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Work Problem

Expert replies
by Gurpinder » Thu Oct 27, 2011 9:37 am
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?
I tried plugging in a smart number for work (12) which screws up the answer.

Why is plugging in a number for work not applicable here?

Can I ever plugin numbers in rate/work problems?
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
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Source: — Problem Solving |

by Anurag@Gurome » Thu Oct 27, 2011 9:43 am
Gurpinder wrote: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?

I tried plugging in a smart number for work (12) which screws up the answer.
Say, the work has 12 parts.
Hence, X does one part in one hour.
Hence, Y does 1/10 of one part in one hour.

Y has to finish 1/3 of the work = 12/3 = 4 parts of the work.

Hence, Y needs 4*10 = 40 hours to finish his work.
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by Gurpinder » Thu Oct 27, 2011 10:53 am
Thanks Anurag.

I dont understand what I am doing wrong. Since we are plugging in a number for work, lets say work = 1.

----r-------t------d
x -- r -----12------2/3

12r=2/3
36r=2
r=1/18 <- rate of x

----r-------t------d
y -- 1/180 ---t------1/3

t/180=1/3
t=60
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by Gurpinder » Thu Oct 27, 2011 11:19 am
Anurag@Gurome wrote:
Say, the work has 12 parts.
Hence, X does one part in one hour.
Hence, Y does 1/10 of one part in one hour.
This is where I am confused. How can x do one part in one hour? X only does 2/3 of the job. So wouldn't you take the amount of work he "actually" did to calculate rate?

Thats how I am getting his rate to be 1/18.
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
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