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# Running at their respective constant rates, machine X takes

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Manager
Joined: 06 Jul 2010
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Running at their respective constant rates, machine X takes  [#permalink]

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Updated on: 30 Sep 2013, 03:09
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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

Originally posted by heyholetsgo on 05 Aug 2010, 02:01.
Last edited by Bunuel on 30 Sep 2013, 03:09, edited 3 times in total.
Edited the question and added the OA.
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Running at their respective constant rates, machine X takes  [#permalink]

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05 Aug 2010, 03:00
49
75

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 produce $$w$$ widgets will be negative 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 - $$2t=12$$ --> $$t=6$$ --> $$\frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}$$.

Some work problems with solutions:
http://gmatclub.com/forum/time-n-work-p ... cal%20rate
http://gmatclub.com/forum/facing-proble ... reciprocal
http://gmatclub.com/forum/what-am-i-doi ... reciprocal
http://gmatclub.com/forum/gmat-prep-ps- ... cal%20rate
http://gmatclub.com/forum/questions-fro ... cal%20rate
http://gmatclub.com/forum/a-good-one-98 ... hilit=rate

Hope it helps.
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15 Aug 2010, 22:12
29
12
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!
##### General Discussion
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05 Aug 2010, 03:16
1
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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05 Aug 2010, 03:39
2
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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Posts: 55609

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05 Aug 2010, 03:57
4
6
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?

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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05 Aug 2010, 04:13
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

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05 Aug 2010, 05:09
3
1
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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15 Aug 2010, 06:00
1
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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Running at their respective constant rates, machine X takes  [#permalink]

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16 Jan 2011, 19:04
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14 Feb 2011, 20:08
7
1
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

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11 Mar 2011, 08:43
1
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]

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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:

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

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]

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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:

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

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]

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21 May 2012, 08:57
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]

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06 Jan 2013, 08:46
3
1
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

Best Regards,
Mansoor

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

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09 Jan 2013, 07:50

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]

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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.

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

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13 Aug 2013, 01:55
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.

Posted from GMAT ToolKit

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]

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13 Aug 2013, 06:13
2
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 26516 times ]

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Re: Running at their respective constant rate, machine X takes 2   [#permalink] 13 Aug 2013, 06:13

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