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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
avatar+272 
+1

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.

 Aug 27, 2021
 #2
avatar+144 
+1

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

 Sep 4, 2021

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