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Running at their respective constant rates, Machine X takes

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Running at their respective constant rates, Machine X takes [#permalink] New post 18 Mar 2014, 00:39
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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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Question: 173
Category: Algebra; Applied problems
Page: 85
Difficulty: 600


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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 18 Mar 2014, 00:39
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SOLUTION

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

For work problems one of the most important thin to know is \(rate*time=job \ done\).

Let the time needed for machine X to produce \(w\) widgets be \(t\) days, so the rate of X would be \(rate=\frac{job \ done}{time}=\frac{w}{t}\);

As "machine X takes 2 days longer to produce \(w\) widgets than machines Y" then time needed for machine Y to produce \(w\) widgets would be \(t-2\) days, so the rate of Y would be \(rate=\frac{job \ done}{time}=\frac{w}{t-2}\);

Combined rate of machines X and Y in 1 day would be \(\frac{w}{t}+\frac{w}{t-2}\) (remember we can sum the rates). In 3 days two machines together produce 5w/4 widgets so: \(3(\frac{w}{t}+\frac{w}{t-2})=\frac{5w}{4}\) --> \(\frac{w}{t}+\frac{w}{t-2}=\frac{5w}{12}\).

\(\frac{w}{t}+\frac{w}{t-2}=\frac{5w}{12}\) --> reduce by \(w\) --> \(\frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}\).

At this point we can either solve quadratic equation: \(5t^2-34t+24=0\) --> \((t-6)(5t-4)=0\) --> \(t=6\) or \(t=\frac{4}{5}\) (which is not a valid solution as in this case \(t-2=-\frac{6}{5}\), the time needed for machine Y to ptoduce \(w\) widgets will be negatrive value and it's not possible). So \(t=6\) days is needed for machine X to produce \(w\) widgets, hence time needed for machine X to produce \(2w\) widgets will be \(2t=12\) days.

OR try to substitute the values from the answer choices. Remember as we are asked to find the time needed for machine X alone to produce \(2w\) widgets then the answer should be \(2t\) among answer choices: E work - \(2t=12\) --> \(t=6\) --> \(\frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}\).

Answer: E.
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Running at their respective constant rates, Machine X takes [#permalink] New post 20 Mar 2014, 00:52
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Time taken by Machine Y = t days

Time taken by Machine X = (t+2) days

Rate of Machine \(X = \frac{w}{t+2}\)
Rate of Machine \(Y = \frac{w}{t}\)

\(\frac{5w}{4}\) widgets produced in 3 days; so \(\frac{5w}{12}\) produced in 1 day

Equation setup will be

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

Solving,

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

t = 4

Time taken by Machine X = t+2 = 6 for w

So for 2w; it will be 12

Answer = E

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Last edited by PareshGmat on 14 Aug 2014, 02:01, edited 1 time in total.
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 23 Mar 2014, 07:13
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SOLUTION

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

For work problems one of the most important thin to know is \(rate*time=job \ done\).

Let the time needed for machine X to produce \(w\) widgets be \(t\) days, so the rate of X would be \(rate=\frac{job \ done}{time}=\frac{w}{t}\);

As "machine X takes 2 days longer to produce \(w\) widgets than machines Y" then time needed for machine Y to produce \(w\) widgets would be \(t-2\) days, so the rate of Y would be \(rate=\frac{job \ done}{time}=\frac{w}{t-2}\);

Combined rate of machines X and Y in 1 day would be \(\frac{w}{t}+\frac{w}{t-2}\) (remember we can sum the rates). In 3 days two machines together produce 5w/4 widgets so: \(3(\frac{w}{t}+\frac{w}{t-2})=\frac{5w}{4}\) --> \(\frac{w}{t}+\frac{w}{t-2}=\frac{5w}{12}\).

\(\frac{w}{t}+\frac{w}{t-2}=\frac{5w}{12}\) --> reduce by \(w\) --> \(\frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}\).

At this point we can either solve quadratic equation: \(5t^2-34t+24=0\) --> \((t-6)(5t-4)=0\) --> \(t=6\) or \(t=\frac{4}{5}\) (which is not a valid solution as in this case \(t-2=-\frac{6}{5}\), the time needed for machine Y to ptoduce \(w\) widgets will be negatrive value and it's not possible). So \(t=6\) days is needed for machine X to produce \(w\) widgets, hence time needed for machine X to produce \(2w\) widgets will be \(2t=12\) days.

OR try to substitute the values from the answer choices. Remember as we are asked to find the time needed for machine X alone to produce \(2w\) widgets then the answer should be \(2t\) among answer choices: E work - \(2t=12\) --> \(t=6\) --> \(\frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}\).

Answer: E.
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 16 May 2014, 06:22
QE: 5t^2 - 14t - 24 = 0, How do we get solution = 4?
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 16 May 2014, 08:32
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gciftci wrote:
QE: 5t^2 - 14t - 24 = 0, How do we get solution = 4?


Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm

Solving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htm

Hope it helps.
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 27 May 2014, 14:16
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Alternate Solution without algebra using POE.

If it takes 3 days to produce 5/4 widgets, it will take 3*(4/5) = 12/5 days to produce 1 widget.

If two machines were working at the same rate it would take, 2*12/5 i.e. approx 5 days days for x to produce 1 widget. Since machine X is slower it will take more than 5 days to produce 1 widget and subsequently more than 10 days to produce 2 widgets. The only answer more than 10 is 12 days. Answer E
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 10 Aug 2014, 01:25
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Back tracking method:
We have to find the number of days it takes Machine X alone to produce 2w widgets.
Answer options are as below:

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12

We are given that X and Y produce (5/4)w in 3 days and also given that X take 2 days more than Y.
So to find corresponding value of X and Y use the answer options.
First half the value of X to find number of days it takes Machine X to complete *W* widgets and find the value of Y by reducing 2 from all answer options.

--> X to complete W widgets
(A) X=2 and Y= 2-2 =0
(B) X=3 and Y= 3-2 =1
(C) X=4 and Y= 4-2 =2
(D) X=5 and Y= 5-2 =3
(E) X=6 and Y= 6-2 =4

Now we have all the values of X and Y. Find out the combination X and Y, whose values will produce (5/4)w for 3 days of work.

3(1/6+1/4)=>5/4-->Option (E) gives this value so correct answer is (E)-->12.
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 20 Oct 2014, 09:39
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.
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 20 Oct 2014, 19:34
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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
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 20 Oct 2014, 20:20
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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.


It doesn't matter whether you take the number of days as
Days taken by machine X = X and
Days taken by machine Y = X-2
or
Days taken by machine X = X+2
Days taken by machine Y = X

Either way, you will get one valid value for X and you will need to manipulate that to get your answer. You need to find the number of days that will be taken by machine X to make 2w widgets.

In first case, you get the equation as \(5X^2 - 34X + 24 = 0\) and get X = 6 as the only valid value (X = 4/5 gives X-2 as negative but number of days cannot be negative).
Machine X takes X days i.e. 6 days to make w widgets. So it will take 12 days to make 2w widgets.

In the second case, you get \(5x^2 - 14x - 24 = 0\)
which is \(5x^2 - 20x + 6x - 24 = 0\)
5x(x - 4) + 6(x - 4) = 0
(x-4)(5x + 6) = 0
x = 4
This is the number of days taken by machine Y to make w widgets. So machine X takes 4+2 = 6 days to make w widgets. It will take 12 days to make 2w widgets.

Either way, your answer will be the same.
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 20 Oct 2014, 20:39
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VeritasPrepKarishma

Thanks a lot. Careless of me to not have noticed such a simple solution to the equation 5x^2 - 14x - 24 = 0
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Running at their respective constant rates, Machine X takes [#permalink] New post 08 Dec 2014, 07:30
VeritasPrepKarishma 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.


It doesn't matter whether you take the number of days as
Days taken by machine X = X and
Days taken by machine Y = X-2
or
Days taken by machine X = X+2
Days taken by machine Y = X

Either way, you will get one valid value for X and you will need to manipulate that to get your answer. You need to find the number of days that will be taken by machine X to make 2w widgets.

In first case, you get the equation as \(5X^2 - 34X + 24 = 0\) and get X = 6 as the only valid value (X = 4/5 gives X-2 as negative but number of days cannot be negative).
Machine X takes X days i.e. 6 days to make w widgets. So it will take 12 days to make 2w widgets.

In the second case, you get \(5x^2 - 14x - 24 = 0\)
which is \(5x^2 - 20x + 6x - 24 = 0\)
5x(x - 4) + 6(x - 4) = 0
(x-4)(5x + 6) = 0
x = 4
This is the number of days taken by machine Y to make w widgets. So machine X takes 4+2 = 6 days to make w widgets. It will take 12 days to make 2w widgets.

Either way, your answer will be the same.



Just wanted to point out that this problem can be solved quite easily without solving a quadratic.

Our equation for x is given by:

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

\(\frac{x-2 + x} {x(x-2)} = \frac{5}{12}\).

\(\frac{2x-2}{x(x-2)} = \frac{5}{12}\).

\(\frac{2(x-1)}{x(x-2)} = \frac{5}{12}\).

\(\frac{(x-1)}{x(x-2)} = \frac{5}{24}\).

We see immediately that by letting \(x=6\) we find our solution of x. Therefore, since we're looking for the value of 2x, we see that the answer must be 12.

Answer: E
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 26 Mar 2015, 00:59
Bunuel wrote:
SOLUTION

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



Any questions similar to this one somewhere? The regular Rate problems are quite easy, but as soon as they start mixing it up, adding variables, I'm clueless as to how to approach things.
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 26 Mar 2015, 03:08
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erikvm wrote:
Bunuel wrote:
SOLUTION

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



Any questions similar to this one somewhere? The regular Rate problems are quite easy, but as soon as they start mixing it up, adding variables, I'm clueless as to how to approach things.


Check 700+ level Work/Rate PS problems HERE and DS problems HERE.

Hope it helps.
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Running at their respective constant rates, Machine X takes [#permalink] New post 26 Mar 2015, 03:49
MensaNumber wrote:
Alternate Solution without algebra using POE.

If it takes 3 days to produce 5/4 widgets, it will take 3*(4/5) = 12/5 days to produce 1 widget.

If two machines were working at the same rate it would take, 2*12/5 i.e. approx 5 days days for x to produce 1 widget. Since machine X is slower it will take more than 5 days to produce 1 widget and subsequently more than 10 days to produce 2 widgets. The only answer more than 10 is 12 days. Answer E



+1 Kudos !!

Have a doubt , why did you discount 'w' while calculating time require to produce 1 widget ?
If it takes 3 days to produce 5/4 widgets, it will take 3*(4/5) = 12/5 days to produce 1 widget.

shldn't it be \(\frac{12}{5w}\) ?
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 10 May 2015, 03:48
Lucky2783 wrote:
MensaNumber wrote:
Alternate Solution without algebra using POE.

If it takes 3 days to produce 5/4 widgets, it will take 3*(4/5) = 12/5 days to produce 1 widget.

If two machines were working at the same rate it would take, 2*12/5 i.e. approx 5 days days for x to produce 1 widget. Since machine X is slower it will take more than 5 days to produce 1 widget and subsequently more than 10 days to produce 2 widgets. The only answer more than 10 is 12 days. Answer E



+1 Kudos !!

Have a doubt , why did you discount 'w' while calculating time require to produce 1 widget ?
If it takes 3 days to produce 5/4 widgets, it will take 3*(4/5) = 12/5 days to produce 1 widget.

shldn't it be \(\frac{12}{5w}\) ?



Hi,

Correction - If it takes 3 days to produce 5/4 widgets, it will take 1/3*(4/5) = 4/15 days to produce 1 widget.

Thanks
Re: Running at their respective constant rates, Machine X takes   [#permalink] 10 May 2015, 03:48
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