Date: 06/10/99 at 08:56:26 From: Ben Dixon Subject: Algebra Problem On Christmas Eve two candles, one of which is one inch longer than the other, are lighted. The longer one is lighted at 4:30 and the shorter one at 6:00. At 8:30 they are both the same length. The longer one burns out at 10:30 and the shorter one at 10:00. How long was each candle originally? Thank you.
Date: 06/13/99 at 03:51:19 From: Doctor Floor Subject: Re: Algebra Problem Hi, Ben, Thanks for your question. We have to assume that the candles shorten at a constant rate when burning. Let us assume that the shorter one burns down S inches per hour, and the longer one L inches per hour. We know that the total length of the longer candle is 6L, because it burns out in 6 hours. We also know that the shorter candle is 4S, as it burns out in 4 hours. Since the longer candle is 1 inch longer than the shorter one, we find the equation: 6L = 4S + 1 .......... We also know that at 8:30 the longer candle has burned for 4 hours, and the shorter one for 2.5 hours. The remaining length of the longer candle is 6L - 4L = 2L at that time, and the length of the shorter candle is 4S - 2.5S = 1.5S. This gives the equation: 2L = 1.5S ............. When we multiply by three we find 6L = 4.5S. Insert this in , to find: 4.5S = 4S + 1 (subtract 4S from both sides) 0.5S = 1 (multiply by 2) S = 2 When we insert this in  you'll find that L = 1.5 (check it). So the shorter candle originally was 4S = 8 inches, and the longer one 6L = 9 inches. I hope this helped. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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