Last visit was: 26 Apr 2026, 07:13 It is currently 26 Apr 2026, 07:13
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
nick1816
User avatar
Retired Moderator
Joined: 19 Oct 2018
Last visit: 12 Mar 2026
Posts: 1,841
Own Kudos:
8,511
 [33]
Given Kudos: 707
Location: India
Posts: 1,841
Kudos: 8,511
 [33]
Consider 3 friends A,B and C work at different speed. When the slowest 09 May 2019, 11:18
3
Kudos
Add Kudos
30
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

54% (03:18) correct 46% (02:52) wrong based on 177 sessions

History

Date
Time
Result
Not Attempted Yet
Consider 3 friends A,B and C work at different speed. When the slowest 2 work together, they take m days to finish the task and when 2 quickest of them work together, they take n days to finish the same task. One of them if work alone would take thrice as much time as it would take when all three work together. How much time it will take when all three work together?

1. \(\frac{3mn}{2(m+n)}\)
2. \(\frac{2mn}{(m+n)}\)
3. \(\frac{4mn}{3(m+n)}\)
4. \(\frac{5mn}{3(m+n)}\)
5. \(\frac{8mn}{3(m+n)}\)
ShowHide Answer
Official Answer
C
Most Helpful Reply
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 26 Apr 2026
Posts: 11,229
Own Kudos:
45,021
 [9]
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
My Reviews
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,229
Kudos: 45,021
 [9]
Consider 3 friends A,B and C work at different speed. When the slowest 10 May 2019, 07:26
2
Kudos
Add Kudos
7
Bookmarks
Bookmark this Post
nick1816
Consider 3 friends A,B and C work at different speed. When the slowest 2 work together, they take m days to finish the task and when 2 quickest of them work together, they take n days to finish the same task. One of them if work alone would take thrice as much time as it would take when all three work together. How much time it will take when all three work together?

1. \(\frac{3mn}{2(m+n)}\)
2. \(\frac{2mn}{(m+n)}\)
3. \(\frac{4mn}{3(m+n)}\)
4. \(\frac{5mn}{3(m+n)}\)
5. \(\frac{8mn}{3(m+n)}\)
Show more

Let them take A, B and C days with A>B>C..
So (1/A)+(1/B)=1/m...(I)
And (1/B)+(1/C)=1/n...(II)
The average days has to be the middle value so B..
Therefore, 1/B =(1/3)(1/A+1/B+1/C)..

Add I and II, so
(1/A)+(1/B)+ (1/B)+(1/C)=1/n+1/m.
(1/A)+(1/B)+ (1/B)+(1/C)-(1/B)=(1/n+1/m)-(1/B)...
(1/A)+(1/B)+(1/C)=(m+n)/mn-(1/3)(1/A+1/B+1/C)..
So =(4/3)(1/A+1/B+1/C)=(m+n)/mn...
1/A+1/B+1/C=(3/4)(m+n)/MN=3(m+n)/4mn..
Since this is one day work, time taken by all three = 4mn/3(m+n)
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 23 Apr 2026
Posts: 16,441
Own Kudos:
79,415
 [7]
Given Kudos: 485
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,441
Kudos: 79,415
 [7]
Consider 3 friends A,B and C work at different speed. When the slowest 10 May 2019, 07:41
3
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
nick1816
Consider 3 friends A,B and C work at different speed. When the slowest 2 work together, they take m days to finish the task and when 2 quickest of them work together, they take n days to finish the same task. One of them if work alone would take thrice as much time as it would take when all three work together. How much time it will take when all three work together?

1. \(\frac{3mn}{2(m+n)}\)
2. \(\frac{2mn}{(m+n)}\)
3. \(\frac{4mn}{3(m+n)}\)
4. \(\frac{5mn}{3(m+n)}\)
5. \(\frac{8mn}{3(m+n)}\)
Show more

Assume values since mn/(m+n) is same in all options. Only multiplication factor is different.
Say their speeds are a, b and c in increasing order. Say a and b together finish in 2 days (= m) and b and c together finish in 1 day (= n)

a + b = 1/2

b + c = 1

a + 2b + c = 3/2

a + b + c = 3/2 - b

Since one of them takes 3 times as much time, he is 1/3rd as slow as all 3 together. So he represents the average of the 3 speeds and hence must be the middle guy, b

3/2 - b = 3b
b = 3/8

a + b + c = 3/2 - 3/8 = 9/8

So all 3 will take 8/9 days. Put m = 2 and n = 1 in options to see that option (C) works.
General Discussion
User avatar
hadimadi
User avatar
Joined: 26 Oct 2021
Last visit: 03 Dec 2022
Posts: 113
Own Kudos:
31
 [1]
Given Kudos: 94
Posts: 113
Kudos: 31
 [1]
Consider 3 friends A,B and C work at different speed. When the slowest 23 Dec 2021, 03:44
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi,

w_s=slowest work rate, w_m=middle work rate, w_h=highest work rate. We are given:

(1) 1/(w_s+w_m)=m -> w_m=(1/m)-w_s
(2) 1/(w_m+w_h)=n -> w_m=(1/n)-w_h
(3) 3w_m=w_s+w_m+w_h -> 2w_m=w_s+w_h

(3) is w_m and not w_s or w_h, because if we assume that the highest work rate would need thrice the time than all three together, that wouldn't work (test algebraically and logically). Taking the slowest work rate, that also wouldn't work, here is more detailed why:

3w_s=w_s+w_m+w_h
3=(w_s+w_m+w_h)/w_s
3=(w_s/w_s)+(w_m/w_s)+(w_h/w_s)
Since w_s<w_m<w_h we get 1 for the first term, something greater than one for the second, and something greater than one for the third term, implying
3=(something greater than 3), and that doesn't work.

Set (1)=(2):

(4) w_h-w_s=(1/n)-(1/m)

Add (3) and (4):
2w_h=(1/n)-(1/m)+w_m
(5) w_h=(1/2n)-(1/2m)+w_m

We add can get w_s from (1) just by rearranging, and adding (5) and rearranged (1) we can eliminate the w_m term in (5), so in total we get:
w_h+w_s=(1/2n)-(1/2m)+(1/m)
(6) w_h+w_s=(m+n)/(2mn)

But from (2) we know that (w_h+w_s)/2=w_m, so adding up (6) and what we know from (2) we get:

w_s+w_m+w_h=[(m+n)/(4mn)]+[(m+n)/(2mn)]
=(3m+n)/(4mn)

That's the combined work rate, to get the time, we we have 1/WR=(4mn)/(3m+n)
User avatar
Sanskar115
User avatar
Joined: 24 Aug 2023
Last visit: 13 Dec 2024
Posts: 24
Own Kudos:
19
 [2]
Given Kudos: 39
Posts: 24
Kudos: 19
 [2]
Consider 3 friends A,B and C work at different speed. When the slowest 13 Oct 2024, 00:05
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Rate A+ Rate B=\(\frac{1}{n}\)
Rate C+ Rate B= \(\frac{1}{m}\)
Rate A+ Rate B + Rate C = \(\frac{1}{x}\)
Rate B = \(\frac{1}{3x}\)
Therefore
\(\frac{1}{n}+\frac{1}{m}=\frac{1}{x}+\frac{1}{3x}\)
\(\frac{(n+m)}{mn}=\frac{4}{3x}\)
\(\frac{ 3(n+m)}{4mn }=\frac{1}{x}\)
x=\(\frac{4mn}{ 3(n+m) }\)
User avatar
Sanskar115
User avatar
Joined: 24 Aug 2023
Last visit: 13 Dec 2024
Posts: 24
Own Kudos:
19
Given Kudos: 39
Posts: 24
Kudos: 19
Consider 3 friends A,B and C work at different speed. When the slowest 13 Oct 2024, 00:05
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Rate A+ Rate B=\(\frac{1}{n}\)
Rate C+ Rate B= \(\frac{1}{m}\)
Rate A+ Rate B + Rate C = \(\frac{1}{x}\)
Rate B = \(\frac{1}{3x}\)
Therefore
\(\frac{1}{n}+\frac{1}{m}=\frac{1}{x}+\frac{1}{3x}\)
\(\frac{(n+m)}{mn}=\frac{4}{3x}\)
\(\frac{ 3(n+m)}{4mn }=\frac{1}{x}\)
x=\(\frac{4mn}{ 3(n+m) }\)
User avatar
wtang123
User avatar
Joined: 16 Apr 2021
Last visit: 25 Jan 2025
Posts: 6
Own Kudos:
3
Given Kudos: 32
Posts: 6
Kudos: 3
Consider 3 friends A,B and C work at different speed. When the slowest 28 Nov 2024, 12:36
Kudos
Add Kudos
Bookmarks
Bookmark this Post
chetan2u
The average days has to be the middle value so B..
Therefore, 1/B =(1/3)(1/A+1/B+1/C)..
Show more
I don't understand why B has to be the middle. It only says either A, B, or C would take 3 times as long as A, B, and C combined
User avatar
hr1212
User avatar
GMAT Forum Director
Joined: 18 Apr 2019
Last visit: 26 Apr 2026
Posts: 929
Own Kudos:
1,338
 [1]
Given Kudos: 2,217
GMAT Focus 1: 775 Q90 V85 DI90
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
GMAT Focus 1: 775 Q90 V85 DI90
Posts: 929
Kudos: 1,338
 [1]
Consider 3 friends A,B and C work at different speed. When the slowest 07 Dec 2024, 14:08
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
wtang123
chetan2u
The average days has to be the middle value so B..
Therefore, 1/B =(1/3)(1/A+1/B+1/C)..
I don't understand why B has to be the middle. It only says either A, B, or C would take 3 times as long as A, B, and C combined
Show more
Let's assume that all three of them can finish work in 1 day,
Therefore, (1/A + 1/B + 1/C) = 1

Now suppose one of them takes 3 times as long, which means 3 days to complete full work and hypothetically let's say it's not B
1/A = 1/3
so, 1/B + 1/C = 2/3

If we divide above fraction in 2 non-equal parts then 1/B < 1/3 and 1/C > 1/3, which results into 1/A becoming the middle fraction of these three values.

Hence, because of our initial assumption (A>B>C) that A takes the maximum time and C takes the least time, 1/B = (1/3)(1/A + 1/B + 1/C) must be true as the central fraction is the only one which can take thrice the time.
Moderators:
Bunuel
Math Expert
109837 posts
Krunaal
Tuck School Moderator
852 posts