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Posted by T.Gracken on October 30, 2002 at 18:52:18:

In Reply to: Length of Time posted by RD on October 30, 2002 at 17:47:44:

: Two candles are in a building. One is red and the other is blue. The candles are the same height, but the red is slightly thicker. The red burns out in 3 hours and the blue 2. After an amount of time the length of the red left unburned was twice that of the blue. A person who was in the building had a watch but it stopped, and there was no other clock in the building. How long was the person there?

I didn't see the other responses until after I typed all this crap, so I'm still posting (like it or not...)

choose any positive time unit and you have an answer that can't be refuted legitimately...

however, some will argue a minimum time is required for the person to be considered "in the building".

Also, it was never mentioned whether the person departed the building, so until further information is provided, some will argue that no answer is correct.

But wait, there's more... (i just feel like being 'a little precise' today)

Also not mentioned was whether the candles are lit at the same time.

O.K. I get it,

...however, (in the spirit of 'somewhat precise')

I will assume you (or the author of the question) intended the question to be,

"Two candles are in a building. One is red and the other is blue. The candles are the same height, but the red is slightly thicker. The red burns out in 3 hours and the blue 2. if the two candles were lit at the same time, and... After an amount of time the length of the red left unburned was twice that of the blue. A person who was in the building when the candles were lit had a watch but it stopped, and there was no other clock in the building. if the person was in the building while the above conditions occured, what is the least amount of time the person was in the building?

O.K. here’s one way to view it:

suppose each candle has a height of “h” (units... whether inches, meters, yards, etc.)

now note that the rate the red candle burns is (1/3)h per hour

and note that the rate the blue candle burns is (1/2)h per hour.

(more math stuff...)

let “T” be the time (in hours) that the candles have been lit.

THEN

the height of the blue candle is: h - (1/2)h*T

and the height of the red candle is: h - (1/3)h*T

[if you do not understand what I wrote above, think clearly about the following...
the height of a candle after it has been lit should equal the starting height minus the rate-candle-burns times the time-candle-burns]

So, since we want the time, T, when the height of the red candle is twice the height of the blue candle we want

two times the height of blue = height of red [after T hours]

so

2[h - (1/2)hT] = h - (1/3)hT

...divide both sides by h to get

2 - T = 1 - T/3

solve for T to get

T = 3/2

meaning the time for the conditions to occur is 1 hour 30 minutes.

So, the person must have been in the building at least one hour, thirty minutes.

• my way... Denis Borris 23:24:17 10/30/02 (0)

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