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04 Jul 2025, 10:08
Kudos
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In 1 minutes, Thor drinks k/m liter, Loki drinks k/n liter. So together they drink 2k liters in t minutes, described as follows:
t = 2k /(k/m + k/n) = 2 / ((n+m)/mn) = 2mn/(n+m)
The liters that Thor drink more than Loki are:
2mn/(m+n) x (k/m - k/n) = (2mnk/(m+n)) x ((n-m)/nm) = 2k x (n-m)/(n+m)
Answer: C
t = 2k /(k/m + k/n) = 2 / ((n+m)/mn) = 2mn/(n+m)
The liters that Thor drink more than Loki are:
2mn/(m+n) x (k/m - k/n) = (2mnk/(m+n)) x ((n-m)/nm) = 2k x (n-m)/(n+m)
Answer: C
04 Jul 2025, 10:19
Kudos
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Let's layout what's given to us:
For Thor:
Work = \(k\)
Time = \(m\)
we can calculate, Rate = \(\frac{k}{m}\)
For Loki:
Work = \(k\)
Time = \(n\)
we can calculate, Rate = \(\frac{k}{n}\)
Also we are told that Loki's rate is slower than Thor's, so we can say, time taken by Loki is greater than Thor's
Hence, \(n>m\)
Now, if they were to work together, they would work at a rate of
\(R_t\) = Rate of Thor + Rate of Loki
Thus,
\(R_t \)= \(\frac{k}{m}\) + \(\frac{k}{n}\) = \(\frac{k(m+n)}{mn}\)
They together drink \(2k\) liters of beer, so, \(W_t = 2k\)
Time taken by both working together, is \(T=\frac{W_t}{R_t}\) =\(\frac{2mn}{(m+n)}\)
Now we know that, both Thor and Loki worked for \(T\) minutes. Let's calculate the beer they drank in that time individually using their individual rates:
Beer drank by Thor: \((Thor's rate)*(T)\) = \(\frac{k}{m}*\frac{2mn}{(m+n)}\) = \(\frac{2kn}{(m+n)}\)
Beer drank by Loki: \((Loki's rate)*(T)\) = \(\frac{k}{n}*\frac{2mn}{(m+n)}\) = \(\frac{2km}{(m+n)}\)
Diff between beer drank by Thor and Loki = Beer drank by Thor - Beer drank by Loki
\(\frac{2kn}{(m+n)}\) - \(\frac{2km}{(m+n)}\) = \(\frac{2k(n-m)}{(m+n)}\)
Answer C.
For Thor:
Work = \(k\)
Time = \(m\)
we can calculate, Rate = \(\frac{k}{m}\)
For Loki:
Work = \(k\)
Time = \(n\)
we can calculate, Rate = \(\frac{k}{n}\)
Also we are told that Loki's rate is slower than Thor's, so we can say, time taken by Loki is greater than Thor's
Hence, \(n>m\)
Now, if they were to work together, they would work at a rate of
\(R_t\) = Rate of Thor + Rate of Loki
Thus,
\(R_t \)= \(\frac{k}{m}\) + \(\frac{k}{n}\) = \(\frac{k(m+n)}{mn}\)
They together drink \(2k\) liters of beer, so, \(W_t = 2k\)
Time taken by both working together, is \(T=\frac{W_t}{R_t}\) =\(\frac{2mn}{(m+n)}\)
Now we know that, both Thor and Loki worked for \(T\) minutes. Let's calculate the beer they drank in that time individually using their individual rates:
Beer drank by Thor: \((Thor's rate)*(T)\) = \(\frac{k}{m}*\frac{2mn}{(m+n)}\) = \(\frac{2kn}{(m+n)}\)
Beer drank by Loki: \((Loki's rate)*(T)\) = \(\frac{k}{n}*\frac{2mn}{(m+n)}\) = \(\frac{2km}{(m+n)}\)
Diff between beer drank by Thor and Loki = Beer drank by Thor - Beer drank by Loki
\(\frac{2kn}{(m+n)}\) - \(\frac{2km}{(m+n)}\) = \(\frac{2k(n-m)}{(m+n)}\)
Answer C.
04 Jul 2025, 10:30
Kudos
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Thor’s Rate: k/m
Loki’s Rate: k/n
Rates sum:
K/m + k/n
Kn/mn + km/mn
K(n+m)/mn
Know, we can use the formula Rate x Time = Work to calculate the time.
Time = Work / Rate
Work = 2k
Time = 2k / (k(n+m)/nm)
Time = ( 2knm/ (k(n+m) )
Time = (2nm)/(n+m)
Thor’s work: Rate x Time
(K/m) * (2nm)/(n+m) = (2kn)/(n+m)
Loki’s work: Rate x Time
(K/n) * (2nm)/(n+m) = (2km)/(n+m)
Thor - Loki:
((2kn)/(n+m)) - ((2km)/(n+m))
(2kn - 2km) / (n+m) = (2k(n-m)) / (n+m)
Answer C
Loki’s Rate: k/n
Rates sum:
K/m + k/n
Kn/mn + km/mn
K(n+m)/mn
Know, we can use the formula Rate x Time = Work to calculate the time.
Time = Work / Rate
Work = 2k
Time = 2k / (k(n+m)/nm)
Time = ( 2knm/ (k(n+m) )
Time = (2nm)/(n+m)
Thor’s work: Rate x Time
(K/m) * (2nm)/(n+m) = (2kn)/(n+m)
Loki’s work: Rate x Time
(K/n) * (2nm)/(n+m) = (2km)/(n+m)
Thor - Loki:
((2kn)/(n+m)) - ((2km)/(n+m))
(2kn - 2km) / (n+m) = (2k(n-m)) / (n+m)
Answer C
