Bunuel wrote:
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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