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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 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
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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
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05 Aug 2010, 02:00
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 \(t2\) days, so the rate of Y would be \(rate=\frac{job \ done}{time}=\frac{w}{t2}\); Combined rate of machines X and Y in 1 day would be \(\frac{w}{t}+\frac{w}{t2}\) (remember we can sum the rates). In 3 days two machines together produce 5w/4 widgets so: \(3(\frac{w}{t}+\frac{w}{t2})=\frac{5w}{4}\) > \(\frac{w}{t}+\frac{w}{t2}=\frac{5w}{12}\). \(\frac{w}{t}+\frac{w}{t2}=\frac{5w}{12}\) > reduce by \(w\) > \(\frac{1}{t}+\frac{1}{t2}=\frac{5}{12}\). At this point we can either solve quadratic equation: \(5t^234t+24=0\) > \((t6)(5t4)=0\) > \(t=6\) or \(t=\frac{4}{5}\) (which is not a valid solution as in this case \(t2=\frac{6}{5}\), 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  \(2t=12\) > \(t=6\) > \(\frac{1}{6}+\frac{1}{62}=\frac{5}{12}\). Answer: E. Some work problems with solutions: http://gmatclub.com/forum/timenworkp ... cal%20ratehttp://gmatclub.com/forum/facingproble ... reciprocalhttp://gmatclub.com/forum/whatamidoi ... reciprocalhttp://gmatclub.com/forum/gmatprepps ... cal%20ratehttp://gmatclub.com/forum/questionsfro ... cal%20ratehttp://gmatclub.com/forum/agoodone98 ... hilit=rateHope it helps.
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Re: Running at their respective constant rates, machine X takes 2 days lon
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15 Aug 2010, 21:12
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 = \(\frac{1}{(x2)}\) 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!




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Re: Running at their respective constant rates, machine X takes 2 days lon
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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^234t+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
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05 Aug 2010, 02:39
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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Re: Running at their respective constant rates, machine X takes 2 days lon
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05 Aug 2010, 02:57
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^234t+24=0 After I found the factors for 24, there is no way getting around trial and error, is there? Factoring Quadratics: http://www.purplemath.com/modules/factquad.htmFor our original question you can also use substitution method: we \(\frac{1}{t}+\frac{1}{t2}=\frac{5}{12}\) 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
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05 Aug 2010, 04:09
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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Re: Running at their respective constant rates, machine X takes 2 days lon
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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
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14 Feb 2011, 19:08
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(y4) = 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.
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Re: Running at their respective constant rates, machine X takes 2 days lon
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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; \(5d^214d24=0\) \(5d^220d+6d24=0\) \((5d+6)(d4)=0\)
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
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06 Jan 2013, 07:46
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 62=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
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Re: Running at their respective constant rates, machine X takes 2 days lon
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13 Aug 2013, 05:13
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:
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Re: Running at their respective constant rates, machine X takes 2 days lon
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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.
Thanks in advance ! Chris



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Re: Running at their respective constant rates, machine X takes 2 days lon
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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.
Thanks in advance ! Chris \(\frac{1}{t}+\frac{1}{t2}=\frac{5}{12}\); \(\frac{(t2)+t}{t(t2)}=\frac{5}{12}\); \(\frac{2t2}{t^22t}=\frac{5}{12}\); \(24t24=5t^210t\); \(5t^234t+24=0\). Hope it's clear.
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Re: Running at their respective constant rates, machine X takes 2 days lon
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17 Aug 2016, 10:57
Tough one! My recommended solution is plugging in 1 for w. Attached is a visual that should help.
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Re: Running at their respective constant rates, machine X takes 2 days lon
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10 Jun 2017, 07:10
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. Answer: E
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Re: Running at their respective constant rates, machine X takes 2 days lon
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Updated on: 12 Jun 2019, 05:37
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 YLet 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 daysIn other words, The two machines together produce 5(12)/4 widgets in 3 daysOr 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 Answer: E Cheers, Brent
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Re: Running at their respective constant rates, machine X takes 2 days lon
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10 Sep 2018, 03:23
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
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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
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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. 64 = 2 and this adhers to our original constraint.




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