Machine A produces pencils at a constant rate of 9000 : PS Archive
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Machine A produces pencils at a constant rate of 9000

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Machine A produces pencils at a constant rate of 9000 [#permalink]

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26 Oct 2005, 21:58
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Machine A produces pencils at a constant rate of 9000 pencils/hour, and machine B products pencils at a constant rate of 7000 pencils per hour. If the two machine together must product 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) 14/3
c) 16/3
d) 6
e) 25/4
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26 Oct 2005, 22:23
Ans: A

We need to find the least working hrs for B, hence we should maximize the working hrs of A.

Max working hrs of A is 8 -> 8 * 9000 = 72000

Remaining pencils = 28000 -> 28000/Working rate of B = 28000/7000
= 4 hours
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31 Oct 2005, 02:12
but the question says each machine together must produce....i think the ans should be E) 25/4
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Re: PS - Prep1 - Two machines working together [#permalink]

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31 Oct 2005, 02:48
If B has to be minimum then A should work for the maximum time.
The original equn is 7000 X + 9000 Y = 100000
Where x is number of hours that A works and Y for B. We want maximum for B, so Y will be 8.
7000x + (9000 * 8) = 100000
=> 7000X = 280000 => x = 4
So machine B must operate for atleast 4 hours. - A
Re: PS - Prep1 - Two machines working together   [#permalink] 31 Oct 2005, 02:48
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