GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 05 Jul 2020, 11:39 ### GMAT Club Daily Prep

#### 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

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.  # Running at their respective constant rates, Machine X takes

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 64949
Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

6
1
135 00:00

Difficulty:   95% (hard)

Question Stats: 58% (03:09) correct 42% (03:12) wrong based on 1203 sessions

### HideShow timer Statistics

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

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

GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

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:
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.

Thank you!

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 64949
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

13
18
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 - $$2t=12$$ --> $$t=6$$ --> $$\frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}$$.

_________________
SVP  Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1706
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

36
2
15
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

Originally posted by PareshGmat on 20 Mar 2014, 00:52.
Last edited by PareshGmat on 14 Aug 2014, 02:01, edited 1 time in total.
##### General Discussion
Math Expert V
Joined: 02 Sep 2009
Posts: 64949
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

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

_________________
Intern  Joined: 13 Feb 2014
Posts: 6
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

1
QE: 5t^2 - 14t - 24 = 0, How do we get solution = 4?
Math Expert V
Joined: 02 Sep 2009
Posts: 64949
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

gciftci wrote:
QE: 5t^2 - 14t - 24 = 0, How do we get solution = 4?

Hope it helps.
_________________
Retired Moderator Joined: 29 Oct 2013
Posts: 245
Concentration: Finance
GPA: 3.7
WE: Corporate Finance (Retail Banking)
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

8
2
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
_________________

My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876
Manager  Joined: 17 Oct 2012
Posts: 59
Location: India
Concentration: Strategy, Finance
WE: Information Technology (Computer Software)
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

3
1
Back tracking method:
We have to find the number of days it takes Machine X alone to produce 2w widgets.

(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.
Intern  Joined: 21 Mar 2014
Posts: 2
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

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.
SVP  Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1706
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

5
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
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10629
Location: Pune, India
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

4
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.

_________________
Karishma
Veritas Prep GMAT Instructor

Intern  Joined: 21 Mar 2014
Posts: 2
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

PareshGmat
VeritasPrepKarishma

Thanks a lot. Careless of me to not have noticed such a simple solution to the equation 5x^2 - 14x - 24 = 0
Manager  B
Joined: 23 May 2013
Posts: 179
Location: United States
Concentration: Technology, Healthcare
GMAT 1: 760 Q49 V45 GPA: 3.5
Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

4
2
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.

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.

Manager  Joined: 26 Feb 2015
Posts: 108
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 64949
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

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.
_________________
Senior Manager  Joined: 07 Aug 2011
Posts: 488
GMAT 1: 630 Q49 V27
Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

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 $$\frac{12}{5w}$$ ?
Manager  B
Joined: 26 Dec 2011
Posts: 115
Schools: HBS '18, IIMA
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

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
Intern  B
Joined: 23 Jul 2015
Posts: 36
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

can someone pelase explain how they got from (1/t) + (1/((t-2)) = 5/12
TO
5t^2-34t+24=0
CEO  G
Joined: 20 Mar 2014
Posts: 2531
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44 GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

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.
_________________
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10629
Location: Pune, India
Running at their respective constant rates, Machine X takes  [#permalink]

### Show Tags

1
jasonfodor wrote:
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.
_________________
Karishma
Veritas Prep GMAT Instructor Running at their respective constant rates, Machine X takes   [#permalink] 26 Jul 2015, 20:36

Go to page    1   2    Next  [ 35 posts ]

# Running at their respective constant rates, Machine X takes  