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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] New post 08 Apr 2008, 23:45
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

65% (02:48) correct 35% (01:45) wrong based on 26 sessions
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] New post 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] New post 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] New post 09 Apr 2008, 10:54
i got it from answer

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] New post 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] New post 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] New post 27 Jun 2014, 01:03
Expert's post
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 \(rate_A*1.1=rate_B\) --> \(\frac{660}{a}*1.1=\frac{660}{a-10}\) --> \(a=110\) --> \(rate_A=\frac{job \ done}{time}=\frac{660}{a}=6\).

Answer: A.

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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