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PS: Widgets and machines.

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Joined: 12 Jun 2006
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PS: Widgets and machines. [#permalink] New post 06 Jun 2007, 17:11



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I'd like to see your solutions to this one. And please explain your process.

Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce (5/4)w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?
Joined: 28 Aug 2006
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 [#permalink] New post 06 Jun 2007, 19:54
Lets say Machine Y makes w widgets in A days
Machine X makes the same w widgets in A+2 days

Together they make w widgets in A(A+2)/(2A+2) days -----1 (Formula for combined workoutput)

Now if together they have completed 5/4w widgets in 3 days => they would have completed w widgets in 12/5 days-------2

Equating 1 &2 we get
A=4 and -6/5(cannot be -ve)

X makes w widgets in A+2 daya=> 6 days
and 2w widgets in 12 days------Answer
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 [#permalink] New post 07 Jun 2007, 00:28
can we avoid quadratic equation somehow?
Can it be solved easier somehow?
This method is too long.
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 [#permalink] New post 07 Jun 2007, 04:15
LEt Y produce w widgets in a days
X shall produce w widgets in a+2 days
X in 1 days shall produce w/a+2 widgets
Y in 1 day shall produce w/a widgets
If 5/4 widgets are produced in 3 days, then in 1 days 5/12 widgets shall be produced
SO if X produces w/a+2 widgest and Y produces w/a widgets in one day, the combined they shall produce w/a+2 + w/a which is equal to 5/12
so setting up the equation
w/a+w/a+2 = 5/12
solving this equation, we get a=4 & -6/5. Since days cannot be in -ve, so ahas to be 4.
Now, if X produces w widgets in a+2=6 days, then it shall produce 2w widgets in 6*2=12 days.
so 12 days in the answer
  [#permalink] 07 Jun 2007, 04:15
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PS: Widgets and machines.

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