1956 AHSME Problems/Problem 19

Problem 19

Two candles of the same height are lighted at the same time. The first is consumed in $4$ hours and the second in $3$ hours. Assuming that each candle burns at a constant rate, in how many hours after being lighted was the first candle twice the height of the second?

$\textbf{(A)}\ \frac{3}{4}\qquad\textbf{(B)}\ 1\frac{1}{2}\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 2\frac{2}{5}\qquad\textbf{(E)}\ 2\frac{1}{2}$

Solution

In $x$ hours, the height of the first candle is $1 - \frac{x}{4}$ and the height of the second candle is $1 - \frac{x}{3}$ Essentially, we are solving the equation $1 - \frac{x}{4} = 2(1 - \frac{x}{3})$ .

$- \frac{x}{4} = 1 - \frac{2x}{3}$ | $-1$

$- 3x = 12 - 8x$ | $\cdot 12$

$5x = 12$ | $+8x$

$x = \frac{12}{5}$ | $/5$

$x = \boxed{\textbf{(D)} \quad 2 \frac{2}{5}}$ hours.

1956 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1956 AHSME Problems/Problem 19

Problem 19

Solution

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