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?

Solution

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

$A$ burns $\frac{1}{4}$ of 12 inches in 1 hour = 3 in/hr.

$B$ burns $\frac{1}{3}$ of 12 inches in 1 hour = 4 in/hr.

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

\begin{aligned} 12 - 3x &=& 2(12 - 4x), \\ 12 - 3x &=& 24 - 8x, \\ 5x &=& 12, \\ x &=& \frac{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 $B(x) = 2 A(x)$ ?