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GMAT Club Olympics 2025 (Day 5): Thor, drinking at a constant rate

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Tanish9102

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Correct Answer: Option C
Rate of Thor= k/m
Rate of Loki= k/n

If we combined both of there rates we get
k/m+k/n= k(1/m+1/n)
=k(n+m/mn)

Now we will find time taken to drink 2k beer
assume total time= t
Total= Rate*time
k(n+m/mn)*t
We need to find for 2k so,
2k= k(n+m/mn)
K will cancel out and we get
t= 2mn/m+n

Now we will find individual amount of beer
Thor rate= k/m
k/m*t= k/m * 2mn/m+n
we get 2kn/m+n

Loki rate= k/n
k/n*t= k/m * 2mn/m+n
we get 2km/m+n

Now well subtract Thor rate from Loki rate
(2kn/m+n) - (2km/m+n)
Final Answer we got is
2k(n-m)/m+n

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Bunuel

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Thor's drinking rate = k/m
Loki's drinking rate = k/n

Since Loki drinks slowly we have
n< m

Total beer drunk by them together = 2k
rate * time = work
Let's say they both take time 't' to finish the beer

=> kt/m + kt/n = 2k
=> t/m + t/n = 2
=> t(n+m)/nm = 2
=> t = 2nm/(n+m)

Thor drank = k/m * 2nm/(n+m)
Loki drank = k/n * 2nm/(n+m)

Thor drank more by
k/m * 2nm/(n+m) - k/n * 2nm/(n+m)
=> 2kn/(n+m) - 2km/(n+m)
=> $\frac{2k(n-m)}{(n+m)}$

IMHO Option C

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Bunuel

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Bunuel

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Thor can finish k litres of beer in m minutes.
Thor's rate = k/m

Loki can finish k litres of beer in n minutes
Loki's rate = k/n

Total Rate: k/m+k/n
Together they drink 2k litres of beer.
Total Time taken: 2k/(k/m+k/n)
= 2nm/n+m

Volume of Beer Thor drinks:= Time * Thor's Rate
= 2nm/n+m * k/m
= 2kn/m+n

Volume of beer Loki drinks = Time * Loki's Rate
= 2nm/n+m * k/n
= 2km/m+n

How much more beer did Thor drink compared to Loki?
= 2kn/m+n - 2km/m+n
= 2k(n-m)/m+n

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In this we can find combined rate of them drinking ie 2k=(k/m+k/n)*T
So T would be 2mn/m+n
Now how much more beer did he drink would be n-m as n is greater than m
So we would get 2k(n-m)/m+n Option C

Bunuel

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IMO Answer is C

R(Thor)= k/m and R(Loki)= k/n, given R(Thor) is greater than R(Loki)

Combined Rate= k/m+k/n= k(m+n)/mn
Time taken for Thor and Loki to drink 2k beer = 2k/combined rate= 2k/k(m+n)/mn= 2mn/(m+n)

In time 2mn/(m+n), Thor would drink- R(Thor)*time=(k/m)*(2mn/m+n)
and Loki would drink - R(Loki)*time=(k/n)*(2mn/m+n)
so difference would be = (k/m)*(2mn/m+n) -(k/n)*(2mn/m+n)= 2k(n-m)/(m+n)

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Updated on: 05 Jul 2025, 08:37

Originally posted by BongBideshini on 04 Jul 2025, 23:48.
Last edited by BongBideshini on 05 Jul 2025, 08:37, edited 1 time in total.

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Rate of Thor, $ r_T=\frac{k}{m}$
Rate of Loki, $ r_L=\frac{k}{n}$

$r_T:r_L$
$=\frac{k}{m}$:$\frac{k}{n}$
$=n:m$

So, Thor drank $\frac{2kn}{m+n}$ Litres and Loki drank $\frac{2km}{m+n}$ Litres
Thor drank $\frac{2kn}{n+m}-\frac{2km}{n+m}$ Litres more
$=\frac{2k(n-m)}{(m+n)}$

IMO, Ans C

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Thor drinks= k/m lit per min
Loki drinks= k/n lit per min
total they consume=2k lit

Their total rate= (k/m)+(k/n) => k(1/m +1/n)

Total time= Total amount/Total rate
=2k/ k(1/m +1/n) => 2mn/(m+n)

Thor drank= rate*time
=(k/m)*(2mn/(m+n)) => 2kn/(m+n)
Similarly,
Loki drank = 2km/(m+n)

Difference in amt = 2kn/(m+n) - 2km/(m+n)
= 2k(n-m)/(m+n)

Option C

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Bunuel

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Thor's rate -> \frac{k}{m}
Loki's Rate-> \frac{k}{n}

Total rate => \frac{k}{m} + \frac{k}{n}
=k\frac{(m+n)}{(mn)}

Total drank = 2k

Total time = \frac{(Total drank)}{ (Total drank)}
=>\frac{(2k)}{(k[fraction](m+n)/(mn)})[/fraction]

= \frac{(2mn)}{(m+n)}

Thor's rate -> \frac{k}{m}* Total time
= \frac{k}{m} * \frac{(2mn)}{(m+n)}

Loki's Rate-> \frac{k}{n} * Total time
=\frac{k}{n} * \frac{(2mn)}{(m+n)}

how much more beer in liters did Thor drink than Loki:
=>> Thor's rate - Loki's rate

= fraction]k/m[/fraction] * \frac{(2mn)}{(m+n)} - \frac{k}{n} * \frac{(2mn)}{(m+n)}

=\frac{(2mn)}{(m+n)}*( \frac{k}{m} - \frac{k}{n} )

=\frac{(2mn)}{(m+n)}*( \frac{(kn-km)}{(mn)} )

=$\frac{2k(n - m)}{m+n}$

ANSWER:
C. $\frac{2k(n - m)}{m+n}$

Gg153

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05 Jul 2025, 00:30

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Rate of Thor: k/m
Rate of Loki: k/n
Let the time they drank for be x minutes.
(k/m + k/n)x = 2k
solving for x:
x = $\frac{2(mn)}{(m+n)}$
Beer by thor - beer by loki:
kx/m - kx/n
= $\frac{2k(n - m)}{m+n}$
Option C.

Bunuel

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05 Jul 2025, 00:33

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Thor drank 2k(n−m)/(m+n) liters more than Loki.

Thor's drinking rate: k/m liters per minute.
Loki's drinking rate: k/n liters per minute.
Together, they finish 2k liters in 2mn/(m+n) minutes.
Thor's total: 2kn/(m+n)liters.
Loki's total: 2km/(m+n) liters.
The difference: 2k(n−m)/(m+n) liters.

Bunuel

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05 Jul 2025, 00:39

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Option C is the answer.

First lets understand the information mentioned in the question and what needs to be answer in the question. So the questions starts by telling us that "there is 'k' liters of beer which Thor can drink in 'm' minutes and Loki can finish the same amount of beer in 'n' minutes when Loki's drinking speed is less than that of Thor which means that m>n". Then the question asks us "that there if '2k' liter of beer and both Thor and Loki are drinking it at their respective speeds than how much more beer in liters did Thor drink than Loki".

Now after understanding the information mentioned in the question lets try solving it but as we can see instead of number all the information in question as well as in the options are given in the form of variable. While solving the question with variable is considerable but it gets really messy and hectic sometimes so lets try solving it by assuming some numbers.

So lets assume the following vales for the given variables, we are taking the values which will result in getting the integer numbers because trying to solve for decimal numbers also tend to confuse a lot of people.

K = 30 =>2K = 60
Thor's Drinking Rate = 3
Time take by Thor (m) = 10
Loki's Drinking Rate = 2
Time take by Loki (n) = 15
Combined drinking rate of both Thor and Loki = (3+2) => 5
Drinking time taken when both of them drink together Thor and Loki = 2K/(m+n) => 60/5 => 12
According to this Thor must have drank 36 liter of beer (3*12) and Loki must have drank 24 liter of beer (2*12).
Which according to the question will get in: 36-24 => Thor drank 12 liter more beer than Loki.

Now lets check our options and see which option given us the same answer.

Option A: (k(m-n))/(2(m+n)), Now lets the assumed values of k, m & n into the equation: (30*(10-15))/2(10+15) => -3, We the getting the answer in negative which mean Thor drank less beer than Loki which is incorrect. Eliminated

Option B: (k(m-n)/(m+n), Now lets the assumed values of k, m & n into the equation: (30*(10-15))/(10+15) => -6, We the getting the answer in negative which mean Thor drank less beer than Loki which is incorrect. Eliminated

Option C: (2k(n-m))/(m+n), Now lets the assumed values of k, m & n into the equation: (60*(15-10))/(10+15) => 12, Bingo it exactly the answer which we were looking for. Selected

Option D: (2k(m-n))/(m+n), Now lets the assumed values of k, m & n into the equation: (60*(10-15))/(10+15) => -12, We the getting the answer in negative which mean Thor drank less beer than Loki which is incorrect. Eliminated

Option E: (2k(m+n))/(n-m), Now lets the assumed values of k, m & n into the equation: (60*(10+15))/(15-10) => 300, This answer is not possible because the total amount of beer is '60' only so it is not possible to drink '300' liter beer from '60' liter. Eliminated

Here we can see only Option B matches the answer so this must be our answer option.

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05 Jul 2025, 00:43

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Drinking rate of Thor (T): k/m (liters per minute)
Drinking rate of Loki (L): k/n (liters per minute)
=> In 1 minute, Thor and Loki can drink: k/m+k/n = k(m+n)/mn (litters)
=> Total time for both to drink 2k litters = 2*mn/(m+n) (minutes)

In 2k litters:
Thor drank: (k/m)*(2mn/(m+n))= 2kn/(m+n) (litters)
Loki drank: (k/n)*(2mn/(m+n)) = 2km/(m+n) (litters)
=> Thor drank more than Loki by: 2kn/(m+n) - 2km/(m+n) = 2k(n-m)/(m+n) (litters)

Answer: C

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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?

Thor's rate = k/m
Loki's rate =k/n

combined rate for Thor and Loki = K(1/m+1/n)

If they drink 2 k liters together the time taken for both of them given that

Total volume= time taken for both of them * combined rate

Time taken is thus

2k= t * k(1/m+1/n)

divide both sides by k

2= t * (1/m+1/n)
t= 2/(1/m+1/n)
t= 2 /((m+n)/(mn)) = 2 *((mn/(m+n))
t=2mn/(m+n)

How much did Thor drink

kt= k/m*t
kt= k/m*2mn/(m+n)
kt =2kn/(m+n)

How much did Loki drink

Kl= k/n*t
Kl= k/n*2mn/(m+n)
kl =2km/(m+n)

How much more did Thor drink than loki

=Kt- Kl

=2kn/(m+n) -2km/(m+n)

=2K(n-m)/(m+n)

The answer is thus C

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M denotes slower rate of drinking, taking more time whereas N denotes faster rate of drinking.So M-N, multiplied with total litres of drinking,
2k(n-m)/m+n

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Total rate

Thor: k liters in m minutes - speed = k/m liters per minute
Loki: k liters in n minutes - speed = k/n liters per minute

Time to finish 2k liters

Combined speed = k/m + k/n = k(n+m)/(mn)
Time to drink 2k liters = 2k ÷ [k(n+m)/(mn)] = 2mn/(m+n) minutes

how much each drinks
In time 2mn/(m+n):

Thor drinks: (k/m) × 2mn/(m+n) = 2kn/(m+n)
Loki drinks: (k/n) × 2mn/(m+n) = 2km/(m+n)

difference
Thor - Loki
= 2kn - 2km
(m+n) (m+n)

= 2k(n-m)
(m+n)

Since Loki is slower, n > m (takes more time), so n-m > 0

Answer: C. 2k(n−m)/(m+n)

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Given,
Thor’s rate of drinking = k/m
Loki’s rate of drinking = k/n
k/m > k/n
Together they drank 2k litre beer, how much more Thor drink in comparison to Loki?

Assume,
Time taken to drink the total 2k litre beer by Thor & Loki is “t”.
Solution:
From above,
2k = (k/m)*t +(k/n)*t
2k = tk(m+n)/mn
t = 2mn/(m+n) .......(i)

Then, the excess beer drank by thor, x
= (k/m)*t – (k/n)*t
= tk(n-m)/mn
Using value of “t” from equation (i),
= [2mn/(m+n)] * k*[(n-m)/mn]
x = 2k(n-m)/(m+n)

Ans : C

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05 Jul 2025, 03:01

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Let k= 6 litres
m = 3 min
n = 6 min

Rate for Thor = 2 litre/min
Rate for Loki = 1 litre/min

for 2k litres, combined rate is 3 litre/min
time taken is 12/3=4 min

In 4 min Thor can drink, 2x4 = 8 litres
" " " Loki " " , 1x4 = 4 litres

Hence Thor drank 4 litre more. Substitute in options. Will show C as the right answer

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05 Jul 2025, 03:24

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Thor’s speed = k ltr in m mins = k / m ltr/min.
Loki’s speed = k ltr in n mins = k / n ltr/min.
Together they drink at a rate of
(k / m) + (k / n) => k(m + n) / (mn) ltr/min.
They keep drinking until they have finished 2k liters.
Time taken = total amount / combined rate
= 2k / (k(m + n)/(mn))
= 2mn / (m + n) mins.
In that time
Thor drinks : (k/m) * 2mn / (m + n) = 2kn / (m + n) ltr.
Loki drinks : (k/n) * 2mn / (m + n) = 2km / (m + n) ltr.
(Thor − Loki) = (2kn / (m + n)) − (2km / (m + n)) -> 2k(n − m) / (m + n).

Answer : C

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05 Jul 2025, 04:22

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Thor's rate of drinking = k liters / m minutes = k/m
Loki's rate of drinking = k liters / n minutes = k/n

When drinking together, effective total rate of Thor + Loki = Thor's rate + Loki's rate = k/m + k/n
Total time taken by both to consume 2k liters
= $\frac {2k} {k/m + k/n} $
= $\frac {2} {1/m + 1/n} $
= $\frac {2mn} {m+n} $

Amt of beer Thor drinks in these total minutes = rate * time = $\frac {k} {m}$ * $\frac {2mn} {m+n} $ = $\frac {2nk} {m+n} $

Amt of beer Loki drinks in these total minutes = rate * time = $\frac {k} {n}$ * $\frac {2mn} {m+n} $ = $\frac {2mk} {m+n} $

Answer = amt of beer thor drinks - amt of beer loki drinks
= $\frac {2nk} {m+n} $ - $\frac {2mk} {m+n} $
= $\frac {2k (n-m)} {m+n} $
= answer C