my way...

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Posted by Denis Borris on October 30, 2002 at 23:24:17:

In Reply to: I know the problem has already been answered, but... posted by T.Gracken on October 30, 2002 at 18:52:18:

My 1st step is usually rewrite the damn thing:

We have a red and a blue candle of same height.
The red burns out in 3 hours, the blue in 2.
Both are lit at the same time.
After an amount of time, the length of the red left
unburned was twice that of the blue.
How long had the candles been burning?

[Now, why problem posers seem to go out of their way to say
something simple in a complicated way, such that it takes
more time to figure out what they're asking than solving the
damn thing is beyond me....burns my ass...]

Let L = Length of candles
Let X = blue unburned length; then 2X = red unburned length
So blue burned = L - X, red burned = L - 2X

red speed = L/3 ; blue speed = L/2

L/3 = (L - 2x) / T ; L/2 = (L - X) / T : took same Time to get where thet are!
SO: L = 4x

Then it's obvious: since red is half-way (2X), then time is 3/2,
or 1.5 hours.

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