Two candles of equal height but …

Let height of each candle = {tex}x\ unit.{/tex}
First candle burns off in 6 hours.
Second candle burns off in 8 hours.
Height of 1st candle after burning for 1 hr = {tex}\frac{x}{6}{/tex}{tex}unit{/tex}
and height of 2nd candle after burning for 1 hr = {tex}\frac{x}{8}{/tex}{tex}unit{/tex}
Let the required time {tex}= y\ hrs{/tex}.
Length of 1st candle burnt after y hrs = {tex}\frac{y \times x}{6}{/tex}{tex}unit{/tex}
Height of 1st candle left = {tex}\left(x-\frac{x y}{6}\right){/tex}
Length of 2nd candle burnt after y hrs = {tex}\left(\frac{y \times x}{8}\right){/tex}{tex}unit{/tex}
Height of 2nd candle left = {tex}\left(x-\frac{x y}{8}\right){/tex}
According to the question,
Height of 1st candle = {tex}\frac{1}{2} \times{/tex}Height of 2nd candle
{tex}\Rightarrow \quad x-\frac{x y}{6}=\frac{1}{2}\left(x-\frac{x y}{8}\right){/tex}
{tex}\Rightarrow \quad x\left(1-\frac{y}{6}\right)=\frac{1}{2} x\left(1-\frac{y}{8}\right){/tex}
{tex}1-\frac{y}{6}=\frac{1}{2}\left(1-\frac{y}{8}\right){/tex}
{tex}\Rightarrow \quad 2-\frac{y}{3}=1-\frac{y}{8}{/tex}
{tex}2 -1 ={/tex} {tex}\frac{y}{3}-\frac{y}{8}{/tex}
1 = {tex}\frac{8 y-3 y}{24}{/tex}
{tex}\Rightarrow{/tex} {tex}24 = 5y{/tex}
{tex}\Rightarrow{/tex} {tex}y ={/tex} {tex}\frac{24}{5}{/tex}
{tex}y = 4.8 hours = 4\ hours\ 48\ minutes.{/tex}