Class 10

Two dice each numbered 1, 2, 3, 4, 5, 6 and 1,1,2,2,3,3 . . .

Question : Two dice each numbered 1, 2, 3, 4, 5, 6 and 1,1,2,2,3,3 respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.

Doubt by Lavanya

Solution :
Total possible outcomes
{

(1,1), (1,1), (1,2), (1, 2), (1,3), (1,3)

(2,1), (2,1), (2,2), (2, 2), (2,3), (2,3)

(3,1), (3,1), (3,2), (3, 2), (3,3), (3,3)

(4,1), (4,1), (4,2), (4, 2), (4,3), (4,3)
(5,1), (5,1), (5,2), (5, 2), (5,3), (5,3)

(6,1), (6,1), (6,2), (6, 2), (6,3), (6,3)
} = 36

P(E) = No. of favourable outcomes / Total number of possible outcomes

No. of outcomes when sum on both the dice is 2 {(1,1), (1,1)} = 2

P(Sum is 2) = 2/36 = 1/18

No. of outcomes when sum on both the dice is 3 {(1,2), (1, 2), (2,1), (2,1)} = 4

P(Sum is 3) = 4/36 = 1/9

No. of outcomes when sum on both the dice is 4 {(1,3), (1,3), (2,2), (2, 2), (3,1), (3,1)} = 6

P(Sum is 4) = 6/36 = 1/6

No. of outcomes when sum on both the dice is 5 {(2,3), (2,3), (3,2), (3, 2), (4,1), (4,1)} = 6

P(Sum is 5) = 6/36 = 1/6

No. of outcomes when sum on both the dice is 6 {(3,3), (3,3), (4,2), (4, 2), (5,1), (5,1)} = 6

P(Sum is 6) = 6/36 = 1/6

No. of outcomes when sum on both the dice is 7 {(4,3), (4,3), (5,2), (5, 2), (6,1), (6,1)} = 6

P(Sum is 7) = 6/36 = 1/6

No. of outcomes when sum on both the dice is 8 {(5,3), (5,3), (6,2), (6, 2)} = 4

P(Sum is 8) = 4/36 = 1/9

No. of outcomes when sum on both the dice is 9 {(6,3), (6,3)} = 2

P(Sum is 9) = 2/36 = 1/18

What is the probability of having exactly 52 sundays in a non leap year . . .

Question : What is the probability of having exactly 52 Sundays in a non leap year?

Doubt by Divya

Solution :

In a non leap year, there are 365 days which can be divided as 52 Full Weeks + 1 Additional day.

This one extra days could be

1.) (Mondyay)

2.) (Tuesday)

3.) (Wednesday)

4.) (Thursday)

5.) (Friday)

6.) (Saturday)

7.) (Sunday)

Total Possible outcomes = 7

In order to have 52 sundays, this extra day should not be (Sunday)

So, favourable outcomes = {(Mondyay), (Tuesday), (Wednesday), (Thursday), (Friday), (Saturday)}= 6

P(52 sundays in a non-leap year) = 6/7

What is the probability that a randomly taken leap year has 52 Sundays . . .

Question : What is the probability that a randomly taken leap year has 52 Sundays ?

Doubt by Divya

Solution :

In a leap year, there are 366 days which can be divided as 52 Full Weeks + 2 Additional days.

These two extra days could be

1.) (Mondyay, Tuesday)

2.) (Tuesday, Wednesday)

3.) (Wednesday, Thursday)

4.) (Thursday, Friday)

5.) (Friday, Saturday)

6.) (Saturday, Sunday)

7.) (Sunday, Monday)

Total Possible outcomes = 7

In order to have 52 sundays, these two extra days should not be Saturday, Sunday) & (Sunday, Monday)

So, favourable outcomes = {(Mondyay, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday)}= 5

P(52 sundays in a leap year) = 5/7

What is the probability of 53 sundays in a non-leap . . .

Question : What is the probability of 53 Sundays in a non-leap year?

Doubt by Lavanya

Solution :

In a non-leap year, we have 365 days
336 days = 52 Full Weeks + 1 Extra Days

This one extra days could be
1.) (Mondyay)
2.) (Tuesday)
3.) (Wednesday)
4.) (Thursday)
5.) (Friday)
6.) (Saturday)
7.) (Sunday)

Total Possible outcomes = 7
Favourable outcomes = {(Sunday)}= 1

P(53 sundays in a leap year) = 1/7

What is the probability of 53 sundays in a leap year . . .

Question : What is the probability of 53 Sundays in a leap year?

Doubt by Lavanya

Solution :

In a leap year, we have 366 days
366 days = 52 Full Weeks + 2 Extra Days

These two extra days could be
1.) (Mondyay, Tuesday)
2.) (Tuesday, Wednesday)
3.) (Wednesday, Thursday)
4.) (Thursday, Friday)
5.) (Friday, Saturday)
6.) (Saturday, Sunday)
7.) (Sunday, Monday)

Total Possible outcomes = 7
Favourable outcomes = {(Saturday, Sunday), (Sunday, Monday)}= 2

P(53 sundays in a leap year) = 2/7

[CBSE 2025] In order to organise, Annual Sports Day, a school prepared an eight lane. . .

In order to organise, Annual Sports Day, a school prepared an eight lane running track with an integrated football field inside the track area as shown below :

The length of the innermost lane of the track is 400 m and each subsequent lane is 7.6 m longer than the preceding lane.

Based on given information, answer the following questions, using the concept of Arithmetic Progression.

(i) What is the length of the 6th lane? [1 Marks]

(ii) How long is the 8th lane than that of the 4th lane? [1 Marks]

(iii) (a) While practicing for a race, a student took one round each in the first six lanes. Find the total distance covered by the student. [2 Marks]

(b) A student took one round each in lane 4 to lane 8. Find the total distance covered by the student. [2 Marks]

Solution :

Here a=400 m

d=7.6 m

(i) What is the length of the 6th lane? Ans : an=a+(n-1)d a6=400+(6-1)7.6 a₆=400+5×(7.6)
a₆=400+38
a₆=438

(ii) How long is the 8th lane than that of the 4th lane? Ans : a₈-a₄ = a+7d-(a+3d) = a+7d-a-3d = 4d = 4(7.6)

= 30.4 m

(iii) (a) While practicing for a race, a student took one round each in the first six lanes. Find the total distance covered by the student.

Ans : S_n=n/2[2a+(n-1)d]
S₆=6/2[2(400)+(6-1)7.6]
S₆=3[800+5×7.6]
S₆=3[800+38]
S₆=3[838]
S₆=2514 m

(iv) (b) A student took one round each in lane 4 to lane 8. Find the total distance covered by the student.
Ans :
S₈=8/2[2(400)+(8-1)7.6]
S₈=4[800+7×7.6]
S₈=4[800+53.2]
S₈=4[853.2]
S₈=3412.8 m

S₃=3/2[2(400)+(3-1)7.6]
S₃=1.5[800+2×7.6]
S₃=1.5[800+15.2]
S₃=1.5[815.2]
S₃=1222.8 m

S₈-S₃ =3412.8-1222.8 =2190 m

[CBSE 2025] In an equilateral triangle of side 10 cm, equilateral triangles of side 1 . . .

In an equilateral triangle of side 10 cm, equilateral triangles of side 1 cm are formed as shown in the figure below, such that there is one triangle in the first row, three triangles in the second row, five triangles in the third row and so on.

Based on the given information, answer the following questions using Arithmetic Progression.

(i) How many triangles will be there in bottom most row?

(ii) How many triangles will be there in fourth row from the bottom?

(iii) (a) Find the total number of triangles of side 1 cm each till 8th row.

(b) How many more number of triangles are there from the 5th row to the 10th row than in the first 4 rows? Show working.

[CBSE 2025] A school is organising a charity run to raise funds for a local hospital. The run . . .

A school is organising a charity run to raise funds for a local hospital. The run is planned as a series of rounds around a track, with each round being 300 metres. To make the event more challenging and engaging, the organisers decided to increase the distance of each subsequent round by 50 metres. For example, the second round will be 350 metres, the third round will be 400 metres and so on. The total number of rounds planned is 10.

Based on the information given above, answer the following questions :

(i) Write the fourth, fifth and sixth term of the Arithmetic Progression so formed. [1 Mark]

(ii) Determine the distance of the 8th round. [1 Mark]

(iii) (a) Find the total distance run after completing all 10 rounds. [2 Marks]

(iii) (b) If a runner completes only the first 6 rounds, what is the total distance run by the runner ? [2 Marks]

The minimum age of children eligible to participate in a painting competition . . .

Question : The minimum age of children eligible to participate in a painting competition is 8 years. It is observed that the age of the youngest boy was 8 years and the ages of the participants, when seated in order of age, have a common difference of 4 months. If the sum of the ages of all the participants is 168 years, find the age of the eldest participant in the painting competition.

CBSE Board Exam 2025

Solution :

Minimum age of children
= 8 years
= 8×12 = 96 months

So,

First Term of the AP (a) = 96 months
Common difference (d) = 4 months

Sum of the ages of all the participants (S_n)
= 168 years
= 168×12
= 2016 months

Let total number of participants be 'n'.

S_n=n/2[2a+(n-1)d]

2016=n/2[2(96)+(n-1)(4)]
2016=n/2[192+4n-4]
2016=n/2[188+4n]
2016=[n/2]×2[94+2n]
2016=n[94+2n]
2016=94n+2n²
2n²+94n-2016=0
2[n²+47n-1008]=0
n²+47n-1008=0
n²+(63-16)n-1008=0
n²+63n-16n-1008=0
n(n+63)-16(n+63)=0
(n+63)(n-16)=0
n+63=0
n=-63
∵n≠-63
So n=-63 is rejected.

n-16=0
n=16

So, there are 16 students.

Now,
a_n=a+(n-1)d
a16=96+(16-1)4
a16=96+15×4
a16=96+60
a16=156
=156 months
=156/12
= 13 years

So, age of eldest child is 156 months or 13 years.

If a=2^2×3^x, b=2^2×3×5, c=2^2×3×7 and . . .

Question : If a=2²×3^x, b=2²×3×5, c=2²×3×7 and LCM (a, b, c) = 3780, then x is equal to

(A) 1
(B) 2
(C) 3
(D) 0

CBSE Sample Question Paper 2025-2026

Solution :

a=2²×3^x

b=2²×3×5

c=2²×3×7

LCM (a,b,c) = 2²×3^x×5×7 — (1)

LCM (a,b,c) = 3780 (Given)
LCM (a,b,c) = 2²×3³×5×7 — (2)

From (1) and (2)
2²×3^x×5×7=2²×3³×5×7
3^x=3³

Clearly x=3

So, (C) 3, would be the correct option.

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