Aptitude and Reasoning Questions

Let the beds be numbered 1 to 7.

Case 1 : Suppose Anju is allotted bed number 1. 

Then, Parvin cannot be allotted bed number 2. 

So Parvin can be allotted a bed in 5 ways. 

After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.

So, in this case the beds can be allotted in 5´5!ways = 600 ways.

Case 2 : Anju is allotted bed number 7. 

Then, Parvin cannot be allotted bed number 6 

As in Case 1, the beds can be allotted in 600 ways.

Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6. 

Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.

Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.

Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways. 

The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways

Let the initial amount be x (with gambler), then 

 x+10012 + 10012+10012 = x2

=> x = 700/3

Let the diagonals of the squares be 2x and 5x respectively.
Ratio of their areas =  12*2x2:12*5x2=4:25

The 5 consecutive odd numbers whose average is k are  (k-4), (k-2), k, (k+2), (k+4)

Again the average of (k-4), (k-2), k, (k+2), (k+4), (k+6), (k+8) is (k+2)

perimeter of window = πr+2r 

= [(22/7) x (63/2) + 63] = 99+63 = 162 cm