anandIIMA

 Profile Answers by anandIIMA 

• Jan 4th, 2010

 

Here, consider there are n number of persons sitting on the table.
Let Ramesh has Rs. x.
Then the person to his right will have (x-1) then to his right will have (x-2) and so on till the person to the left of ramesh have Rs (x-(n-1))

Here it is said that x gives Re 1 to his right and he in turns give Rs2 to his right and the process continuess untill we get a person who cannot give to his right i.e he is let with no money.

In the 1st round person to the right will be left with Rs x-2, to his right will have x-3 and so on and the person to the left of Ramesh have Rs (x-n)
And now Ramesh has Rs(x +n-1).

In the 2nd round person to the right will be left with Rs x-3, to his right will have x-4
and so on and the person to the left of Ramesh have Rs (x-(n+1))
And now ramesh has Rs(x + 2(n-1))

Let this process continues for r rounds
Then
person to the right of Ramesh has (x-r-1)
person to the left of Ramesh has (x-(n+r-1))
and Ramesh has (x + r(n-1))

Now we have a condition that we stop when we have a person who has 0 sum
Here person to the left of Ramesh has Rs.0
i.e x=n+r-1

Also ramesh has 4 times the sum of the person to his right
i.e x+r(n-1) = 4(x-r-1)
i.e 3x -4 = 3r  + nr
=> 3(n+r-1) - 4 = 3r + nr
=> 3n = nr +7

Now if the process completes in 1 round then r=1
we get 2n = 7
this can not be the number of persons

Now when r=2
we get n=7
=> x = 8

Thus Ramesh has Rs. 8 and the person to his right will have 7,6,5,4,3,2

Proof:
In the 1st Round
Ramesh's right = 6
Ramesh's left = 1
Ramesh = 14

In the 2nd round
Ramesh's right = 5
Ramesh's left = 0
Ramesh = 20

and 20 = 4(5)