# CBSE Sample Papers for Class 10 Maths Paper 9

## CBSE Sample Papers for Class 10 Maths Paper 9

These Sample Papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 9

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections – A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
Write down the decimal expansion of rational number $$\\ \frac { 14588 }{ 625 }$$

Question 2.
Check whether the following equation is a quadratic equation : (x – 2) (x + 1) = (x – 1) (x + 5).

Question 3.
In figure, O is the centre of a circle, AB is a chord and AP is the tangent at A. If ∠AOB = 120°, then find ∠BAP.

Question 4.
If 2 cosec² θ (1 – cos θ) (1 + cos θ) = k + 2, then find the value of k.

Question 5.
Find the area of a sector of a circle with radius 4 cm, if angle of the sector is 60°. (Use π = 3.14)

Question 6.
A bag contains cards numbered from 1 to 50. A card is drawn from the bag. Find the probability that the number on this card is divisible by 3 and 5.

SECTION-B

Question 7.
Show that 7√5 is irrational.

Question 8.
How many two-digit number are divisible by 5 ?

Question 9.
Find the point on the X-axis which is equidistant from (2, – 5) and (- 2, 9).

Question 10.
If cos A = $$\\ \frac { 3 }{ 5 }$$ then calculate sin A and cot A.

Question 11.
Prove the identity : $$\frac { cosA }{ 1+sinA } +\frac { 1+sinA }{ cosA } =2secA$$

Question 12.
Find the area of a quadrant of a circle whose circumference is 44 cm. (Use $$\pi =\frac { 22 }{ 7 }$$)

SECTION-C

Question 13.
Show that any positive odd integer is of the form (6q + 1) or (6q + 3) or (6q + 5) where q is some integer.

Question 14.
Solve for x and y : $$\frac { 10 }{ x+y } +\frac { 2 }{ x-y } =4\quad and\quad \frac { 15 }{ x+y } -\frac { 5 }{ x-y } =-2$$

OR

Solve for x : $$\frac { 1 }{ x+1 } +\frac { 3 }{ 5x+1 } =\frac { 5 }{ x+4 } ;x\neq -1,-\frac { 1 }{ 5 } ,-4$$

Question 15.
Find the roots of the equation 2x² + x – 4 = 0 by the method of completing the square.

OR

If ad ≠ bc, then prove that the equation (a² + b²) x² + 2(ac + bd)x + (c² + d²) = 0 has no real roots.

Question 16.
For what value of n are the nth terms of two A.Ps : 63, 65, 67, … and 3,10,17, … equal ?

OR

The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 400. Find the number of terms and the common difference of the A.P.

Question 17.
Prove the identity : (cosec A – sin A) (sec A – cos A) = $$\frac { 1 }{ tanA+cotA }$$

Question 18.
In figure, PQ = 12 cm, PR = 9 cm and O is the radius of the circle. Find the area of the shaded region. (Use π = 3.14)

OR

In figure, are shown two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre O and radius OP while arc PBQ is a semi-circle drawn on PQ as diameter with centre M. If OP = PQ = 10 cm show that area of shaded region $$25\left( \sqrt { 3 } -\frac { \pi }{ 6 } \right)$$ cm²

Question 19.
Prove that the parallelogram circumscribing a circle is a rhombus.

Question 20.
In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C, is intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

Question 21.
A card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:

1. a king of black colour,
2. a face card,
3. queen of hearts.

Question 22.
The given distribution shows the marks, out of 50 obtained by 100 students in a test:

Find the mean and median

SECTION-D

Question 23.
Find the 40th term of the A.P. 8,10,12, …, if it has a total of 40 terms and hence find the sum of its last 10 terms.

Question 24.
The total cost of a certain length of a piece of cloth is Rs 100. If the piece was 5 m longer and each metre of cloth costs Rs 1 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per metre ?

OR

A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how many days will B take to finish it ?

Question 25.
Find the coordinates of the points of trisection of the line segment joining (4, – 1) and (- 2, – 3).

OR

In what ratio does the point $$\left( \frac { 24 }{ 11 } ,y \right)$$ divide the line segment joining the points P(2, – 2) and Q(3, 7) ? Also find the value of y.

Question 26.
Prove that the lengths of the tangents drawn from external point to a circle are equal.

Question 27.
Construct a triangle ABC in which AB = 8 cm, BC = 10 cm and AC = 6 cm. Then construct another triangle whose sides are $$\\ \frac { 3 }{ 5 }$$ of the corresponding sides of ∆ABC.

Question 28.
The angles of depression of the top and bottom at a 12 m tall building from the top of a multi-storeyed building are 30° and 60° respectively. Find the height of the multi-storey building.

OR

An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and 60° respectively. Find the width of the river. (Use √3 = 1.732)

Question 29.
A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of base of the cone is 21 cm and its volume is $$\\ \frac { 2 }{ 3 }$$ of the volume of the hemisphere. Calculate the height of the cone and the surface area of the toy. (Use $$\pi =\frac { 22 }{ 7 }$$)

Question 30.
A box contains cards bearing numbers from 5 to 69. If one card is drawn at random from the box, find the probability that it bears :
(i) a one-digit number,
(ii) a number divisible by 5,
(iii) an odd number less than 30,
(iv) a prime number between 50 to 69.

SOLUTIONS

SECTION-A

Solution 1.
Let
$$x=\frac { 14588 }{ 625 } =\frac { 14588 }{ { 5 }^{ 4 } }$$
= $$\frac { 14588\times { 2 }^{ 4 } }{ { 2 }^{ 4 }\times { 5 }^{ 4 } } =\frac { 14588\times { 2 }^{ 4 } }{ 10^{ 4 } }$$
= $$\frac { 233408 }{ { 10 }^{ 4 } }$$
= 23.3408
So, decimal expression is 23.3408

Solution 2.
We have,
(x – 2)(x + 1) = (x -1) (x + 5)
x² – 2x + x – 2 = x² + 5x – x – 5
x² – x – 2 = x² + 4x – 5
– 5x + 3 = 0
Since, the above equation is not of the form ax² + bx + c = 0
Therefore, the given equation is not a quadratic equation.

Solution 3.
Given
∠AOB = 120° [Radii]
Since AO = OB
∴ ∆AOB is an isosceles triangle.
=> ∠OAB = ∠OBA [cpct]
∴ In ∆ABOB
∠OAB + ∠AOB + ∠OBA =180° [Angle sum property]
=> 2∠OAB + 120° = 180°
=> ∠OAB = ∠OBA = 30°
Now, ∠OAP = 90° [∴ Tangent AP ⊥ radius OA]
Thus, ∠BAP = ∠OAP – ∠OAB
= 90° – 30° = 60°
Hence, ∠BAP = 60°. Ans.

Solution 4.
We have,
2 cosec² θ(1- cos θ) (1 + cos θ) = k + 2
=> 2 cosec² θ (1 – cos² θ) = k + 2 [∵ (a + b)(a – b) = a²-b²)]
=> 2 cosec² θ sin² θ = k + 2 [∵ 1 – cos² θ = sin² θ]
=> $$2\times \frac { 1 }{ { sin }^{ 2 }\theta } \times { sin }^{ 2 }\theta =k+2$$
=> 2 = k + 2
=> k = 0 Ans.

Solution 5.
Given sector is OAPB.
Radius of circle, r = 4 cm
Angle of sector, θ = 60°
Now,

Hence, the area of the sector is 8.37 cm².

Solution 6.
Total number of possible outcomes = 50 = n(S)
Let E be the event that the number on the card drawn is divisible by both 3 and 5.
∴ E = {15, 30,45}
∴ n(E) = 3
∴ $$P(E)=\frac { n(E) }{ n(S) } =\frac { 3 }{ 50 }$$

SECTION-B

Solution 7:
Let us assume that 7√5 is rational.
Therefore, we can find co-prime a and b(≠ 0) such that
7√5 = $$\\ \frac { a }{ b }$$
or
√5 = $$\\ \frac { a }{ 7b }$$
Now, since a, b and 7 are integers, therefore $$\\ \frac { a }{ 7b }$$ is rational and so √5 is rational, which is not true as √5 is irrational.
Hence, we conclude that 7√5 is irrational.

Solution 8:
Two-digit numbers that are divisible by 5 are
10,15,20,…, 90,95.
It is an A.P. with a = 10, d = 15 – 10 = 5 and an = 95
We know an = a + (n – 1 )d
=> 95 = 10 + (n – 1)5
=> 85 = (n – 1)5
=> 17 = n – 1
=> n =18
So, there are 18 two-digit numbers that are divisible by 5.

Solution 9:
Let the required point on the X-axis be (a, 0).
Then, according to question,
Distance of (a, 0) and (2, – 5) = Distance of (a, 0) and ( – 2, 9)
=> $$\sqrt { { (2-a) }^{ 2 }+{ (-5-0) }^{ 2 } } =\sqrt { { (-2-a) }^{ 2 }+{ (9-0) }^{ 2 } }$$
Squaring on both sides, we get
(2 – a)² + 5² = (- 2 – a)² + 9²
=> 4 + a² – 4a + 25 = 4 + a² + 4a + 81
=> – 8a = + 56
=> a = – 7
∴ ( – 7, 0) is the point which is equidistant from (2, – 5) and (- 2, 9).

Solution 10:
Let ABC be a triangle, then
$$cosA=\frac { AB }{ AC } =\frac { 3 }{ 5 }$$
Let AB = 3k and AC = 5k

Solution 11:
L.H.S = $$\frac { cosA }{ 1+sinA } +\frac { 1+sinA }{ cosA }$$
$$\frac { { cos }^{ 2 }A+{ (1+sinA) }^{ 2 } }{ (1+sinA)cosA }$$

Solution 12:
Let r be the radius of the circle,
Given
circumference = 2πr = 44 cm

SECTION-C

Solution 13:
Let k be any integer.
Using Euclid’s division lemma with k and 6, we get
k = 6q + r, where 0 ≤ r ≤ 6
For r = 0
k = 6q i.e., even number
For r = 1
k = 6q + 1 i.e., odd number
For r = 2
k = 6q + 2 i.e., even number
For r = 3
k = 6q + 3 i.e., odd number
For r = 4
k = 6q + 4 i.e., even number
For r = 5
k = 6q + 5 i.e., odd number
Therefore, any positive odd integer is of the form (6q + 1) or (6q + 3) or (6q + 5).

Solution 14:
Given equations are,
$$\frac { 10 }{ x+y } +\frac { 2 }{ x-y } =4$$
and $$\frac { 15 }{ x+y } -\frac { 5 }{ x-y } =-2$$

Solution 15:
Given equation is,
2x² + x – 4 = 0
On multiplying the equation by 1/2 ,we get

Solution 16:
First A.P. is 63, 65, 67, …
Here, a = 63
d = 65 – 63 = 2
.’. nth term, an = a + (n – 1)d
= 63 + (n – 1) 2
= 63 + 2n – 2
= 61 + 2n

Solution 17:
L.H.S = (cosec A – sin A) (sec A – cos A)
= $$\left( \frac { 1 }{ sinA } -sinA \right) \left( \frac { 1 }{ cosA } -cosA \right)$$

Solution 18:
Since, the angle in a semi-circle is a right angle
∠RPQ = 90°
Thus, PQR represents a right angled triangle
where PR = 9 cm, PQ = 12 cm
In ∆PQR,using Pythagoras theorm

Solution 19:
Consider a parallelogram ABCD circumscribing a circle. Sides AB, BC, CD and AD touch the circle at points P, Q, R and S respectively.

Since the lengths of the tangents drawn from an external point to a circle are equal.
AP =AS …(i)
PB =BQ …(ii)
DR =DS …(ii)
CR = CQ …(iv)
On adding all equations, we get
(AP + PB) + (DR + CR) = (AS + DS) + (BQ + CQ)
=> AB + CD = AD + CB
=> 2AB = 2BC [∵ in a parallelogram, opposite sides are equal]
=> AB =BC
Since, adjacent sides of a parallelogram are equal
∴ ABCD is a rhombus.

Solution 20:
Given : XY and X’Y’ are parallel.
Tangent AB is another tangent which touches the circle at C.

Solution 21:
Total number of outcomes = 52 = n(S).
(i) Let E denote the event of getting a king of black colour.
There are two cards in this category i.e., king of club and king of spade.

Solution 22:

SECTION-D

Solution 23:
The given A.P. is 8,10,12,…
So, a = 8, d = 10 – 8 = 2.
We know that an = a + (n – 1 )d
So, 40th term = a40 = 8 + 39 x 2
= 8 + 78
= 86
Hence, the 40th term of the given A.P. is 86.
Now, since the A.P. has a total of 40 terms. So, to find the sum of last n terms, we take

Hence, the sum of last 10 terms of the given A.P. is 770.

Solution 24:
Let the length of the piece of the cloth be x m.
Total cost of the piece of cloth = Rs 100
Then cost per metre = Rs $$\\ \frac { 100 }{ x }$$
New length = (x + 5) m
Since the cost of the piece of cloth remains unchanged

Solution 25:
We have, A(4, -1) and B(- 2, – 3)
Let P and Q be the points of trisection of A and B.
i.e., AP = PQ = QB

Solution 26:
Given, PA and PB are two tangents drawn from an external point P to the circle C(0, r).
Construction: Join OA and OB.
To prove : PA = PB
Proof : Since, tangent at any point to a circle is perpendicular to the
radius through the point of contact
PA ⊥ OA and PB ⊥ OB

Solution 27:
Steps of the construction :
(1) Draw a line segment BC = 10 cm.
(2) Taking B point as centre and radius 8 cm, draw an arc.
(3) Similarly, taking point C as centre and radius equal to 6 cm, draw an arc. Let the two arcs intersect at point A.
(4) Join AB, AC.
Thus, ∆ABC is obtained.
(5) Draw a ray BX making an acute angle with line BC on the opposite side of vertex A.
(6) Mark 5 points B1 ,B2 ,B3, B4 and B5 on ray BX such that
BB1 = B1B2 = B2B3 = B3B4 = B4B5.
(7) Join CB5.
(8) Through B3, draw a line parallel to CB5 intersecting BC at point C.
(9) Draw a line parallel to line AC through C’ intersecting AB at A’.
Thus, ∆A’BC’ is obtained whose sides are $$\\ \frac { 3 }{ 5 }$$ of corresponding sides of ∆ABC.

Solution 28:
Let AB be the tall building of height 12 m and CD be the multi-storey building.
Let CD = h m
Given: ∠DBE = 30° and ∠DAC = 60°
Here, AB = CE = 12 m
DE = CD – CE = (h – 12)m

Solution 29:
Let
h = height of the cone
Radius of the cone = Radius of hemisphere
= 21 cm

Solution 30:
Numbers of possible outcomes = 69 – 5 + 1 = 65
n(S) = 65
(i) Let A be the event of getting one digit number.
Favourable outcomes = {5,6,7,8,9}
n(A) = 5

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