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# Six machines, all working at the same steady rate, can

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Intern
Joined: 14 Oct 2004
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Six machines, all working at the same steady rate, can [#permalink]

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22 Apr 2005, 08:40
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Six machines, all working at the same steady rate, can finish a particular job in 1 hour. If we wanted to finish the same job in 40 minutes, how many identical machines would we have to add?

A. 1
B. 2
C. 3
D. 6
E. 9
Intern
Joined: 14 Nov 2004
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22 Apr 2005, 08:55
6 Machines take 60 mins.
therefore the number of machines needed to finish the job in 40 mins =

(60/40)*6=9 machines.

number of extra machines required=9-6=3.
Manager
Joined: 22 Apr 2005
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22 Apr 2005, 11:31
More formal solution

J - job
M - average rate of a single machine

J/(6 * M) = 1
J/((6+x) * M) = 2/3

J = 6M * 1

6M * 1 / ((6+x) * M) = 2/3

6 / (6+x) = 2/3

2 * (6+x) = 18

x = 3
Senior Manager
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24 Apr 2005, 08:20
60 mins job requires = 6 machines
1 min job would require = 60*6 machines ( as number of m/c would be more)
40 min job would require = 60*6/40 = 9 m/c

No of machines required = 9-6 =3
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ash
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I'm crossing the bridge.........

VP
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24 Apr 2005, 18:42
Agree with 3

6 machines during 1h -> 6h -> 360mn

x*40 = 360
x = 9

9-6 = 3
VP
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Location: Taiwan
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25 Apr 2005, 04:48

agree with the above-mentioned workings.
25 Apr 2005, 04:48
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# Six machines, all working at the same steady rate, can

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