document.write( "Question 884541: I have 2 candles with equal lengths. Both candles are lit at the same time. Candle A burns out at 4 hours. Candle B burns out at 3 hours. At what time is the slower burning candle twice as long as the faster burning candle? \n" ); document.write( "

Algebra.Com's Answer #534411 by geetha_rama(94) $\"\"$ $\"About$

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Let x be the length of two candles. s1 and s2 rate of burning of A & B candles respectively.
\n" ); document.write( "For A, x= 4s1 => s1 = x/4
\n" ); document.write( "For B s2 = x/3
\n" ); document.write( "Let t be the time at which candles A is twice as much as candle B height
\n" ); document.write( "At time t, using the formula distance = speed * time
\n" ); document.write( "candle A: $\"t=%28x-y%29%2Fs1\"$ -> eq A
\n" ); document.write( "For candle B $\"t=%28x-%28y%2F2%29%29%2Fs2\"$
\n" ); document.write( "Equating above two we get
\n" ); document.write( " $\"%28x-y%29%2Fs1=%282x-y%29%2F2s2\"$
\n" ); document.write( "substitute for s1, s2
\n" ); document.write( " $\"%28x-y%29%2F%28x%2F4%29+=+%282%2Ax-y%29%2F%282%2A%28x%2F3%29%29\"$
\n" ); document.write( "=> $\"x=%285%2F2%29%2Ay\"$
\n" ); document.write( "Substituting for x,s1 intermns of y in eq A
\n" ); document.write( " $\"t+=+%28x-y%29%2Fs1\"$
\n" ); document.write( "We get t = 2.4 hrs \n" ); document.write( "

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