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Work problem

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by crackgmat007 » Wed May 06, 2009 9:38 am
Q: Working alone at its constant rate, machine K took 3 hours to produce 1/4of the units produced
last Friday. Then machine M started working and the two machines, working simultaneously at
their respective constant rates, took 6 hours to produce the rest of the units produced last Friday.
How many hours would it have taken machine M, working alone at its constant rate, to produce all
of the units produced last Friday?
A. 8
B. 12
C. 16
D. 24
E. 30

OA – D

Total hours of work is 12 (1/4 is 3 hours, hence 1 is 12 hours). Balance work is 9 hours, which is completed by M & K in 6 hours. Can someone explain how do I go from here?
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by DanaJ » Wed May 06, 2009 10:12 am
You're almost there.

Since it took machine K 3 hours to complete 1/4 of the job, it would indeed take it 12 hours to do it alone.
Since it took both machines 6 hours to complete the remaining 3/4 of the job, it would take them both 4/3 * 6 = 8 hours to complete the whole thing.

Now all you need to do is apply that "work formula"

1/k + 1/m = 1/(mk) - where k = time it takes k to complete the job alone; m = time it takes m to complete the job alone; mk = time it takes both machines to do the job working together.

1/12 + 1/m = 1/8

1/m = 1/8 - 1/12 = 1/24 ------------ m = 24.
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by crackgmat007 » Wed May 06, 2009 11:10 am
got it...thanks much
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by Bhattu » Fri May 08, 2009 4:20 am
I went a bit more basic on this.

K took 3 hrs to do 1/4 of the job, then K took 6 hrs to do 1/2 the job. Therefore K did 3/4 of the work in 9 Hrs and M did 1/4 of the work in 6 hrs.

If M did 1/4 in 6hrs, it would take M 6*4 hours to complete the job.

I would go for the earlier solution as it sticks to the tried and tested formula for work problems
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