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Word problem

by Aman verma » Thu Feb 04, 2010 11:24 am
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.
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by ajith » Thu Feb 04, 2010 11:37 am
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
Last edited by ajith on Thu Feb 04, 2010 12:03 pm, edited 2 times in total.
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by Aman verma » Thu Feb 04, 2010 12:01 pm
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
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.
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by ajith » Thu Feb 04, 2010 12:09 pm
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.
Original post edited to include brackets - will take care in the future.

After the first man goes the vendor has x/2 -1/2 chocolates left = (x-1)/2

Now to the second customer vendor sells half of what he has and half a box [Half of what he has = 1/2( (x-1)/2) = (x-1)/4] = (x-1)/4 +1/2

now after that vendor has nothing, so vendor must have given the second customer all he has, hence

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