Two candles of the same height are lighted at the same time. The first is consumed in $8$ hours and the second in $6$ hours. Assuming that each candle burns at a constant rate, in how many hours after being lighted, the ratio between the height of first and second candles becomes $2:1$?

The correct option is C: $4$ hours $48$ minutes

Given: For first candle (A), Time taken to consumed $=8$ hours

and, For the second candle(B), Time taken to be consumed $=6$ hours

We have to find the time, when ratio of the height of first to the height of second $=2:1$

i.e, Height of candle A $=$ 2$\times$ Height of candle B

Let us consider the height of the two candles as $24cm,$ which is the LCM of the time taken to be consumed i.e $8$ hours and $6$ hours respectively.

Let ${}^{\prime }{x}^{\prime }$ the time taken at which candle A is twice the height of candle B.

In $8$ hours the height of candle A consumed is $24cm.$

In ${}^{\prime }{x}^{\prime }$ hours the height of candle A consumed will be $\frac{24}{8}\times x=3xcm$

Height of candle A in ${}^{\prime }{x}^{\prime }$ hour time $=24-3xcm\dots \dots \left(i\right)$

In $6$ hours the height of candle B consumed is $24cm.$

In ${}^{\prime }{x}^{\prime }$ hours the height of candle B consumed will be $\frac{24}{6}\times x=4xcm$

Height of candle B in ${}^{\prime }{x}^{\prime }$ hours time $=24-4xcm\dots \dots \left(ii\right)$

After time ${}^{\prime }{x}^{\prime }$ hours, we have

$\frac{\text{Height of Candle A}}{\text{Height of Candle B}}=\frac{2}{1}$

$\Rightarrow \frac{24-3x}{24-4x}=\frac{2}{1}$ [from $eq.\left(i\right)$ and $\left(ii\right)$]

$\Rightarrow 1(24-3x)=2(24-4x)$

$\Rightarrow 24-3x=48-8x$

$\Rightarrow -3x+8x=48-24$

$\Rightarrow 5=24x$

$\Rightarrow x=\frac{24}{5}=4\frac{4}{5}$ hours $=4$ hours $+\frac{4}{5}\times 60$ minutes $=4$ hours $48$ minutes

Hence, after $4$ hours 48 minutes, the ratio of height of first candle (A) to the second candle will be $2:1$