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# Machine A and Machine B can produce 1 widget in 3 hours

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Manager
Joined: 05 May 2005
Posts: 81
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Machine A and Machine B can produce 1 widget in 3 hours [#permalink]  18 May 2007, 16:44
Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

a) 1/2

b) 2

c) 3

d) 5

e) 6
Manager
Joined: 02 May 2007
Posts: 152
Followers: 1

Kudos [?]: 5 [0], given: 0

Re: Word Problem - machines [#permalink]  18 May 2007, 19:26
above720 wrote:
Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

a) 1/2

b) 2

c) 3

d) 5

e) 6

(E) it is

A = hours it takes machine A to produce 1 widget on its own
B = hours it takes machine B to produce 1 widget on its own

1/A + 1/B = 1/3
=> 3B + 3A = AB (1)

2/A + 1/B = 1/2
=> 4B + 2A = AB (2)

(1) & (2) => A = B

=> A = 6
Director
Joined: 26 Feb 2006
Posts: 905
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Re: Word Problem - machines [#permalink]  18 May 2007, 20:29
above720 wrote:
Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

a) 1/2

b) 2

c) 3

d) 5

e) 6

= inverse of (1/2 - 1/3) = inverse (1/6) = 6 hours.
Current Student
Joined: 22 Apr 2007
Posts: 1097
Followers: 5

Kudos [?]: 21 [0], given: 0

A+B = 1/3 => 3A + 3B = 1 => 6A + 6B = 2
2A+B = 1/2 => 4A + 2B = 1 => 12A + 6B = 3

Together, 6A = 1 => A = 1/6. So A takes 6 hours to produce the widget
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