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Re: Running at their respective constant rates, Machine X takes 2 days [#permalink]
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Vegita wrote:
avigutman

Loved your approach to this question. 😀 A lot of critical thinking involved rather than going a quadratic route! Would it please be possible for you to explain how you came up with your statement "Product of individual times divided by sum of times gives us the total time of the combined workers"?

P.S: I tested your statement and it works but curious to know how it is derived.

Sure thing, Vegita. Here's an excerpt from my book.
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Re: Running at their respective constant rates, Machine X takes 2 days [#permalink]
Bunuel wrote:
chrishhaantje wrote:
Can somebody please explain the following:

How does this algebra work: 1/t + 1/t+2 = 5/12 is translated to 5t^2 + 34t - 24.

Thanks in advance ! Chris


\(\frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}\);

\(\frac{(t-2)+t}{t(t-2)}=\frac{5}{12}\);

\(\frac{2t-2}{t^2-2t}=\frac{5}{12}\);

\(24t-24=5t^2-10t\);

\(5t^2-34t+24=0\).

Hope it's clear.


Bunuel

Why can't you do

(w/x)+ w/(x+2)=(5/12)*w

Essentially, why do you need to subtract the 2 versus adding the two? I let x = machine y's time and x+2 = machine X's time because it takes machine X two more days than machine Y. When I went ahead and tried to solve though, I could not get the answer into the same quadratic equation because there were w's and x's whereas the w's canceled out when you do w/(x-2) compared to w/(x+2). Thank you for your help.
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Re: Running at their respective constant rates, Machine X takes 2 days [#permalink]
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woohoo921


Your approach is fine. The w's should all cancel out of your equation. In fact, you can cancel them directly from the form you have right now!

(w/x)+ w/(x+2)=(5/12)*w
(1/x) + 1/(x+1)= 5/12

From there the math is much the same as for Bunuel's solution. You should end up here:

5x^2 -14x - 24 = 0
(5x+6)(x-4) = 0

Positive solution for x = 4, so X's actual time is x+2 = 6.

However, notice that you are setting yourself up for trouble here. If you represent Machine Y as "x" and Machine X as "x+2," you are far more likely to end up providing the wrong answer after all that calculation! Two valuable principles here:
1) Choose variables that match what they represent. You were really solving for Y here, so call it y.
2) When possible, solve directly for what you want. The question is about X's time, so set up and solve for that.
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Re: Running at their respective constant rates, Machine X takes 2 days [#permalink]
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Quick approximation:

First, get a value for how long the 2 machines together would take to produce 2w widgets. We're told that it takes 3 days to produce 5w/4. That means they will take 3(4/5) = 12/5 days to produce w widgets and 2(12/5) = 24/5 days to produce w widgets. So basically, together the 2 machines take almost 5 days to produce 2w widgets.

Now we know that X is slower than Y. If they equally fast, one machine alone would take twice as long, or almost 10 days, to produce 2w widgets. However, we know that X is considerably slower than Y. So it will take more than 10 days. E.

(This would definitely be a weird approach to invent on the fly to solve this one problem, but you might be surprised how often this kind of approximation will get you to--or near--the right answer.)
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Re: Running at their respective constant rates, Machine X takes 2 days [#permalink]
KarishmaB wrote:
Baten80 wrote:
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

Please show the whole calculation.


Together they make 5/4 w widgets in 3 days. So they make w widgets in 3*4/5 = 12/5 days.

1/(t+2) + 1/t = 5/12

Now, calculating this is really cumbersome so try to plug in options to get to the answer.
If machine X takes 12 days to produce 2w widgets, it would take 6 days to make w widgets, t would be 4.

1/6 + 1/4 = 5/12
It works so t = 4.

Answer is 12.


Why do you put 12/5 in the first lane adn change it to 5/12 in the second?

Thanks!
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Re: Running at their respective constant rates, Machine X takes 2 days [#permalink]
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This solution is relying on the idea that rate and time are reciprocal. It's the same reason that if they make 5/4 w widgets per day, they take 4/5 of a day to make w widgets.

In the part you're asking about, 12/5 days is the time to make w widgets. If they take 12/5 days to make w, then they do 5/12 of w per day. The rates 1/t and 1/(t+2) must add to this combined rate.

Thib33600 wrote:
KarishmaB wrote:
Baten80 wrote:
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

Please show the whole calculation.


Together they make 5/4 w widgets in 3 days. So they make w widgets in 3*4/5 = 12/5 days.

1/(t+2) + 1/t = 5/12

Now, calculating this is really cumbersome so try to plug in options to get to the answer.
If machine X takes 12 days to produce 2w widgets, it would take 6 days to make w widgets, t would be 4.

1/6 + 1/4 = 5/12
It works so t = 4.

Answer is 12.


Why do you put 12/5 in the first lane adn change it to 5/12 in the second?

Thanks!
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Re: Running at their respective constant rates, Machine X takes 2 days [#permalink]
 
Bunuel wrote:
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

­Let us assume, Y takes n days to produce w widgets. Then X will take (n+2) days to produce w widgets.
Efficiency of Y=1/n, efficiency of X=1/(n+2). So total efficiency to produce w widgets is 1/n + 1/(n+2)=(2n+2)/(n(n+1)).

Take the reciprocal of the total efficiency, we will get total days i.e. (n(n+1))/(2n+2). This is the time to produce w widgets ... (i)

Given 5w/4 widgets are produced in 3 days, then w widgets are produced in 3*w*4/(5*w)=12/5 days ... (ii)

(i) and (ii) represent the same, let's equate them. \((n(n+1))/(2n+2)\)=12/5
We get 5\(n^2\)-14n-24=0. Solve it for the positive value, n will be 4.

Therefore, X takes 4+2=6 days to produce w widgets. Then for 2w widgets, X will take 6*2=12 days. Option (E) is correct.
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