2420.51 – Glenice Burns her Candles


Glenice has a passion for candles. She has two of the same length. Each burns at a constant rate but they are made of different types of wax so that one lasts 4 hours and the other lasts 3. When should Glenice light her candles so that at midnight one stub is exactly twice the length of the other?

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Solution

Assume for convenience (but without loss of generality) that the two candles (AA and BB) are 12 inches tall, twelve being the least common multiple of 3 and 4.

AA burns 14\dfrac{1}{4} of 12 inches in 1 hour, or 3 inhr\dfrac{3 \ \inch}{\hr}. BB burns 13\dfrac{1}{3} of 12 inches in 1 hour, or 4 inhr\dfrac{4 \ \inch}{\hr}.

Let A(x)A(x) and B(x)B(x) be the length of the candles after xx hours. Then A(x)=123xA(x) = 12 - 3x and B(x)=124xB(x) = 12 - 4x. Since AA is the slower burning candle, we want to find xx so that A(x)=2B(x)A(x) = 2 B(x). Solving,

123x=2(124x)123x=248x5x=12x=125\begin{aligned}12 - 3x &= 2(12 - 4x) \\12 - 3x &= 24 - 8x \\5x &= 12 \\x &= \dfrac{12}{5}\end{aligned}

Since a fifth of an hour is 12 minutes, twelve-fifths is 144 minutes.

One hundred and forty-four minutes before midnight is 9:36 PM. That is when Glenice should light her candles.

Extra Challenge: What happens mathematically if we try to make B(x)=2A(x)B(x) = 2 A(x)?