Answer / sasidhar

Let the initial height of the candles be equal to 1 unit. If the candles have burned for x hours for the first candle to be half the height of the second candle, we can write

(x/4)=((4-x)/4)


(x/6)=((6-x)/6)

Height of the first candle left = (1/2)Height of the second candle


((4-x)/4)=((6-x)/6)

x=1 hr

so ramesh light the candles before=total length of thin candle-remaining length of thin candle)=4-1=3hr

Answer / guest

2 hours ago

Answer / sarala

1 hour answer,how to know correct or wrong is the my answer

Answer / sandhya

let L be the length of candles and t be the time(hrs) he
litted the candles.

after t hrs the propotions of loss is same...

so..
for thicker 6-t/4-t =(L-x/2)/(L-x/4)

then u get t as 2Hrs

Answer / palaniappan

let x be time at witch RAMESH lighted the candles
since thicker one last for 6hrs &thinner for 2hrs lesser
than thicker
so,
6=2x
x=6/2
=>x=3hrs

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