Bryan had some apples. He put 1/3 of the apples and another 4 into basket A, 1/3 of the remaining apples and another 3 into basket B, and 1/3 of the rest and another 3 into basket C. In the end, he had 3 apples left for himself. How many apples did he have at first?
+313 Let Bryan had x apples.
He put 1/3 of the apples and another 4 into basket A, i.e.
\( ({x \over 3}+4){}{}\)apples in basket A.
So, remaining apples are
x - (x/3 + 4)=\(( {2x \over 3}- 4).{}{}\)
Now, he put1/3 of the remaining apples and another 3 into basket B, i.e.
1/3 (2x/3 - 4) + 3 = (2x/9 - 4/3 + 3) = (2x/9 + 5/3)
apples in basket B.
This time remaining apples are
(2x/4 - 4) - (2x/9 + 5/3) = (4x/9 - 17/3).
Now, he put 1/3 of the rest and another 3 into basket C, i.e.
1/3 (4x/9 - 17/3) + 3 = (4x/27 + 10/9) apples in basket C.
So, the remaining apples are
(4x/9 - 17/3) - (4x/27 + 10/9) = (8x/27 - 61/9).
But in the end, he had 3 apples left for himself, i.e.
(8x/27 - 61/9) = 3
⇒ (8x/27) = 3 + 61/9 =88/9
⇒ x = 88/9 * 27/8 = 33
So,
33 apples he have at first.
+164 Suppose that he has x apples at first
basket * A
1/3x+4
basket B: 1/3(x - 1/3x - 4) + 3
= 2/9x + 5/3
basket C: 1/3(x - 1/3x - 4 - 2/9x - 5/3) + 3
= 4/28x + 10/9
all apples:
1/3x + 4 + 2/9x + 5/3 + 4/27x + 10/9 + 3=x
{ look question carefully} 19/27x + 88/9 = x
8/27x = 88/9
{ careful calculation} x = 33
So the answer is 33
