Table of Contents

CBSE Class 10 Maths board paper consists of questions from the NCERT Mathematics books of the Class 10. The types of questions and their marking scheme can be fully understood by going through the NCERT Previous Year question papers of the class 10 Maths subject. For this, the CBSE Question Papers for Class 10 Maths are provided here for students to help in their exam preparation.

Along with the question papers, we have also provided students with the solutions to these questions. It will help students in understanding the approach of answer writing in the Mathematics exam. The previous year question paper of the Mathematics class 10 board exam will help students get familiar with the marking scheme and the overall difficulty level of the exam.

## Class 10 Maths Previous Year Question Papers with Solutions

Class 10th Maths Previous Year Question Paper is worth solving for students appearing in the CBSE Class 10th Examination 2024. Solving the Class 10th Maths Previous Year Question Paper will give students the confidence to solve mathematical problems and also practicing the Class 10th Maths Previous Year Question Paper on daily basis will reduce the stress of CBSE Class 10th Exam 2024.

## NCERT previous year question papers class 10 Maths with solutions

If students will solve the Class 10th Maths Previous Year Question Paper then they will analyze by themselves which topics are important for their CBSE Class 10th Exam 2024 and which types of questions are repeated every year. On this page, we have given Class 10th Maths Previous Year Question Paper with solutions, by solutions students will get the idea of answer writing style which can fetch extra marks in maths paper.

## PYQ Class 10 Maths

Mathematics is all about practice. The more students practice, the more command they will get over the subject. Solving the CBSE Previous Year Question Papers of Class 10 Maths before the board exams will give students an idea about the question paper pattern, marking scheme, and difficulty level of exams. PYQ Class 10 Maths for Board Exams 2024, Click on the below Link to check all papers

**Class 10 Maths Sample Paper 2024 (Standard and Basic)**

**Class 10 Maths MCQ Question with Answers**

**Class 10 Maths Important Question with Answers**

## Previous Year Question Paper Class 10 Maths NCERT PDF 2023

Last Year The CBSE Class 10 Maths Test was held on March 21, 2023. Maths is all about practice. Moreover, they will also get thorough with the types of questions asked in Maths. The CBSE designs these CBSE Class 10 Maths Question Papers based on the syllabus. As a result, we must optimize our performance on the Class 10 Mathematics Exam right now. by clicking on the below link.

Check: CBSE Class 10 Maths Paper 2023

## Previous Year Question Paper Class 10 Maths NCERT PDF 2022

**Section A**

**Q. Present weight of Saket is 100 kg. He has decided to take a weight loss challenge, during his challenge he can only reduce 2.5 kg in each month. After how many months his weight will be 70 kg?**

**Answer:**

As here. a = 100

d = – 2.5 (As his weight is decreasing by 2.5 kg in each month).

So, an = 70

As, an = a + (n-1)d

=> 70 = 100 + (n -1) * (-2.5)

=> 70 = 100 -2.5n +2.5

=> 2.5 n = 102.5 -70

=> 2.5 = 32.5

=> n=32.52.5 = 13

**Q. Determine the factors of the equation by using the factored form method.**

y2-23y+112=0

**Answer:**

=>y2-23y+112=0

y2-16y-7y+112=0

y(y-16)-7(y-16)=0

y=+16, +7

**Q. A tangent PQ at a point P of a circle of radius 4 cm meets a line through the centre O at a point Q so that OQ = 5 cm. Length PQ is**

**Answer:**

Triangle OPQ is a right – angled triangle

Therefore,

OQ2 = PQ2 + OP2

PQ2 = OQ2 – OP2

PQ2 = OQ2– OP2

PQ2 = 52– 32

PQ2 = 25-9

PQ2 = 16

PQ =4 cm

**Q. A cube has a volume of 64 cm3 then, what is the length of the cube?**

**Answer:**

Volume of cube = a3

**=> **64 = a3

=> a = 364

=> a = 4 cm.

**Q.The daily minimum steps climbed by a man during a week were as under:**

Monday |
Tuesday |
Wednesday |
Thursday |
Friday |
Saturday |

30 |
35 |
24 |
26 |
32 |
28 |

**Find the mean of the steps.**

**Answer:**

** **Number of steps climbed in a week: 30,35,24,26,32,28.

So, we get,

Mean = sum of observation (steps) / total no of observations

=30 + 35 + 24 + 26 + 32 + 286

=1756

= 29.17

**OR**

**Q. A coin and a dice is tossed together. What will be the probability of getting a tail and a prime number?**

**Answer:**

Since the probability of getting tail = 1/2

As when dice is thrown,

We get numbers from 1 to 6 i.e. 1, 2, 3, 4, 5, 6

So, prime numbers are = 2, 3, 5 = 3

The probability of getting a prime number = 3/6 = 1/2

So, the probability of getting tail and a prime number will be = 1/2 *1/2 =1/4

Q. **Find two numbers whose sum is 27 and product is 182**.

**Answer:**

Let the number is *x*

Then equation will be

=> x2+ 27*x *+ 182 =0

=> x2 + 14*x *+ 13*x *+ 182 =0

=> *x*(*x*+14) + 13 (*x*+14) = 0

=> (*x*+14) (*x*+13)

=> *x* = 14,13

**Section B**

**Q. Which is correct sigma notation of the AP:101, 97, 93, 89, ….?**

**Answer:**

AP is 101, 97, 93, 89 ,….

Here, a = 101 and other terms are continuously decreasing with a common difference of 4

So, we can write “a” i.e. 101 as (105 – 4*1) = (105 – 4*n)

So, correct sigma notation will be,

**OR**

**Q. Present weight of Saket is 100 kg. He has decided to take a weight loss challenge, during his challenge he can only reduce 2.5 kg in each month. After how many months his weight will be 70 kg?**

**Answer:**

As here, a = first term = 3 kg

d = common difference = 2.5 (As his weight is increasing by 2.55 kg in every year)

So, after 15 years, means value of n = 15

Weight of baby will be:

= a15 = a + (n-1)d

= 3+ (15-1)×2.5

=>3+ 14×2.5

=>3 + 35 = 38 kg.

**Q. In triangle ABC, ****∠****C = 90°, BC = 24 cm and AC = 7 cm. Determine the value of SinA * CotB.**

**Answer:**

As here,

Hypotenuse2= 242 + 72

=> Hypotenuse = 25 cm

=> SinA = opposite side/ hypotenuse = 24/25

=> TanB =opposite side/adjacent = 7/24

=>SinA * CotB = 24/25 * 7/24 = 7/25

**Q. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.**

**Answer:**

fix and calculate the value of and

Calculate the unknown frequency

Mean

f = 402

f = 20

**Q. A box contains 4 white and 8 yellow balls. Two balls are taken out at random, without replacement. If the first ball is white, then what is the probability that the second ball is also white?**

**Answer:**

Number of white balls = 4, Number of yellow balls = 8

Number of balls taken out at random = 2

Total balls = 4 + 8 = 12

Required probability = 4/12 * 3/11 = 1/11

**Section C**

**Q. What will be the length of AD and BC of the given figure?**

**Answer:- **

Given that,

ABC is a triangle,

D is a point on the side BC of ΔABC

=> ∠ADC=∠BAC

By using properties of similar triangle,

=> AB/AD = BC/AC = AC/DC

=> 8/AD = 6/3

=> AD = 4cm

Now, BC/6 = 6/3

=> BC = 12cm

**Q. In the given figure, if ****∠****PQR = 30°, find the value of ****∠****POR.**

**Answer:**

Here, OP ⊥ QP and OR ⊥ QR

∴ ∠OSR = ∠ORQ = 90°

In triangle OPQR, ∠POR + ∠OPQ + ∠PQR + ∠ORQ = 360°

∠POR + 90° + 30° + 90° = 360°

∠POR = 360° – 210° = 150°

**OR**

**Q. In the given figure, PQ and PR are tangents to the circle with centre O such that ****∠****QPR = 45°, then ****∠****QOR is equal to: **

**Answer:**

In quadrilateral PQOR

∠PQO + ∠QOR + ∠ORP + ∠QPR = 360°

⇒ 90° + ∠QOR + 90° + 45° = 360°

⇒ ∠QOR = 360° – 225° = 135°

## Previous Year Question Papers PYQ Class 10 Maths– 2021

**Section A**

**Q. Determine the factors of the equation by using the factored form method.**

**3x****2**** – 10x + 8 = 0**

**Answer:**

=>3x2 – 10x + 8 = 0

3x2 – 6x – 4x + 8 = 0

3x(x – 2) – 4(x – 2) = 0

x = 2, 4/3

Q. **What will be the approximate least and greatest value of “y” in the given equation:(**y-7)2=484?

**Answer:**

y-72=484

=> y -72 = 222

=>y = 22+7 and -22 +7

=> y = 29, -15

**Q. If the distance between points (5, 2) and (1, a) is 5, then find the value of a, If (a < 0).**

**Answer:**

Distance = {(5-1)2+2-a2}

=>5= {(4)2+2-a2}

=>25 = 16+2-a2

=>25-16 = 2-a2

=>9=2-a2

As value of a is negative.

=>2-a=3

=>a = 2-3 = -1

**Q. A circle with its center at O (x,11) has a radius of 26 units. Point P (3,-13) lies on the circle.Also, ***x***> 10. Find x.**

**Answer:**

**As **O is at (x, 11) and p (3, – 13)

=> OP = x-32+(11--132)

=> OP = (x-3)^2+24^2

=>OP = (x-3)^2+576

=>262 = x-32+576

=>676 = x-32 +576

=>x-32= 100

=>(x-3) = 10

=>x = 13 (As x > 10)

**Q. What will be the sum of the first 21 terms of the AP 4, 7, 10, …..?**

**Answer:**

AP = 4, 7, 10, ……

Here, a = 4

d = 7-4 = 3

Number of terms = n = 21

So, sum of first 21 terms of AP will be Sn =n2 (2a+(n-1)d)

=>Sn =212 * [2*4 +(21-1)*3]

=>Sn =212 * [8+60]

=>Sn =212 * 68

=>Sn = 21*34 =714

**Q. Mohan bought a toy which is packed in box to gift his Sohan on his birthday. The dimension of the box is 6cm x 6cm x 6cm. What size of gift paper is required to wrap the gift?**

**Answer:**

The dimension of box is 6cm x 6cm x 6cm** **

Curved surface area of cube = 6 x a2

= 6x 62

= 216 m2

**Q. Find the mode of the following items.**

**1,2,2,5,6,6,4,2,4,0,4,4,6,5,3**

**Answer:**

On arranging the items in ascending order, we get:

0,1,2,2,2,3,4,4,4,4,5,5,6,6,6

As we can see 4 occurs the maximum number of times.

Therefore, the mode of the given items = 4

**Q. ****A bag contains a red ball, a blue ball, a yellow ball and an orange ball, all the balls being of the same size. Priti takes out a ball from the bag without looking into it. What is the probability that she takes out the**

**(i) yellow ball? (ii) orange ball?**

**Answer:**

Priti takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them.

Let be the event ‘the ball taken out is yellow ‘and be the event ‘the ball taken out is orange”

Now, the number of possible outcomes .

(i) The number of outcomes favourable to the event .

So,

Similarly, (ii)

**Section B**

**Q. If the mid point of line AB is (7, 1) determine the coordinates of B if A (9.5, -6.5).**

**Answer:**

Given, (x1, y1) = (9.5, -6.5)

B = (x2, y2)

According to the formula we can find the midpoint (x, y):

(x, y) = [x1 + x22,y1 + y22]

(7, 1) = [9.5+ x22,-6.5+y22]

=> 9.5+x2=14

=> x2 = 5.5

And, y2 = -6.5+y2 = 2

=> y2=8.5

**Q. In the figure given below, ****∠****PTS = ****∠****QPR, PS || QR, PS : QR = 7 : 9 and PR = 36 cm, then find the length of TR.**

**Answer:- **

∠PTS = ∠QPR

PS || QR

PS : QR = 8 :10

PR = 36 cm

Rule of similarity AA(Angle-Angle)

∆TSP and ∆PQR are similar.

Now, PS/QR = PT/PR

⇒ 7/9 = PT/36

⇒ PT = 28 cm

Now, TR = PR – PT

= 36 – 28 = 8 cm

**Q. A box contains lemon and banana, if there are 13 bananas in the box and the probability of selecting 2 bananas from the box is 1/10, then find the number of lemons in the box?**

**Answer:**

The number of bananas in the box =13

Let the number of lemons in the box =X

13*12X+13X+12=110

156X2+13X+12X+156=110

1560=X2+25X+156

X2+25X-1404=0

X2-27X+52X-1404=0

X(X-27)+52(X-27)=0

X=27, -52 (Neglecting the negative values)

Number of lemons in the box =27

Q. **The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify?**

Number of wickets |
20 – 60 |
60 – 100 |
100 – 150 |
150 – 250 |
250 – 350 |
350 – 450 |

Number of bowlers |
7 |
5 |
16 |
12 |
2 |
3 |

**Answer:**

Here, the class size varies, and the xi,s are large. Let us still apply the step deviation method with a = 200 and h = 20. Then, we obtain the data as in Table

Number of wickets taken | Number of bowlers
( fi ) |
xi | di = xi – 200 | ui = di20 | ui fi |

20 – 60 | 7 | 40 | –160 | –8 | –56 |

60 – 100 | 5 | 80 | –120 | –6 | –30 |

100 – 150 | 16 | 125 | –75 | –3.75 | –60 |

150 – 250 | 12 | 200 | 0 | 0 | 0 |

250 – 350 | 2 | 300 | 100 | 5 | 10 |

350 – 450 | 3 | 400 | 200 | 10 | 30 |

Total | 45 | -106 |

So,

Therefore,

This tells us that, on an average, the number of wickets taken by these 45 bowlers in one-day cricket is 152.89.

**Section C**

**Q. Find the roots of the following quadratic equation by completing the square method. 14 x2+23 x-30=0.**

**Answer;**

x2 + 13x – 17 = 0

Comparing to general equation:

ax2 +bx +c = 0

=> a = 1, b = 13, c = -17

=> Roots = [-b ± √(b² – 4ac)]/2a

**Q. What is the length of the radius for the given figure?**

**Answer:**

Here, B is the point of tangency.

So,

Subtract both sides by 2500,

=> r2+3900 = r2+100r

Cancelling r2

=> 3900 = 100r

=> r = 39 ft

**OR**

**What is the radius of the circle given below?**

**Answer:-**

Here, AB is a tangent and BC is the radius

∴ BC ⊥ AB

Now, in triangle CBA,

⇒ CA2= AB2 + CB2

⇒ 302 = 262 + OT2

⇒ 900 – 676 = OT2

⇒ 224 = OT2

⇒ 14.96 cm = OT

**Q. If the value of 25Sinθ = 24, then what will be the value of Sinθ * Cosθ * Cotθ?**

**Answer:**

As, sin θ = Perpendicular/Hypotenuse

25sin θ = 24

=>sin θ = 24/25

=> Perpendicular = 24, Hypotenuse = 25

\[=>Base=\sqrt{625-576}=7\]

cos θ = Base/Hypotenuse = 7/25

Cotθ = Base/Perpendicular = 7/24

=>Sinθ * Cosθ * Cotθ

=>[24/25 * 7/25 ] * 7/24

=>49/625

**Q. A juice seller was serving his customers using glasses as shown in fig. below. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find the apparent capacity of the glass and its actual capacity. (Use π = 3.14.)**

**Answer:**

Since the inner diameter of the glass and height ,

the apparent capacity of the glass

But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass.

i.e., it is less by

So, the actual capacity of the glass =

## Previous Year Question Papers PYQ Class 10 Maths – 2019

**Section A**

**Q. If Mahesh were younger by 5 years than what he really is. The square of his age would have been 11 more than five times of his present age. What is his present age?**

**Answer:**

Let x be the present age of Mahesh

Then,

=> x-52 = 5x+11

=> x2-10x+25 =5x+11

=> x2-15x+14=0

=>x2-14x-x+14 =0

=>x(x-14)-1(x-14)=0

So,

x=14 or x=1

As age can’t be negative so, now age of Mahesh is 14.

**Q. Find the midpoint of a line whose endpoints are (11, 13) and (9, 11).**

**Answer:**

Given, (x1, y1) = (11, 13)

(x2, y2) = (9, 11)

According to the formula we can find the midpoint (x, y):

(x, y) = [x1 + x22,y1 + y22]

(x, y) = [11 + 92,13 + 112]

= (10, 12)

**Q. Disha has two dairy milk and 3 perk chocolates. What will be the probability of taking two chocolates, one of dairy milk and one of perk?**

**Answer**

Dairy milk = 2

=> Probability of taking one dairy milk P(A) = 1/2

Perk = 3

=> Probability of taking one perk P(B) = 1/3

=> P(A and B)=P(A∩B)=P(A)⋅P(B) = 1/3*1/2 = 1/6

**OR**

**Q. Find the value of y from the following observations if these are already arranged in ascending order. The Median is 63.**

**20, 24, 42, y , y + 2, 73, 75, 80, 99**

**Answer:**

As the number of observations made is odd, so the median will be the middle term, i.e. y + 2.

Therefore,

y + 2 = 63

y = 63 – 2 = 61

**Q. What will be the approximate least and greatest value of “x” in the given equation?**

x+42 -15 =0?

**Answer:**

x+42 -15 =0

=>x+42 = 15

=>(x+4)^2 = 152

=>x = √15 -4 and √15+4

As, √15 = 3.8729

=> x = 3.873 – 4 and 3.873 + 4 => -0.127 and 7.873

**OR**

Q. **The earning for the month of January of Mr. Tiwari is Rs. 1000. If his earning gets increased by Rs. 125 every month, then what will be his earning in the month of August?**

**Answer:**

As here, a = first term = 1000

d = common difference = 125 (As earning is increased by Rs. 125 in every month)

So, for August, value of n will be = 7

=> Earning of August will be

= a7 = a + (n-1)d

= 1000+ (7-1)×125

=>1000+ 6×125

=>1000 + 750 = Rs. 1750

Q. **A cylinder shaper container has a height of 5m and its radius is 2m. How much water can a container store?**

**Answer:**

Volume of cylinder

= πr2h

= π×2×2×5

= 62.83 m3 ≃63 m3

The container can contain 63 m3 of water .

**Q. The weight of a new born baby is 3 kg after every year his weight is increasing by 2.5 kg. What will be his weight when he will be 16 years old?**

As here, a = first term = 3 kg

d = common difference = 2.5 (As his weight is increasing by 2.55 kg in every year)

So, after 15 years, means age when he will be 16 years old,

So, value of n = 15

Weight of baby will be:

= a15 = a + (n-1)d

= 3+ (15-1)×2.5

=>3+ 14×2.5

=>3 + 35 = 38 kg.

**Section B**

**Q. If the value of Sin**θ** = 63/65, then what will be the value of Cos**θ** * Tan**θ**?**

**Answer:**

As, sin θ = Perpendicular/Hypotenuse = 63/65

=> Perpendicular =63 , Hypotenuse = 65

=> Base = 652–632 = 16

=> Cos θ = Base/ Hypotenuse = 16/65

tan θ = Perpendicular/Base = 63/16

=>Cos θ * Tan θ = 16/65 * 63/16 = 65/63

** Q. ****In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number.**

**Answer:**

Get the values of a and h

a = 57

h = 53 – 50 = 3

Calculate the mean

Mean (x) = a + h × *∑f**i**u*i*∑f**i*

x = 57 + 3×25400

x = 57 + 316

x = 57 + 0.19

x = 57.19

**Q. A bag contains 3 red, 5 yellow and 4 blue balls. If three balls are drawn at random, then find the probability that balls are yellow?**

**Answer:**

Required probability = 5C3/12C3

= 5 * 4 * 3/12 * 11 * 10

= 1/22

**OR**

**Q. Yugali is playing a cricket match and a game of football; the probability of winning cricket is 0.4 while the probability of winning football is 0.6. What will be the probability that she will win both matches assuming that the result of the matches is independent?**

**Answer:**

Probability of winning cricket = 0.4

Probability of winning football = 0.6

Probability of winning cricket and football = 0.4*0.6 = 0.24

**Section C**

**Q. Find the roots of the following quadratic equation by completing the square method. 14 x2+23 x-30=0 .**

**Answer:**

To convert the coefficient of the first term in 14x2, multiply the equation by 14.

Result is,

Take the square root of both sides of the equation.

Result is, 14 x+232=472

The roots of the equation are given by,

** Q. In the given figure, S is the point of tangency. What will be the radius of the circle T?**

**Answer:- **

Given that,

S is the point of tangency,

So

Subtract both sides by 1296,

**OR**

**Q. What is the value of x for the given figure?**

**Answer:- **

Here, US is a tangent to point R at S,

UV is a tangent to point R at V

Tangents are congruent,

Therefore,

SV = VU

**Q. If the value of 17Sinθ = 8, then what will be the value of (2Secθ + 4Tanθ)/ Cosec θsinθ?**

**Answer: **

As, sin θ = Perpendicular/Hypotenuse

17sin θ = 8

=>sin θ = 8/17

=> Perpendicular = 8, Hypotenuse = 17

=> sec θ = Hypotenuse/Base = 13/12

tan θ = Perpendicular/Base = 5/12

Cosecθ = 17/8

(2Secθ + 4Tanθ)/ Cosec θsinθ

=>136+531

=>136+53

=>13+106 =236

**Q. 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 16 cm by 10 cm by 3.7 cm. The radius of each of the depressions is 0.6 cm and the depth is 1.5 cm. Find the volume of the wood in the entire stand.**

**Answer:**

The dimensions of the cuboid are16 cm by 10 cm by 3.7 cm.

So, Volume of the cuboid is = l × b × h = 16 × 10 × 3.7 = 592 cm3

Radius of the conical depression (r) = 0.6 cm

Height of the conical depression (h) = 1.5 cm

As the pen stand made of wood is in the shape of a cuboid with four conical depressions,

Volume of wood

= volume of cuboid – 4 (volume of cone)

= 592 – (4 × 13 × 227 × 0.6 × 0.6 × 1.5)

= 592 – 2.26

= 589.74 cm3

So, the Volume of wood = 589.74 cm3

### Previous Year Question Paper PYQ Class 10- All Subjects