[Case Study] The students of a school decided to beautify the school . . .

Question : 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. 

(a) How much distance did she cover in placing 6 flags on either side of centre point? [1 Marks]

(b) Represent above information in Arithmetic progression. [1 Marks]

OR

How much distance did she cover in completing this job and returning to collect her books?

(c) What is the maximum distance she travelled carrying a flag? [2 Marks]

Doubt by Prateek 

Solution : 

Total Flags = 27 

Distance between two flags = 2 m
Position of Ruchi = Middle Flag
Here
n=27 (odd)
Middle Flag = (n+1)^th/2
=(27+1)^th/2
=28^th/2
=14^th

So Ruchi is standing at the position of 14^th flag

and there are 13 flags on the both the side of her position. 

Distance covered by Ruchi in putting the first flag nearest to her position and then returning back on one side only = 2(2) = 4 m 

Distance covered by Ruchi in putting the second flag nearest to her position and then returning back on one side only = 2(2+2) 
=2(4)
= 8 m 

Distance covered by Ruchi in putting the third flag nearest to her position and then returning back on one side only = 2(2+2+2)
=2(6)
= 12 m 

and so on. 

So the required AP would be . . .

4, 8, 12 . . .

Here
a=4
d=a2-a1
d=8-4
d=4

(a) n=6

S_n=n/2[2a+(n-1)d
S₁₃=6/2[2(4)+(6-1)4]
S₁₃=3[8+(5)4]
S₁₃=3[8+20]
S₁₃=3[28]
S₁₃=3[28]
S₁₃=84

The distance covered by Ruchi in placing 6 flags on either side of centre point is 84 m

(b) The required AP would be 4, 8, 12 . . .

OR

Distance covered in putting all 13 flags on one side

S_n=n/2[2a+(n-1)d
S₁₃=13/2[2(4)+(13-1)4]
S₁₃=13/2[8+(12)4]
S₁₃=13/2[8+48]
S₁₃=13/2[56]
S₁₃=13[28]
S₁₃=364

Total distance covered by Ruchi in putting the flags on both the sides = 2×364
= 728 m

(c) Maximum distance Ruchi covered while carrying the flag will be the distance covered in putting the last flag = 2[13]

=26 m 

The students of a school decided to beautify the school . . .

Question : 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? What is the maximum distance she travelled carrying a flag?

Doubt by Prateek 

Solution : 

Total Flags = 27 
Distance between two flags = 2 m
Position of Ruchi = Middle Flag
Here
n=27 (odd)
Middle Flag = (n+1)^th/2
=(27+1)^th/2
=28^th/2
=14^th

So Ruchi is standing at the position of 14^th flag

and there are 13 flags on the both the side of her position. 

Distance covered by Ruchi in putting the first flag nearest to her position and then returning back on one side only = 2(2) = 4 m 

Distance covered by Ruchi in putting the second flag nearest to her position and then returning back on one side only = 2(2+2) 
=2(4)
= 8 m 

Distance covered by Ruchi in putting the third flag nearest to her position and then returning back on one side only = 2(2+2+2)
=2(6)
= 12 m 

and so on. 

So the required AP would be . . .

4, 8, 12 . . .

Here
a=4
d=a2-a1
d=8-4
d=4
n=13

S_n=n/2[2a+(n-1)d
S₁₃=13/2[2(4)+(13-1)4]
S₁₃=13/2[8+(12)4]
S₁₃=13/2[8+48]
S₁₃=13/2[56]
S₁₃=13[28]
S₁₃=364

Total distance covered by Ruchi in putting the flags on both the sides = 2×364
= 728 m

Maximum distance Ruchi covered while carrying the flag will be the distance covered in putting the last flag = 2[13]
=26 m 

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]

OR

(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. 

OR

(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]

OR

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