The answer is C for this question
You set it up as a y=mx+b problem 150=55x+40 then you subtract 40 from 150 making 110=55x so x=2
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Base on my own understanding what should be done is that three stacks will not work, because it wont be a symmetric arrangement and also one will be left out. So, two stacks of five each would be better and easy to carry. But the stacks should be arranged in such a manner that the lengths will be parallel to each other and not in-line which would increase the length making it comparatively very long. its easier to hold a (2*8.5,11,2*5=17,11,17) compact box, because it will be easy to carry a long and not heavy.
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so yea what was the answer again
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