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kevincan
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Two cardboard boxes and four wooden boxes are loaded onto a Updated on: 06 Apr 2026, 01:50
Originally posted by kevincan on 31 Mar 2026, 03:05.
Last edited by kevincan on 06 Apr 2026, 01:50, edited 1 time in total.
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69% (03:29) correct 31% (03:49) wrong based on 62 sessions

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Two cardboard boxes and four wooden boxes are loaded onto a plane. The cardboard boxes account for 1/8 of the total weight. One cardboard box weighs 3/5 as much as the other. Among the wooden boxes, the heaviest box accounts for 1/3 of the total weight of the wooden boxes. What is the ratio of the weight of the heaviest box to that of the lightest cardboard box?

A. 7 : 1
B. 64 : 3
C. 21 : 4
D. 56 : 9
E. 9 : 2
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GMATcracker101
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Two cardboard boxes and four wooden boxes are loaded onto a 31 Mar 2026, 11:49
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Hello, I understand that we take the heaviest wooden box : the lightest cardboard box. Can there be no instance of lightest wooden box?
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Lorenzini
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Two cardboard boxes and four wooden boxes are loaded onto a 05 Apr 2026, 12:06
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Whats the easiest way to do this sort of word problems?
Do I just pick a Total Weight number that is easily divisible by 8?
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AUGINBERG
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Two cardboard boxes and four wooden boxes are loaded onto a 05 Apr 2026, 15:15
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Let the two cardboard boxes be C1 and C2(C1 being the lightest box), and the 4 wooden boxes be W1, W2, W3 and W4, with W1 being the heaviest box.
Consider C1 + C2 + W1 + W2 + W3 + W4 = T (T here being the total weight)

It is given that C1 + C2 = T/8 and C2 = 5/3 * C1,
Therefore, C1 = (3T)/64 ....... (1)

Now, W1 + W2 + W3 + W4 = (7T) / 8 (Total weight (T) - Cardboard boxes(T/8))
W2 + W3 + W4 = ((7T) / 8) - W1
It is given that W1 = (W1 + W2 + W3 + W4) / 3
Hence, W1 will now be equal to (7T) / 24..... (2)

Therefore (2)/(1) = 56/9

If there's an easier way to solve this, please do let me know.
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