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Running at their respective constant rates, machine X takes : GMAT Problem Solving (PS) - Page 2

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Re: Running at their respective constant rate, machine X takes 2 [#permalink]
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 1003 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]
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]
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?

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