# 1967 AHSME Problems/Problem 27

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## Problem

Two candles of the same length are made of different materials so that one burns out completely at a uniform rate in $3$ hours and the other in $4$ hours. At what time P.M. should the candles be lighted so that, at 4 P.M., one stub is twice the length of the other?

$\textbf{(A) 1:24}\qquad \textbf{(B) 1:28}\qquad \textbf{(C) 1:36}\qquad \textbf{(D) 1:40}\qquad \textbf{(E) 1:48}$

## Solution

If the candles both have length $\ell$, then the candle that burns in $3$ hours has a stub of $\ell$ at $0$ minutes, and a stub of $0$ at $180$ minutes. Since the candle burns at a constant rate (i.e. linearly), the stub length of this candle $t$ minutes after being lit is $f(t) = \frac{\ell}{180}(180 - t)$, since $f(0) = \ell$ and $f(180) = 0$.

Similarly, for the second candle that burns out in $240$ minutes, $g(t) = \frac{\ell}{240}(240 - t)$

Since the first candle burns out faster, if the two candles are lighted simultaneously, it will always have a shorter stub. The problem asks for when $g(t) = 2f(t)$. Solving this equation gives:

$\frac{\ell}{240}(240 - t) = 2\frac{\ell}{180}(180 - t)$

$240 - t = \frac{480}{180}(180 - t)$

$240 - t = 480 - \frac{480}{180}t$

$\frac{8}{3} t - t = 480 - 240$

$t = \frac{3}{5} \cdot 240$

$t = 144$

So, the second candle will have a stub twice as big as the first candle $144$ minutes after they are both lit. If we want this to happen at $4$ PM, the candles have to be lit $144$ minutes earlier, or $2$ hours and $24$ minutes earlier. This is at $\text{1:36 PM}$, which is option $\fbox{C}$