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# Running at their respective constant rates, machine X takes 2 days lon

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Running at their respective constant rates, machine X takes 2 days lon  [#permalink]

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Updated on: 30 Sep 2019, 03:01
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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

Originally posted by heyholetsgo on 05 Aug 2010, 01:01.
Last edited by Bunuel on 30 Sep 2019, 03:01, edited 4 times in total.
Edited the question and added the OA.
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Re: Running at their respective constant rates, machine X takes 2 days lon  [#permalink]

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05 Aug 2010, 02:00
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103

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 produce $$w$$ widgets will be negative 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:
http://gmatclub.com/forum/time-n-work-p ... cal%20rate
http://gmatclub.com/forum/facing-proble ... reciprocal
http://gmatclub.com/forum/what-am-i-doi ... reciprocal
http://gmatclub.com/forum/gmat-prep-ps- ... cal%20rate
http://gmatclub.com/forum/questions-fro ... cal%20rate
http://gmatclub.com/forum/a-good-one-98 ... hilit=rate

Hope it helps.
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Re: Running at their respective constant rates, machine X takes 2 days lon  [#permalink]

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15 Aug 2010, 21:12
42
21
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!
##### General Discussion
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Re: Running at their respective constant rates, machine X takes 2 days lon  [#permalink]

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05 Aug 2010, 02:16
1
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)
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Re: Running at their respective constant rates, machine X takes 2 days lon  [#permalink]

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05 Aug 2010, 02:39
2
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
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Posts: 65384
Re: Running at their respective constant rates, machine X takes 2 days lon  [#permalink]

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05 Aug 2010, 02:57
7
7
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.
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Re: Running at their respective constant rates, machine X takes 2 days lon  [#permalink]

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05 Aug 2010, 04:09
5
2
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..
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Posts: 65384
Re: Running at their respective constant rates, machine X takes 2 days lon  [#permalink]

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16 Jan 2011, 18:04
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Re: Running at their respective constant rates, machine X takes 2 days lon  [#permalink]

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14 Feb 2011, 19:08
8
1
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

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

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11 Mar 2011, 07:43
2
1
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"
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Re: Running at their respective constant rates, machine X takes 2 days lon  [#permalink]

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06 Jan 2013, 07:46
3
3
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
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Re: Running at their respective constant rates, machine X takes 2 days lon  [#permalink]

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13 Aug 2013, 05:13
4
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

my style of solution ,which is pretty common:
Attachments

work widgets.png [ 30.15 KiB | Viewed 47667 times ]

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

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02 Apr 2014, 06:02
Can somebody please explain the following:

How does this algebra work: 1/t + 1/t+2 = 5/12 is translated to 5t^2 + 34t - 24.

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

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02 Apr 2014, 06:54
chrishhaantje wrote:
Can somebody please explain the following:

How does this algebra work: 1/t + 1/t+2 = 5/12 is translated to 5t^2 + 34t - 24.

$$\frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}$$;

$$\frac{(t-2)+t}{t(t-2)}=\frac{5}{12}$$;

$$\frac{2t-2}{t^2-2t}=\frac{5}{12}$$;

$$24t-24=5t^2-10t$$;

$$5t^2-34t+24=0$$.

Hope it's clear.
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Re: Running at their respective constant rates, machine X takes 2 days lon  [#permalink]

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17 Aug 2016, 10:57
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Top Contributor
1
Tough one! My recommended solution is plugging in 1 for w. Attached is a visual that should help.
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Screen Shot 2016-08-17 at 11.55.33 AM.png [ 151.2 KiB | Viewed 16678 times ]

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

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10 Jun 2017, 07:10
2
1
heyholetsgo wrote:
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

We are given that running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y does. If we let t = the time in days that it takes machine Y to produce w widgets, then t + 2 = the time in days that it takes machine X to produce w widgets. Furthermore, we can say the following:

rate of machine Y = w/t

rate of machine X = w/(t + 2)

We are given that the machines produce 5w/4 widgets in 3 days. Since work = rate x time and each machine works for 3 days, we first calculate the work done by machine X and machine Y individually.

work of machine Y = (w/t) x 3 = 3w/t

work of machine X = w/(t + 2) x 3 = 3w/(t + 2)

Since the machines work together to produce 5w/4 widgets, we can sum their work and set that sum to 5w/4.

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

Divide the entire equation by w and we have:

3/t + 3/(t + 2) = 5/4

Now, multiplying the entire equation by 4t(t + 2) to eliminate the denominators in the equation, we obtain:

3[4(t + 2)] + 3(4t) = 5[t(t + 2)]

12t + 24 + 12t = 5t^2 + 10t

5t^2 - 14t - 24 = 0

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

t = -6/5 or t = 4

Since t cannot be negative, t must equal 4. That is, it takes machine Y 4 days to produce w widgets. Thus, it will take machine X 6 days to produce w widgets and 12 days to produce 2w widgets.

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

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Updated on: 12 Jun 2019, 05:37
Top Contributor
heyholetsgo wrote:
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

Another approach is to assign a nice value to the job (w)

Let's say that w = 12.

GIVEN: Running at their respective constant rates, machine X takes 2 days longer to produce 12 widgets than machine Y
Let t = time for machine Y to produce 12 widgets
So, t+2 = time for machine X to produce 12 widgets

RATE = output/time

So, machine X's RATE = 12 widgets/(t + 2 days) = 12/(t+2) widgets per day
And machine Y's RATE = 12 widgets/(t days) = 12/t widgets per day

The two machines together produce 5w/4 widgets in 3 days
In other words, The two machines together produce 5(12)/4 widgets in 3 days
Or the two machines together produce 15 widgets in 3 days
This means the COMBINED RATE = 5 widgets per day

So, we can write: 12/(t+2) + 12/t = 5
Multiply both sides by (t+2)(t) to get: 12t + 12t + 24 = 5(t+2)(t)
Simplify: 24t + 24 = 5t² + 10t
Rearrange: 5t² - 14t - 24 = 0
Factor to get: (5t + 6)(t - 4) = 0
So, EITHER t = -6/5 OR t = 4
Since the time cannot be negative, it must be the case that t = 4

If t = 4, then it takes Machine Y 4 days to produce 12 widgets
And it takes Machine X 6 days to produce 12 widgets

How many days would it take machine X alone to produce 2w widgets?
In other words, how many days would it take machine X alone to produce 24 widgets? (since w = 12)

If it takes Machine X 6 days to produce 12 widgets, then it will take Machine X 12 days to produce 24 widgets

Cheers,
Brent
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Originally posted by BrentGMATPrepNow on 05 Mar 2018, 09:14.
Last edited by BrentGMATPrepNow on 12 Jun 2019, 05:37, edited 1 time in total.
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Re: Running at their respective constant rates, machine X takes 2 days lon  [#permalink]

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10 Sep 2018, 03:23
1
heyholetsgo wrote:
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

Let $$w = 60$$ widgets, implying that $$\frac{5}{4}w = 75$$ widgets.
We can PLUG IN THE ANSWERS, which represent X's time to produce 2w widgets.
When the correct answer choice is plugged in, X and Y will produce 75 widgets in 3 days.

D: 10
Here, X can produce 2w widgets in 10 days.
Thus, the time for X to produce w widgets = 5 days.

Since X takes 5 days to produce w=60 widgets, X's rate $$= \frac{60}{5} = 12$$ widgets per day.
Since X takes 2 days longer than Y to produce w widgets, Y's time to produce w widgets = 3 days.
Since Y takes 3 days to produce w=60 widgets, Y's rate $$= \frac{60}{3} = 20$$ widgets per day.

Combined rate for X and Y = 12+20 = 32 widgets per day.
Work produced by X and Y in 3 days = 3*32 = 96 widgets.

Here, X and Y are working TOO FAST.
Implication:
X must take LONGER to produce 2w widgets, with the result that X and Y will work more SLOWLY.

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

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28 May 2020, 20:09
A super easy way would be to assume w = 1 widget.
Hence X would take 6 days ( Y= 4 days, 2 days more for X) to make one widget and for 2 widgets, 12 days. hence that's the answer.

However, we do need to check whether our assumption of w = 1 is correct by checking in the condition of working together.

5w / 4 = 10*3/24 ....... work together rate is 1/6 + 1/4, LCM is 10/24
by solving you do get w = 1 therefore our assumption was correct.

Hence, 12 days is the answer.

Bunuel is this correct approach ?
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Re: Running at their respective constant rates, machine X takes 2 days lon  [#permalink]

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06 Jul 2020, 05:45
Let the rate of X be x; the rate of Y be y

Some posters have already shared that number plugging is a better approach. I agree, but I'd like to add another step that will save time by avoiding back and forth between the numbers.

We’ve been given that:
3= $$\frac{5W}{4(x+y)}$$
Therefore, I can say that 12 = $$\frac{5W}{(x+y)}$$

Now, let’s say that x+y =5 so that the 5 cancels out. Therefore, W must be 12.

With this, we’ve said that the sum of the rates must be 5 and the difference between the time taken by x and y to do 12 units of work must be 2 days. Pick x as 2...there time taken by X = 6. Therefore, Y has to be 3 and time taken by Y will be 4.
6-4 = 2 and this adhers to our original constraint.
Re: Running at their respective constant rates, machine X takes 2 days lon   [#permalink] 06 Jul 2020, 05:45