Machine A and machine B are each used to manufacture 660 spr : GMAT Problem Solving (PS)
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Machine A and machine B are each used to manufacture 660 spr

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Machine A and machine B are each used to manufacture 660 spr [#permalink]

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08 Apr 2008, 23:45
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Question Stats:

68% (02:33) correct 32% (02:50) wrong based on 47 sessions

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Machine A and machine B are each used to manufacture 660 sprockets. It takes machine A 10 hours longer to produce 660 sprockets than machine B. Machine B produces 10 percent more sprockets per hour than machine A. How many sprockets per hour does machine A produces?

A. 6
B. 6.6
C. 60
D. 100
E. 110

OPEN DISCUSSION OF THIS QUESTION IS HERE: machine-a-and-machine-b-are-each-used-to-manufacture-98696.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 27 Jun 2014, 01:03, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: Machine A and machine B are each used to manufacture 660 spr [#permalink]

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09 Apr 2008, 00:09
I think the correct answer is A.

Machine A produces at a speed of A sp/hour and B at a speed of B sp/hour.
so, 660/A=(660/B)+10 and B=1,1 A--->1,1*660=660+11A--->A=6, so answer A is correct
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Re: Machine A and machine B are each used to manufacture 660 spr [#permalink]

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09 Apr 2008, 08:43
Speed of A = $$\frac{660}{n + 10}$$ sockets/hour

Speed of B = $$\frac{660}{n}$$ sockets/hour

Speed of B = 10 % of Speed of A

--> $$\frac{660}{n} = 1.1*\frac{660}{n + 10}$$ --> $$\frac{1}{n} = 1.1*\frac{1}{(n + 10)}$$ --> $${n+10} = 1.1*{n}$$ ---> $${n} = 100$$ ---> Hence speed of A = $$\frac {660}{110} = 6 sockets/hr$$
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Re: Machine A and machine B are each used to manufacture 660 spr [#permalink]

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09 Apr 2008, 10:54

by testing those
and it is not take much time
and i think it is took less time than by equation
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Re: Machine A and machine B are each used to manufacture 660 spr [#permalink]

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09 Apr 2008, 23:36
Thanks guys,
OA is A
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Re: Machine A and machine B are each used to manufacture 660 spr [#permalink]

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26 Jun 2014, 19:50
neelesh wrote:
Speed of A = $$\frac{660}{n + 10}$$ sockets/hour

Speed of B = $$\frac{660}{n}$$ sockets/hour

Speed of B = 10 % of Speed of A

--> $$\frac{660}{n} = 1.1*\frac{660}{n + 10}$$ --> $$\frac{1}{n} = 1.1*\frac{1}{(n + 10)}$$ --> $${n+10} = 1.1*{n}$$ ---> $${n} = 100$$ ---> Hence speed of A = $$\frac {660}{110} = 6 sockets/hr$$

HOW DO YOU GET FROM 1.1*\frac{660}{n + 10}[/m] --> $$\frac{1}{n} = 1.1*\frac{1}{(n + 10)}$$ where does the 660 go?
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Re: Machine A and machine B are each used to manufacture 660 spr [#permalink]

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27 Jun 2014, 01:03
sondenso wrote:
Machine A and machine B are each used to manufacture 660 sprockets. It takes machine A 10 hours longer to produce 660 sprockets than machine B. Machine B produces 10 percent more sprockets per hour than machine A. How many sprockets per hour does machine A produces?

A. 6
B. 6.6
C. 60
D. 100
E. 110

Let time needed for machine A to produce 660 sprockets be $$a$$ hours, then the rate of machine A would be $$rate_A=\frac{job \ done}{time}=\frac{660}{a}$$ sprockets per hour;

As "it takes machine A 10 hours longer to produce 660 sprockets than machine B" then time needed for machine B to produce 660 sprockets be $$a-10$$ hours and the rate of machine B would be $$rate_B=\frac{job \ done}{time}=\frac{660}{a-10}$$ sprockets per hour;

As "machine B produces 10 percent more sprockets per hour than machine A" then $$rate_A*1.1=rate_B$$ --> $$\frac{660}{a}*1.1=\frac{660}{a-10}$$ --> $$a=110$$ --> $$rate_A=\frac{job \ done}{time}=\frac{660}{a}=6$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: machine-a-and-machine-b-are-each-used-to-manufacture-98696.html
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Re: Machine A and machine B are each used to manufacture 660 spr   [#permalink] 27 Jun 2014, 01:03
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