'Work' Word Problems Made Easy : Quantitative

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'Work' Word Problems Made Easy

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

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10 Jan 2015, 23:13

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

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11 Jan 2015, 10:58

asocialnot wrote:

Bunuel,

I have a question regarding Example 2.

Quote:

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

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11 May 2015, 23:49

SivaKumarP wrote:

MBAhereIcome wrote:

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

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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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30 Jul 2015, 08:23

JRAppz wrote:

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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30 Jul 2015, 08:26

Bunuel wrote:

JRAppz wrote:

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

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30 Jul 2015, 08:32

JRAppz wrote:

Bunuel wrote:

JRAppz wrote:

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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30 Jul 2015, 08:38

JRAppz wrote:

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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30 Jul 2015, 09:27

JRAppz wrote:

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

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01 Oct 2018, 02:41

dave13 wrote:

sriharimurthy wrote:

Imp: Ensure that you always have your eye on the units. When you are flipping a quantity, know why.

A and B finish a task in 24 mins. If work done has to be 1 (complete work), their rate of work is (1/24)th of the work per min. (It is usually the case that we consider work to be 1 work)

Work = Rate*Time
1 work = Rate* 24 mins
Rate = (1/24)th of work per min

Similarly, A's rate of work = (1/60)th of work per min

Since rates are additive, we get that B's rate must be (1/40)th of work per min.

So B does (1/40)th of the work in a min while B does only (1/60)th of the work. We are given that this difference of (1/40)th work and (1/60)th work is 5 pages.
So 1 complete work is 600 pages.
Now the rate of work can be expressed in terms of pages/min too.
B prints (1/40)*600 = 15 pages/min
A prints (1/60)*600 = 10 pages/min
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