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# 'Work' Word Problems Made Easy

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Current Student
Joined: 20 Mar 2014
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Concentration: Finance, Strategy
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30 Jul 2015, 09:38
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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Joined: 08 Jul 2010
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30 Jul 2015, 10:27
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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Current Student
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02 Jan 2016, 07:03
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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20 Mar 2016, 01:13
rainbooow wrote:
sriharimurthy wrote:

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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09 May 2016, 12:59
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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Joined: 22 Jun 2016
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02 Oct 2016, 05:43
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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23 Mar 2017, 22:30
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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13 May 2017, 15:35
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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Joined: 04 Aug 2014
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09 Mar 2018, 01:26
rainbooow wrote:
sriharimurthy wrote:

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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