Question 165043
let c="length of candles"


a candle that burns for h hours will burn t/h of the length in time t
__  the REMAINING length will be c[1-(t/h)]


the slower burning candle will be the longest


c[1-(t/5)]=3c[1-(t/4)] __ dividing by c and distributing __ 1-(t/5)=3-(3t/4)


adding (3t/4)-1 __ (3t/4)-(t/5)=2 __ multiplying by LCD (20) __ 15t-4t=40 __ 11t=40 __ t=40/11