Two candles of the same length (challenge problem)
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26 Sep 2021, 09:01
To solve this one, let's try and turn into math what we know:
Firstly, we'll call the faster burning candle \(A\) and the other \(B\), and we'll let \(x\) be the rate of melting of the candle \(A\) and \(y\) be the rate of melting for candle \(B\).
You can think of the candles melting as an analogy to someone running a race: as time goes on, the full length of the candle (race track) is decreasing. In other words, the length of the candle is jointly proportional to the rate at which it melts and the time that passes.
So we'll use \(d = rt\) to help us solve this, with \(d\) being the length of the candle.
We know that \(d = x(3)\) and \(d = y(4)\), so this leads to:
\( 3x = 4y\)
Which we'll use later.
Now, to find out how long it'll take for one to be half the length of the other, we can think about it like this:
Given the rates of each, there must be some time \(t\) that satisfies:
\(d - yt = 2(d - xt)\)
Since from a logical standpoint, Candle \(A\) burns faster than Candle \(B\), the scenario where this is true must have that the slower burning candle ends up being double the length of the faster burning one.
Essentially, \(xt\) and \(yt\) represent the rates multiplied by the time we're looking for, which will give us the length that melted for Candle \(A\) and likewise, the length that melted for Candle \(B\), after time \(t\). We subtract that from the total length \(d\) which gives us the remaining length of each candle. From here, one candle will be double the length of the other, hence one side of the equality being multiplied by 2.
Now, with the equation above, we have a system of equations:
\( 3x = 4y\)
\(d - yt = 2(d - xt)\)
Let's simplify with the second equation:
\(d - yt = 2d - 2xt\)
\(2xt - yt = d\)
Now:
\( 3x = 4y\) is the same as \(y = \frac{3}{4}x\), so:
\(2xt - (\frac{3}{4}x)t = d\)
\(t(2x - \frac{3}{4}x) = d\)
\(t(\frac{5}{4}x) = d \)
Now \(d = 3x\), and rearranging for \(x\) we get \(x = \frac{d}{3}\), so:
\(t(\frac{5}{4} * \frac{d}{3}) = d \)
We can now divide both sides by \(d\):
\(t(\frac{5}{4} * \frac{1}{3}) = 1\)
\(t(\frac{5}{12}) = 1\)
\(t = \frac{12}{5}\)
Since \(t\) is in hours, we have \(\frac{12}{5}\) being equivalent to 2.4 hours, or 2 hours and 24 minutes.
Therefore, if we light both candles 2 hours and 24 minutes before 4pm, Candle \(B\) will end up being double the length of Candle \(A\) at 4pm.
So the answer is lighting both candles at 1:36pm