2 persons have the same no of marbles at starting.after that 1st person gains 20 marbles from 2nd..n later 1st loses 2/3rd of wat he had... now 2nd has 4 times as many marbles as 1st...wat r the initial no of marbles tat each of them had?

This question is related to Infosys Interview

Interview Candidate
Jul 22nd, 2006
8
2055

Showing Answers 1 - 8 of 8 Answers

Vidhya

Jul 22nd, 2006

Originally both had 300 marbles.
Let x be the original number of marbles both are having.
Given,
A B
x+20 = x-20
x+20-(2x/3) = x-20+(2x/3)
4[x+20-(2x/3)]=x-20+(2x/3)
By solving it, x = 300.

Bala

Jul 23rd, 2006

Is that answer is correct. Please make note of the poin that 1st person looses 2/3 of the marbles what he had. At the second round the first person has x+20. At the third stage the 1st person should have (X+20)-2/3(X+20), so as per my suggestion A BNo of Marbles X XA gains 20 marbles from B X+20 X-20B gains 2/3 of marbles from A (X+20)-2/3(X+20) (X-20)+2/3(X+20)Now 4[(X+20)-2/3(X+20)]=(X-20)+2/3(X+20)From this i got answer as each has 100ANS-100

Gopalakrishnan

Jul 23rd, 2006

Even I got the answer as 100. I think Bala is correct.

Bala

Jul 24th, 2006

Sorry my answers are messed up. I have arranged one by one, here it isA has X and B has X. A gains 20 marbles from B, so a has X+20 and B has X-20. B gains 2/3 of marbles from A , so A has(X+20)-2/3(X+20) and B has (X-20)+2/3(X+20)Now 4[(X+20)-2/3(X+20)]=(X-20)+2/3(X+20)From this i got answer as each has 100ANS-100