Please help me figure out this one: Two candles each have length of 10 inches. but they arenot identical except for the lengths. they are lit at the sametime. One candle burning at a constant rate burns out in 4 hourswhile other one burning at a constant rate burns out in 5 hours.After how many hours will the length of the 5 hour candle be3-times the length of the four hour candle? Thanks!! Please help me figure out this one: Two candles each have length of 10 inches. but they arenot identical except for the lengths. they are lit at the sametime. One candle burning at a constant rate burns out in 4 hourswhile other one burning at a constant rate burns out in 5 hours.After how many hours will the length of the 5 hour candle be3-times the length of the four hour candle? Thanks!! Solution One candle burns at the rate of 10in/5hr or 2 in/hr so if t ismeasured in hours the length the candle left is d=10-2t. Forcandle number 2 it burns at 10/4=5/2 in/hr so the legnth of thecandle left is e=10-(5/2)t. So the length of the first candled is 3 times the length of the second candle e, sod=3e. So 10-2t=30-(15/2)t, so -20=(4/2)t- (15/2)tso -20=-(11/2)t Thus 40=11t so t=40/11 hrs. (or about 3.64 hrs).

1. Please help me figure out this one:
Two candles each have length of 10 inches. but they arenot identical except for the lengths. they
are lit at the sametime. One candle burning at a constant rate burns out in 4 hourswhile other one
burning at a constant rate burns out in 5 hours.After how many hours will the length of the 5
hour candle be3-times the length of the four hour candle?
Thanks!!
Please help me figure out this one:
Two candles each have length of 10 inches. but they arenot identical except for the lengths. they
are lit at the sametime. One candle burning at a constant rate burns out in 4 hourswhile other one
burning at a constant rate burns out in 5 hours.After how many hours will the length of the 5
hour candle be3-times the length of the four hour candle?
Thanks!!
Solution
One candle burns at the rate of 10in/5hr or 2 in/hr so if t ismeasured in hours the length the
candle left is d=10-2t. Forcandle number 2 it burns at 10/4=5/2 in/hr so the legnth of thecandle
left is e=10-(5/2)t. So the length of the first candled is 3 times the length of the second candle e,
sod=3e. So 10-2t=30-(15/2)t, so -20=(4/2)t- (15/2)tso -20=-(11/2)t
Thus 40=11t so t=40/11 hrs. (or about 3.64 hrs)