Aman verma wrote:Q: A man sells chocolates which are in 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 a 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 ?

a) 2

b) 3

c) 4

d) 3.5

e) Can't be determined

Can anybody solve this algebraically ? I will appreciate if anybody could provide the solution for this and the underlying algebra.

ajith wrote:

Aman verma wrote:Q: A man sells chocolates which are in 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 a 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 ?

a) 2

b) 3

c) 4

d) 3.5

e) Can't be determined

Can anybody solve this algebraically ? I will appreciate if anybody could provide the solution for this and the underlying algebra.

Assumptions: chocolate boxes are additive (half box + half box = full box)

Say initially they had x boxes

First man purchases x/2+1/2

Now x/2 -1/2 is left

Second person purchases x-1/4 +1/2

Nothing is left after that

x-1/4 +1/2 =x-1/2

x-1/4 =1/2

x-1 = 2

x=3

Aman verma wrote:
Thanks Ajit. But the notation you used is confusing. Please use brackets for the numerators for example : for x-1/4 use
(x-1)/4 otherwise it appears as if x - (1/4) .Also please mention how you arrived at(x-1)/4 +1/2 for the purchase made by the second person.