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# 'Work' Word Problems Made Easy

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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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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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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

Originally posted by dagmat on 01 Jul 2010, 02:42.
Last edited by dagmat on 28 Jul 2010, 04:16, edited 1 time in total.
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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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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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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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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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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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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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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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Hi,

Can sombody please solve this wit solution for this question.

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

Regards,
Rrsnathan.
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rrsnathan wrote:
Hi,

Can sombody please solve this wit solution for this question.

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

Regards,
Rrsnathan.

Check here: it-takes-printer-a-4-more-minutes-than-printer-b-to-print-98479.html

Hope it helps.
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Just 3 hours back, literally I was at zero confidence to start with work rate problems, but after going through your notes and solving 5 or 6 problems, I have solved 80 to 90% difficulty level problems like a pro, solutions are also matching closely to Bunuel's. Really work rate is not that as much tough as I was thinking before. Reciprocate the hours to find the rate of work per day or per minute that's it half battle won-> "master key to every work rate problem".

Thanks allot.
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Bunuel wrote:
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.

Hi Bunuel,

For #2, I set-up my equation as:

1/3 + 1/A = 1/(45/23)

Since we already know the rate of the 5 craftsmen and the total hours needed for the 2 groups to finish the job. However, I arrived at a different answer for A (45/8). I can't seem to figure out where it went wrong, hope you can help.

Thanks.
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pauc wrote:
Bunuel wrote:
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.

Hi Bunuel,

For #2, I set-up my equation as:

1/3 + 1/A = 1/(45/23)

Since we already know the rate of the 5 craftsmen and the total hours needed for the 2 groups to finish the job. However, I arrived at a different answer for A (45/8). I can't seem to figure out where it went wrong, hope you can help.

Thanks.

You should get the same answer: 1/3 + 1/A = 1/(45/23) --> A=45/8.
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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 = $$\frac{1}{24}-\frac{1}{60}=\frac{1}{40}$$.

[i]‘B prints 5 pages a minute more than printer A’[/i] 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.

Is there any other way to tackle this part. I am not able to understand it.
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earnit wrote:
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 = $$\frac{1}{24}-\frac{1}{60}=\frac{1}{40}$$.

[i]‘B prints 5 pages a minute more than printer A’[/i] 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.

Is there any other way to tackle this part. I am not able to understand it.

Check here: working-together-printer-a-and-printer-b-would-finish-the-100221.html