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Running at their respective constant rates, machine X takes [#permalink]
05 Aug 2010, 01:01

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55% (medium)

Question Stats:

57% (03:27) correct
42% (03:05) wrong based on 290 sessions

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]
21 May 2012, 07:35

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]
21 May 2012, 07:50

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]
21 May 2012, 07:57

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]
06 Jan 2013, 07:46

2

This post received KUDOS

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]
09 Jan 2013, 06:50

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]
12 Aug 2013, 20:28

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]
13 Aug 2013, 00:55

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.