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Quant Strategy: Solving Certain Work Rate Problems using the LCM

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Quant Strategy: Solving Certain Work Rate Problems using the LCM  [#permalink]

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New post 13 Jun 2017, 21:36
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Hello GMATClubbers,

I have recently completed my GMAT and wanted to share an approach I used to solve work rate problems with you all. This forum was really helpful during my preparation and I guess sharing some of the methods I used while solving certain problems is a way I would like to give back to the forum. Hope you all find this useful.

Before I begin, I'd like to make it clear that this is not a method I invented. I came across this a few years ago while studying for some other entrance and found it useful for GMAT as well. I'm sure there was no copyright and hence I'm quite sure I'm not violating one in sharing this method. I;m not being able to find the source now but will definitely look for it and edit the post with it if I find it.

Work rate problems can be a tad bit nasty and messed up at times. It is all good as long as just A and B are at it and you just have one equation and one variable. It can get tricky if there are 3 or more entities involved and they all work for different periods of time. The usual approach to such problems includes the use of variables and fractions which leave plenty of scope for mistakes. The method I use is primarily aimed at avoiding these possibilities.

Let’s call the method I will be discussing the LCM method. The basic principle of this method is to take what all is known, find their LCM and take that to be the total units of work to be done.

Sounds confusing? What better way to clarify than the use of examples.

Example 1:
X can do a work in 5 days and Y can do it in 15 days, In how many ways can they together complete the work?
Standard solution: X can do 1/5 work in one day and Y can do 1/15.
Working together, X and Y can do 1/5+1/15=4/15 of the work. So to complete 1 unit of work they will take 15 days.
LCM Method solution:The LCM of 5 and 15 is 15. So lets assume the total work to be 15 units.
So in 1 day, X does 3 units of work and B does 1 unit. Working together, they can do 4 units of work in 1 day.
Therefore total time is 15/4 days.

This is a simple example and so the new method does not seem to be a great improvement over the standard one. So let's apply it to a complicated one.

Example 2:
X, Y and Z can complete a piece of work in 15 days. The three of them start working together. After 2 days, X leaves. Y and Z work for 10 more days, after which Y also leaves. Z completes the rest of the work in 40 days. If Z had worked alone, he would have taken 75 days to complete the job. In how many days can X alone complete the job?
Standard solution: I wont get into the complications of variables and fractions and multiple equations. I'll leave that for now.
LCM method solution: LCM of 15, 2, 10, 40 and 75 is 600. So let the total work to be done be 600 units.
From the information in the question, Z can do 8 units of work per day. And (X+Y+Z) can together do 40 units per day.
Now lets form a single equation for the full scenario
(X+Y+Z)*2+(Y+Z)*10+Z*40 = 600
80+(Y+Z)*10+320=600
This gives Y+Z as 20 units.
Therefore, X's one day;s work = (X+Y+z) - (Y+Z) = 40 - 20 = 20
So X alone will take 600/20 = 30 days to complete the work alone

Example 3:
There are 3 pipes X, Y and Z attached to a tank. X and Y can fill the tank alone in 20 minutes and 30 minutes respectively while Z can empty it alone in 45 minutes. If all three of them are opened together, in how much time will the tank be full?
LCM Method solution: Let total units of work be 180 (LCM of 20, 30 and 45)
So X and Y can do 9 and 6 units of work per minute while Z can do 4 units.
Working together, total work per minute = 9 + 6 - 4 = 11 units.
So total time taken will be 180/11 minutes

Guess it seems more straightforward now. Let's apply this to a few GMAT questions picked up from this forum itself.

Example 4:
Working at their respective constant rates, Paul, Abdul and Adam alone can finish a certain work in 3, 4, and 5 hours respectively. If all three work together to finish the work, what fraction of the work will be done by Adam?
(A) 1/4
(B) 12/47
(C) 1/3
(D) 5/12
(E) 20/47
LCM Method solution: LCM of 3,4 and 5 is 60. So let the total amount of work be 60 units.
So based on the information give, Paul, Abdul and Adam can do 20 units, 15 units and 12 units of work per hour respectively.
So together, they will do (20+15+12) or 47 units of work in one hour.
Since Adam will do 12 units of this, fraction of work done by Adam will be 12/47. Option (B)

Example 5:
Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A’s speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?
(A) 1/2
(B) 2
(C) 3
(D) 5
(E) 6
LCM Method solution: LCM of 3 and 2 is 6. So let total work done be 6 units.
This means A and B can do 2 units of work per hour.
If A's speed is doubled, they do 3 units of work per hour.
So the extra unit of work is done simply because A's speed is doubled. This implies A's initial speed was 1 unit per hour.
Since total work to be done is 6 units, A will take 6 hours. Option (E)

Example 6:
A pool can be filled in 4 hours and drained in 5 hours. The valve that fills the pool was opened at 1:00 pm and some time later the drain that empties the pool was also opened. If the pool was filled by 11:00 pm and not earlier, when was the drain opened?
A. at 2:00 pm
B. at 2:30 pm
C. at 3:00 pm
D. at 3:30 pm
E. at 4:00 pm
LCM Method solution: LCM of 4 and 5 is 20. So let total work done be 20 units.
Valve to fill the pool will do 5 units of work in 1 hour when it is open alone.
When both valves are open, total work done is (5-4)=1 unit per hour.
Now the pool was filled in 10 hours. Let the valve be open alone for h hours.
Then, 5 h + 1 x (10-h) = 20
4h = 10 or h = 2.5
Thus the drain was open 2.5 hrs after 1:00 pm at 3:30 pm. Option (D)

I hope these examples are enough to explain the method. Obviously it is not applicable to all work rate problems. Especially for problems that have variables in the question and/or options such as 'X takes t hours more than Y', 'Find the total time in terms of x and y' and so on.

If you need help in applying this method to any specific problem feel free to post here and I can give it a shot.

Cheers and good luck!!!
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GMAT 1: 750 (Q50; V40; IR 8; AWA 5.5)
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Re: Quant Strategy: Solving Certain Work Rate Problems using the LCM  [#permalink]

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New post 14 Jun 2017, 01:56
I was also initially struggling with speed with work related problems and then came across this somewhere in a Gmatclub solution.
Note that if finding LCM is time consuming, multiplying all the numbers in the problem will work just as fine and can save some time if the problem involves difficult numbers.
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Re: Quant Strategy: Solving Certain Work Rate Problems using the LCM  [#permalink]

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New post 14 Jun 2017, 02:04
nishantd88 wrote:
Hello GMATClubbers,

I have recently completed my GMAT and wanted to share an approach I used to solve work rate problems with you all. This forum was really helpful during my preparation and I guess sharing some of the methods I used while solving certain problems is a way I would like to give back to the forum. Hope you all find this useful.

Before I begin, I'd like to make it clear that this is not a method I invented. I came across this a few years ago while studying for some other entrance and found it useful for GMAT as well. I'm sure there was no copyright and hence I'm quite sure I'm not violating one in sharing this method. I;m not being able to find the source now but will definitely look for it and edit the post with it if I find it.

Work rate problems can be a tad bit nasty and messed up at times. It is all good as long as just A and B are at it and you just have one equation and one variable. It can get tricky if there are 3 or more entities involved and they all work for different periods of time. The usual approach to such problems includes the use of variables and fractions which leave plenty of scope for mistakes. The method I use is primarily aimed at avoiding these possibilities.

Let’s call the method I will be discussing the LCM method. The basic principle of this method is to take what all is known, find their LCM and take that to be the total units of work to be done.

Sounds confusing? What better way to clarify than the use of examples.

Example 1:
X can do a work in 5 days and Y can do it in 15 days, In how many ways can they together complete the work?
Standard solution: X can do 1/5 work in one day and Y can do 1/15.
Working together, X and Y can do 1/5+1/15=4/15 of the work. So to complete 1 unit of work they will take 15 days.
LCM Method solution:The LCM of 5 and 15 is 15. So lets assume the total work to be 15 units.
So in 1 day, X does 3 units of work and B does 1 unit. Working together, they can do 4 units of work in 1 day.
Therefore total time is 15/4 days.

This is a simple example and so the new method does not seem to be a great improvement over the standard one. So let's apply it to a complicated one.

Example 2:
X, Y and Z can complete a piece of work in 15 days. The three of them start working together. After 2 days, X leaves. Y and Z work for 10 more days, after which Y also leaves. Z completes the rest of the work in 40 days. If Z had worked alone, he would have taken 75 days to complete the job. In how many days can X alone complete the job?
Standard solution: I wont get into the complications of variables and fractions and multiple equations. I'll leave that for now.
LCM method solution: LCM of 15, 2, 10, 40 and 75 is 600. So let the total work to be done be 600 units.
From the information in the question, Z can do 8 units of work per day. And (X+Y+Z) can together do 40 units per day.
Now lets form a single equation for the full scenario
(X+Y+Z)*2+(Y+Z)*10+Z*40 = 600
80+(Y+Z)*10+320=600
This gives Y+Z as 20 units.
Therefore, X's one day;s work = (X+Y+z) - (Y+Z) = 40 - 20 = 20
So X alone will take 600/20 = 30 days to complete the work alone

Example 3:
There are 3 pipes X, Y and Z attached to a tank. X and Y can fill the tank alone in 20 minutes and 30 minutes respectively while Z can empty it alone in 45 minutes. If all three of them are opened together, in how much time will the tank be full?
LCM Method solution: Let total units of work be 180 (LCM of 20, 30 and 45)
So X and Y can do 9 and 6 units of work per minute while Z can do 4 units.
Working together, total work per minute = 9 + 6 - 4 = 11 units.
So total time taken will be 180/11 minutes

Guess it seems more straightforward now. Let's apply this to a few GMAT questions picked up from this forum itself.

Example 4:
Working at their respective constant rates, Paul, Abdul and Adam alone can finish a certain work in 3, 4, and 5 hours respectively. If all three work together to finish the work, what fraction of the work will be done by Adam?
(A) 1/4
(B) 12/47
(C) 1/3
(D) 5/12
(E) 20/47
LCM Method solution: LCM of 3,4 and 5 is 60. So let the total amount of work be 60 units.
So based on the information give, Paul, Abdul and Adam can do 20 units, 15 units and 12 units of work per hour respectively.
So together, they will do (20+15+12) or 47 units of work in one hour.
Since Adam will do 12 units of this, fraction of work done by Adam will be 12/47. Option (B)

Example 5:
Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A’s speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?
(A) 1/2
(B) 2
(C) 3
(D) 5
(E) 6
LCM Method solution: LCM of 3 and 2 is 6. So let total work done be 6 units.
This means A and B can do 2 units of work per hour.
If A's speed is doubled, they do 3 units of work per hour.
So the extra unit of work is done simply because A's speed is doubled. This implies A's initial speed was 1 unit per hour.
Since total work to be done is 6 units, A will take 6 hours. Option (E)

Example 6:
A pool can be filled in 4 hours and drained in 5 hours. The valve that fills the pool was opened at 1:00 pm and some time later the drain that empties the pool was also opened. If the pool was filled by 11:00 pm and not earlier, when was the drain opened?
A. at 2:00 pm
B. at 2:30 pm
C. at 3:00 pm
D. at 3:30 pm
E. at 4:00 pm
LCM Method solution: LCM of 4 and 5 is 20. So let total work done be 20 units.
Valve to fill the pool will do 5 units of work in 1 hour when it is open alone.
When both valves are open, total work done is (5-4)=1 unit per hour.
Now the pool was filled in 10 hours. Let the valve be open alone for h hours.
Then, 5 h + 1 x (10-h) = 20
4h = 10 or h = 2.5
Thus the drain was open 2.5 hrs after 1:00 pm at 3:30 pm. Option (D)

I hope these examples are enough to explain the method. Obviously it is not applicable to all work rate problems. Especially for problems that have variables in the question and/or options such as 'X takes t hours more than Y', 'Find the total time in terms of x and y' and so on.

If you need help in applying this method to any specific problem feel free to post here and I can give it a shot.

Cheers and good luck!!!


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Re: Quant Strategy: Solving Certain Work Rate Problems using the LCM  [#permalink]

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New post 28 Jun 2017, 10:18
hi, I'm curious if this applies to all types of work problems, so my question is how do you solve this problem using the lcm method:

"Twelve identical machines, running continuously at the same constant rate, take 8 days to complete a shipment. How many additional machines, each running at the same constant rate, would be needed to reduce the time required to complete a shipment by two days?"

I already know the answer as it was discussed in another forum, but I'm just wondering how the approach would be based on your method.
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Re: Quant Strategy: Solving Certain Work Rate Problems using the LCM  [#permalink]

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New post 06 Jul 2017, 19:25
giangalang wrote:
hi, I'm curious if this applies to all types of work problems, so my question is how do you solve this problem using the lcm method:

"Twelve identical machines, running continuously at the same constant rate, take 8 days to complete a shipment. How many additional machines, each running at the same constant rate, would be needed to reduce the time required to complete a shipment by two days?"

I already know the answer as it was discussed in another forum, but I'm just wondering how the approach would be based on your method.


Let the total work to be done be 24 units (LCM of 8,12 and 2)

So currently 3 units of work are done per day.

In the new setup 12 units will have to be done.

To do 4 times the work, you will need 4 times the machines i.e. total 48 machines.

So 36 additional machines needed.

However, the simpler way is to just say that the work has to be completed in 1/4th the time and hence would need 4 times the initial number of machines.
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Re: Quant Strategy: Solving Certain Work Rate Problems using the LCM  [#permalink]

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