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31 Jul 2025, 08:54
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Thor, drinking at a constant rate, can finish \(k\) liters of beer in \(m\) minutes, while Loki, drinking at a slower constant rate, can finish the same \(k\) liters in \(n\) minutes. If they start drinking together and consume \(2k\) liters in total, in terms of \(m\), \(n\), and \(k\), how much more beer in liters did Thor drink than Loki?
A. \(\frac{k(m - n)}{2(m+n)}\)
B. \(\frac{k(m - n)}{m+n}\)
C. \(\frac{2k(n - m)}{m+n}\)
D. \(\frac{2k(m - n)}{m+n}\)
E. \(\frac{2k(m + n)}{n-m}\)
A. \(\frac{k(m - n)}{2(m+n)}\)
B. \(\frac{k(m - n)}{m+n}\)
C. \(\frac{2k(n - m)}{m+n}\)
D. \(\frac{2k(m - n)}{m+n}\)
E. \(\frac{2k(m + n)}{n-m}\)
31 Jul 2025, 08:54
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Official Solution:
Thor, drinking at a constant rate, can finish \(k\) liters of beer in \(m\) minutes, while Loki, drinking at a slower constant rate, can finish the same \(k\) liters in \(n\) minutes. If they start drinking together and consume \(2k\) liters in total, in terms of \(m\), \(n\), and \(k\), how much more beer in liters did Thor drink than Loki?
A. \(\frac{k(m - n)}{2(m+n)}\)
B. \(\frac{k(m - n)}{m+n}\)
C. \(\frac{2k(n - m)}{m+n}\)
D. \(\frac{2k(m - n)}{m+n}\)
E. \(\frac{2k(m + n)}{n-m}\)
Thor can drink \(k\) liters in m minutes: \(rate_{Thor} = \frac{\text{amount}}{\text{time}} = \frac{k}{m}\) liters per minute.
Loki can drink \(k\) liters in n minutes: \(rate_{Loki} = \frac{\text{amount}}{\text{time}} = \frac{k}{n}\) liters per minute.
Together, to consume \(2k\) liters at the combined rate of \(\frac{k}{m} + \frac{k}{n}\) liters per minute, they'd take:
\(time = \frac{\text{amount}}{\text{combined rate}} = \frac{2k}{\frac{k}{m} + \frac{k}{n}} = \frac{2k}{k(\frac{1}{m} + \frac{1}{n})} = \frac{2}{\frac{1}{m} + \frac{1}{n}} = \frac{2mn}{m + n}\) minutes.
Therefore, the difference in amounts consumed would be:
{Time} * {Rate of Thor} − {Time} * {Rate of Loki} =
\(= \frac{2mn}{m + n} * \frac{k}{m} - \frac{2mn}{m + n} * \frac{k}{n} = \frac{2k(n - m)}{m + n}\) liters.
Answer: C
Thor, drinking at a constant rate, can finish \(k\) liters of beer in \(m\) minutes, while Loki, drinking at a slower constant rate, can finish the same \(k\) liters in \(n\) minutes. If they start drinking together and consume \(2k\) liters in total, in terms of \(m\), \(n\), and \(k\), how much more beer in liters did Thor drink than Loki?
A. \(\frac{k(m - n)}{2(m+n)}\)
B. \(\frac{k(m - n)}{m+n}\)
C. \(\frac{2k(n - m)}{m+n}\)
D. \(\frac{2k(m - n)}{m+n}\)
E. \(\frac{2k(m + n)}{n-m}\)
Thor can drink \(k\) liters in m minutes: \(rate_{Thor} = \frac{\text{amount}}{\text{time}} = \frac{k}{m}\) liters per minute.
Loki can drink \(k\) liters in n minutes: \(rate_{Loki} = \frac{\text{amount}}{\text{time}} = \frac{k}{n}\) liters per minute.
Together, to consume \(2k\) liters at the combined rate of \(\frac{k}{m} + \frac{k}{n}\) liters per minute, they'd take:
\(time = \frac{\text{amount}}{\text{combined rate}} = \frac{2k}{\frac{k}{m} + \frac{k}{n}} = \frac{2k}{k(\frac{1}{m} + \frac{1}{n})} = \frac{2}{\frac{1}{m} + \frac{1}{n}} = \frac{2mn}{m + n}\) minutes.
Therefore, the difference in amounts consumed would be:
{Time} * {Rate of Thor} − {Time} * {Rate of Loki} =
\(= \frac{2mn}{m + n} * \frac{k}{m} - \frac{2mn}{m + n} * \frac{k}{n} = \frac{2k(n - m)}{m + n}\) liters.
Answer: C
