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

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13 Aug 2013, 05:13

1

This post received KUDOS

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 2288 times ]

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 - \(2t=12\) --> \(t=6\) --> \(\frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}\).

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 - \(2t=12\) --> \(t=6\) --> \(\frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}\).

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 - \(2t=12\) --> \(t=6\) --> \(\frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}\).

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.

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 - \(2t=12\) --> \(t=6\) --> \(\frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}\).

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.

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 - \(2t=12\) --> \(t=6\) --> \(\frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}\).

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 - \(2t=12\) --> \(t=6\) --> \(\frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}\).

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

Can somebody please help me understand the mistake in my calculation?!

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.

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?

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

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04 Apr 2015, 14:46

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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.

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
_________________

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?

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

Please advise where am I going wrong
_________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

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.

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

gmatclubot

Re: Running at their respective constant rates, machine X takes
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21 Jul 2016, 06:52

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