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Knewton - Rate and work

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Knewton - Rate and work

by nikhilkatira » Wed Jul 14, 2010 9:03 pm
Eight workers from Company A can paint 7 homes in 84 hours. Working together, three workers from Company A and five workers from company B can paint 9 homes in 96 hours. If each worker from company A works at one constant rate, each worker from Company B works at another constant rate, and each home requires the same amount of work to paint, how many workers from Company B are required to paint 9 homes in 60 hours?

(A)12

(B)15

(C)18

(D)20

(E)30
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Nikhil H. Katira
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by kvcpk » Wed Jul 14, 2010 11:56 pm
Let work perhour of a worker of A be a
Total manhours from A = 84*8a = 672a.
So 672a hours needed for 7 homes.

Let work per hour of B be b
(3a+5b)*96 hours are needed for 9 homes.
=288a+480b hours are needed for 9 homes.
Here, 288*7/672 homes can be built by 3a workers. = 3 homes.
Therefore 480b hours are needed for bulding 6 homes by B workers.
multiply by 1.5
720b hours are needed for bulding 9 homes.

Let the number of workers needed be X
Xb*60 hours are needed for 9 homes.
xb*60=720b
x=12

pick A.

What is OA?
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by ez2dj86 » Thu Jul 15, 2010 12:23 am
Let A be the work productivity for a worker in company A and B for that of company B.

8A can paint 7 homes in 84 hours.
8A=7/84
A=1/96 meaning one worker at company A can finish painting 1/96 of the house per hour.

3A+5B can paint 9 homes in 96 hours.
Substituting A=1/96 to this equation,
3(1/96)+5B=9/96
B=1/80 meaning one worker at company B can finish painting 1/80 of the house per hour.

So in 60 hours, one worker from company B can paint 60*(1/80)=3/4 of the house.

So to paint 9 homes, you need 9/(3/4)=12 workers
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