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Running at their respective constant rates, Machine X takes

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Senior Manager
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Joined: 06 Dec 2016
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Re: Running at their respective constant rates, Machine X takes [#permalink]

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New post 22 Apr 2017, 05:25
Engr2012 wrote:
jasonfodor wrote:
can someone pelase explain how they got from (1/t) + (1/((t-2)) = 5/12
TO
5t^2-34t+24=0


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

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

Cross multiplying and rearranging terms, you get ,

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

Hope it helps.



Note, if you do it this way below, you are going to get the same answer.

1/x + 1/x+2 = 5/12
2x+2/x^2 + 2x = 5/12
12(2x + 2) = 5(x^2 + 2x)
24x + 24 = 5x^2 + 10x
5x^2 - 14x - 24 = 0
(5x + 6)(x-4)

x + 2
4 + 2
6

Which is answer C

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Re: Running at their respective constant rates, Machine X takes [#permalink]

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New post 05 Jun 2017, 16:23
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

1. Do all the calculations for 2w widgets
2. It takes x 4 days more to produce 2w widgets'
3. Together they take 4.8 days to produce 2w widgets
4. 1/y + 1/x=1/4.8 or 1/y + 1/(y+4) =5/24, y=8 . Therefore y+4=12.
X takes 12 days to produce 2w widgets
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Re: Running at their respective constant rates, Machine X takes [#permalink]

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New post 08 Jun 2017, 16:06
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Expert's post
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


We can let x = the number of days it takes Machine X to produce w widgets and y = the number of days it takes Machine Y to produce w widgets. Thus, x = y + 2 or y = x - 2. Furthermore, the rate of Machine X is w/x and the rate of Machine Y is 1/y = w/(x - 2). We are given that they can produce (5/4)w widgets in 3 days. Thus, we have:

3(w/x) + 3[w/(x - 2)] = (5/4)w

3w/x + 3w/(x - 2) = (5/4)w

Multiplying both sides by 4x(x - 2), we have:

12w(x - 2) + 12wx = 5wx(x - 2)

12(x - 2) + 12x = 5x(x - 2)

12x - 24 + 12x = 5x^2 - 10x

5x^2 - 34x + 24 = 0

(5x - 4)(x - 6) = 0

x = ⅘ or x = 6

However, x can’t be ⅘; if it were, y would be negative. Thus, x must be 6. Since it takes Machine X 6 days to produce w widgets, it will take 12 days to produce 2w widgets.

Answer: E
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Running at their respective constant rates, Machine X takes [#permalink]

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New post 11 Oct 2017, 12:37
Set the number of widgets to a preferred number. I choose 20.

(1/y + 2) + (1/y) + 1/3 = 5/4(20) = 25
2/6+y + 3/6+y + 6/6+y = 25
11/25 = 6 + y
11/25 - 6/1 = y
5/25 = 1/5 = y = machine y takes 5 days to make 25, therefore it takes 4 days to make 20
machine x takes two days more than y to produce 20 widgets; therefore, it takes machine x 2 + 4 = 6 days for 20 widgets

6 days for 20 widgets, 12 days for 40 widgets or 2w
therefore, answer (E) 12 days

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Running at their respective constant rates, Machine X takes   [#permalink] 11 Oct 2017, 12:37

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