SOLUTION: 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

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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?
Found 2 solutions by skartikey, geetha_rama:
Answer by skartikey(21) About Me  (Show Source):
You can put this solution on YOUR website!
let length of both candles are 1. After t hours, length remaining of first candle would be 1-t/4 and length remaining of second candle would be 1-t/3. According to the problem:
2(1-t/3)=1-t/4
=>2-2t/3=1-t/4
=>1=5t/12
t=2.4 hours

Answer by geetha_rama(94) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the length of two candles. s1 and s2 rate of burning of A & B candles respectively.
For A, x= 4s1 => s1 = x/4
For B s2 = x/3
Let t be the time at which candles A is twice as much as candle B height
At time t, using the formula distance = speed * time
candle A: t=%28x-y%29%2Fs1 -> eq A
For candle B t=%28x-%28y%2F2%29%29%2Fs2
Equating above two we get
%28x-y%29%2Fs1=%282x-y%29%2F2s2
substitute for s1, s2
%28x-y%29%2F%28x%2F4%29+=+%282%2Ax-y%29%2F%282%2A%28x%2F3%29%29
=>x=%285%2F2%29%2Ay
Substituting for x,s1 intermns of y in eq A
t+=+%28x-y%29%2Fs1
We get t = 2.4 hrs