Question

Two candles of the same height are lighted at the same timee of the 1st is consumed in 4 hours and the second is 3 hours assuming that each candle burns at constant rate in how many hours after being lighted was the first candle twice the height of second.

Answer 1

For candle A time taken for $100\%$ bearning = 4 hour
So in 1 hour it burns for $25\%\left(\frac{100}{4}\right)$
So, after 1 hr. $\frac{25}{100}$ i.e., $\frac{1}{4}$ of it has burn 1
Hence, after x hours, amount burnt = $\frac{x}{4}$
So, amount left =$1-\frac{x}{4}=\frac{4-x}{4}$
Simply for candle B, let no assume that its height is half of candle A after x hours.
So, after x hours, part vemacing =$1-\frac{x}{3}=\frac{3-x}{3}$
According to question,
$\frac{4-x}{4}=1\left(\frac{3-x}{3}\right)$
i.e, Height of candle A=2$\times$ Height of candle B.
$\therefore 12-3x=24-8x$
$\Rightarrow 5x=12$
$x=\frac{12}{5}$
$=2.4$
So, after 2.4 hours, the height of candle 4 is twice of that of candle B.