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27 Nov 2009, 16:18
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This post is a part of [GMAT MATH BOOK]

created by: sriharimurthy
edited by: bb, walker, Bunuel

--------------------------------------------------------

NOTE: In case you are not familiar with translating word problems into equations please go through this post first : word-problems-made-easy-87346.html

What is a ‘Work’ Word Problem?

• It involves a number of people or machines working together to complete a task.
• We are usually given individual rates of completion.
• We are asked to find out how long it would take if they work together.

Sounds simple enough doesn’t it? Well it is!

There is just one simple concept you need to understand in order to solve any ‘work’ related word problem.

The ‘Work’ Problem Concept

STEP 1: Calculate how much work each person/machine does in one unit of time (could be days, hours, minutes, etc).

How do we do this? Simple. If we are given that A completes a certain amount of work in X hours, simply reciprocate the number of hours to get the per hour work. Thus in one hour, A would complete $$\frac{1}{X}$$ of the work. But what is the logic behind this? Let me explain with the help of an example.

Assume we are given that Jack paints a wall in 5 hours. This means that in every hour, he completes a fraction of the work so that at the end of 5 hours, the fraction of work he has completed will become 1 (that means he has completed the task).

Thus, if in 5 hours the fraction of work completed is 1, then in 1 hour, the fraction of work completed will be (1*1)/5

STEP 2: Add up the amount of work done by each person/machine in that one unit of time.

This would give us the total amount of work completed by both of them in one hour. For example, if A completes $$\frac{1}{X}$$ of the work in one hour and B completes $$\frac{1}{Y}$$ of the work in one hour, then TOGETHER, they can complete $$\frac{1}{X}+\frac{1}{Y}$$ of the work in one hour.

STEP 3: Calculate total amount of time taken for work to be completed when all persons/machines are working together.

The logic is similar to one we used in STEP 1, the only difference being that we use it in reverse order. Suppose $$\frac{1}{X}+\frac{1}{Y}=\frac{1}{Z}$$. This means that in one hour, A and B working together will complete $$\frac{1}{Z}$$ of the work. Therefore, working together, they will complete the work in Z hours.

Advice here would be: DON'T go about these problems trying to remember some formula. Once you understand the logic underlying the above steps, you will have all the information you need to solve any ‘work’ related word problem. (You will see that the formula you might have come across can be very easily and logically deduced from this concept).

Now, lets go through a few problems so that the above-mentioned concept becomes crystal clear. Lets start off with a simple one :

Example 1.
Jack can paint a wall in 3 hours. John can do the same job in 5 hours. How long will it take if they work together?

Solution:
This is a simple straightforward question wherein we must just follow steps 1 to 3 in order to obtain the answer.

STEP 1: Calculate how much work each person does in one hour.
Jack → (1/3) of the work
John → (1/5) of the work

STEP 2: Add up the amount of work done by each person in one hour.
Work done in one hour when both are working together = $$\frac{1}{3}+\frac{1}{5}=\frac{8}{15}$$

STEP 3: Calculate total amount of time taken when both work together.
If they complete $$\frac{8}{15}$$ of the work in 1 hour, then they would complete 1 job in $$\frac{15}{8}$$ hours.

Example 2.
Working, independently X takes 12 hours to finish a certain work. He finishes 2/3 of the work. The rest of the work is finished by Y whose rate is 1/10 of X. In how much time does Y finish his work?

Solution:
Now the only reason this is trickier than the first problem is because the sequence of events are slightly more complicated. The concept however is the same. So if our understanding of the concept is clear, we should have no trouble at all dealing with this.

‘Working, independently X takes 12 hours to finish a certain work’ This statement tells us that in one hour, X will finish $$\frac{1}{12}$$ of the work.

‘He finishes 2/3 of the work’ This tells us that $$\frac{1}{3}$$ of the work still remains.

‘The rest of the work is finished by Y whose rate is (1/10) of X’ Y has to complete $$\frac{1}{3}$$ of the work.

‘Y's rate is (1/10) that of X‘. We have already calculated rate at which X works to be $$\frac{1}{12}$$. Therefore, rate at which Y works is $$\frac{1}{10}*\frac{1}{12}=\frac{1}{120}$$.

‘In how much time does Y finish his work?’ If Y completes $$\frac{1}{120}$$ of the work in 1 hour, then he will complete $$\frac{1}{3}$$ of the work in 40 hours.

So as you can see, even though the question might have been a little difficult to follow at first reading, the solution was in fact quite simple. We didn’t use any new concepts. All we did was apply our knowledge of the concept we learnt earlier to the information in the question in order to answer what was being asked.

Example 3.
Working together, printer A and printer B would finish a task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A?

Solution:
This problem is interesting because it tests not only our knowledge of the concept of word problems, but also our ability to ‘translate English to Math’

‘Working together, printer A and printer B would finish a task in 24 minutes’ This tells us that A and B combined would work at the rate of $$\frac{1}{24}$$ per minute.

‘Printer A alone would finish the task in 60 minutes’ This tells us that A works at a rate of $$\frac{1}{60}$$ per minute.

At this point, it should strike you that with just this much information, it is possible to calculate the rate at which B works: Rate at which B works = $$\frac{1}{24}-\frac{1}{60}=\frac{1}{40}$$.

‘B prints 5 pages a minute more than printer A’ This means that the difference between the amount of work B and A complete in one minute corresponds to 5 pages. So, let us calculate that difference. It will be $$\frac{1}{40}-\frac{1}{60}=\frac{1}{120}$$

‘How many pages does the task contain?’ If $$\frac{1}{120}$$ of the job consists of 5 pages, then the 1 job will consist of $$\frac{(5*1)}{\frac{1}{120}} = 600$$ pages.

Example 4.
Machine A and Machine B are used to manufacture 660 sprockets. It takes machine A ten hours longer to produce 660 sprockets than machine B. Machine B produces 10% more sprockets per hour than machine A. How many sprockets per hour does machine A produce?

Solution:
The rate of A is $$\frac{660}{t+10}$$ sprockets per hour;
The rate of B is $$\frac{660}{t}$$ sprockets per hour.

We are told that B produces 10% more sprockets per hour than A, thus $$\frac{660}{t+10}*1.1=\frac{660}{t}$$ --> $$t=100$$ --> the rate of A is $$\frac{660}{t+10}=6$$ sprockets per hour.

As you can see, the main reason the 'tough' problems are 'tough' is because they test a number of other concepts apart from just the ‘work’ concept. However, once you manage to form the equations, they are really not all that tough.

And as far as the concept of ‘work’ word problems is concerned – it is always the same!

DS work problems to practice: search.php?search_id=tag&tag_id=46
PS work problems to practice: search.php?search_id=tag&tag_id=66

Feel free to discuss any question or doubt you might have in this post.
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Click below to check out some great tips and tricks to help you deal with problems on Remainders!
http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html

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11 May 2010, 23:49
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Can someone help me solve this DS problem on 'Work'

A group of 5 craftsmen, working together at the same rate, can finish a job in 3 hours. How long will it take a group of 4 apprentices working together to do the same job?

(1) Each apprentice works at 2/3 the rate of a craftsman.
(2) The 5 craftsmen and the 4 apprentices working together will take 45/23 hours to finish the job.
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15 May 2010, 21:18
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JoyLibs wrote:
Can someone help me solve this DS problem on 'Work'

A group of 5 craftsmen, working together at the same rate, can finish a job in 3 hours. How long will it take a group of 4 apprentices working together to do the same job?

(1) Each apprentice works at 2/3 the rate of a craftsman.
(2) The 5 craftsmen and the 4 apprentices working together will take 45/23 hours to finish the job.

x= time 4 apprentices need to do a job and what is x?
1.
1/(3*5) = (2/3)* {1/(4*x)}-->x is specific. suff

2.

1/(3*5) + 1/(4*x) = 23/45 --->x is specific, suff

D
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01 Jul 2010, 02:42
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Quote:
Example 4.
Machine A and Machine B are used to manufacture 660 sprockets. It takes machine A ten hours longer to produce 660 sprockets than machine B. Machine B produces 10% more sprockets per hour than machine A. How many sprockets per hour does machine A produce?

Solution 4.
You wont come across many problems tougher than this on the GMAT. But as we will see, even the toughest of problems can be solved with relative ease if we employ the concept discussed above.

‘It takes machine A ten hours longer to produce 660 sprockets than machine B’
Let machine A produce 660 sprockets in ‘x’ hours.
Therefore, machine B will produce 660 sprockets in ‘x – 10’ hours.

With this information, we can calculate the amount of work machine A and B do per hour respectively.
Rate at which machine A works = [1/x] per hour
Rate at which machine B works = [1/(x – 10)] per hour

‘Machine B produces 10% more sprockets per hour than machine A’
If machine A produces [1/x] sprockets an hour, then machine B will produce (1/x) +(10/100)*(1/x) = (11/10x)

But we already know that rate at which machine B works = [1/(x – 10)] per hour. Therefore, equating it to (11/10x) we get the following equation:

(11/10x) = 1/(x-10) → x = 110 hours

‘How many sprockets per hour does machine A produce?’
If in 110 hours A produces 660 sprockets,
Then in 1 hour it will produce (660*1)/110 = 6

Another way in which it can be solved:
Let 'x' be the no. of hours taken by B to complete the work, therefore A takes x+10 hours
Also, Let 'y' be the rate of work (sprockets/hr) for A, therefore, for B it becomes 1.1y ((110/100) * y) since B produces 10% more.
Equation then becomes: 1.1y * x = 660 (B)
y * (x+10)= 660 (A)
Since both =660, 1.1y * x = y * (x+10)
Solving the equation gives us x=100, y=6

Last edited by dagmat on 28 Jul 2010, 04:16, edited 1 time in total.
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02 Aug 2010, 21:26
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Can someone explain what the 1 represents in the examples above like 1*1 or 5*1 or 1/3*1? Is this number of fraction of jobs complete x number of hours? I'm confused on this aspect of it.

For example if we say it takes 8 hours to paint a house, why do we say (1*1)/8 and not just 1/8?
How would this change if we said it takes 13 hours to paint 2 houses? would it be (1*2)/13. I guess I don't understand where the second 1 comes from. Is there ever a case where it is not 1* a number and is instead a different number?
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30 Dec 2010, 11:18
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Bunuel,

Can you help me with this question?

A group of 5 craftsmen, working together at the same rate, can finish a job in 3 hours. How long will it take a group of 4 apprentices working together to do the same job?
(1) Each apprentice works at 2/3 the rate of a craftsman.
(2) The 5 craftsmen and the 4 apprentices working together will take 45/23 hours to finish the job.

Answer choice seems to be D.
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30 Dec 2010, 12:07
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resh924 wrote:
Bunuel,

Can you help me with this question?

A group of 5 craftsmen, working together at the same rate, can finish a job in 3 hours. How long will it take a group of 4 apprentices working together to do the same job?
(1) Each apprentice works at 2/3 the rate of a craftsman.
(2) The 5 craftsmen and the 4 apprentices working together will take 45/23 hours to finish the job.

Answer choice seems to be D.

Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.

$$time*speed=distance$$ <--> $$time*rate=job \ done$$.

So, if we say that the rate of a craftsmen is $$x$$ job/hours and the rate of an apprentice is $$y$$ job/hour then we'll have $$(5x)*3=job=(4y)*t$$ --> $$(5x)*3=(4y)*t$$. Question: $$t=\frac{15x}{4y}=?$$

(1) Each apprentice works at 2/3 the rate of a craftsman --> $$y=\frac{2}{3}x$$ --> $$\frac{x}{y}=\frac{3}{2}$$ --> $$t=\frac{15x}{4y}=\frac{45}{8}$$ hours. Sufficient.

(2) The 5 craftsmen and the 4 apprentices working together will take 45/23 hours to finish the job --> as the 5 craftsmen need 3 hours to do the job then in 45/23 hours they'll complete (45/23)/3=15/23 rd of the job (15 parts out of 23) so the rest of the job, or 1-15/23=8/23 (8 parts out of 23) is done by the 4 apprentices in the same amount of time (45/23 hours): $$\frac{5x}{4y}=\frac{15}{8}$$ --> $$\frac{x}{y}=\frac{3}{2}$$, the same info as above. Sufficient.

Check this for more: word-translations-rates-work-104208.html
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30 Dec 2010, 13:09
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Bunuel,

Thank you so much! Your ability to explain things so concisely and with such depth is amazing.

Thanks for your help! I'm a huge fan.
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22 Apr 2011, 07:17
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I do not know if it as been posted -

If X can do a set of work in m # days/hr/time unit and Y can do the same work in n days/hr/time unit independently then both can do it only M*N/M+N time unit.

You can do cross checks on this- I have found it useful ...
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06 Jul 2011, 00:04
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I want to clarify example 4

Rate at which machine A works wont be = [660/x] per hour
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27 Aug 2011, 13:02
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this is a great post but, i'd suggest to use either of the below formulas depending on what the question asks.

time taken = A*B/A+B ............. where A,B is the respective time of each person/machine.
work done = W1 + W2 ............... where W1,W2 is the respective work done by each person/machine.
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Awesome post.
Don't know if this has been dealt with in a separate post but thought i might add.
A formula i sometimes use when 3 variables are tested:

(P1×T1)/J1= (P2×T2)/J2

P1 = the number of workers (or machines) in the 1st situation.
P2 = the number of workers (or machines) in the 2nd situation.
T1 = the number of time units (hours, days, etc) in the 1st situation.
T2 = the number of time units (hours, days, etc) in the 2nd situation.
J1 = the number of jobs done in the 1st situation.
J2 = the number of jobs done in the 2nd situation.
note: Be sure to convert time into a single unit (hours or minutes)

Example:
It takes 7 people 8 hours to build 3 cars. How many hours will it take 2 people to build 1 car?
7x8/3 = 2xT2/1
T2 = 9.33 or 9 hours 20 minutes.
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26 Oct 2011, 11:05
Great resource!

Question. For example 2, I tried plugging in a smart number for work (12) which screws up the answer.

Why is plugging in a number for work not applicable here?

Can I ever plugin numbers in rate/work problems?
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03 Dec 2011, 09:08
sriharimurthy ,

Thanks for your post. I have this question. Have checked the solutions to this problem online and found them quite complicated.Is there an easier solution to the question without getting involved in quadratics?
The question is:
It takes Printer A 4 more minutes than printer B to print 40 pages. Working together, the two printers can print 50 pages in 6 minutes. how long will it take printer A to print 80 pages?
a. 12 b. 18 c. 20 d. 24 e. 30

Thanks
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04 Jan 2012, 20:52
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Super Quick Way Follows.....

If Machine A produces 100, in the same amount of time Machine B produces 110
Thus, if Machine A produces 600, in the same amount of time Machine B produces 660.---(A)
It is given that Machine A takes 10 hours more than that of Machine B to produce 660---(B)
From the statements A & B, we can conclude that in a given amount of time where B can produce 660, since A can produce only 600, to produce the remaining 60 it takes another 10 hours.

So A takes 10 hours to produce 60 sprockets, thus making it 6 sprockets in 1 hour...

GMATPASSION wrote:
dagmat wrote:
Quote:
Example 4.
Machine A and Machine B are used to manufacture 660 sprockets. It takes machine A ten hours longer to produce 660 sprockets than machine B. Machine B produces 10% more sprockets per hour than machine A. How many sprockets per hour does machine A produce?

Solution 4.
You wont come across many problems tougher than this on the GMAT. But as we will see, even the toughest of problems can be solved with relative ease if we employ the concept discussed above.

‘It takes machine A ten hours longer to produce 660 sprockets than machine B’
Let machine A produce 660 sprockets in ‘x’ hours.
Therefore, machine B will produce 660 sprockets in ‘x – 10’ hours.

With this information, we can calculate the amount of work machine A and B do per hour respectively.
Rate at which machine A works = [1/x] per hour
Rate at which machine B works = [1/(x – 10)] per hour

‘Machine B produces 10% more sprockets per hour than machine A’
If machine A produces [1/x] sprockets an hour, then machine B will produce (1/x) +(10/100)*(1/x) = (11/10x)

But we already know that rate at which machine B works = [1/(x – 10)] per hour. Therefore, equating it to (11/10x) we get the following equation:

(11/10x) = 1/(x-10) → x = 110 hours

‘How many sprockets per hour does machine A produce?’
If in 110 hours A produces 660 sprockets,
Then in 1 hour it will produce (660*1)/110 = 6

Another way in which it can be solved:
Let 'x' be the no. of hours taken by B to complete the work, therefore A takes x+10 hours
Also, Let 'y' be the rate of work (sprockets/hr) for A, therefore, for B it becomes 1.1y ((110/100) * y) since B produces 10% more.
Equation then becomes: 1.1y * x = 660 (B)
y * (x+10)= 660 (A)
Since both =660, 1.1y * x = y * (x+10)
Solving the equation gives us x=100, y=6

Well done. Another fast way of tackling this problem.

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05 Jan 2012, 03:56
Quick Solution Follows.....

Given that if A produces 100, in the same time B will produce 110
=>if A produces 600, in the same time B will produce 660---(A)
Given that A takes 10 hours more than B to produce 660---(B)
From A & B, since the time in which B produces 660, A can produce only 600 to produce the remaining 60 sprockets A takes 10 more hours.

Hence, A produces 60 sprockets in 10 hrs => 6 sprockets in 1 hr

GMATPASSION wrote:
dagmat wrote:
Quote:
Example 4.
Machine A and Machine B are used to manufacture 660 sprockets. It takes machine A ten hours longer to produce 660 sprockets than machine B. Machine B produces 10% more sprockets per hour than machine A. How many sprockets per hour does machine A produce?

Solution 4.
You wont come across many problems tougher than this on the GMAT. But as we will see, even the toughest of problems can be solved with relative ease if we employ the concept discussed above.

‘It takes machine A ten hours longer to produce 660 sprockets than machine B’
Let machine A produce 660 sprockets in ‘x’ hours.
Therefore, machine B will produce 660 sprockets in ‘x – 10’ hours.

With this information, we can calculate the amount of work machine A and B do per hour respectively.
Rate at which machine A works = [1/x] per hour
Rate at which machine B works = [1/(x – 10)] per hour

‘Machine B produces 10% more sprockets per hour than machine A’
If machine A produces [1/x] sprockets an hour, then machine B will produce (1/x) +(10/100)*(1/x) = (11/10x)

But we already know that rate at which machine B works = [1/(x – 10)] per hour. Therefore, equating it to (11/10x) we get the following equation:

(11/10x) = 1/(x-10) → x = 110 hours

‘How many sprockets per hour does machine A produce?’
If in 110 hours A produces 660 sprockets,
Then in 1 hour it will produce (660*1)/110 = 6

Another way in which it can be solved:
Let 'x' be the no. of hours taken by B to complete the work, therefore A takes x+10 hours
Also, Let 'y' be the rate of work (sprockets/hr) for A, therefore, for B it becomes 1.1y ((110/100) * y) since B produces 10% more.
Equation then becomes: 1.1y * x = 660 (B)
y * (x+10)= 660 (A)
Since both =660, 1.1y * x = y * (x+10)
Solving the equation gives us x=100, y=6

Well done. Another fast way of tackling this problem.

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06 Nov 2012, 03:19
Hi,
thank you for this.
but I still dont understand something in Q4.

it should be:

M............JOB.............Time..............Rate
A............660.............t+10.............660/t+10
B............660..............t..................660/t

and now to use the 10%...
i still cant get this to work.
can anyone help?
thanks
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07 Nov 2012, 05:53
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roygush wrote:
Hi,
thank you for this.
but I still dont understand something in Q4.

it should be:

M............JOB.............Time..............Rate
A............660.............t+10.............660/t+10
B............660..............t..................660/t

and now to use the 10%...
i still cant get this to work.
can anyone help?
thanks

The rate of A is 660/(t+10) sprockets per hour;
The rate of B is 660/t sprockets per hour.

We are told that B produces 10% more sprockets per hour than A, thus 660/(t+10)*1.1=660/t --> t=100 --> the rate of A is 660/(t+10)=6 sprockets per hour.

Hope it's clear.
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15 Feb 2013, 07:24
sriharimurthy wrote:

We are told that B produces 10% more sprockets per hour than A, thus $$\frac{660}{t+10}*1.1=\frac{660}{t}$$ --> $$t=100$$ --> the rate of A is $$\frac{660}{t+10}=6$$ sprockets per hour.

Can someone explain to me how to get t=100? from this equation?

Thank you!
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