Running at their respective constant rates, machine X takes : GMAT Problem Solving (PS)
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 08 Dec 2016, 22:37

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Running at their respective constant rates, machine X takes

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 06 Jul 2010
Posts: 75
Followers: 1

Kudos [?]: 141 [6] , given: 12

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

### Show Tags

05 Aug 2010, 01:01
6
KUDOS
57
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

57% (03:23) correct 43% (03:18) wrong based on 992 sessions

### HideShow timer Statistics

Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 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
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Sep 2013, 02:09, edited 3 times in total.
Edited the question and added the OA.
Manager
Joined: 16 Apr 2010
Posts: 221
Followers: 4

Kudos [?]: 117 [0], given: 12

Re: Work Problem [#permalink]

### Show Tags

05 Aug 2010, 01:41
Hello,

The equivalent Time
1/T = 1/T1 + 1/T2 or T=(T1*T2)/(T1+T2)
T=(5/4)/3=5/12

Let A and A+2 be the required times
T=(T1*T2)/(T1+T2)
5/12 = (A*(A+2))/(A+(A+2))
Now you can solve for A

The answer is 2A.

regards,
Jack
Math Expert
Joined: 02 Sep 2009
Posts: 35927
Followers: 6853

Kudos [?]: 90069 [18] , given: 10413

Re: Work Problem [#permalink]

### Show Tags

05 Aug 2010, 02:00
18
KUDOS
Expert's post
31
This post was
BOOKMARKED
Please post full questions with answer choices.

Running at their respective constant rate, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 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 - $$2t=12$$ --> $$t=6$$ --> $$\frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}$$.

Some work problems with solutions:
time-n-work-problem-82718.html?hilit=reciprocal%20rate
facing-problem-with-this-question-91187.html?highlight=rate+reciprocal
what-am-i-doing-wrong-to-bunuel-91124.html?highlight=rate+reciprocal
gmat-prep-ps-93365.html?hilit=reciprocal%20rate
a-good-one-98479.html?hilit=rate

Hope it helps.
_________________
Manager
Joined: 06 Jul 2010
Posts: 75
Followers: 1

Kudos [?]: 141 [0], given: 12

Re: Work Problem [#permalink]

### Show Tags

05 Aug 2010, 02:16
Thanks a lot, this was very helpful! I know how to factorize but as soon as it looks like this I can"t handle it anymore 5t^2-34t+24=0 After I found the factors for 24, there is no way getting around trial and error, is there?

And is this a formula to compute the times 2 machines need to finish the same job? I always thought the times of the machines do NOT add up in work problems..
1/T = 1/T1 + 1/T2 or T=(T1*T2)/(T1+T2)
Manager
Joined: 16 Apr 2010
Posts: 221
Followers: 4

Kudos [?]: 117 [1] , given: 12

Re: Work Problem [#permalink]

### Show Tags

05 Aug 2010, 02:39
1
KUDOS
1/T = 1/T1 + 1/T2 or T=(T1*T2)/(T1+T2)

Yes, you can use the above formula but you have to not that:

T is the time needed to complete a job J by members 1 and 2
T1 is the time needed to complete a job J in time T1
T2 is the time needed to complete a job J in time T2
Math Expert
Joined: 02 Sep 2009
Posts: 35927
Followers: 6853

Kudos [?]: 90069 [2] , given: 10413

Re: Work Problem [#permalink]

### Show Tags

05 Aug 2010, 02:57
2
KUDOS
Expert's post
4
This post was
BOOKMARKED
heyholetsgo wrote:
Thanks a lot, this was very helpful! I know how to factorize but as soon as it looks like this I can"t handle it anymore 5t^2-34t+24=0 After I found the factors for 24, there is no way getting around trial and error, is there?

For our original question you can also use substitution method: we $$\frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}$$ and we know that answer would be $$2t$$. Try to substitute half of the values listed in the answer choices and you'll see that answer choice E will work.

heyholetsgo wrote:
And is this a formula to compute the times 2 machines need to finish the same job? I always thought the times of the machines do NOT add up in work problems..
1/T = 1/T1 + 1/T2 or T=(T1*T2)/(T1+T2)

If:
Time needed for A to complete the job =A hours;
Time needed for B to complete the job =B hours;
Time needed for C to complete the job =C hours;
...
Time needed for N to complete the job =N hours;

Then if time needed for all of them working simultaneously to complete the job is $$T$$, then: $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+..+\frac{1}{N}=\frac{1}{T}$$ (General formula).

For two and three entities (workers, pumps, ...):

General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:

Given that $$a$$ and $$b$$ are the respective individual times needed for $$A$$ and $$B$$ workers (pumps, ...) to complete the job, then time needed for $$A$$ and $$B$$ working simultaneously to complete the job equals to $$T_{(A&B)}=\frac{a*b}{a+b}$$ hours, which is reciprocal of the sum of their respective rates ($$\frac{1}{a}+\frac{1}{b}=\frac{1}{t}$$).

General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:

$$T_{(A&B&C)}=\frac{a*b*c}{ab+ac+bc}$$ hours.

Hope it helps.
_________________
Manager
Joined: 06 Jul 2010
Posts: 75
Followers: 1

Kudos [?]: 141 [0], given: 12

Re: Work Problem [#permalink]

### Show Tags

05 Aug 2010, 03:13
Thanks guys, this is very good stuff:)

Edit: @ jakolik: I don't think your method works because 5/12 is the rate and not the time. Even if we took the reciprocal for the time, as both machines together produce 5/4w instead of 1w, we can't really apply the formula, can we?

Let A and A+2 be the required times
T=(T1*T2)/(T1+T2)
5/12 = (A*(A+2))/(A+(A+2))
Now you can solve for A

The answer is 2A.
Manager
Joined: 18 Feb 2010
Posts: 174
Schools: ISB
Followers: 8

Kudos [?]: 196 [1] , given: 0

Re: Work Problem [#permalink]

### Show Tags

05 Aug 2010, 04:09
1
KUDOS
heyholetsgo wrote:
Running at their respective constant rate, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 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 ideas how to solve this problem? I have three rows in my chart but somehow I can't properly solve it...

THX!

Let total widgets be 600. (for calculation simplicity)

machine days widgets per day output
x d+2 600 600/(d+2)
y d 600 600/d
x+y 3 750# we will find out

# 5w/4 = 750

now we know
3 (daily output of x + daily output of y) = 750

this gives

3 (600/(d+2) + 600/d) =750

solve for d = 4

now for 600 widgets x takes d+2 i.e. 6 days...

So for 2*600 = 1200 widgets x would take 6*2 = 12 days...

simple maths !!! isnt it...

Hope this helps..
_________________

CONSIDER AWARDING KUDOS IF MY POST HELPS !!!

Intern
Joined: 15 Aug 2010
Posts: 2
Followers: 0

Kudos [?]: 1 [1] , given: 4

Re: Work Problem [#permalink]

### Show Tags

15 Aug 2010, 05:00
1
KUDOS
Hi,

I tried a different method to solve the problem, but somehow I couldn't. This was what I came up with so far:

I was trying to apply:Time = job done / rate

Let X = X widgets/day,
Y = Y widgets/day

1)w/X = (w/Y) + 2
2) 5w/4 = (X + Y)*3
From 1), Y = wX/(w - 2X) 3)
I plugged 3) into 2), but I came out with something really ugly and got stuck.

Is there anyone able to solve these two equations, or is there anything wrong with my method? Thank you.
Manager
Joined: 22 Oct 2009
Posts: 242
GMAT 1: 760 Q49 V44
GPA: 3.88
Followers: 6

Kudos [?]: 87 [11] , given: 1

Re: Work Problem [#permalink]

### Show Tags

15 Aug 2010, 21:12
11
KUDOS
6
This post was
BOOKMARKED
Hey everyone,

While I definitely agree that Bunuel's method is the safest for finding the solution, I hate dealing with algebra when it's not necessary, so I found that this problem is really easy if you just plug in numbers.

Try w = 1!
$$w=1$$
Rate of x = $$\frac{1}{(x)}$$ widgets per day
Rate of y = $$\frac{1}{(x-2)}$$ widgets per day

Now take the answer choices and plug them in, add the fractions, and then multiply by three (since you're looking for 3 days of production) and see which answer gives you $$\frac{5}{4}$$.

I always start at C when plugging in numbers:

C: $$\frac{1}{4} + \frac{1}{2} = \frac{3}{4}$$

$$\frac{3}{4}*3$$ is not equal to $$\frac{5}{4}$$, so we move on to D

D: $$\frac{1}{5} + \frac{1}{3} = \frac{8}{15}$$

$$\frac{8}{15}*3$$ is not equal to $$\frac{5}{4}$$, but we're getting closer, so we move on to E

E: $$\frac{1}{6} + \frac{1}{4} = \frac{5}{12}$$

$$\frac{5}{12} * 3$$ is equal to $$\frac{5}{4}$$, so this is the answer!

Hope this helps all you people who hate to factor!
Math Expert
Joined: 02 Sep 2009
Posts: 35927
Followers: 6853

Kudos [?]: 90069 [1] , given: 10413

Re: Advanced Rate Problem QR#173 PS [#permalink]

### Show Tags

16 Jan 2011, 18:04
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
SVP
Joined: 16 Nov 2010
Posts: 1672
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 33

Kudos [?]: 507 [4] , given: 36

Re: Work Problem [#permalink]

### Show Tags

14 Feb 2011, 19:08
4
KUDOS
1
This post was
BOOKMARKED
Let machine y take y days for w widhets

1 day - w/y widgets by machine Y

1 day - w/(y+2) widgets by machine X

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

=> 12/y + 12/(y+2) = 5

=> 12y + 24 + 12y = 5y^2 + 10y

=> 5y^2 - 14y - 24 = 0

=> 5y^2 - 20y + 6y - 24 = 0

=> 5y(y - 4) + 6(y-4) = 0

=> y = 4

=> Machine x takes 4+2 = 6 days for w widgets, and 6 * 2 = 12 days for 2w widgets

So answer is E.
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2021
Followers: 162

Kudos [?]: 1670 [0], given: 376

Re: 173. ps QR. [#permalink]

### Show Tags

11 Mar 2011, 07:43
Looks like we got this problem in the forum before;

Anyway;

X
d+2 days-> w widgets
1 day-> w/(d+2) widgets
3 days-> 3w/(d+2) widgets

Y
d days-> w widgets
1 day-> w/d widgets
3 days-> 3w/d widgets

So; widgets by X and Y in 3 days;

3w/(d+2) + 3w/d = (5/4)w
Cancel out w from both sides;

$$\frac{3}{d+2}+\frac{3}{d} = \frac{5}{4}$$

Upon solving;
$$5d^2-14d-24=0$$
$$5d^2-20d+6d-24=0$$
$$(5d+6)(d-4)=0$$

d= -6/5
or
d=4

-ve is not possible as number of days
so; d=4

X makes w widgets in d+2=4+2=6 days
X makes 2w widgets in twice the time i.e. 6*2=12 days

Ans: "E"
_________________
Intern
Joined: 13 Jan 2012
Posts: 4
Followers: 0

Kudos [?]: 0 [0], given: 0

Running at their respective constant rate, machine X takes 2 [#permalink]

### Show Tags

21 May 2012, 07:35
Hi,

I came across this problem in the quantitative review. I am struggling with a part of the books solution, a part I feel would be useful to know.

How does one get from
$$12x (x-2) ( \frac{1}{x} + \frac{1}{x-2} )= 12x(x-2)(\frac{5}{12})$$

to

$$12 [(x-2)+x] = 5x(x-2)$$

my instinct is to multiply the 12x by x-2 resulting in 12x squared - 24x and then foil the two binomials and multply the terms on the right resulting in this more complicated equation:

$$12x + \frac{12x^2}{x-2}-24 - \frac{24x}{x-2} = 5x^2 - 10x$$

I want to know how to simplify problems like this that involve multiplying a monomial X binomial X Binomial(that has a 1 binomial denominator).

I would like to know how to handle this type of simplification I feel its important to know in algebra.

Any help is greatly appreciated!!

-Josh
Manager
Joined: 19 Mar 2012
Posts: 174
Location: United States
Concentration: Finance, General Management
GMAT 1: 750 Q50 V42
GPA: 3.69
WE: Analyst (Mutual Funds and Brokerage)
Followers: 7

Kudos [?]: 46 [0], given: 13

Re: Running at their respective constant rate, machine X takes 2 [#permalink]

### Show Tags

21 May 2012, 07:50
joshuaRome wrote:
Hi,

I came across this problem in the quantitative review. I am struggling with a part of the books solution, a part I feel would be useful to know.

How does one get from
$$12x (x-2) ( \frac{1}{x} + \frac{1}{x-2} )= 12x(x-2)(\frac{5}{12})$$

to

$$12 [(x-2)+x] = 5x(x-2)$$

my instinct is to multiply the 12x by x-2 resulting in 12x squared - 24x and then foil the two binomials and multply the terms on the right resulting in this more complicated equation:

$$12x + \frac{12x^2}{x-2}-24 - \frac{24x}{x-2} = 5x^2 - 10x$$

I want to know how to simplify problems like this that involve multiplying a monomial X binomial X Binomial(that has a 1 binomial denominator).

I would like to know how to handle this type of simplification I feel its important to know in algebra.

Any help is greatly appreciated!!

-Josh

A general rule I try to follow for simplification is trying to get rid of the most complicated parts of the equation first.

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

So starting with the left side of the equation, first thing I'd work on is getting rid of that "x" in the denominator in $$( \frac{1}{x} + \frac{1}{x-2} )$$. I simplified that to (x-2+x)/[(x)(x-2)]. From there, (x-2) in the denominator cancels out with the 2nd term of the equation, and the (x) in the denominator cancels out with the 12x term, so we get $$12 [(x-2)+x]$$.

Now looking at the right side of the equation:$$12x(x-2)(\frac{5}{12})$$, we just have the 12x cancel out with the 12, leaving us with (5x)(x-2)
Math Expert
Joined: 02 Sep 2009
Posts: 35927
Followers: 6853

Kudos [?]: 90069 [0], given: 10413

Re: Running at their respective constant rate, machine X takes 2 [#permalink]

### Show Tags

21 May 2012, 07:57
joshuaRome wrote:
Hi,

I came across this problem in the quantitative review. I am struggling with a part of the books solution, a part I feel would be useful to know.

How does one get from
$$12x (x-2) ( \frac{1}{x} + \frac{1}{x-2} )= 12x(x-2)(\frac{5}{12})$$

to

$$12 [(x-2)+x] = 5x(x-2)$$

my instinct is to multiply the 12x by x-2 resulting in 12x squared - 24x and then foil the two binomials and multply the terms on the right resulting in this more complicated equation:

$$12x + \frac{12x^2}{x-2}-24 - \frac{24x}{x-2} = 5x^2 - 10x$$

I want to know how to simplify problems like this that involve multiplying a monomial X binomial X Binomial(that has a 1 binomial denominator).

I would like to know how to handle this type of simplification I feel its important to know in algebra.

Any help is greatly appreciated!!

-Josh

Merged your question with an earlier discussion of the same problem.

Check this post for the solution: running-at-their-respective-constant-rate-machine-x-takes-98599.html#p759876 and this one for the theory: running-at-their-respective-constant-rate-machine-x-takes-98599.html#p759894

As for your question: $$12x (x-2) ( \frac{1}{x} + \frac{1}{x-2} )= 12x(x-2)(\frac{5}{12})$$ --> reduce by $$12x(x-2)$$: $$\frac{1}{x}+\frac{1}{x-2}=\frac{5}{12}$$ --> $$\frac{x-2+x}{x(x-2)}=\frac{5}{12}$$ --> cross-multiply: $$24x-24=5x^2-10x$$ --> $$5x^2-34x+24=0$$ --> $$(x-6)(5x-4)=0$$ --> $$x=6$$ or $$x=\frac{4}{5}$$

Hope it's clear.
_________________
Manager
Joined: 05 May 2012
Posts: 67
Location: India
Concentration: General Management, Finance
GMAT 1: 670 Q48 V34
GMAT 2: 720 Q49 V40
Followers: 5

Kudos [?]: 24 [2] , given: 16

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

### Show Tags

06 Jan 2013, 07:46
2
KUDOS
Hi Drik,

a wise approach here would be to plugin numbers.

together they make 5/4 widgets in 3 days => widget in 3*4/5 = 2.4 days
so 2.4 is our target value. we will plug value of X for w widgets and calculate Y and thereby no of days they take together to produce w widgets.
This should match with our target value of 2.4

so lets put X=6 then Y would be 6-2=4
together they can do in 6*4/(6+4)=2.4 days

This is exactly our target value.

Beware X would take 6 days to produce w widgets. Question asked for 2w so X will be 2*6=12

Best Regards,
Mansoor

Please consider kudos if you find it useful
Manager
Joined: 28 Feb 2012
Posts: 115
Concentration: Strategy, International Business
Schools: INSEAD Jan '13
GPA: 3.9
WE: Marketing (Other)
Followers: 0

Kudos [?]: 41 [0], given: 17

Re: Running at their respective constant rate, machine X takes 2 [#permalink]

### Show Tags

09 Jan 2013, 06:50
Bunuel could you please comment!

I have one question, can we accept it as GMAT-like (i mean the wording and the numbers) question. I am asking because normally i was not facing with calculations where you have equations like 5A^2-34A+24=0, to be honest i have calculated everything up to this point and then thought that i was doing something wrong (because i normally simplify such equations to A^2-34/5A+24/5=0 then try to find factors). But until here i have already spent 2 min then i start doing again and faced the same equation again. Then i thought that i did something principally wrong.
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Senior Manager
Joined: 14 Jul 2013
Posts: 295
Location: India
Concentration: Marketing, Strategy
GMAT 1: 690 Q49 V34
GMAT 2: 670 Q49 V33
GPA: 3.6
WE: Brand Management (Retail)
Followers: 13

Kudos [?]: 183 [0], given: 130

Re: Running at their respective constant rate, machine X takes 2 [#permalink]

### Show Tags

12 Aug 2013, 20:28
Whats wrong with the below mentioned approach. I know its wrong but cant get my head whats wrong. X number of days taken by x Y number of days taken by Y.

1/x - 1/y = 1/2
1/x + 1/y = 5/12

I got the right ones explained earlier just want to know whats wrong with this one.

Posted from GMAT ToolKit
_________________

Cheers
Farhan

My Blog - Student for Life ( Oxford MBA)

Math Expert
Joined: 02 Sep 2009
Posts: 35927
Followers: 6853

Kudos [?]: 90069 [0], given: 10413

Re: Running at their respective constant rate, machine X takes 2 [#permalink]

### Show Tags

13 Aug 2013, 00:55
farhanc85 wrote:
Whats wrong with the below mentioned approach. I know its wrong but cant get my head whats wrong. X number of days taken by x Y number of days taken by Y.

1/x - 1/y = 1/2
1/x + 1/y = 5/12

I got the right ones explained earlier just want to know whats wrong with this one.

Posted from GMAT ToolKit

Not clear what are you doing there.

Given: running at their respective constant rate, machine X takes 2 days longer to produce w widgets than machines Y.

Now, if x and y are the number of days for machines X and Y to produce w widgets, respectively, then it should be x-y=2.
_________________
Re: Running at their respective constant rate, machine X takes 2   [#permalink] 13 Aug 2013, 00:55

Go to page    1   2   3    Next  [ 42 posts ]

Similar topics Replies Last post
Similar
Topics:
86 Running at their respective constant rates, Machine X takes 20 18 Mar 2014, 00:39
4 Running at their respective constant rates, machine X takes 2 days lon 9 12 May 2011, 09:33
13 Working at their respective constant rates, printing machine X, Y, and 6 17 Feb 2011, 14:11
7 Machines X and Y produce bottles at their respective constant rates. 12 29 Aug 2010, 19:43
9 Running at their respective constant rate, machine X takes 2 19 22 Mar 2009, 20:53
Display posts from previous: Sort by

# Running at their respective constant rates, machine X takes

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.