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Burning Candles

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   ..........[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   .............[2]

When we multiply by three we find 6L = 4.5S. Insert this in [1], to 

   4.5S = 4S + 1   (subtract 4S from both sides)
   0.5S = 1        (multiply by 2)
      S = 2

When we insert this in [2] 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   
Associated Topics:
Middle School Algebra
Middle School Word Problems

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