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Running at their respective constant rates, Machine X takes

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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
1
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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Re: Running at their respective constant rates, Machine X takes  [#permalink]

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New post 09 May 2018, 13:56
PareshGmat wrote:
muhtasimhassan wrote:
Hi,

I am new here and I have solved this math correctly. However, here's my problem:

If I consider the number of days required to produce 'W' widgets as 'X' and 'X-2' for X and Y respectively, I end up with the equation 5t^2 - 34t + 24 which yields me 2 solutions, one of which is correct (x=6)

However, when I take 'X' and 'Y' as 'X+2' and 'X', which are the same as above, because either ways, X is taking 2 days longer, I end up with the equation 5x^2 - 14x - 24 which has no solutions (I hope I'm not making some stupid mistake here)

Can someone please clarify this and correct me where I'm wrong.


Refer my post above; I did in the same way.....

As far as solving the equation, refer below.....

\(5x^2 - 14x - 24 = 0\)

\(5x^2 - 20x + 6x - 24 = 0\)

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

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

x = 4 (Ignore the -ve equation)

x+2 = 6



pushpitkc
here cant fugure out what numbers are multiplied to get -24 and add up to -14. i wanna do factorization as i did in this post, https://gmatclub.com/forum/how-many-rea ... l#p1999687 but simply these figures dont come to my mind :) and i dont want to use lengthy way through formula b^2 -4ac

is it some kind of new way of factorization ? :)

\(5x^2 - 14x - 24 = 0\)

\(5x^2 - 20x + 6x - 24 = 0\) how to understand this ?

\(5x(x-4) + 6(x-4) = 0\) and this :)

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

how we get x = 4 ?

see ya tomorrow :) thanks!
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Running at their respective constant rates, Machine X takes  [#permalink]

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New post 10 May 2018, 00:16
1
dave13 wrote:
PareshGmat wrote:
muhtasimhassan wrote:
Hi,

I am new here and I have solved this math correctly. However, here's my problem:

If I consider the number of days required to produce 'W' widgets as 'X' and 'X-2' for X and Y respectively, I end up with the equation 5t^2 - 34t + 24 which yields me 2 solutions, one of which is correct (x=6)

However, when I take 'X' and 'Y' as 'X+2' and 'X', which are the same as above, because either ways, X is taking 2 days longer, I end up with the equation 5x^2 - 14x - 24 which has no solutions (I hope I'm not making some stupid mistake here)

Can someone please clarify this and correct me where I'm wrong.


Refer my post above; I did in the same way.....

As far as solving the equation, refer below.....

\(5x^2 - 14x - 24 = 0\)

\(5x^2 - 20x + 6x - 24 = 0\)

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

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

x = 4 (Ignore the -ve equation)

x+2 = 6



pushpitkc
here cant fugure out what numbers are multiplied to get -24 and add up to -14. i wanna do factorization as i did in this post, https://gmatclub.com/forum/how-many-rea ... l#p1999687 but simply these figures dont come to my mind :) and i dont want to use lengthy way through formula b^2 -4ac

is it some kind of new way of factorization ? :)

\(5x^2 - 14x - 24 = 0\)

\(5x^2 - 20x + 6x - 24 = 0\) how to understand this ?

\(5x(x-4) + 6(x-4) = 0\) and this :)

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

how we get x = 4 ?

see ya tomorrow :) thanks!



Hi dave13

Unfortunately, there will be times when you will have to use the lengthy method :(

Here is a simple workaround which can be used for this problem.
We need two numbers whose sum is -14 and product is -24*5 = -120

Prime-factorizing 120, we get \(2^3 * 3 * 5\) and there are 4*4 or 16 factors possible.
Listing down the factors, we will have 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120. From these
factors we need to find a combination which gives a sum of -14, which is possible when x=-20 and 6

However, for bigger numbers where is it not possible to list down the factors, it will be difficult
to solve a problem using this method. I would recommend that you solve some problems using
this formula just in case you can't use the above-given method.

Hope this helps you!
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New post 10 May 2018, 08:13
hey pushpitkc :)

will you join on my bumpy ride to understand a few questions regarding my soluton in the post above :)

In my solution above I get the roots 4 and -1.2 are these roots correct ? :? whereas you get x=-20 and 6 through factorization..why ?

Also when you say "We need two numbers whose sum is -14 and product is -24*5 = -120"

\(5y^2 -14y-24=0\) and considering that \(5y^2\) is A \(-14y\) is B and \(-24\) is C

so i can conclude you mulptiply C by A ---> -24 *5 to find the numbers that add up -14 ---> can you please remind the formula.

And if equation were like this \(y^2 -14y-24=0\) (without 5) so i simply would multiply -24 by 1 ? right ?

many thanks and have great weekend :)

p.s looks like I accidently deleted the solution I posted yesterday...
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Re: Running at their respective constant rates, Machine X takes  [#permalink]

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New post 11 May 2018, 09:59
1
dave13 wrote:
hey pushpitkc :)

will you join on my bumpy ride to understand a few questions regarding my soluton in the post above :)

In my solution above I get the roots 4 and -1.2 are these roots correct ? :? whereas you get x=-20 and 6 through factorization..why ?

Also when you say "We need two numbers whose sum is -14 and product is -24*5 = -120"

\(5y^2 -14y-24=0\) and considering that \(5y^2\) is A \(-14y\) is B and \(-24\) is C

so i can conclude you mulptiply C by A ---> -24 *5 to find the numbers that add up -14 ---> can you please remind the formula.

And if equation were like this \(y^2 -14y-24=0\) (without 5) so i simply would multiply -24 by 1 ? right ?

many thanks and have great weekend :)

p.s looks like I accidently deleted the solution I posted yesterday...


Hey dave13

The roots(in the previous solution) are x = 4 and \(5x = -6\) -> \(x = -\frac{6}{5} = -1.2\)
You have solved the problem correctly by using the formula and arrived at the right solution.

Have a great weekend!
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Re: Running at their respective constant rates, Machine X takes  [#permalink]

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New post 15 Aug 2018, 08:11
Running at their respective constant rates, Machine X takes 2 dazys 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

It can be very easily solved using efficiency method.

So the question says, X+Y together produces 1.25 widgets in 3 days
Thus, efficiency of X and Y to produce 1.25 widget in 3 days= 3/1.25=41.6% ----(eqn 1)

Now we are asked to find the days taken by X to produce 2w widgets.
Lets go to the answers.
Consider C: 8 days.
thus, X takes 8 days to produce 2w widgets==>4 days to produce w widgets
thus X's efficiency= 1/4=25%
So Y efficiency would be (41.6-25)=16.66% . ---(from eqn 1)
This implies: X would take 4 days (i.e 100/25=4) and Y would take 6 days (i.e 100/16.66=6)
but the question says the opposite: X should take 2 more days.

So now consider option E: 12 days.
So 12 days for 2w widgets.
So 6 days for w widget.

So X's efficiency= 100/6=16.66% = 6 days (i.e 100/16.66=6)
==> Y's efficiency= 41.66-16.66=25%= 4 days (i.e 100/25=4)
So this is correct as X would take 6 days and Y would take 4 days; So 6-4=2 days.
Hence option E is the correct answer.
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Re: Running at their respective constant rates, Machine X takes  [#permalink]

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New post 16 Aug 2018, 03:04
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 # days required for machine Y to produce w widgets be y days

Hence rate of machine Y = w/y widgets per day

Hence rate of machine X = w/y+2 widgets per day

given that both machines together can produce 5w/4 widgets in 3 days

Hence rate at which both machines working together = (5w/4)/3

We get w/(y+2) + w/y = (5w/4)/3

solving this for y we get, y = 4 days

Hence # of days machine X takes to produce w widgets = 4 + 2 = 6 days

Therefore # of days machine X takes to produce 2w widgets = 12 days


Answer E.



Thanks,
GyM
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Re: Running at their respective constant rates, Machine X takes &nbs [#permalink] 16 Aug 2018, 03:04

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