# CBSE Sample Papers for Class 10 Maths Paper 2

## CBSE Sample Papers for Class 10 Maths Paper 2

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 2.

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.
What are the coordinates of the centroid of the triangle ABC if A(2,3), B(5, 8) and C(2, 1) ? [1]

Question 2.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. Find the radius of the circle. [1]

Question 3.
Express 4 as a sum of two irrational numbers.  [1]

Question 4.
Check if the system 3x – y – 8 = 0, 4x – 12y + 16 = 0 has infinite solution.  [1]

Question 5.
Check if x = 1 is a solution of 3x2 – 2x – 1 = 0.  [1]

Question 6.
∆PQR is right angled at Q. If tan R = $$\frac { 24 }{ 7 }$$, find sec R and sin R.  [1]

SECTION-B

Question 7.
If the points A(1, 2), O(0, 0) and C(a, b) are collinear, then find the relationship between
‘a’ and ‘b’.  [2]

Question 8.
There is a circular path around a sports field. Sania takes 24 minutes to drive one round of the field, while Ravi takes 16 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point ?  [2]

Question 9.
What is the probability that a number selected from the numbers 1, 2, 3,…., 25 is a prime number, when each of the given numbers is equally likely to be selected ?  [2]

Question 10.
If a number x is chosen at random from the numbers -2,-1, 0, 1, 2, what is the probability that x2 < 2 ? [2]

Question 11.
Find the values of for which x2 – 2x + k = 0 has real roots.  [2]

Question 12.
Write the value of a30 – a10 for the A.P. 4, 9, 14, 19, 24,….  [2]

SECTION-C

Question 13.
If the point A(2, – 4) is equidistant from P(3, 8) and Q(- 10, y), find the value of y. Also find the distance PQ. [3]

Question 14.
If ∆ABC ~ ∆DEF and AL and DM are their corresponding angle bisector segments then show that
$$\frac { AL }{ DM } =\frac { AB }{ DE } =\frac { BC }{ EF } =\frac { AC }{ DF }$$  [3]

Question 15.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.  [3]

Question 16.
Find the LCM of 396, 429 and 561 by prime factorization method.  [3]

Question 17.
The diameters of the front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that rear wheel will make to cover the distance which the front wheel covers in 140 revolutions.  [3]
OR
In the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region.

Question 18.
A rectangular water tank of base 11 m x 6 m contains water up to a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.  [3]
OR
A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.  [3]

Question 19.
Find the zeroes of the quadratic polynomial x2 + 13x + 30, and verify the relationship between the zeroes and the coefficients.  [3]

Question 20.
The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum ?  [3]
OR
How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3 ?

Question 21.

Question 22.
Find median for the following data :  [3]

 Classes 10-29 30-49 50-69 70-89 90 -109 Frequency 18 4 6 4 8

SECTION-D

Question 23.
Construct a triangle ABC in which AC = AB = 4.5 cm and ∠A = 90°, then constuct a triangle similar to ∆ABC with its sides equal to ($$\frac { 5 }{ 4 }$$)th of the corresponding sides of ∆ABC.  [4]

Question 24.
The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metre towards the foot of the tower to a point B, the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from the
point A.  [4]
OR
As observed from the top of a lighthouse, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 45°. Determine the distance travelled by the ship during the period of observation. [4]

Question 25.
In figure, P is the mid-point of BC and Q is the mid-point of AP. If BQ when produced meets AC at R, prove that RA = $$\frac { 1 }{ 3 }$$CA  [4]

Question 26.
A milk container of height 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of ? 22 per litre which the container can hold.  [4]
OR
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.

Question 27.
In a rectangle if length is increased and breadth is reduced by 2 metre, the area is reduced by 28 sq. m. If length is reduced by 1 metre and breadth is increased by 2 metre, the area increased by 33 sq. m. Find the length and breadth of the rectangle.  [4]
OR
Out of group of swans, $$\frac { 7 }{ 2 }$$ times the square root of the total number are playing on the shore of a pond. The remaining two are swimming in water. Find the total number of swans.  [4]

Question 28.
The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books ?  [4]

Question 29.
Show $$\frac { { cot }^{ 2 }A }{ 1+cosecA } +1=cosec A$$  [4]

Question 30.
Find the value of p, if the mean of the following distribution is 7.5.  [4]

 X 3 5 7 9 11 13 f 6 8 15 p 8 4

SOLUTIONS
Section – A

Here A(2,3) B(5,8) and C(2,1).

Let, radius of circle OP = r
∵ Tangent to a circle is perpendicular to the radius
∴ In ∆POQ, using Pythagoras theorem

OQ² = PQ² + OP²
25² = 24² + r²
⇒ r² = 25² – 24²
⇒ r² = (25 + 24)(25 – 24)
⇒ r² = 49
⇒ r = 7 cm

We know that (2 + √3) and (2 – √3) are irrational numbers
(2 + √3) + (2 – √3)
2 + √3 + 2 – √3 = 4

Given system of equations is,
3x – y – 8 =0
and 4x – 12y + 16 = 0
For infinite solution

We have
3x² – 2x – 1 = 0
If x = 1,
Then,
3 × (1)2 – 2 × 1 – 1 = 0
3 – 2 – 1 = 0
3 – 3 = 0
0 = 0
L.H.S = R.H.S
Hence, x = 1 is a solution of the given equation.

We have,
tan R =$$\frac { 24 }{ 7 }$$
From ∆PQR, tan R = $$\frac { PQ }{ RQ }$$
In ∆PQR, PR² = PQ² + RQ²
PR² = 24² + 7²
PR² = 576+ 49
PR² = 625
PR = √625
PR 25

sec R = $$\frac { PR }{ RQ } =\frac { 25 }{ 7 }$$
sin R = $$\frac { PQ }{ PR } =\frac { 24 }{ 25 }$$

SECTION-B

We have, A( 1, 2), O(0, 0) and C(a, b) are collinear.
Since, the given points are collinear, therefore, area of triangle formed by them is zero.
i.e., ar (∆AOB) = 0
⇒ $$\frac { 1 }{ 2 }$$ x1(y2 – y3) + x2(y3 – yx) + x3{yx – y2) 1=0
⇒ $$\frac { 1 }{ 2 }$$ 1(0 – b) + 0(b – 2) + a(2 – 0) i =0
⇒ – b + 2a = 0
⇒ b = 2a.

Given, Sania takes 24 minutes for one round and Ravi takes 16 minutes for one round. It means that Ravi takes lesser time than Sonia to drive one round of field.
Now,
Required time = L.C.M. of the time taken by Sonia and Ravi for completing one round
= L.C.M. of 24 and 16
∵ 24 = 23 x 3 and 16 = 24
∴ L.C.M. = 24 x 3 = 16 x 3 = 48
Hence, Ravi and Sonia will meet each other at the starting point after, 48 minutes.

Numbers are 1, 2, 3,…., 25.
∴ Total numbers n(S) = 25
Prime numbers between 1 to 25 are 2, 3, 5, 7, 11, 13, 17, 19, 23
⇒ Total prime numbers = n(E) = 9
∴ P(Prime Number) = $$\frac { n(E) }{ n(S) } \quad \frac { 9 }{ 25 }$$

Given numbers are -2,-1, 0, 1, 2
⇒ Total numbers = 5
Here, (-2)2 = 4, (-1)2 = 1, 02 = 0, 12 = 1, 22 = 4
For x = – 1, 0, 1  x< 2
⇒ Favourable outcomes = 3
p(x² < 2) = $$\frac { Favourable\quad outcomes }{ Total\quad outcomes } = \frac { 3 }{ 5 }$$

Given equation is,
x² – 2x + k = 0
Here a = 1, b = -2, c = k
We know, D = b2 – 4ac
⇒ D = (-2)2 – 4 × 1 × k
⇒ D = 4 – 4k

For real roots,
D =0
⇒ 4 – 4 k =0
⇒ – 4k = – 4
⇒ k = $$\frac { -4 }{ -4 }$$
⇒ k = 1

Given A.P. is, 4, 9, 14, 19, 24
Here, a = 4
d = 9 – 4=5
We know
an = a + (n – 1) d
a30 = 4 + (30 – 1) 5
⇒ a30 = 4 + 29 x 5
⇒ a30 = 4 + 145
⇒ a30 = 149
And an = a + (10 – 1) d = 4 + (30 -1 )5
= 4 + (9 x 5) = 4 + 45
⇒ a30 = 49
∴ a30 – a10 = 149 – 49
= 100

SECTION-C

AP =AQ
⇒ AP2 = AQ2
Let (x1, y1) = (2, – 4)
(x2, y2) = (3, 8)
(x3, y3) = (-10, y)
Then,
(x2 – x1)2 + (y2 – y1)2 = (x3 – x1 )2 + (y3 – x1)2
(3 – 2)2 + (8 + 4)2 = (-10 – 2)2 + (y + 4)2
(1)2 + (12)2 = (12)2 + (y + 4)2
1 + 144 = 144 + (y + 4)2
1 = (y + 4)2
±1 = y+4
⇒ y = -1 – 4 or y = 1 – 4
⇒ y=-5 or y = -3

Given : A circle with centre at O. AB and CD are tangents drawn at the end of diameter.
To prove : AB || CD
Proof : We know that a tangent at any point of a circle is perpendicular to the radius throught the point of contact.
∴ OM ⊥ AB and ON ⊥ CD

⇒ ∠OMB = 90°
∠OMB = 90°
∠OND = 90°
∠ONC = 90°
∠OMA = ∠OND
∠OMB = ∠ONC
These are the pair of alternate interior angles.
Since, alternate angles are equal, the line AB and CD are parallel to each other.
i.e, AB || CD

Prime factorisation of the given numbers are,
396 = 2 × 2 × 3 × 3 × 11 = 2² × 3² × 11
429 = 3 × 143
561 = 3 × 187
∴ LCM (396, 429, 561) = 2² × 3² × 143 × 187 × 11 = 10589436

Given, Diameter of front wheel = 80 cm
∴ Radius, r = $$\frac { 80 }{ 2 }$$ = 40 cm
and Diameter of rear wheel = 2 m
∴ Radius, R = $$\frac { 2 }{ 2 }$$ = 1 m = 100 cm
Number of revolutions covered by front wheel = 140.
Let the number of revolutions covered by rear wheel be n.
According to the question,
Distance covered by real wheel in n revolutions = Distance covered by front wheel in 140 revolutions
⇒ n x 2πR = 140 x 2πr
n × R = 140 x r
n x 100 = 140 x 40
n = $$\frac { 140\times 40 }{ 100 }$$ = 56
Thus, the number of revolutions covered by rear wheel is 56.
OR
Given : Radius, OA = 3.5 cm
and OD = 2 cm
Area of shaded region = Area of quadrant OACB – Area of ∆DOB

Hence, area of shaded region is 6-125 cm2.

Given : Length of cuboid, l = 6 m
Breadth of cuboid, b = 11 m
Height of cuboid, h = 5 m
Radius of cylinder, r = $$\frac { 3.5 }{ 10 }$$ m = $$\frac { 7 }{ 2 }$$ m
Let height of water level in the cylindrical tank be H m
Since, water in rectangular tank is transferred to cylindrical tank.
∴ Volume of cuboid = Volume of cylinder of height H
l ×b × h =πr²H
6 x 11 x 5 = $$\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 }$$ × H
$$\frac { 30\times 2 }{ 7 }$$ = H
H = $$\frac { 60 }{ 7 }$$ m = 8.57 m.
OR
Given : Diameter of the copper rod = 1 cm
So, Radius (r) of the copper rod = 1/2 cm
Length of copper rod, h = 8 cm
Length of wire, H = 18 m = 18 x 100 cm
Let R be the radius of the cross-section (circular) of the wire.
We know that volume of cylinder = πr²h Now, according to the question
Now, according to the question,

Let p(x) = x² + 13x + 30
= x² + 10x + 3x + 30
= x(x + 10) + 3(x + 10)
=(x + 3) (x + 10)
⇒ x + 3 = 0; x + 10 = 0
⇒ x = – 3, x = – 10
Let α = – 3
and β = – 10
Now, α + β = $$-\frac { b }{ a }$$
– 3 – 10 = $$-\frac { (-13) }{ 1 }$$
– 13 = – 13
and αβ = $$\frac { c }{ a }$$
(-3) (-10) = $$\frac { 30 }{ 1 }$$
30 = 30
Thus, the releationship between the zeroes and coefficients is verified.

Given : First term of A.P., a = 17
Last term, l = 350
Common difference, d = 9
We know,
an = a + (n – 1) d
∵ 350 is the nth term of given A.P.,
∵ 350 = 17 + (n – 1) 9
350 – 17 = (n – 1) 9
333 = (n – 1) 9
$$\frac { 333 }{ 9 }$$ = n – 1
37 = n – 1
37 + 1 =n
38 = n
We know that sum of n terms of A.P is given as,
Sn = $$\frac { n }{ 2 }$$[a + l]
⇒ S38 = $$\frac { 38 }{ 2 }$$[17 + 350]
= 19 [367] = 6973
n = 38 and Sn = 6973.
OR
The first number greater than 10 which when divided by 4 leaves a remainder 3 is 11.
So, the next number will be 11 + 4 = 15
The other numbers will be 15 + 4 = 19, 19 + 4 = 23….
Thus, A.P. = 11, 15, 19, 23 …..
The last term of this A.P. will be 299.
We have to find n. i.e., number of terms in the A.P.
We know,
a + (n – 1 )d = an
⇒ 11 + (n – 1)4 =299
⇒ 11 + 4n – 4 = 299
⇒ 7 + 4n =299
⇒ 4n = 299 – 7
⇒ 4n = 292
⇒ n = $$\frac { 292 }{ 4 }$$
⇒ n = 73
∴ numbers lie between 10 and 300 which divided by 4 leaves a remainder 3.

 Class Frequency (/) c.f. 10-29 18 18 30-49 4 22 50-69 6 28 70-89 4 32 90 -109 8 40

N = 40 ⇒ $$\frac { N }{ 2 }$$ = 20.
Frequency just greater than 20 is 22.
So, 30 – 49 will be median class.
Here, l = 30
ƒ = 4
c ƒ = 18
h = 9

SECTION-D

Steps of construction :

• Draw a line segment AB = 4-5 cm
• Construct ∠YAB = 90°
• Taking A as centre and radius 4-5 cm, draw an arc intersecting AY at C.
• Join BC. Thus, ∆ABC is obtained. Y
• Draw any ray AX making acute angle with AB on the side opposite to the vertex A.
• Mark 5 points namely A1, A2, A3, A4, A5 on AX such that, AA1 = A1A2 = A2A3 = A3 A4 = A4 A5
• Join A4 to B and draw a line through A5 parallel to A4B intersecting extended AB at B’
• Draw a line B’C’ parallel to BC to intersect AY at C’.

Thus, ∆AB’C’ is the required triangle whose sides are $$\frac { 5 }{ 4 }$$ times of the corresponding sides of ∆ABC.

Let CD be the tower of height y m.
Let BD = x.
In ∆CAD, we have

tan 30° = $$\frac { CD }{ AD }$$
$$\frac { 1 }{ \sqrt { 3 } }$$ = $$\frac { y }{ 20+x }$$
20 + x = y√3
x = y√3 – 20
In ∆CBD, we have
tan 60°= $$\frac { CD }{ BD }$$
√3 = $$\frac { y }{ x }$$
⇒ x√3 = y
Putting the value of y in equation (i), we get
x = (x√3) (√3) – 20
x = 3x – 20
20 = 3x – x
20 = 2x
$$\frac { 20 }{ 2 }$$ = x
10 = x
Putting the value of x in equation (ii), we get
10√3 = y
⇒ y = 10 × 1.732 = 1.732
Hence, distamce of the foot of tower from point A
= 20 + x = 20 + 10 = 30 m
and Height of tower = 17.32 m

OR

Let A and B be the two positions of the ship. Let d be the distance travelled by ship during the period of observation i.e. AB = d metres.
Period of observation i.e. AB = d meters.
Let the observer be at O, the top of the light house PO.
It is given that PO = 100 m and the angles of depression from O of A and B are 30° and 45° respectively.
∴ ∠OAP = 30° and ∠OBP = 45°

In ∆OPA, we have
tan 45° = $$\frac { OP }{ BP }$$
⇒ 1 = $$\frac { 100 }{ BP }$$
⇒ BP = 100 m
In ΔOPA, we have
⇒ tan 30° = $$\frac { OP }{ AP }$$
⇒ $$\frac { 1 }{ \sqrt { 3 } }$$ = $$\frac { 100 }{ d+BP }$$
⇒ d + BP = 100√3
⇒d + 100 = 100√3 [∵ BP = 100 m]
⇒ d = 100√3 – 100
⇒ d = 100(√3 – 1)
= 100(1.732 – 1) = 73.2 m
Hence, the distance travelled by the ship from A to B is 73.2 m

Given : ∆ABC in which P is the mid-point of BC, Q is the mid-point of AP, such that BQ produced meets AC at R.
To prove : RA = $$\frac { 1 }{ 3 }$$ CA.

Construction : Draw PS || BR, meeting AC at S.
Proof : In ∆BCR, P is the mid-point of BC and PS || BR.
∴ By basic proportionality theorem, S is the mid-point of CR.
⇒ CS = SR
Similarly, in ∆APS, Q is the mid-point of AP and QR || PS.
∴ R is the mid-point of AS.
⇒ AR=RS
From equations (i) and (ii), we get
AR =RS = SC .
⇒ AC =AR +RS +SC = 3 AR
AR = $$\frac { 1 }{ 3 }$$ AC = $$\frac { 1 }{ 3 }$$ CA

Given :
Radii of upper end of frustum, R = 20 cm
Radii of lower end of frustum, r = 8 cm
Height of frustum, h = 16 cm

We know that
1 cm3 = 0.001 litre
⇒ Volume of container = 10.46 litre
Cost of 1 litre milk = ₹ 22
⇒ Total cost = 22 × 10.46 = ₹ 230.12
OR
Let r cm be the radius and h cm be height of the cone .
Then r = 0.7 cm and h = 2.4 cm
let r1 cm be the radius, l cm be the slant height and h1 cm be the height of the cone.
Then r1 = 0.7 cm and h1 = 2.4 cm

Now, total surface area of the remaining solid
= C.S.A. of cylinder + C.S.A. of cone + Area of base of the cylinder
= 2πrh + πr1l + πr2
= 2πrh + πr1 + πr2   [∵ r = r1]
= [ 2 × $$\frac { 22 }{ 7 }$$ × 0.7 × 2.4 + $$\frac { 22 }{ 7 }$$ × 0.7 × 2.5 + $$\frac { 22 }{ 7 }$$ × 0.7 × 0.7 ] cm2
= ( 10.56 + 5.5 + 1.54 )cm2 = 17.6 cm2

Let Length of rectangle = x
and Breadth of rectangle = y
∴ Area of rectangle = xy
According to question,
(x + 2)(y – 2) = xy – 28
and (x – 1) (y + 2) = xy + 33
From equation (i), we have
xy – 2x + 2y – 4 = xy – 28
– 2x + 2y = – 28 + 4
– 2x + 2y = – 24
2x – 2y = 24
x – y =12
x =12 + y
From equation (ii), we have
(x -1) (y + 2) = xy + 33
xy + 2x – y – 2 = xy + 33
2x – y =33 + 2
2x – y =35
Putting the value of x from equation (iii) to equation (iv)
2(12 + y) – y =35
24 + 2y – y =35
24 + y = 35
y = 35 – 24 =11
Putting the value y in equation (i), we get
x = 12 + 11 = 23
Hence, Length of rectangle = 19 m
Breadth of rectangle = 11 m
OR
Let the total number of swans be x2.
Then, number of swans playing at the pond shore
= $$\frac { 7 }{ 2 } \sqrt { { x }^{ 2 } } =\frac { 7 }{ 2 } x$$
Number of swans playing in water = 2
Total number of swans = No. of swans playing on the shore + No.of swans playing in water
⇒ x² = $$\frac { 7 }{ 2 }$$x + 2
⇒ 2x² = 7x + 4
⇒ 2x² – 7x – 4 = 0
⇒ 2x² – 8x + x – 4 = 0
⇒ 2x(x – 4) + (x – 4) = 0
⇒ (2x + 1) + (x – 4) = 0
⇒2x + 1 = 0 or x – 4 = 0
⇒ x = $$\frac { -1 }{ 2 }$$ or x = 4
But x = $$\frac { -1 }{ 2 }$$ is not feasible. Thus, x = 4
Hence, number of swans = x² = 4² = 16.

Let A be the middle-most point and 13 flags are fixed at points A1,A2,A3,…. A13, to the left of A and remaining to the right of A.
Distance travelled for fixing and coming back to A for
(i) first flag = (2 + 2)m = 4m = a
(ii) second flag = (4 + 4) m = 8 m = a2
(iii) third flag = (6 + 6) m = 12 m = a3
∴ a1, a2, a3…., form an AP in which
a = 4, a2 = 8, a3 = 12……n = 13
and common difference, d =8 – 4 = 4
S13 = $$\frac { 13 }{ 2 }$$[2 x 4 + (13-1) 4]
= $$\frac { 13 }{ 2 }$$(8 + 48) = $$\frac { 13 }{ 2 }$$ × 56 = 364
∴ Distance travelled to fix 13 flags to the left of A = 364 m.
Similarly, the distance travelled to fix remaining 13 flags to right of A = 364 m
∴ Total distance travelled = 2 × 364 m = 728 m.

 xi ƒi ƒixi 3 6 18 5 8 40 7 15 105 9 V 9p 11 8 88 13 4 52 N=∑ƒi = 41 + p ∑ƒixi= 303 + 9p

We have, ∑fi = 41 + p, ∑fixi = 303 + 9p
But Mean = 7.5
∴ Mean = $$\frac { \Sigma { f }_{ i }{ x }_{ i } }{ \Sigma { f }_{ i } }$$
⇒ 7.5 = $$\frac { 303+9p }{ 41+p }$$
⇒ 7.5 × (41 + p) = 303 + 9p
⇒ 307.5 + 7.5p = 303 + 9p
⇒ 9p – 7.5p = 307.5 – 303
⇒ 1.5p = 4.5
⇒ p = 3

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