Posted by **carolyn** on Monday, September 15, 2008 at 11:27am.

Rachel loves to color. One of her fat crayons wears down to nothing after 15 hours and her skinny crayon wears down to nothing after 6 hours.. One day she starts coloring at 10:30am with two crayons one fat one skinny with the same height. At what time will the skinny crayon be 1/2 height of fat crayon

- crayon riddle 8th grade -
**bobpursley**, Monday, September 15, 2008 at 11:39am
The amount of crayon left is the original length-amount used.

The amount used depends on time. Now if one assumes both were the same original length (not stated in problem), then

lengthskinny=orglength-time*origlength/6

lengthfat=origlength-time*origlength/15

so if lenghtskinny=1/2 lengthfat, then

1/2lengthfat=Lo(1-time/6) and

lengthfat=Lo(1-time/15)

I would divide the first equation by the second...

1/2= (1-time/6)/(1-time/15)

multiply both sides by 2(1-time/15)

and it ought to solve quickly.

Nice problem: compliments to the originator.

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