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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
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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
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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.


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???
Re: 'Work' Word Problems Made Easy   [#permalink] 13 May 2017, 14:35

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