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Difficulty: 805+ Level,   Work and Rate Problems,                                 
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Re: Running at their respective constant rates, Machine X takes 2 days [#permalink]
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heyholetsgo wrote:
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


We are given that running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y does. If we let t = the time in days that it takes machine Y to produce w widgets, then t + 2 = the time in days that it takes machine X to produce w widgets. Furthermore, we can say the following:

rate of machine Y = w/t

rate of machine X = w/(t + 2)

We are given that the machines produce 5w/4 widgets in 3 days. Since work = rate x time and each machine works for 3 days, we first calculate the work done by machine X and machine Y individually.

work of machine Y = (w/t) x 3 = 3w/t

work of machine X = w/(t + 2) x 3 = 3w/(t + 2)

Since the machines work together to produce 5w/4 widgets, we can sum their work and set that sum to 5w/4.

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

Divide the entire equation by w and we have:

3/t + 3/(t + 2) = 5/4

Now, multiplying the entire equation by 4t(t + 2) to eliminate the denominators in the equation, we obtain:

3[4(t + 2)] + 3(4t) = 5[t(t + 2)]

12t + 24 + 12t = 5t^2 + 10t

5t^2 - 14t - 24 = 0

(5t + 6)(t - 4) = 0

t = -6/5 or t = 4

Since t cannot be negative, t must equal 4. That is, it takes machine Y 4 days to produce w widgets. Thus, it will take machine X 6 days to produce w widgets and 12 days to produce 2w widgets.

Answer: E
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Re: Running at their respective constant rates, Machine X takes 2 days [#permalink]
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heyholetsgo wrote:
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

An alternative approach—For some reason, many test-takers seem hesitant to alter information given in PS problems, but they have no trouble doing so in DS problems. In this question, we can reduce the confusing rate in the middle of the problem to one that is a much more relatable production of w widgets.

Take the original rate—

\(\frac{5w}{4}\) in 3 days

Divide both parts by 5/4— w in

\(\frac{12}{5}\) (or 2.4) days

Since the question asks us about the slower machine producing twice the number of widgets, we can deduce that the answer must be greater than 4 (since 2 * 2.4 is already nearly 5). Get rid of (A). At this point, we can test the answers in the following manner. I like to start in the middle of the original five options with either (B) or (D), since, if I get lucky, I will either choose the answer directly or know which option, greater than (D) or less than (B), must be the answer without doing additional work. I would start with (D), given that, again, the question is asking about the slower machine at twice the production.

a) If I assume that Machine X takes 10 days to produce 2w widgets, then it produces w widgets in 5 days.
b) If Machine X takes 5 days to produce w widgets, then Machine Y takes 5 - 2, or 3 days to produce the same w widgets.

Putting everything together, the daily production of Machine X + Machine Y should equal the known production rate of w widgets. Just invert the rates we have worked out so far to get the daily production.

\(\frac{1}{5}w+\frac{1}{3}w=\frac{5}{12}w\)

Divide out the w

\(\frac{1}{5}+\frac{1}{3}=\frac{5}{12}\)

\(\frac{3}{15}+\frac{5}{15}=\frac{5}{12}\)

\(\frac{8}{15}≠\frac{5}{12}\)

In fact,

\(\frac{8}{15}>\frac{5}{12}\)

One fraction is greater than half, while the other is less than half, so we know the calculated production of Machine X is too high. It must be slower, so the answer has to be 12. It would not take long to run the numbers:

c) X: 6 days, Y: 4 days

\(\frac{1}{6}w+\frac{1}{4}w=\frac{5}{12}w\)

\(\frac{1}{6}+\frac{1}{4}=\frac{5}{12}\)

\(\frac{2}{12}+\frac{3}{12}=\frac{5}{12}\)

\(\frac{5}{12}=\frac{5}{12}\)

Yes, a little knowledge of work rates helps. But algebra and logic can help you reach the answer with certainty without too much effort or time, and keeping calm under pressure separates a 50-51 test-taker from someone with the same mathematical skill who balks at seeing a familiar concept presented in a less familiar way.

Good luck with your studies.

- Andrew
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Re: Running at their respective constant rates, Machine X takes 2 days [#permalink]
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heyholetsgo wrote:
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


Another approach is to assign a nice value to the job (w)

Let's say that w = 12.

GIVEN: Running at their respective constant rates, machine X takes 2 days longer to produce 12 widgets than machine Y
Let t = time for machine Y to produce 12 widgets
So, t+2 = time for machine X to produce 12 widgets

RATE = output/time

So, machine X's RATE = 12 widgets/(t + 2 days) = 12/(t+2) widgets per day
And machine Y's RATE = 12 widgets/(t days) = 12/t widgets per day

The two machines together produce 5w/4 widgets in 3 days
In other words, The two machines together produce 5(12)/4 widgets in 3 days
Or the two machines together produce 15 widgets in 3 days
This means the COMBINED RATE = 5 widgets per day

So, we can write: 12/(t+2) + 12/t = 5
Multiply both sides by (t+2)(t) to get: 12t + 12t + 24 = 5(t+2)(t)
Simplify: 24t + 24 = 5t² + 10t
Rearrange: 5t² - 14t - 24 = 0
Factor to get: (5t + 6)(t - 4) = 0
So, EITHER t = -6/5 OR t = 4
Since the time cannot be negative, it must be the case that t = 4

If t = 4, then it takes Machine Y 4 days to produce 12 widgets
And it takes Machine X 6 days to produce 12 widgets

How many days would it take machine X alone to produce 2w widgets?
In other words, how many days would it take machine X alone to produce 24 widgets? (since w = 12)

If it takes Machine X 6 days to produce 12 widgets, then it will take Machine X 12 days to produce 24 widgets

Answer: E

Cheers,
Brent

Originally posted by BrentGMATPrepNow on 05 Mar 2018, 10:14.
Last edited by BrentGMATPrepNow on 12 Jun 2019, 06:37, edited 1 time in total.
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Re: Running at their respective constant rates, Machine X takes 2 days [#permalink]
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heyholetsgo wrote:
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


Let \(w = 60\) widgets, implying that \(\frac{5}{4}w = 75\) widgets.
We can PLUG IN THE ANSWERS, which represent X's time to produce 2w widgets.
When the correct answer choice is plugged in, X and Y will produce 75 widgets in 3 days.

D: 10
Here, X can produce 2w widgets in 10 days.
Thus, the time for X to produce w widgets = 5 days.

Since X takes 5 days to produce w=60 widgets, X's rate \(= \frac{60}{5} = 12\) widgets per day.
Since X takes 2 days longer than Y to produce w widgets, Y's time to produce w widgets = 3 days.
Since Y takes 3 days to produce w=60 widgets, Y's rate \(= \frac{60}{3} = 20\) widgets per day.

Combined rate for X and Y = 12+20 = 32 widgets per day.
Work produced by X and Y in 3 days = 3*32 = 96 widgets.

Here, X and Y are working TOO FAST.
Implication:
X must take LONGER to produce 2w widgets, with the result that X and Y will work more SLOWLY.

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Re: Running at their respective constant rates, Machine X takes 2 days [#permalink]
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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.

Let's assume that total work = \(W\) units.

Time taken by machine X alone to complete W widgets = \(t\) days

Time taken by machine Y alone to complete W widgets = \(t-2\) days

Per day work of machine X = \(\frac{W}{t}\)

Per day work of machine Y = \(\frac{W}{(t-2)}\)

if the two machines together, they produce 5W/4 widgets in 3 days
That means they both together can produce 5W widgets in 12 days or W widgets in \(12/5\) days.
Since Machine X and Y together can complete W widgets in 12/5 days, per day work of machine X and Y together =\( \frac{W}{(12/5)} \)= \(\frac{5W}{12}\)

Per day work of machine X + Per day work of machine Y = Per day work of machine X and Y together

\(\frac{W}{t }\)+ \(\frac{W}{(t-2)}\) = \(\frac{5W}{12}\)

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

At this point you can either solve the quadratic equation or plugin the values of t from answer options and crosscheck. I prefer the second approach though.

#1: Solving Quadratic equation
\( \frac{1}{t} + \frac{1}{(t-2)} = \frac{5}{12}\)

\(\frac{(t-2 +t)}{(t^2-2t)} = \frac{5}{12}\)

\(12(2t-2) = 5(t^2 -2t)\)

\(24t-24 = 5t^2 - 10 t\)

\(5t^2 -34t+ 24 = 0\)

Product = \(5* 24\) = \(5* 6*4 \) Sum = \(-34\)

Find two numbers m,n where the product = 5* 6*4 and the sum = -34
m= -30 n = -4
we can directly find the quadratic equation without factorization by using the formula, Roots are \(\frac{-m}{a}\) and \(\frac{ -n}{a}\) where a is the coefficient of \(x^2\) in the quadratic equation.
t= -(30)/5 = 6 or -(-4)/5 = 4/5

Since machine Y takes t-2 days to complete W widgets , we can eliminate 4/5 as t-2 will not be negative.

Therefore, Machine X days 6 days to complete W widgets.
Here the question is how many days would it take machine X alone to produce 2W widgets i.e 6*2 = 12 days.
Option E is the answer.

#2 Approach 2 : Using answer options

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

The question here is how many days would it take machine X alone to produce 2W widgets.
So time taken by machine X to produce W widgets i.e. t = Answer option/2

Let's start with Option C: 8

That is time taken machine X alone to produce 2W widgets = 8 days
So t= time taken by machine X to produce W widgets = 8/2 =4 days
Substituting value of t and crosschecking the equation:
1/t + 1/(t-2) = 5/12

1/4 + 1/2 ≠ 5/12 Option C is eliminated. 1/2 itself is greater 5/12 that means value of t should be more than 8. Option A,B are eliminated .

Option D: 10

t= 10/2 = 5

1/5 + 1/3 ≠ 5/12 Option D is eliminated. You can also eliminate the options by cross checking the LCM of the denominators whether it's 12 or not.

That means you are left with only one option i.e. Option E: 12 days would be the answer.
Crosschecking:
t = 12/2 = 6
1/6 + 1/4 = 5/12

Ureeka!!

Hope it helps!
Clifin J Francis,
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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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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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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: https://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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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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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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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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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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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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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:
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Time taken by Machine Y = t days

Time taken by Machine X = (t+2) days

Rate of Machine \(X = \frac{w}{t+2}\)
Rate of Machine \(Y = \frac{w}{t}\)

\(\frac{5w}{4}\) widgets produced in 3 days; so \(\frac{5w}{12}\) produced in 1 day

Equation setup will be

\(\frac{w}{t+2} + \frac{w}{t} = \frac{5w}{12}\)

Solving,

\(5t^2 - 14t - 24 = 0\)

t = 4

Time taken by Machine X = t+2 = 6 for w

So for 2w; it will be 12

Answer = E

Originally posted by PareshGmat on 20 Mar 2014, 01:52.
Last edited by PareshGmat on 14 Aug 2014, 03:01, edited 1 time in total.
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Re: Running at their respective constant rates, Machine X takes 2 days [#permalink]
Can somebody please explain the following:

How does this algebra work: 1/t + 1/t+2 = 5/12 is translated to 5t^2 + 34t - 24.

Thanks in advance ! Chris
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chrishhaantje wrote:
Can somebody please explain the following:

How does this algebra work: 1/t + 1/t+2 = 5/12 is translated to 5t^2 + 34t - 24.

Thanks in advance ! Chris


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

\(\frac{(t-2)+t}{t(t-2)}=\frac{5}{12}\);

\(\frac{2t-2}{t^2-2t}=\frac{5}{12}\);

\(24t-24=5t^2-10t\);

\(5t^2-34t+24=0\).

Hope it's clear.
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