The probability of getting an odd and a prime number on . . .

Question : The probability of getting an odd and a prime number on throwing an unbiased die is :
(A) 1/3
(B) 2/3
(C) 1/6
(D) 1/2

Doubt by Lavanya 

Solution : 
Total possible outcomes when a die is thrown 
{1,2,3,4,5,6} = 6

No. on the die which is both odd and prime 
{3, 5} = 2

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

P(Odd Prime Number) = 2/6 = 1/3

Hence, (A) 1/3 would be the correct option.

A hollow cube of internal edge 22cm is filled with spherical marbles of . . .

Question : A hollow cube of internal edge 22cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that 1/8 space of the cube remains unfilled. Then the number of marbles that the cube can accomodate is 

(A) 142296 

(B) 142396 

(C) 142496

(D) 142596

Doubt by Veer

Solution : 

Side of cube (a) = 22 cm 

Diameter of spherical marble (d) = 0.5 cm 
Radius of spherical marble (r)
= 0.5/2
= 5/20
=1/4 cm

Let number of marbles cube can accomodate be x

ATQ = 
Volume of Cube - Volume of empty space = x × Volume of each marble 
Volume of cube - ⅛ (Volume of cube) = x × Volume of each marble 

Volume of cube (1-⅛) = x × Volume of each marble 

⅞ Volume of cube = x × Volume of each marble 

[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