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# Machine A and Machine B are each used to manufacture 660

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Manager
Joined: 05 Feb 2007
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Machine A and Machine B are each used to manufacture 660 [#permalink]  26 Apr 2008, 12:55
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0% (00:00) correct 0% (00:00) wrong based on 4 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 produce?
Senior Manager
Joined: 19 Apr 2008
Posts: 325
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Kudos [?]: 44 [0], given: 0

Re: PS: Rates/Machines [#permalink]  26 Apr 2008, 16:11
assume hours that B takes is H , and number of machines that A produces in an hour is M

work --------hours------------machines

A 660 = (H+10) * M
B 660 = H * 1.1M

solving this we get

A 660 = 110 * 6
B 660 = 100 * 6.6

So A produces 6 m/c , is the answer correct?
Current Student
Joined: 28 Dec 2004
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Location: New York City
Schools: Wharton'11 HBS'12
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Kudos [?]: 181 [0], given: 2

Re: PS: Rates/Machines [#permalink]  26 Apr 2008, 17:13
giantSwan 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 produce?

this is how i did..

1.1=rb*t; 1=ra*t
1.1ra=rb

660=rb*t
660=ra(t+10)
660=1.1ra*t
t=660/1.1ra you can right 1.1 as 110/100; 660*100/110*ra =600/ra
660=ra(600/ra +10)
660-600=10ra
ra=60/10 ra=6 sockets per hour
Manager
Joined: 26 Aug 2005
Posts: 60
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Kudos [?]: 4 [0], given: 0

Re: PS: Rates/Machines [#permalink]  27 Apr 2008, 11:02
this is how i solved it.

rate_a = 660/(t+10)
rate_b = 660/t
rate_b = rate_a*1.1

solve for t and you get t=100.
rate_a = 660/(t+10) = 660/110 = 6.
Re: PS: Rates/Machines   [#permalink] 27 Apr 2008, 11:02
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