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Running at their respective constant rates, machine X takes
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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, 02:01.
Last edited by Bunuel on 30 Sep 2013, 03:09, edited 3 times in total.
Edited the question and added the OA.




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Running at their respective constant rates, machine X takes
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05 Aug 2010, 03:00
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: Work Problem
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15 Aug 2010, 22:12
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: Work Problem
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05 Aug 2010, 03:16
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: Work Problem
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05 Aug 2010, 03:39
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: Work Problem
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05 Aug 2010, 03:57
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: Work Problem
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05 Aug 2010, 04:13
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.



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Re: Work Problem
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05 Aug 2010, 05:09
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: Work Problem
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15 Aug 2010, 06:00
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.



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Running at their respective constant rates, machine X takes
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Re: Work Problem
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14 Feb 2011, 20:08
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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11 Mar 2011, 08:43
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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Running at their respective constant rate, machine X takes 2
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21 May 2012, 08:35
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 \(12x (x2) ( \frac{1}{x} + \frac{1}{x2} )= 12x(x2)(\frac{5}{12})\)
to
\(12 [(x2)+x] = 5x(x2)\)
my instinct is to multiply the 12x by x2 resulting in 12x squared  24x and then foil the two binomials and multply the terms on the right resulting in this more complicated equation:
\(12x + \frac{12x^2}{x2}24  \frac{24x}{x2} = 5x^2  10x\)
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



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Re: Running at their respective constant rate, machine X takes 2
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21 May 2012, 08:50
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 \(12x (x2) ( \frac{1}{x} + \frac{1}{x2} )= 12x(x2)(\frac{5}{12})\)
to
\(12 [(x2)+x] = 5x(x2)\)
my instinct is to multiply the 12x by x2 resulting in 12x squared  24x and then foil the two binomials and multply the terms on the right resulting in this more complicated equation:
\(12x + \frac{12x^2}{x2}24  \frac{24x}{x2} = 5x^2  10x\)
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. \(12x (x2) ( \frac{1}{x} + \frac{1}{x2} )= 12x(x2)(\frac{5}{12})\) 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 \(( \frac{1}{x} + \frac{1}{x2} )\). I simplified that to (x2+x)/[(x)(x2)]. From there, (x2) 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 \(12 [(x2)+x]\). Now looking at the right side of the equation:\(12x(x2)(\frac{5}{12})\), we just have the 12x cancel out with the 12, leaving us with (5x)(x2)



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Re: Running at their respective constant rate, machine X takes 2
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21 May 2012, 08:57
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 \(12x (x2) ( \frac{1}{x} + \frac{1}{x2} )= 12x(x2)(\frac{5}{12})\)
to
\(12 [(x2)+x] = 5x(x2)\)
my instinct is to multiply the 12x by x2 resulting in 12x squared  24x and then foil the two binomials and multply the terms on the right resulting in this more complicated equation:
\(12x + \frac{12x^2}{x2}24  \frac{24x}{x2} = 5x^2  10x\)
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: runningattheirrespectiveconstantratemachinextakes98599.html#p759876 and this one for the theory: runningattheirrespectiveconstantratemachinextakes98599.html#p759894As for your question: \(12x (x2) ( \frac{1}{x} + \frac{1}{x2} )= 12x(x2)(\frac{5}{12})\) > reduce by \(12x(x2)\): \(\frac{1}{x}+\frac{1}{x2}=\frac{5}{12}\) > \(\frac{x2+x}{x(x2)}=\frac{5}{12}\) > crossmultiply: \(24x24=5x^210x\) > \(5x^234x+24=0\) > \((x6)(5x4)=0\) > \(x=6\) or \(x=\frac{4}{5}\) Hope it's clear.
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Re: Running at their respective constant rates, Machine X takes.
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06 Jan 2013, 08:46
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 rate, machine X takes 2
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09 Jan 2013, 07:50
Bunuel could you please comment! I have one question, can we accept it as GMATlike (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^234A+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^234/5A+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.
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Re: Running at their respective constant rate, machine X takes 2
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12 Aug 2013, 21:28
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
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Re: Running at their respective constant rate, machine X takes 2
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13 Aug 2013, 01:55
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 ToolKitNot 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 xy=2.
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Re: Running at their respective constant rate, machine X takes 2
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13 Aug 2013, 06:13
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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