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

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05 Aug 2010, 02:01

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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.

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

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^2-34t+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)

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

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^2-34t+24=0 After I found the factors for 24, there is no way getting around trial and error, is there?

For our original question you can also use substitution method: we \(\frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}\) 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:

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

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...

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 = \(\frac{1}{(x-2)}\) 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!

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

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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 \(12x (x-2) ( \frac{1}{x} + \frac{1}{x-2} )= 12x(x-2)(\frac{5}{12})\)

to

\(12 [(x-2)+x] = 5x(x-2)\)

my instinct is to multiply the 12x by x-2 resulting in 12x squared - 24x and then foil the two binomials and multply the terms on the right resulting in this more complicated equation:

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

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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 \(12x (x-2) ( \frac{1}{x} + \frac{1}{x-2} )= 12x(x-2)(\frac{5}{12})\)

to

\(12 [(x-2)+x] = 5x(x-2)\)

my instinct is to multiply the 12x by x-2 resulting in 12x squared - 24x and then foil the two binomials and multply the terms on the right resulting in this more complicated equation:

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 \(( \frac{1}{x} + \frac{1}{x-2} )\). I simplified that to (x-2+x)/[(x)(x-2)]. From there, (x-2) 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 \(12 [(x-2)+x]\).

Now looking at the right side of the equation:\(12x(x-2)(\frac{5}{12})\), we just have the 12x cancel out with the 12, leaving us with (5x)(x-2)

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

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21 May 2012, 08:57

Expert's post

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 \(12x (x-2) ( \frac{1}{x} + \frac{1}{x-2} )= 12x(x-2)(\frac{5}{12})\)

to

\(12 [(x-2)+x] = 5x(x-2)\)

my instinct is to multiply the 12x by x-2 resulting in 12x squared - 24x and then foil the two binomials and multply the terms on the right resulting in this more complicated equation:

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

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06 Jan 2013, 08:46

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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 6-2=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

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

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09 Jan 2013, 07:50

Bunuel could you please comment!

I have one question, can we accept it as GMAT-like (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^2-34A+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^2-34/5A+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. _________________

If you found my post useful and/or interesting - you are welcome to give kudos!

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

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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.

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

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13 Aug 2013, 01:55

Expert's post

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.

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