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Re: Work Problem
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30 Sep 2013, 01:35
Bunuel wrote: 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 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  \(2t=12\) > \(t=6\) > \(\frac{1}{6}+\frac{1}{62}=\frac{5}{12}\). Answer: E. Some work problems with solutions: timenworkproblem82718.html?hilit=reciprocal%20ratefacingproblemwiththisquestion91187.html?highlight=rate+reciprocalwhatamidoingwrongtobunuel91124.html?highlight=rate+reciprocalgmatprepps93365.html?hilit=reciprocal%20ratequestionsfromgmatpreppracticeexampleasehelp93632.html?hilit=reciprocal%20rateagoodone98479.html?hilit=rateHope 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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Re: Work Problem
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04 Oct 2013, 19:47
Bunuel wrote: 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 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  \(2t=12\) > \(t=6\) > \(\frac{1}{6}+\frac{1}{62}=\frac{5}{12}\). Answer: E. Some work problems with solutions: timenworkproblem82718.html?hilit=reciprocal%20ratefacingproblemwiththisquestion91187.html?highlight=rate+reciprocalwhatamidoingwrongtobunuel91124.html?highlight=rate+reciprocalgmatprepps93365.html?hilit=reciprocal%20ratequestionsfromgmatpreppracticeexampleasehelp93632.html?hilit=reciprocal%20rateagoodone98479.html?hilit=rateHope 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 'x2'. 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 'x2'? 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
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14 Oct 2013, 04:59
Time take by X = t days Time take by Y = (t2) days
1/X = w/t and 1/Y = w/(t2)
(w/t+w/(t2)) X 3 = 5w/4 Simplifying we get :
(t1)/t(t2) = 5/24 Instead of solving the quadratic eqn substitute values for t. Clearly t=6
So, Rate * Time = Work or w/6 * T = 2w or T= 12



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Re: Work Problem
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17 Oct 2013, 04:17
bulletpoint wrote: Bunuel wrote: 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 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  \(2t=12\) > \(t=6\) > \(\frac{1}{6}+\frac{1}{62}=\frac{5}{12}\). Answer: E. Some work problems with solutions: timenworkproblem82718.html?hilit=reciprocal%20ratefacingproblemwiththisquestion91187.html?highlight=rate+reciprocalwhatamidoingwrongtobunuel91124.html?highlight=rate+reciprocalgmatprepps93365.html?hilit=reciprocal%20ratequestionsfromgmatpreppracticeexampleasehelp93632.html?hilit=reciprocal%20rateagoodone98479.html?hilit=rateHope 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 'x2'. 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 'x2'? 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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Re: Work Problem
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17 Oct 2013, 10:20
Bunuel wrote: 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 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  \(2t=12\) > \(t=6\) > \(\frac{1}{6}+\frac{1}{62}=\frac{5}{12}\). Answer: E. Some work problems with solutions: timenworkproblem82718.html?hilit=reciprocal%20ratefacingproblemwiththisquestion91187.html?highlight=rate+reciprocalwhatamidoingwrongtobunuel91124.html?highlight=rate+reciprocalgmatprepps93365.html?hilit=reciprocal%20ratequestionsfromgmatpreppracticeexampleasehelp93632.html?hilit=reciprocal%20rateagoodone98479.html?hilit=rateHope it helps. There must be a quicker way to solve these, doing that math took me almost 6 minutes



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Re: Work Problem
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17 Oct 2013, 10:41
AccipiterQ wrote: Bunuel wrote: 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 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  \(2t=12\) > \(t=6\) > \(\frac{1}{6}+\frac{1}{62}=\frac{5}{12}\). Answer: E. Some work problems with solutions: timenworkproblem82718.html?hilit=reciprocal%20ratefacingproblemwiththisquestion91187.html?highlight=rate+reciprocalwhatamidoingwrongtobunuel91124.html?highlight=rate+reciprocalgmatprepps93365.html?hilit=reciprocal%20ratequestionsfromgmatpreppracticeexampleasehelp93632.html?hilit=reciprocal%20rateagoodone98479.html?hilit=rateHope 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 \(\frac{1}{t}+\frac{1}{t2}=\frac{5}{12}\) quite quickly and then substitute.
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Re: Work Problem
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17 Oct 2013, 12:33
Bunuel wrote: It's a matter of practice. One can get \(\frac{1}{t}+\frac{1}{t2}=\frac{5}{12}\) quite quickly and then substitute.
Adding on to what Bunuel said, If one observes carefully, we can break down the given equation as : \(\frac{1}{t}+\frac{1}{t2}=\frac{(3+2)}{12}\) = \(\frac{3}{12} +\frac{2}{12} = \frac{1}{4}+\frac{1}{6}\) and thus, on comparison, t=6.
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Re: Running at their respective constant rates, machine X takes
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17 Nov 2013, 11:12
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
Can somebody please help me understand the mistake in my calculation?!
Thanks in advance!!



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Re: Running at their respective constant rate, machine X takes 2
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20 Nov 2013, 13:29
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 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. 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/t2...so couldn't we skip the early steps and jump right to the 1/t+1/(t2)=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
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02 Apr 2014, 07:02
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
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Re: Running at their respective constant rates, machine X takes
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05 Apr 2015, 04:30
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 k2 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
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27 Sep 2015, 20:24
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
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17 Feb 2016, 13:10
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 \(t2\) days, so the rate of Y would be \(rate=\frac{job \ done}{time}=\frac{w}{t2}\);
what is the reasoning behind chosing \(t2\) 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
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16 Apr 2016, 19:15
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)14t24=0 Please advise where am I going wrong
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Re: Running at their respective constant rates, machine X takes
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21 Jun 2016, 14:42
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
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21 Jul 2016, 07:52
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
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17 Aug 2016, 11:57
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
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27 Sep 2016, 09:53
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
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27 Sep 2016, 09:53



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