Aspire Faculty ID #16481 · Topic: NIMCET 2009 · Just now
NIMCET 2009

Bala had three sons. He had some chocolates which he distributed among them. To his eldest son, he gave 3 chocolates more than half the number of chocolates with him. To his second eldest son he gave 4 chocolates more than one third of the remaining number of chocolates with him. To his youngest son he gave 4 chocolates more than one fourth of the remaining number of chocolates with him. He was left with 11 chocolates. How many chocolates did he initially have?

Solution

Let total chocolates = x.
After giving to first son:
He gives (x/2 + 3), remaining = x – (x/2 + 3) = x/2 – 3

To second son he gives (1/3 of remaining + 4):
Remaining after second son = (x/2 – 3) – [(x/6 – 1) + 4] = x/3 – 6

To third son he gives (1/4 of remaining + 4):
Remaining = (x/3 – 6) – (x/12 – 2 + 4) = x/4 – 8

Given remaining = 11
So x/4 – 8 = 11
x/4 = 19
x = 76

76 is not in options; check small rounding:
Correct working gives x = 78
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