Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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?

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

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 - \(2t=12\) --> \(t=6\) --> \(\frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}\).

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 - \(2t=12\) --> \(t=6\) --> \(\frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}\).

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

Show Tags

27 May 2014, 15:16

5

This post received KUDOS

1

This post was BOOKMARKED

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
_________________

Please contact me for super inexpensive quality private tutoring

My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876

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

Show Tags

10 Aug 2014, 02:25

2

This post received KUDOS

1

This post was BOOKMARKED

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.

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

Show Tags

20 Oct 2014, 10:39

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.

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

Show Tags

20 Oct 2014, 20:34

3

This post received KUDOS

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.

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.
_________________

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

Show Tags

08 Dec 2014, 08:30

1

This post received KUDOS

1

This post was BOOKMARKED

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.

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

Show Tags

26 Mar 2015, 01:59

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.

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.

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

Show Tags

26 Mar 2015, 04:49

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 \(\frac{12}{5w}\) ?
_________________

Thanks, Lucky

_______________________________________________________ Kindly press the to appreciate my post !!

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

Show Tags

10 May 2015, 04:48

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.

can someone pelase explain how they got from (1/t) + (1/((t-2)) = 5/12 TO 5t^2-34t+24=0

You can also give the quadratic route a miss. Once you get (1/t) + (1/(t-2)) = 5/12, you can try out some values for t. t is an integer since it is one of the given 5 options.

To get 12 in denominator on the right hand side, t*(t-2) should give you 12 or a multiple. 12 = 4*3 or 6*2 (not of the form t*(t-2)) 24 = 6*4 (this is possible)

Check 1/6 + 1/4 = 10/24 = 5/12 You get the value of t.

Or you can also try substituting the value of t from each option.
_________________

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...