Math Fraction

Question

A person sells one more than half of the number of total orange to the first customer, one more than the one third of remaining oranges to the second customer and one more than one fifth  of the remaining oranges to the third customer. At last he finds that three oranges are left with him. The total number of oranges he has initially is :

Answer 1

There are X oranges

1st customer: (1/2)x + 1

Then the remaining is X - [(1/2)x+1] = X - (1/2)x - 1 = (1/2)x - 1

2nd customer : (1/3) [ (1/2)x - 1 ] + 1 =

(1/6)x - 1/3 + 1 =

(1/6) x + 2/3

Then the remaining is : (1/2)x - 1 - [ (1/6) x + 2/3] =

(1/2)x - 1 - (1/6)x - 2/3 =

(1/3)x - 5/3

3rd customer: (1/5) [ (1/3)x - 5/3 ] + 1 =

(1/15)x - 1/3 + 1 =

(1/15) x + 2/3

the remaining is : (1/3)x - 5/3 - [ (1/15) x + 2/3] =

(1/3)x - 5/3 - (1/15)x - 2/3 =

(4/15)x - 7/3

3 = (4/15)x - 7/3

45 = 4x - 35

80 = 4x

x = 20

20 oranges

1st customer purchased (1/2)(20)+1 = 11, so 20-11 = 9 remain

2nd customer purchased (1/3)9 + 1 = 3+1 = 4, so 9-4 = 5 remain

3rd customer purchased (1/5)(5)+1 = 1+1= 2, so 5-2 = 3 remain

20 oranges