Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations - NIMCET

Previous Year Question (PYQs)



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


Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More


Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Ask Your Question or Put Your Review.

loading...