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After all transfers the sand contained in each box is equal to 120 gms.

=> 4/5*C = 120 => C = 150 and C/5 = 30

and, 2/3*B = 120 => B = 180 and B/3 = 60 => C = 150 - 60 = 90

and, 5/6*A + C/5 = 120 => 5/6*A + 30 = 120 => A = 108 => B = 162

So, A = 108, B = 162, and C = 90 at the beginning.
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Bunuel
The total amount of sand contained in Boxes A, B and C is 360 grams. First, 1/6 of the sand from Box A is poured into Box B. After that, 1/3 of the sand from Box B is poured into Box C. Finally, 1/5 of the sand from Box C is poured into Box A. At the end, each box has an equal amount of sand. How much sand did each box A, B and C, respectively have at the beginning?

A. 96, 168, 96

B. 102, 168, 90

C. 108, 156, 96

D. 108, 162, 90

E. 114, 162, 84



Solution:

We can let a, b and c be the amount of sand in boxes A, B and C originally. We can create the equations (notice after sand is transferred between the boxes, each box at the end has an equal amount of same, i.e., 360/3 = 120 grams of sand) :

5a/6 + [c + (b + a/6) * 1/3] * 1/5 = 120

(b + a/6) * 2/3 = 120

and

[c + (b + a/6) * 1/3] * 4/5 = 120

Multiplying the second equation by 3/2, we have:

b + a/6 = 180

Substituting this in the third equation, we have:

[c + 180 * 1/3] * 4/5 = 120

c + 60 = 150

c = 90

Substituting 90 for c and 180 for b + a/6 in the first equation, we have:

5a/6 + [90 + 180 * 1/3] * 1/5 = 120

5a/6 + 150 * 1/5 = 120

5a/6 = 90

a = 90 * 6/5 = 108

Since only choice C has a = 108 and c = 90, the correct answer must be D (we will leave it to the reader to show that b = 162).

Answer: D
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