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SSWTtoKellogg
Joined: 06 Mar 2024
Last visit: 26 Apr 2026
Posts: 90
Given Kudos: 20
Location: India
Schools: INSEAD Jan '27
GMAT Focus 1: 595 Q83 V78 DI77 
GMAT Focus 2: 645 Q87 V79 DI79
GPA: 8.8
05 Jul 2025, 04:39
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In total 2K Liter, Thor would have drunk more than Loki. Let's assume, Thor would have drunk more than K Liter in x min.
Then we need; kx/m - kx/n =?
We know both together have drunk kx/m + kx/n = 2k => x = 2mn/(m+n)
Putting x value in above expression, answer is [2k(n−m)]/(m+n)
Answer is C.
Then we need; kx/m - kx/n =?
We know both together have drunk kx/m + kx/n = 2k => x = 2mn/(m+n)
Putting x value in above expression, answer is [2k(n−m)]/(m+n)
Answer is C.
05 Jul 2025, 04:45
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Let drinking rate of Thor and Loki be \(r_t & r_l\) respectively.
Then from the details given,
\(r_t = k/m\) litres/minute;\( r_l = k/n\) litres/minute
The total rate \(R = k/n + k/m = k(n+m)/mn\) litres/minute
Time taken to drink \(2K\) litres at combined rate, \(T = 2K * mn/k(n + m)\); \(t = 2mn/(n + m)\) minutes.
The question asks us to find out how much more beer in liters did Thor drink than Loki, let it be v
Therefore,
\(v = r_t * T - r_l * T\)
= \(T(r_t - r_l)\)
Substituting & rearranging we get,
\(v = 2K*(n-m) /(n+m)\)
So the correct option is Option C
Then from the details given,
\(r_t = k/m\) litres/minute;\( r_l = k/n\) litres/minute
The total rate \(R = k/n + k/m = k(n+m)/mn\) litres/minute
Time taken to drink \(2K\) litres at combined rate, \(T = 2K * mn/k(n + m)\); \(t = 2mn/(n + m)\) minutes.
The question asks us to find out how much more beer in liters did Thor drink than Loki, let it be v
Therefore,
\(v = r_t * T - r_l * T\)
= \(T(r_t - r_l)\)
Substituting & rearranging we get,
\(v = 2K*(n-m) /(n+m)\)
So the correct option is Option C
05 Jul 2025, 04:54
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Question Recap:
Thor can finish k liters in m minutes → Rate of Thor = k/m liters/min
Loki can finish k liters in n minutes → Rate of Loki = k/n liters/min
They drink together and finish 2k liters in total
We are to find: How much more beer did Thor drink than Loki, in terms of m, n, and k?
---
Step 1: Let’s find the time they drink together
Let the total time they drank together be t minutes.
Since both drink at constant rates, total beer drunk = rate × time for each.
So,
Thor drinks = (k/m) × t liters
Loki drinks = (k/n) × t liters
Total consumed = 2k liters
⇒ (k/m)·t + (k/n)·t = 2k
Factor t:
> t·(k/m + k/n) = 2k
t·k(1/m + 1/n) = 2k
t·(1/m + 1/n) = 2
t = 2 / (1/m + 1/n)
Now simplify:
> 1/m + 1/n = (m + n)/(mn)
⇒ t = 2 / ((m + n)/mn) = 2mn / (m + n)
---
Step 2: Use t to calculate how much each drank
Thor's consumption:
> Thor drank = (k/m) × t = (k/m) × (2mn)/(m + n) = (2k·n)/(m + n)
Loki's consumption:
> Loki drank = (k/n) × t = (k/n) × (2mn)/(m + n) = (2k·m)/(m + n)
---
Step 3: Find the difference
> Thor – Loki = (2k·n)/(m + n) – (2k·m)/(m + n)
= 2k(n – m)/(m + n)
Final Answer: (C)
Thor can finish k liters in m minutes → Rate of Thor = k/m liters/min
Loki can finish k liters in n minutes → Rate of Loki = k/n liters/min
They drink together and finish 2k liters in total
We are to find: How much more beer did Thor drink than Loki, in terms of m, n, and k?
---
Step 1: Let’s find the time they drink together
Let the total time they drank together be t minutes.
Since both drink at constant rates, total beer drunk = rate × time for each.
So,
Thor drinks = (k/m) × t liters
Loki drinks = (k/n) × t liters
Total consumed = 2k liters
⇒ (k/m)·t + (k/n)·t = 2k
Factor t:
> t·(k/m + k/n) = 2k
t·k(1/m + 1/n) = 2k
t·(1/m + 1/n) = 2
t = 2 / (1/m + 1/n)
Now simplify:
> 1/m + 1/n = (m + n)/(mn)
⇒ t = 2 / ((m + n)/mn) = 2mn / (m + n)
---
Step 2: Use t to calculate how much each drank
Thor's consumption:
> Thor drank = (k/m) × t = (k/m) × (2mn)/(m + n) = (2k·n)/(m + n)
Loki's consumption:
> Loki drank = (k/n) × t = (k/n) × (2mn)/(m + n) = (2k·m)/(m + n)
---
Step 3: Find the difference
> Thor – Loki = (2k·n)/(m + n) – (2k·m)/(m + n)
= 2k(n – m)/(m + n)
Final Answer: (C)
