Question

A man sells chocolates that come in boxes. Either full boxes or half a box of chocolates can be bought from him. A customer comes and buys half the number of boxes the seller has plus half a box. A second customer comes and buys half the remaining number of boxes plus half a box. After this, the seller is left with no chocolates box. How many chocolates boxes did the seller have before the

Answer 1

Answer:

3

Step-by-step explanation:

Let the man has total boxes = x

No. of chocolates, first customer bought = $\frac{x}{2} + \frac{1}{2}$

Remaining boxes = $x-\frac{x+1}{2}$ = $\frac{x-1}{2}$

No. of chocolates, second customer bought = $\frac{x-1}{4} +\frac{1}{2} =\frac{x+1}{4}$

Now,

$\frac{x+1}{2}+\frac{x+1}{4}=x$

or, $\frac{2x+2+x+1}{4} =x$

or, $3x+3=4x$

or, x = 3