Glenice has two candles of the same length, but made of different materials so that one lasts 4 hours and the other lasts 3 (each burns at a constant rate). At what time should Glenice light both candles so that at 4:00 pm one stub is exactly twice the length of the other?
Let’s assume (without loss of generality) that the candles are each 30 cm tall – about a foot.
A burns 1/240 of 30 cm in 1 minute = 30/240 = 1/8 cm/min.
B burns 1/180 of 30 cm in 1 minute = 30/180 = 1/6 cm/min.
After x minutes, A has 30 – x/8 cm remaining
and B has 30 – x/6 cm remaining.
A burns 1/240 of 30 cm in 1 minute = 30/240 = 1/8 cm/min.
B burns 1/180 of 30 cm in 1 minute = 30/180 = 1/6 cm/min.
After x minutes, A has 30 – x/8 cm remaining
and B has 30 – x/6 cm remaining.
We want what’s remaining of A to be twice what’s remaining of B:
30 – x/8 = 2(30 – x/6) = 60 – x/3
x/3 – x/8 = 30
(8x – 3x)/24 = 30
5x = 30 · 24
x = 6 · 24
x = 144 min.
x/3 – x/8 = 30
(8x – 3x)/24 = 30
5x = 30 · 24
x = 6 · 24
x = 144 min.
Now 144 minutes before 4:00 pm is 1:36 pm. Light ’em up at 1:36, Glenice, and tell us what’s happening at 4:00.