Brian has some boxes of paper clips.

Question

Brian has some boxes of paper clips. Some boxes hold 10 clips and some hold 100. He has some paper clips left over. He has three more boxes with 100 paper clips than he has boxes with 10 paper clips. He has two fewer paper clips left over than he has numbers of boxes with 100 paper clips. What number of paper clips could he have?

Answer 1

This is an odd one, partly because it doesn't have only one answer. First, let's organize. You have three elements:

Boxes of 10     : X           because we don't know how many it has, and the next element is compared to it.

Boxes of 100   : X + 3     because we have three more of these than the 10s

Loose              : X + 3 -2 because it said we have two fewer than the 100s

 

But how many paper clips does Brian have? Now it gets easier -- but still odd. Simply multiply the size of each element by its variable expression:

Boxes of 10      : 10*X          = 10X

Boxes of 100    : 100 (X+3)   = 100X + 300

Loose               : 1 (X+1)      = X + 1                       If you total everything, combining like terms, you get:

                                           = 111X + 301              We have no other clues about the value of X, so it can be any

natural number (not zero, because the problem implies that all of the boxes DO have clips). That means you can plug in one or more for X and get yet another final answer. I would start with X = 1, of course...