Two candles A and B of the same height are lighted at the same instant. A is consumed in $4$ h while B in $3$ h. Assume each candle burns at a constant rate. In how many hours after. being lighted was A twice the height of B?

1 h 48 m i n

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2 h 24 m i n

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2 h 18 m i n

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1 h 58 m i n

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Solution

The correct option is B $2 h 24 m i n$
Let us consider the height of candles A and B to be $12$ m, which is the LCM of time taken for them to be consumed, i.e., $4$ hand $3$ h, respectively.
Let x be the time taken at which candle A is twice the height of candle B.
Height of candle A in x time is $12 - 4 x .$
Height of candle B in x time is $12 - 3 x .$
In x time, height of candle $B = 2 \times$ height of candle A. Therefore,
$12 - 3 x = 2 \times (12 - 4 x)$
or $12 - 3 x = 24 - 8 x$
or $8 x - 3 x = 24 - 12$
or $5 x = 12$
or $x = \frac{12}{5} = 2 \frac{2}{3}$
$∴ 2 h \frac{2}{5} \times 60 m i n = 2 h 24 m i n$

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