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

and B has 30 –

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 remainingand B has 30 –

*x*/6 cm remaining.We want what’s remaining of A to be twice what’s remaining of B:

30 –

(8

5

*x*/8 = 2(30 –*x*/6) = 60 –*x*/3*x*/3 –*x*/8 = 30(8

*x*– 3*x*)/24 = 305

*x*= 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.