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# Running at their respective constant rates, machine X takes

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Senior Manager
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GMAT 1: 710 Q50 V36
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30 Sep 2013, 00:35
Bunuel 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

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.

Considerable amount of calculation involved, do you think this is a GMAT question, although question was not so difficult.
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30 Sep 2013, 02:12
honchos wrote:
Bunuel 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

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.

Considerable amount of calculation involved, do you think this is a GMAT question, although question was not so difficult.

It's a GMAT Prep question, so representative of "real" questions you can see on the test.

Also, it's OK to spend a bit more time than 2 minutes on tough questions.
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04 Oct 2013, 18:47
Bunuel 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

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.

Bunuel,

For the above question, I defined the time for X to produce w widgets to be 'x+2', and subsequently set the time for Y to produce w widgets to 'x', instead of setting time for X to complete to be 'x' and then set Y time to complete to be 'x-2'. However, when I proceed to solve the equation which is set up so that w/t+2 + w/t = 5/12w, my answer becomes t= -6/5 or 4, which is different to the actual answer.

What I dont understand is why can't I set time for x to be 'x+2'and why do I have to set x to be 'x'and then Y to be 'x-2'? Just trying to understand the logic of setting up the equation as you mentioned.
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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14 Oct 2013, 03:59
Time take by X = t days
Time take by Y = (t-2) days

1/X = w/t and 1/Y = w/(t-2)

(w/t+w/(t-2)) X 3 = 5w/4
Simplifying we get :

(t-1)/t(t-2) = 5/24
Clearly t=6

So, Rate * Time = Work
or w/6 * T = 2w or T= 12
Math Expert
Joined: 02 Sep 2009
Posts: 52431

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17 Oct 2013, 03:17
bulletpoint wrote:
Bunuel 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

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.

Bunuel,

For the above question, I defined the time for X to produce w widgets to be 'x+2', and subsequently set the time for Y to produce w widgets to 'x', instead of setting time for X to complete to be 'x' and then set Y time to complete to be 'x-2'. However, when I proceed to solve the equation which is set up so that w/t+2 + w/t = 5/12w, my answer becomes t= -6/5 or 4, which is different to the actual answer.

What I dont understand is why can't I set time for x to be 'x+2'and why do I have to set x to be 'x'and then Y to be 'x-2'? Just trying to understand the logic of setting up the equation as you mentioned.

You can do this way too. 4 hours is the time for Y to produce w widgets, thus the time for X to produce w widgets is t+2=6 hours and to produce 2w widgets is 12 hours.

Hope it's clear.
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17 Oct 2013, 09:20
Bunuel 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

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.

There must be a quicker way to solve these, doing that math took me almost 6 minutes
Math Expert
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Posts: 52431

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17 Oct 2013, 09:41
AccipiterQ wrote:
Bunuel 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

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.

There must be a quicker way to solve these, doing that math took me almost 6 minutes

It's a matter of practice. One can get $$\frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}$$ quite quickly and then substitute.
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17 Oct 2013, 11:33
Bunuel wrote:

It's a matter of practice. One can get $$\frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}$$ quite quickly and then substitute.

Adding on to what Bunuel said, If one observes carefully, we can break down the given equation as :

$$\frac{1}{t}+\frac{1}{t-2}=\frac{(3+2)}{12}$$ = $$\frac{3}{12} +\frac{2}{12} = \frac{1}{4}+\frac{1}{6}$$ and thus, on comparison, t=6.
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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17 Nov 2013, 10:12
I am trying to solve it using the following formula:

Days per widget x # of widgets = Total number of days

I get an incorrect answer and despite multiple reviews cannot understand where the mistake is.

Machine Y produced w widgets in x days so x/w widgets a day. Machine X produced w widgets in x+2 days so x+2/w widgets a day. Together the machines produced 12/5w a day (3 divided by 5w/4). Therefore:

x/w+(x+2)/w=12/5w -> (x+2+x)/w=12/5w -> simplifying for w -> (x+2+x)/1=12/5
5(2x+2)=12
10x+10=12
10x=2
X=1/5
X+2=11/5

2(x+2)=Days required to produce 2w=22/5

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

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20 Nov 2013, 12:29
Bunuel wrote:
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.

I had an idea here, maybe you could tell me if this makes sense:

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.

So we know that two machines combine to produce 5w/4 widgets in 3 days, so per day they're producing 5/12 of the job combined, now we know that the rates are going to be 1/t and 1/t-2...so couldn't we skip the early steps and jump right to the 1/t+1/(t-2)=5/12? It would cut out about 30 seconds of setup and work if that could apply to other problems, yes?
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Re: Running at their respective constant rates, machine X takes  [#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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Joined: 02 Sep 2009
Posts: 52431
Re: Running at their respective constant rates, machine X takes  [#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  [#permalink]

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05 Apr 2015, 03:30
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

Solved it by options and worked backwards.

it is given that if y takes k days to produce w widgets then x would take k-2 days.

in the answer choices we are given x's time to produce 2w widgets.
half the answer choice(s) and that number would give the number of days x takes to produce w widgets. and that number minus 2 would give number of days taken by y to produce w widgets. after that work out the number of widgets they together can produce in 3 days....the answer choice which gives 5w/4 is correct.

E does that.
if x takes 12 days for 2w widgets.
then x would take 6 days for w widgets (and in 3 days it will produce w/2 widgets)

y will take 4 days for w widgets (and in 3 days it will produce 3w/4 widgets)

x and y together will produce w/2 + 3w/4 = 5w/4 widgets in 3 days.

QED :p
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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27 Sep 2015, 19:24
hi all
this problem can be solved in many ways but to solve it in < 2 minutes -you may apply this technique.
Assume any number of days (Say 4) that Y takes for producing W gadgets or ( 1 unit of work)
hence, x will take 4+2 = 6 days
Take LCM of 6 & 4 =12
therefore x takes 6 days to produce 12 units of work or 2 units/day
similarly y takes 4 days to produce 12 units of work or 3 units/day.
Total units produces by them = 5 units/day
therefore in 3 days they will produce =15 units.
now the total work i.e 15= 5w/4 => w= 12=>2w=24
now since x was producing 2 units /day therefore to produce 24 units it requires total work/production per day =>24/2=12
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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17 Feb 2016, 12:10
Bunuel 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

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 $$rate=\frac{job \ done}{time}=\frac{w}{t-2}$$;

what is the reasoning behind chosing $$t-2$$ days instead of $$t+2$$ days, where $$t+2$$ is the rate of machine X?
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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16 Apr 2016, 18:15
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

Can we solve this by below equation:

1/t + 1/t+2 = 5/12

am not able to solve further 5(t^2)-14t-24=0

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

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21 Jun 2016, 13:42
I would plug numbers for w, you quickly realise that it's either D or E. E happens to work. Solving the GMAT algebraically is suicide - i've tried it before.
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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21 Jul 2016, 06:52
For Y:
time taken by y to produce w widgets => Ty= t days
rate of y * time of y = work done by Y => Ry * Ty = w widgets => Ry = (w/t )widgets/days
Now X:
time taken by x is 2 days more than y => Tx = t + 2 days
Rx = w/(t+2) [widget/days]
Now X & Y together:
Rxy = Rx + Ry = w/(t+2) + (w/t) = 2w(t+1)/t(t+1) --- take LCM and solve.

Now Rxy * Time taken by XY tog = 5w/4 widgets ---> time taken by X and Y = 3hrs
[2w(t+1)/t(t+2)] * 3 = 5w/4
on solving you get =>t =4
Thus it takes Y = 4 days and X = 2+4= 6 days
=> w widgets by X in 6 days
1 widget by X in 6/w days [direct variation]
Thus 2w widgets by X in ----> 12 days
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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17 Aug 2016, 10:57
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Tough one! My recommended solution is plugging in 1 for w. Attached is a visual that should help.
Attachments

Screen Shot 2016-08-17 at 11.55.33 AM.png [ 151.2 KiB | Viewed 1409 times ]

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

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27 Sep 2016, 08:53
1
I got the values of x=6 and x=4/5 by quadratic equation. We use 6 to solve the question. Why can't we use 4/5
Is there an explanation for this?
Re: Running at their respective constant rates, machine X takes &nbs [#permalink] 27 Sep 2016, 08:53

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