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Machine A produces pencils at a constant rate of 9,000 pencils per hou

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Machine A produces pencils at a constant rate of 9,000 pencils per hou [#permalink] New post 04 Apr 2007, 16:09
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Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines to gather must produce 100,000 pencils and if each 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 2/3
C. 5 1/3
D. 6
E. 6 1/4
[Reveal] Spoiler: OA
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou [#permalink] New post 04 Apr 2007, 22:11
Ax + Bx = 100,000
x <= 8

Since it's asking for the least amount of time for B, A must work to its maximum.

100,000-(9000*8) = 28,000
28,000/7000 = 4

A.

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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou [#permalink] New post 05 Apr 2007, 21:04
Total target = 100, 000

Since machine A is faster, and max is 8 hours, we shall allow machine A to orun for 8 hours. It shall thus produce 9000 * 8 = 72, 000 pencils.

We are short by 28, 000, machine B can work for 28K/7K = 4 hours to do it.

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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou [#permalink] New post 05 Apr 2007, 22:19
9,000 x 8 hours + 7,000 x X hours = 100,000
72,000 + 7,000 x X hours = 100,000
X hours =28,000 / 7,000 = 4

A it is
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou [#permalink] New post 05 Apr 2007, 23:51
nitinneha wrote:
Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines togather must produce 100,000 pencils and if each 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 2/3
C)5 1/3
D)6
E)6 1/4


B should operate for least time so A must run for 8 hours and produce 72K. So remaining 28K will be produced by B in 4 hours
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou [#permalink] New post 27 Jul 2010, 12:48
Machine A produces 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 each 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 and 2/3
C. 5 and 1/3
D. 6
E. 6 and 1/4

Last edited by Bunuel on 08 Mar 2015, 05:54, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou [#permalink] New post 27 Jul 2010, 13:16
Expert's post
Machine A produces 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 each 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 and 2/3
c) 5 and 1/3
d) 6
e) 6 and 1/4

To minimize the time that machine B must operate we must maximize the time machine A can operate, so make it operate 8 hours. In 8 hours machine A will produce 8*9,000=72,000 pencils, so 100,000-72,000=28,000 pencils are left to produce, which can be produced by machine B in 28,000/7,000=4 hours.

Answer: A.
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou [#permalink] New post 08 Mar 2015, 03:00
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou   [#permalink] 08 Mar 2015, 03:00
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