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
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.