Two candles of equal height but different thickness are lighted.the fi

1.	Two candles of equal height but different thickness are lighted.the first burns of in 6 hrs
Answer» Let height of each candle = {tex}x\\ unit.{/tex}First candle burns off in 6 hours.Second candle burns off in\xa08 hours.Height of 1st candle after burning for\xa01 hr =\xa0{tex}\\frac{x}{6}{/tex}{tex}unit{/tex}and height of 2nd candle after burning for\xa01 hr\xa0=\xa0{tex}\\frac{x}{8}{/tex}{tex}unit{/tex}Let the required time {tex}= y\\ hrs{/tex}.Length of 1st candle burnt after y hrs =\xa0{tex}\\frac{y \\times x}{6}{/tex}{tex}unit{/tex}Height of 1st candle left =\xa0{tex}\\left(x-\\frac{x y}{6}\\right){/tex}Length of 2nd candle burnt after y hrs =\xa0{tex}\\left(\\frac{y \\times x}{8}\\right){/tex}{tex}unit{/tex}Height of 2nd candle left =\xa0{tex}\\left(x-\\frac{x y}{8}\\right){/tex}According to the question,Height of 1st candle =\xa0{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}\xa0{tex}\\frac{y}{3}-\\frac{y}{8}{/tex}1 =\xa0{tex}\\frac{8 y-3 y}{24}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}24 = 5y{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}y ={/tex}\xa0{tex}\\frac{24}{5}{/tex}{tex}y = 4.8 hours = 4\\ hours\\ 48\\ minutes.{/tex}

Two candles of equal height but different thickness are lighted.the first burns of in 6 hrs