# 1967 AHSME Problems/Problem 27

## 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 $\fbox{C}$

## See also

 1967 AHSME (Problems • Answer Key • Resources) Preceded byProblem 26 Followed byProblem 28 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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