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# Running at their respective constant rates, machine X takes

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Senior Manager
Joined: 05 Jun 2008
Posts: 289
Running at their respective constant rates, machine X takes  [#permalink]

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24 Sep 2008, 07:40
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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?
A. 4
B. 6
C. 8
D. 10
E. 12

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Manager
Joined: 20 Mar 2008
Posts: 147
Location: USA

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24 Sep 2008, 10:22
y takes n days for W widgets Per day -> 1/n
x takes n+2 days W widgets Per day -> 1/n+2

Together these produce 5/4 W widgets in 3 days or 5/12 W per day

Per Day Rate of y + Per day rate of x = 5/12 W
1/n + 1/n+2 = 5/12 W

n=4 satisfies this ( No of days Y takes to produce W)

We have couple of more steps for the answer

No of days X takes is n+2 = 6
For 2W =12

E
Director
Joined: 23 May 2008
Posts: 739

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24 Sep 2008, 16:40
Quote:
Per Day Rate of y + Per day rate of x = 5/12 W
1/n + 1/n+2 = 5/12 W

n=4 satisfies this ( No of days Y takes to produce W)

can you go through the steps you did to get to n=4? I ended up factoring the quadratic equation to get 4 as a possible n, but this method took forever
Manager
Joined: 20 Mar 2008
Posts: 147
Location: USA

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25 Sep 2008, 09:23
bigtreezl wrote:
Quote:
Per Day Rate of y + Per day rate of x = 5/12 W
1/n + 1/n+2 = 5/12 W

n=4 satisfies this ( No of days Y takes to produce W)

can you go through the steps you did to get to n=4? I ended up factoring the quadratic equation to get 4 as a possible n, but this method took forever

When I wrote the original explanation i plugged in couple of numbers to test where n would satisfy the equation and got to 4 .... did not do this calculation. However, this may be counterproductive at times. There is no sure shot way that I am aware of ... we need to go through the lengthy process of solving the quadratic equation.

I forgot to put W on LHS in the original explanation... taking W off from both sides

1/n + 1/n+2 = 5/12

2n+2/ n^2+2n= 5/12

5n^2- 14n -24 =0

5n^2- 20n + 6n -24 =0

5n( n-4) + 6 (n-4)=0

(5n+6)(n-4)=0

n=-6/5 or n=4... by eliminating -ve.. we get n=4

I know that you have done this already but for the completeness of this topic, I am putting the steps here.

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Re: PS-Work Problem &nbs [#permalink] 25 Sep 2008, 09:23
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