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by Vikas Mishra » Thu Nov 17, 2016 2:53 am
Of 128 boxes of oranges each box contains at least 120 and at most 144 oranges. The number of boxes containing the same number of oranges is at the least(i.e minimum possible) is:

a)5

b)6

c)103

d)112

e)118

OA[spoiler]b)6[/spoiler]

How do students feel about the OA ? Is the OA correct ? Somehow I don't think the OA to be correct. I am getting [spoiler]c)103[/spoiler] as the answer. So, please give your views and comments and relevant calculations. Thanks
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by [email protected] » Thu Nov 17, 2016 9:26 am
Hi Vikas Mishra,

Since each box of oranges contains from 120 to 144 oranges (inclusive), there are 144 - 120 + 1 = 25 possibilities per box (re: 120 oranges, 121 oranges... 144 oranges). The question asks us to consider how many "duplicates" we could have at the MINIMUM.

With 128 boxes, but only 25 possibilities, there will clearly be many duplicates, so we have to 'spread out' the options as much as we can to find the minimum number of boxes that would share the same number. Imagine working your way through the options, one at a time....

1st crate with 120
1st crate with 121
1st crate with 122
....
1st crate with 144
2nd crate with 120
2nd crate with 121
Etc.

This cycle would repeat until you ran out of boxes, and would end with...

6th crate with 120
6th crate with 121
6th crate with 122

Thus, we would end up with 6 crates (at the minimum) that have duplicate values.

Final Answer: B

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by Matt@VeritasPrep » Fri Nov 25, 2016 3:34 pm
This uses the Pigeonhole Principle. (Check this link before reading the rest of my explanation, unless you're already familiar with the PP.)

These problems are much easier once you're able to identify the pigeons and the pigeonholes. Here, the pigeonholes are the possible number of oranges in each box (120, 121, ..., 144), and the pigeons are the number of boxes we have (128).

Placing the pigeons into the pigeonholes, we find that 128 / 25 = 5, with 3 left over. So if we spread out the pigeons as much as possible, we'll have five in each pigeonhole, with three left to place. Those three pigeons must each go into a pigeonhole that already has at least five pigeons, so we will have SIX pigeons somewhere.
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