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04 Jul 2025, 08:00
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C
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Difficulty:
55%
(hard)
Question Stats:
73% (02:32) correct
27%
(02:37)
wrong
based on 409
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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. \(\frac{2k(m + n)}{n-m}\)
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04 Jul 2025, 08:30
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Bunuel
GMAT Club Official Explanation:
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.
General Discussion
04 Jul 2025, 08:12
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Information given:
- Thor: k liters in m minutes
- Loki: k liters in n minutes
- Together: 2k liters
Question:
- How much more beer did Thor drink than Loki?
Solution:
- Combined rate: k/m + k/n = k(m+n)/mn
- Total time: 2k / (k(m+n)/mn) = 2mn/(m+n)
- Thor's amount: (k/m) * time = 2kn/(m+n)
- Loki's amount: 2km/(m+n)
- Difference: Thor - Loki = 2k(n-m)/(n+m) = 2k(m-n)/(m+n)
Answer: D, 2k(m-n)/(m+n)
- Thor: k liters in m minutes
- Loki: k liters in n minutes
- Together: 2k liters
Question:
- How much more beer did Thor drink than Loki?
Solution:
- Combined rate: k/m + k/n = k(m+n)/mn
- Total time: 2k / (k(m+n)/mn) = 2mn/(m+n)
- Thor's amount: (k/m) * time = 2kn/(m+n)
- Loki's amount: 2km/(m+n)
- Difference: Thor - Loki = 2k(n-m)/(n+m) = 2k(m-n)/(m+n)
Answer: D, 2k(m-n)/(m+n)
Bunuel
