some mangoes are kept in a bucket.A goes there ,eats one mango and takes one-third of the remaining.then B goes there ,eats one and take one-third of remaining.now C goes there and do the same.finally A-B-C all goes there,throws one mango and divides the remaining into 3 equal parts and all three of them take one part each.now no mangoes are left.how many mangoes were initially in the bucket at the starting?

Asked by architsrivastava02 | 19th Aug, 2010, 06:34: PM

Expert Answer:

Let M mangoes be there initially.
A takes (M - 1)/3 mangoes, and remaining are 2(M - 1)/3
B takes (2(M - 1)/3 - 1) /3 = (2M - 5)/9 and remaining are 2(2(M - 1)/3 - 1)) /3 = 2(2M - 5)/9
C takes [2(2M - 5)/9 - 1]/3 = (4M - 19)/27  and remaining are, 2(4M - 19)/27.
Now if they throw one from the remaining, they are left with,
2(4M - 19)/27 - 1 = (8M - 65)/27, which they divide into three parts and takes one each.
Now the, if we sum total,
Mangoes with A + B + C + 3 which they ate + 1 which they threw + remaining,  we should get M.
(M - 1)/3 + (2M - 5)/9 + (4M - 19)/27 + 4 + (8M - 65)/27 = M

Answered by  | 19th Aug, 2010, 10:29: PM

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