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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 30 Jul 2015, 08:38
1
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 02 Jan 2016, 06:03
hi can anyone help me with this concept please? i get confused why concept employed in Bunuel's example contradicts with the example given in this thread?

m30-184568.html#p1618937
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'Work' Word Problems Made Easy  [#permalink]

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New post 20 Mar 2016, 00:13
rainbooow wrote:
sriharimurthy wrote:
‘Work’ Word Problems Made Easy

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!


660/(t+10)X1.1=660/t

Multiply 660 by 1.1 you get 726.
The equation becomes 726/(t+10)=660/t
Now cross multiply the equation 726t=660t+6600
726t-660t=66T.
so the equation takes the form 66t=6600 or t= 6600/66= 100
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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 09 May 2016, 11:59
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?
in this question solution
We are told that B produces 10% more sprockets per hour than A, thus 660t+10∗1.1=660t660t+10∗1.1=660t --> t=100t=100 --> the rate of A is 660t+10=6660t+10=6 sprockets per hour.


how did 1.1 come ..plz help
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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 02 Oct 2016, 04:43
sondenso wrote:
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


I am just wondering shoudn't the first statement be be (2/3)*1/(3*5)={1/(4*x)} ?
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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 23 Mar 2017, 21:30
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.



From stem: 5(1/C)=1/3, where C is the number hours taken by one craftsman to complete the job. 1/C=1/15, so each craftsman takes 15h to complete the job.

1) (2/3)*(1/15)=2/45=rate of one apprentice
rate of 4 apprentices is therefore 8/45. The apprentices take 45/8 h. Suff.

2) 5(1/C) + 4(1/A) = 23/45
1/3 + 4/A = 23/45
4/A= 23/45 - 1/3 = 69/135 - 45/135 = 24/135 = 8/45 = 45/8 h Suff.
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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 13 May 2017, 14:35
some quantitative questions asks to find the total least amount of time, can sb pls help me with this, thanks.
For example, if A starts every 30 min, and it takes 1h20 to finish ride A.
B starts every 45 min, and it takes 50 min to fnish ride B
.......
what is the least total amount of time?

How to solve this question in less than 2 min???
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'Work' Word Problems Made Easy  [#permalink]

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New post 09 Mar 2018, 00:26
rainbooow wrote:
sriharimurthy wrote:
‘Work’ Word Problems Made Easy

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!


Bumping this question up: How do we quickly solve 660/(t+10)*1.1=660/t --> t=100?

I understand we can multiply both sides by t(t+10) to get rid of the denominators, but solving this way involves some serious computations (which I would take me several minutes, I'm afraid...). Is there a faster way?
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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 09 Sep 2018, 19:10
Ziggygee11 - I think the Answer to your question is 24.

This is how i solved it:

40/t+4 + 40/t = 50/6

Solving this you get t =8

So A' rate = 10/3

For 80 pages it will be 80*3/10 = 24
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'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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Re: 'Work' Word Problems Made Easy  [#permalink]

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New post 01 Oct 2018, 02:41
1
1
dave13 wrote:
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 :)


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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Re: 'Work' Word Problems Made Easy &nbs [#permalink] 01 Oct 2018, 02:41

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