right ∆ PAB, we have
Tan θ = \(\frac{\mathrm{AB}}{\mathrm{PB}}\)
Tan 45° = \(\frac{\mathrm{AB}}{\mathrm{PB}}\)
⇒ 1 = \(\frac{\mathrm{AB}}{\mathrm{PB}}\)
∴ AB = PB
AB = PC + BC
∴ PB = PC + BC
Since, AB = DC and BC = AD = 8m
We get, PC + BC = PB
PC + 8 = PB
8 + \(\sqrt{3}\) DC = AB = DC

So, the height of the multi-storyed building is PC + 8 = PB.
⇒ 4(\(\sqrt{3}\) + 1) + 8 = PB = 4(3 + \(\sqrt{3}\)) m.
The distance between the two buildings is also 4(3 + \(\sqrt{3}\))m.

Question 32.
a) The distribution below gives the weights of 30 students of a class. Find the median weight of the students. (Ch.No-13)

(OR)
b) If the sum of first 7 terms of an A.P is 49 and that of 17 terms is 289, find the sum of first n terms. (Ch.No-5)
Solution:
a)

(OR)

b) Given that
S₇ = 49
S₁₇ = 289
We know that,

Question 33.
a) Solve the following pair of linear equations graphically. (Ch.No-3)
2x + y – 5 = 0
3x – 2y – 4 = 0
(OR)
b) Form the pair of linear equations in the following situation and find their solution
graphically. (Ch.No-3)
3 pens and 4 pencils together coast Rs. 44 whereas 4 pens and 3 pencils together cost Rs. 47.
Solution:
a) Given linear equations are
2x + y – 5 = 0 ……. (1)
3x – 2y – 4 = 0 …… (2)
From eq (1), we have,

From eq (2), we have,


It intersets the point (2,1)
∴ Hence, the solution of a equations are x = 2 & y = 1

(OR)

b) Let the cost of pen is x
The cost of pencil is y
3x + 4y = 44 ….. (1)
4x + 3y = 47 ….. (2)
From eq (1), we have

From eq (2), we have


It intersects the points (8, 5)
∴ Hence the solution of a equations are x = 8, & y = 5
∴ Hence the cost of pen is ₹ 8
The cost of pencil is ₹ 5