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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?

Aug 26, 2021

#1
+313
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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)

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.

Aug 27, 2021
#2
+164
+1

Suppose that he has x apples at first

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