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Machine pencils

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Machine pencils

by Xbond » Sat Dec 12, 2009 11:55 am
Hi

I don't understand anything how to resolve this one. OA is A

Q70)Machine A produced pencils at a constant rate of 9,000 pencils per hour, and machine B produces pencils at a constant rate of 7,000 pencils per hour. If the two machines together must produce 100,000 pencils and if machine can operate for at
most 8 hours, what is the least amount of time, in hours, that machine B must operate ?

A) 4
B) 4,66
C) 5,33
D) 6
E) 6,25
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by heshamelaziry » Sat Dec 12, 2009 1:00 pm
Xbond wrote:Hi

I don't understand anything how to resolve this one. OA is A

Q70)Machine A produced pencils at a constant rate of 9,000 pencils per hour, and machine B produces pencils at a constant rate of 7,000 pencils per hour. If the two machines together must produce 100,000 pencils and if machine can operate for at
most 8 hours, what is the least amount of time, in hours, that machine B must operate ?

A) 4
B) 4,66
C) 5,33
D) 6
E) 6,25
This question is about minimizing and maximizing values. Since the question is asking you to put B to work for the least posssible hours, you need to put A to work for the most possible hours. So, since each machine can't work more than 8 hours per day, A will work 8 hours to produce 9000 * 8 = 72,000 pencils. Now, B must produce 28,000 pencils and it produces 7,000/hour, means that B will have to work 4 hours.

To become familiar with this concept, search November articles by Stacey from MGMAT, I think she posted 3 very good questions, in 2 articles, that illustrate this concept.
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by valleeny » Mon Dec 14, 2009 6:47 am
To have Machine B work the least, we have to maximise the capacity of Machine A, hence let A work for 8 hours.

In 8 hours, Machine A produces 9000x8=72000. Total left = 100000-72000 = 28000 to be produced by B.

So to produce 28000 B needs 28000/7000 hours. = 4 hours
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by Xbond » Mon Dec 14, 2009 12:10 pm
Many thks
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