A man sells chocolates which are in the boxes. Only either full box or half a box of chocolates can be purchased from him. A customer comes and buys half the number of boxes which the seller had plus half box more. A second customer comes and purchases half the remaining number of boxes plus half a box. After this the seller is left with no chocolate boxes. How many chocolate boxes the seller had initially?

The equation would be like.
Let no. Of chocolates boxes be x
Therefore, according to question
((x/2)+(1/2)) for first customer
(1/2(x-((x/2)+1/2))+1/2) for second customer
Total should be x
((x/2)+(1/2))+(1/2(x-((x/2)+1/2))+1/2)=x
Solving for x we get 3

The best way to go through the options.
Let there are initially 3 boxes then,
The 1st customer gets = (3/2)+(1/2) = 2. (অর্থাৎ ১ম ক্রেতার কাছে যে অর্ধেক সংখক বক্স বিক্রি করেছিল তা ছাড়াও অর্ধেক বক্স চকলেট বেশি ছিল)
Remaining boxes = 3-2 = 1.
2nd customer gets = (1/2)+(1/2) = 1.
so remain 1 - 1 = 0(No chocolate left)
So, option 2 is correct.

Go backwards. i.e; from the no. of boxes after sale.
let 'x' be the no. of boxes brought by the customers.

Therefore no. of boxes with seller before customer 1 purchases
↑
boxes bought by customer 1 : x + 1/2
↑
boxes bought by customer 2: x + 1/2
↑
no. of boxes left with the seller after sale = 0.

when x=1, no. of boxes = 3
when x=2, no. of boxes = 5 (not available in option )
Hence, answer is option(B) = 3.

Solution:

Let there be x no. of boxes.

If the first customer gets half of the total number of boxes (x/2) + a half box (1/2)
The total number of boxes the first customer gets is x/2+1/2 = (x+1)/2

If the second customer gets half of the remaining
{[x-(x+1/2)]/2} + a half box
The total number of boxes the second customer gets is {[x-(x+1/2)]/2} + 1/2 = (x+1)/4

The total no. of boxes = The boxes bought by first customer + The boxes bought by second customer
x = [(x+1)/2] + [(x+1)/4]
x = (3x+3)/4
4x = 3x+3
x = 3

By
L.S. Theodre Jackson.
BMHSS.