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There are two candles with the thick one half as high as the thin one.
The thin candle can be consumed in one hour,
and thick one can be consumed in two hours after each is lighted.
If they are lighted at the same time, how long in minutes will they be the same height?
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Let h be the initial height of the shorter (thick) candle;
then the initial height of the longer (thin) candle is 2h.
The rate of burning is 2h units of length per hour for the thin candle and h units of length for the thick candle.
Now I write equations for remaining heights as functions of time
= h - h*t
= 2h - 2h*t
where "t" is the time, in hours.
We want to find "t" when = .
It gives us THIS equation
h - h*t = 2h - 2h*t.
Simplify. First, cancel the common factor "h" in both sides
1 - t = 2 - 2t.
Next,
2t - t = 2 - 1
t = 1.
ANSWER. They will be of the same height in 1 hour (when both will exaust).
Solved.
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Now, when the solution is completed algebraically, we (you and me) do understand (finally) that the answer is OBVIOUS
and could be predicted MENTALLY without any calculations.
So, in this sense, it is an ENTERTAINMENT problem.
With some educational value in it . . .
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Candle problems are nice special class of problems.
See the lesson
- Burning candles
in this site to wider your horizon.
You will find many similar solved problems there.
Actually, candle problems are a special class of Travel & Distance problems.
If you are experienced in solving Travel & Distance problems, you will easily solve any candle problem.