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Number of small boxes in a large box: Easy way of solving

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Number of small boxes in a large box: Easy way of solving

by crackgmat007 » Mon Mar 16, 2009 6:33 pm
Is there an easy way to solve this problem? Please state the steps. Thanks.

What is the maximum number of rectangular boxes, each measuring 2" by 3" by 5" that can be packed into a rectangular packing box measuring 18" by 19 " by 35", if all of the smaller boxes are aligned in the same direction?

320
350
378
410
450

Answer is 378
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by crackthetest » Mon Mar 16, 2009 6:58 pm
Align in this pattern:

Bigger box : 35 : 19 : 18
smaller box: 5 : 3 : 2
counts: 7: 6 : 9 ( multipl 7*6*9 = 378)

IMO. this could be shortest.
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by lav » Tue Mar 17, 2009 5:17 am
hi
can u pls explain more
the alignment

Bigger box : 35 : 19 : 18
smaller box: 5 : 3 : 2

can be done in 6 ways ... how to decide/judge that he alignment taken above will give " maximum number of rectangular boxes" ????
Kid in Verbal :(
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by crackgmat007 » Tue Mar 17, 2009 9:37 am
[quote="lav"]hi
can u pls explain more
the alignment

Bigger box : 35 : 19 : 18
smaller box: 5 : 3 : 2

can be done in 6 ways ... how to decide/judge that he alignment taken above will give " maximum number of rectangular boxes" ????[/quote]

I actually computed for all 6 combinations and found that 378 is the max number of boxes by ignoring the decimals and not rounding off. It took some time to do the calculations but looks like there is no easier way (any thoughts guys!)

You get 378 for 2 combinations, one is as stated above and other one is as below:

Bigger box : 35 : 19 : 18
smaller box: 5 : 2 : 3 (7 : 9 : 6)
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by Stuart@KaplanGMAT » Tue Mar 17, 2009 12:00 pm
lav wrote:hi
can u pls explain more
the alignment

Bigger box : 35 : 19 : 18
smaller box: 5 : 3 : 2

can be done in 6 ways ... how to decide/judge that he alignment taken above will give " maximum number of rectangular boxes" ????
To maximize the number of boxes, we want to minimize the wasted space in the big container. Accordingly, the closer a fit we can get, the better.

Arraging the sides as suggested, we have two exact fits (35:5 and 18:2) and one side that's off by 1 (19:3... 18:3 would be a perfect fit).

If we flip the last two and arrange as follows:

35 : 19 : 18
5 : 2 : 3

We also minimize the wasted space.

In the first example our # of boxes would be 7*6*9; in the second our # of boxes would be 7*9*6. Of course, both give us the same answer to the question.
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