Two candles of the same height are lighted at | vedclass

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(D) Let the initial height of each candle be $x$ and the time elapsed be $T$ hours.The rate of burning for the first candle is $\frac{x}{8}$ per hour.

Vedclass · Accepted Answer

$4 	ext{ h } 48 	ext{ m}$

Vedclass · Answer

(D) Let the initial height of each candle be $x$ and the time elapsed be $T$ hours.The rate of burning for the first candle is $\frac{x}{8}$ per hour.The rate of burning for the second candle is $\frac{x}{6}$ per hour.After $T$ hours,the remaining height of the first candle is $x - \frac{Tx}{8} = x(1 - \frac{T}{8})$.The remaining height of the second candle is $x - \frac{Tx}{6} = x(1 - \frac{T}{6})$.According to the problem,the ratio of the heights is $2:1$:$\frac{x(1 - T/8)}{x(1 - T/6)} = \frac{2}{1}$$\frac{1 - T/8}{1 - T/6} = 2$$1 - \frac{T}{8} = 2(1 - \frac{T}{6})$$1 - \frac{T}{8} = 2 - \frac{T}{3}$Rearranging the terms to solve for $T$:$\frac{T}{3} - \frac{T}{8} = 2 - 1$$\frac{8T - 3T}{24} = 1$$\frac{5T}{24} = 1$$T = \frac{24}{5} = 4.8 	ext{ hours}$.Since $0.8 	ext{ hours} = 0.8 	imes 60 	ext{ minutes} = 48 	ext{ minutes}$,the time is $4 	ext{ hours } 48 	ext{ minutes}$.

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