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

indu

• Jul 25th, 2006

 

x+20=x-20

 (x+20)-2/3(x+20)=(x-20)+2/3(x+20)

 4[(x+20)-23(x=20)]=(x-20)+2/3(x+20)

ans--100--      

bharat

• Jul 26th, 2006

 

there is always a confusion to understand the question because of had or hasi think as it is had so answer is 300

satyam dubey

• Aug 31st, 2006

 

according to my opinion bala is correct

johnjps

 Profile Answers by johnjps 

• Jun 2nd, 2008

 

4[x+20-(2x/3)]=x-20+(2x/3)
x=100