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The first candle burns off in 8 hours and the second in 10 hours. How long until the first candle will be half the height of the second?

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In $n$ hours, $1^{\text {st }}$ candle burns $\frac{n}{8}$ th of its height and in $n$h.

$2^{\text {nd }}$ candle burns $\frac{n}{10}$ th of its height.

Now, the ratios of their respective heights after burning will be $1-\frac{n}{8}: 1-\frac{n}{10}$.

The ratio of height wanted $\rightarrow \frac{1}{2}: 1$.

So, the time taken,
\begin{aligned} & \frac{\frac{8-n}{84}}{\frac{10-n}{10^{5}}}=\frac{1}{2} \Rightarrow \frac{5(8-n)}{4(10-n)}=\frac{1}{2} \\ \Rightarrow & 40-5 n=2(10-n) \Rightarrow 40-5 n=20-2 n . \end{aligned}
$\Rightarrow 3 n=20 \Rightarrow n=\frac{20}{3}=6 \frac{2}{3}$ hours.

Thus, it will take $6 \frac{2}{3}$ hours for the first candle to reach half the height of second.
by Diamond (58,513 points)

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