A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were there in his stock initially? [This Question was asked as TITA]
- 60 toffees
- 62 toffees
- 61 toffees
- 64 toffees
The Correct Option is B
Approach Solution - 1
Given that the person is left with no toffees after distributing them to the fifth student.
Also given that to each student the person gave one more than half the number of toffees at that stage.
For these types of problems, better we go for backward calculation. If the person had not given \(1\) extra toffee, he would have left with that toffee.
This represents that he had \(2\) toffees at that stage. In the previous stage i.e in 4th stage he should have \((2+1)×2\) i.e \(6\) toffees in the third stage, he should have \((6+1)×2\) i.e. \(14\) toffees.
In the second stage, he should have \((14+1)×2\) i.e. \(30\) toffees.
In the first stage, he should have \((30+1)×2\) i.e. \(62\) toffees.
Hence he initially had \(62\) toffees.
Approach Solution -2
Let, initially the number of toffees be \(64x\).
First child gets \(32x+1\) and \(32x-1\) are left.
Second child gets \(16x+\frac 12\) and \(16x-\frac 32\) are left.
Third child gets \(8x+\frac 14\) and \(8x-\frac 74\) are left.
Fourth child gets \(4x+\frac 18\) and \(4x-\frac {15}{8}\) are left
Fifth child gets \(2x+\frac {1}{16}\) and \(2x-\frac {31}{16}\) are left.
Given that,
\(2x-\frac {31}{16}=0\)
\(⇒ 2x=\frac {31}{16}\)
\(⇒ x=\frac {31}{2 \times 16}\)
\(⇒ x=\frac {31}{32}\)
We assumed that Initially the number of chocolates,
\(=64x\)
\(=64\times \frac {31}{32}\)
\(=62\) toffees
So, the correct option is (B): \(62\) toffees.
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