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'Work' Word Problems Made Easy

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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 12 Oct 2014, 15:12
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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New post 12 Oct 2014, 23:04
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
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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 10 Jan 2015, 23:13
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Bunuel,

I have a question regarding Example 2.

Quote:
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:

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

‘He finishes 2/3 of the work’ This tells us that 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 of the work.

‘Y's rate is (1/10) that of X‘. We have already calculated rate at which X works to be . Therefore, rate at which Y works is .

‘In how much time does Y finish his work?’ If Y completes of the work in 1 hour, then he will complete of the work in 40 hours.


Why isn't the rate of work for X calculated as follows:

12hrs ---- 2/3 work
1 hr --- 1/18 work

Therefore, rate of work for Y should be (1/18)*(1/10) = (1/180).

1/180 work --- 1hr
1/3 work --- 180*(1/3) = 60hrs.

Please do clarify.

Thanks.
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New post 11 Jan 2015, 10:58
asocialnot wrote:
Bunuel,

I have a question regarding Example 2.

Quote:
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:

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

‘He finishes 2/3 of the work’ This tells us that 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 of the work.

‘Y's rate is (1/10) that of X‘. We have already calculated rate at which X works to be . Therefore, rate at which Y works is .

‘In how much time does Y finish his work?’ If Y completes of the work in 1 hour, then he will complete of the work in 40 hours.


Why isn't the rate of work for X calculated as follows:

12hrs ---- 2/3 work
1 hr --- 1/18 work

Therefore, rate of work for Y should be (1/18)*(1/10) = (1/180).

1/180 work --- 1hr
1/3 work --- 180*(1/3) = 60hrs.

Please do clarify.

Thanks.


We are told that X takes 12 hours to finish a certain job, not 2/3 of the job. So:
12 hours = 1 job, not 2/3 of the job.
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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 05 May 2015, 17:15
MBAhereIcome wrote:
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.


Can anyone help me how to use this formula for work problems?
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New post 11 May 2015, 23:49
SivaKumarP wrote:
MBAhereIcome wrote:
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.


Can anyone help me how to use this formula for work problems?


Hi SivaKumarP,

The formula is an extension of the basic concept of work rate problem. Let me explain it with an example.

Assume two persons A & B working alone who complete a work \('W'\) in \(5\) and \(6\) days respectively. We need to find the time taken by A & B working together to complete the same work \(W\).

We know the amount of work to be done (i.e. \(W\)) and are asked to find the time taken. From the work rate equation Work = Rate * Time, the only variable remaining is the rate, so let's find out the rates of persons A & B and put in the equation to calculate the time taken.

Rates of work done by A & B
Work done by A = \(W\)
Time taken by A to do W amount of work = \(5\) days

Since Rate = Work/time

Hence rate of work done by A = \(\frac{W}{5}\)

Similarly rate of work done by B = \(\frac{W}{6}\)

Putting the values of rates of A & B in the Work Rate equation to find out the time taken when both A & B work together:

Work = Rate * Time

\(W = (\frac{W}{5} + \frac{W}{6}) * t\) (since A & B would work for the same time 't' to complete the work when working together)

\(W = \frac{11W}{30} * t\) i.e. \(t = \frac{30}{11}\) days = \(\frac{(5*6)}{(5+6)}\) days

Time taken when A & B work together to do the same work = Time taken by A * Time taken by B/(Time taken by A + Time taken by B)


On similar lines, if we need to find the work done by A & B working together for 30 days, we can calculate it using the rate of work of A & B working together which is \(\frac{11W}{30}\).

Work done by A & B working together for 30 days = \(\frac{11W}{30} * 30 = 11W = 6W + 5W\) i.e. work done by A & B respectively working alone for 30 days.

Work done by A & B together in time t = Work done by A in time t + Work done by B in time t


Hope its clear!

Regards
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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 30 Jul 2015, 08:07
Can someone help with this question?
If it takes 4 machines, working at the same constant rate, 2 hours to complete 2 jobs, how long will it take 3 machines, working at the same constant rate, to complete 3 jobs?"

The answer is 4 hours but I have no idea how to arrive at the answer, hopefully someone can help.
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New post 30 Jul 2015, 08:23
JRAppz wrote:
Can someone help with this question?
If it takes 4 machines, working at the same constant rate, 2 hours to complete 2 jobs, how long will it take 3 machines, working at the same constant rate, to complete 3 jobs?"

The answer is 4 hours but I have no idea how to arrive at the answer, hopefully someone can help.


STEP BY STEP:

4 machines, 2 hours to complete 2 jobs;
1 machine, 8 hours to complete 2 jobs;
1 machine, 4 hours to complete 1 job;
3 machines, 4/3 hours to complete 1 job;
3 machines, 4 hours to complete 3 jobs.
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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 30 Jul 2015, 08:26
Bunuel wrote:
JRAppz wrote:
Can someone help with this question?
If it takes 4 machines, working at the same constant rate, 2 hours to complete 2 jobs, how long will it take 3 machines, working at the same constant rate, to complete 3 jobs?"

The answer is 4 hours but I have no idea how to arrive at the answer, hopefully someone can help.


STEP BY STEP:

4 machines, 2 hours to complete 2 jobs;
1 machine, 8 hours to complete 2 jobs;
1 machine, 4 hours to complete 1 job;
3 machines, 4/3 hours to complete 1 job;
3 machines, 4 hours to complete 3 jobs.


Bunuel - Could you possibly structure that in an algebraic manner?
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New post 30 Jul 2015, 08:32
JRAppz wrote:
Bunuel wrote:
JRAppz wrote:
Can someone help with this question?
If it takes 4 machines, working at the same constant rate, 2 hours to complete 2 jobs, how long will it take 3 machines, working at the same constant rate, to complete 3 jobs?"

The answer is 4 hours but I have no idea how to arrive at the answer, hopefully someone can help.


STEP BY STEP:

4 machines, 2 hours to complete 2 jobs;
1 machine, 8 hours to complete 2 jobs;
1 machine, 4 hours to complete 1 job;
3 machines, 4/3 hours to complete 1 job;
3 machines, 4 hours to complete 3 jobs.


Bunuel - Could you possibly structure that in an algebraic manner?


Rate*Time = Job done.

(4*rate of one machine)*2 = 2;

rate of one machine = 1/4 job/hour.

Rate*Time = Job done.

(3*rate of one machine)*x = 3;

(3*1/4)*x = 3;

x = 4 hours.
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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 30 Jul 2015, 08:38
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JRAppz wrote:
Can someone help with this question?
If it takes 4 machines, working at the same constant rate, 2 hours to complete 2 jobs, how long will it take 3 machines, working at the same constant rate, to complete 3 jobs?"

The answer is 4 hours but I have no idea how to arrive at the answer, hopefully someone can help.


You can follow the method mentioned by Bunuel above .

The only thing you need to remember is that the work rate problems are similar to distance-speed-time wherein Distance is similar to job, speed is similar to rate and time is constant.

Thus, as distance = speed X time

Job or work = rate x time

Additionally, as with more machines, the total time required will reduce for the same amount of work, combined rates of 'n' machines will be n*rate of 1 machine.
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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 30 Jul 2015, 09:27
JRAppz wrote:
Can someone help with this question?
If it takes 4 machines, working at the same constant rate, 2 hours to complete 2 jobs, how long will it take 3 machines, working at the same constant rate, to complete 3 jobs?"

The answer is 4 hours but I have no idea how to arrive at the answer, hopefully someone can help.


CONCEPT: \(\frac{(Machine_Power * Time)}{Work} = Constant\)

i.e. \(\frac{(M_1 * T_1)}{W_1} = \frac{(M_2 * T_2)}{W_2}\)

i.e. \(\frac{(4 * 2))}{2} = \frac{(3 * T_2)}{3}\)

i.e. \(T_2 = 4\)
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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 29 Sep 2018, 06:09
sriharimurthy 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 = \(\frac{1}{24}-\frac{1}{60}=\frac{1}{40}\).

‘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.




pushpitkc, Bunuel VeritasKarishma

if in denominator always is given time (unless is reversed ) \(\frac{1}{120}\) then how can 120 be amount of work ? what does 5*1 mean ? and why are we dividing by 1/120 to get WORK DONE


as per FORMULA ---> WORK = RATE * TIME


pls explain :)
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New post 25 Oct 2019, 01:52
Can someone please explain this in Example 2:
‘In how much time does Y finish his work?’ If Y completes 1/120 of the work in 1 hour, then he will complete 1/3 of the work in 40 hours.
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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 19 May 2020, 07:59
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.

Answer: D.

Check this for more: http://gmatclub.com/forum/word-translat ... 04208.html


VeritasKarishma

Can you pls explain me why in statement 2 we are dividing combined time with individual time and also if there is any easier method of solving this

Thanks in advance!
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Re: 'Work' Word Problems Made Easy   [#permalink] 19 May 2020, 07:59

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