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Running at their respective constant rates, machine X takes

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

A. 4
B. 6
C. 8
D. 10
E. 12
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Sep 2013, 02:09, edited 3 times in total.
Edited the question and added the OA.
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Re: Work Problem [#permalink] New post 05 Aug 2010, 01:41
Hello,

The equivalent Time
1/T = 1/T1 + 1/T2 or T=(T1*T2)/(T1+T2)
T=(5/4)/3=5/12

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

The answer is 2A.

regards,
Jack
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Re: Work Problem [#permalink] New post 05 Aug 2010, 02:00
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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 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}.

Answer: E.

Some work problems with solutions:
time-n-work-problem-82718.html?hilit=reciprocal%20rate
facing-problem-with-this-question-91187.html?highlight=rate+reciprocal
what-am-i-doing-wrong-to-bunuel-91124.html?highlight=rate+reciprocal
gmat-prep-ps-93365.html?hilit=reciprocal%20rate
questions-from-gmat-prep-practice-exam-please-help-93632.html?hilit=reciprocal%20rate
a-good-one-98479.html?hilit=rate

Hope it helps.
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Re: Work Problem [#permalink] New post 05 Aug 2010, 02:16
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)
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Re: Work Problem [#permalink] New post 05 Aug 2010, 02:39
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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
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Re: Work Problem [#permalink] New post 05 Aug 2010, 02:57
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heyholetsgo wrote:
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?

Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm

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:

T_{(A&B&C)}=\frac{a*b*c}{ab+ac+bc} hours.

Hope it helps.
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Re: Work Problem [#permalink] New post 05 Aug 2010, 03:13
Thanks guys, this is very good stuff:)

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

The answer is 2A.
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Re: Work Problem [#permalink] New post 05 Aug 2010, 04:09
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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

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...

simple maths !!! isnt it...

Hope this helps..
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Re: Work Problem [#permalink] New post 15 Aug 2010, 05:00
Hi,

I tried a different method to solve the problem, but somehow I couldn't. This was what I came up with so far:

I was trying to apply:Time = job done / rate

Let X = X widgets/day,
Y = Y widgets/day

1)w/X = (w/Y) + 2
2) 5w/4 = (X + Y)*3
From 1), Y = wX/(w - 2X) 3)
I plugged 3) into 2), but I came out with something really ugly and got stuck.

Is there anyone able to solve these two equations, or is there anything wrong with my method? Thank you.
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Re: Work Problem [#permalink] New post 15 Aug 2010, 21:12
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Hey everyone,

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!
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Re: Advanced Rate Problem QR#173 PS [#permalink] New post 16 Jan 2011, 18:04
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Merging similar topics.

Similar problems:
work-problem-98599.html#p759876
hours-to-type-pages-102407.html?hilit=answer%20choices%20or%20solve%20quadratic%20equation.%20R

Theory about work problems:
word-translations-rates-work-104208.html?hilit=printer#p812628

Hope it helps.
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Re: Work Problem [#permalink] New post 14 Feb 2011, 19:08
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Let machine y take y days for w widhets

1 day - w/y widgets by machine Y

1 day - w/(y+2) widgets by machine X


3w/y + 3w/(y+2) = 5w/4

=> 12/y + 12/(y+2) = 5

=> 12y + 24 + 12y = 5y^2 + 10y

=> 5y^2 - 14y - 24 = 0

=> 5y^2 - 20y + 6y - 24 = 0

=> 5y(y - 4) + 6(y-4) = 0

=> y = 4

=> Machine x takes 4+2 = 6 days for w widgets, and 6 * 2 = 12 days for 2w widgets

So answer is E.
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Re: 173. ps QR. [#permalink] New post 11 Mar 2011, 07:43
Looks like we got this problem in the forum before;

Anyway;

X
d+2 days-> w widgets
1 day-> w/(d+2) widgets
3 days-> 3w/(d+2) widgets

Y
d days-> w widgets
1 day-> w/d widgets
3 days-> 3w/d widgets

So; widgets by X and Y in 3 days;

3w/(d+2) + 3w/d = (5/4)w
Cancel out w from both sides;

\frac{3}{d+2}+\frac{3}{d} = \frac{5}{4}

Upon solving;
5d^2-14d-24=0
5d^2-20d+6d-24=0
(5d+6)(d-4)=0

d= -6/5
or
d=4

-ve is not possible as number of days
so; d=4

X makes w widgets in d+2=4+2=6 days
X makes 2w widgets in twice the time i.e. 6*2=12 days

Ans: "E"
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Running at their respective constant rate, machine X takes 2 [#permalink] New post 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:

12x + \frac{12x^2}{x-2}-24 - \frac{24x}{x-2} = 5x^2 - 10x


I want to know how to simplify problems like this that involve multiplying a monomial X binomial X Binomial(that has a 1 binomial denominator).

I would like to know how to handle this type of simplification I feel its important to know in algebra.

Any help is greatly appreciated!!

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

12x + \frac{12x^2}{x-2}-24 - \frac{24x}{x-2} = 5x^2 - 10x


I want to know how to simplify problems like this that involve multiplying a monomial X binomial X Binomial(that has a 1 binomial denominator).

I would like to know how to handle this type of simplification I feel its important to know in algebra.

Any help is greatly appreciated!!

-Josh


A general rule I try to follow for simplification is trying to get rid of the most complicated parts of the equation first.

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

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

12x + \frac{12x^2}{x-2}-24 - \frac{24x}{x-2} = 5x^2 - 10x


I want to know how to simplify problems like this that involve multiplying a monomial X binomial X Binomial(that has a 1 binomial denominator).

I would like to know how to handle this type of simplification I feel its important to know in algebra.

Any help is greatly appreciated!!

-Josh


Merged your question with an earlier discussion of the same problem.

Check this post for the solution: running-at-their-respective-constant-rate-machine-x-takes-98599.html#p759876 and this one for the theory: running-at-their-respective-constant-rate-machine-x-takes-98599.html#p759894

As for your question: 12x (x-2) ( \frac{1}{x} + \frac{1}{x-2} )= 12x(x-2)(\frac{5}{12}) --> reduce by 12x(x-2): \frac{1}{x}+\frac{1}{x-2}=\frac{5}{12} --> \frac{x-2+x}{x(x-2)}=\frac{5}{12} --> cross-multiply: 24x-24=5x^2-10x --> 5x^2-34x+24=0 --> (x-6)(5x-4)=0 --> x=6 or x=\frac{4}{5}

Hope it's clear.
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Re: Running at their respective constant rates, Machine X takes. [#permalink] New post 06 Jan 2013, 07:46
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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

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

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

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Not 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 x-y=2.
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Re: Running at their respective constant rate, machine X takes 2   [#permalink] 13 Aug 2013, 00:55
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