Posted by carolyn on .
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,
The amount of crayon left is the original lengthamount used.
The amount used depends on time. Now if one assumes both were the same original length (not stated in problem), then
lengthskinny=orglengthtime*origlength/6
lengthfat=origlengthtime*origlength/15
so if lenghtskinny=1/2 lengthfat, then
1/2lengthfat=Lo(1time/6) and
lengthfat=Lo(1time/15)
I would divide the first equation by the second...
1/2= (1time/6)/(1time/15)
multiply both sides by 2(1time/15)
and it ought to solve quickly.
Nice problem: compliments to the originator.