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# Running at their respective constant rates, machine X takes2

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Running at their respective constant rates, machine X takes2 [#permalink]  29 Mar 2008, 08:23
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Running at their respective constant rates, machine X takes2 days longer to produce w widgets than machine Y. At these rates, if the 2 machines together produce 5w/4 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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Re: PS: widgets [#permalink]  29 Mar 2008, 08:39
E - 12 days.

$$\frac {3}{N+2} + \frac {3}{N} = \frac {5}{4}$$

Solving for N we get, $$N = 4$$

w widgets are made by machine X in 6 days (4 + 2) . So 2w widgets will take 12 days
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Re: PS: widgets [#permalink]  05 Apr 2008, 14:18
How did you get that N=4 without solving the equation?
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Re: PS: widgets [#permalink]  05 Apr 2008, 14:41
i think plug number is faster in this case, otherwise u have to solve quadratic eqn
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Re: PS: widgets [#permalink]  05 Apr 2008, 16:48
Expert's post
another way: logical reasoning.

1. "machine X takes 2 days longer to produce w widgets than machine Y" - machine X works slowly that machine Y
2. If machine Y work simultaneously, both machine would produce >4w widgets (Y faster than X)
3. So, 5w/4 in 3 days and >4w in ----> >9.6 days
4. in ~10 day Y can produce ~ 4w*2/(10-2) ~ w more widgets than X
5. therefore, answer has to be close to 3*(4w+w)/ (5w/4) = 12

It took ~30sec in my mind and is one of fast ways for solving work problems along with back-solving.
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Re: PS: widgets [#permalink]  05 Apr 2008, 20:22
walker wrote:
another way: logical reasoning.

1. "machine X takes 2 days longer to produce w widgets than machine Y" - machine X works slowly that machine Y
2. If machine Y work simultaneously, both machine would produce >4w widgets (Y faster than X)
3. So, 5w/4 in 3 days and >4w in ----> >9.6 days
4. in ~10 day Y can produce ~ 4w*2/(10-2) ~ w more widgets than X
5. therefore, answer has to be close to 3*(4w+w)/ (5w/4) = 12

It took ~30sec in my mind and is one of fast ways for solving work problems along with back-solving.

Honestly, I still not catch up you logic flow. So I use the traditional one. Anyway, Do you mind using "back-solving" method for this?

netcaesar wrote:
How did you get that N=4 without solving the equation?

X is the time machine X works w job
Y is the time machine Y works w job

X =Y+2 --> Y=X-2
(w/X) + (w/(X-2) = 5w/12

5X^2 -34X + 24 =0
X=6
Machine X do w job in 6 hours, so 2w, corresponding time is 12 hours
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Re: PS: widgets [#permalink]  08 Apr 2008, 06:24
Machine Y takes N days for w widgets => for 2w widgets Y will take 2N days
Machine X takes N+2 days for w widgets => for 2w widgets X will take 2N+4 days

Together:
5/4w widgets in 3 days
1w widget in 3/(5/4) or 3*4/5 days
2w widgets in 2*3*4/5 = 24/5

1/(2N) + 1/(2N+4) = 5/24

(2N + 4 + 2N)/[(2N)(2n+4)] = 5/24
4(N+1)/4(N^2 + 2N) = 5/24

24(N+1) = 5(N^2 + 2N)
5N^2 + 8N - 24 = 0
=> N = 4
So Y takes 4 days and X takes 4+2 = 6 days to complete w widgets
12 (E) for 2w widgets.
Re: PS: widgets   [#permalink] 08 Apr 2008, 06:24
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