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Running at their respective constant rates, machine X takes

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

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New post 08 Jun 2017, 00:09
Bunuel wrote:
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.

Answer: E.



Hi Bunuel,

Could you please explain how do you solve the quadratic in one step ? I don't understand how can we factorize it with a coefficient greater than 1 on the \(x^2\).
When trying to solve with the quadratic formula, it involves big numbers with lot of calculation and requires lot of time.

Thank you so much for your help,
Sincerly yours,
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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New post 10 Jun 2017, 07:10
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  [#permalink]

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New post 03 Oct 2017, 17:25
Bunuel wrote:
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.






thank you for your help
but
shouldn't we use the same formula for rate??!!!
the reciprocal of rates shouldn't be added together?
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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New post 03 Oct 2017, 19:56
soodia wrote:
Bunuel wrote:
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.






thank you for your help
but
shouldn't we use the same formula for rate??!!!
the reciprocal of rates shouldn't be added together?


We ARE adding the rates. A, B and C above are times. The rate is reciprocal of time, so 1/A, 1/B and 1/C are rates.
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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New post 04 Oct 2017, 18:59
nickimonckom wrote:
Bunuel wrote:
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.

Answer: E.



Hi Bunuel,

Could you please explain how do you solve the quadratic in one step ? I don't understand how can we factorize it with a coefficient greater than 1 on the \(x^2\).
When trying to solve with the quadratic formula, it involves big numbers with lot of calculation and requires lot of time.

Thank you so much for your help,
Sincerly yours,



Hi...

we can factorize by taking the product of a and c...so 5*24=120

So 30*4=120

5x^2-30x-4x+24=0
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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New post 29 Oct 2017, 21:14
I solved this way.
What am i making mistake?
Let x,y be the time for X and Y machines.
(w/x)-(w/y)=w/2 ---------1 eqn.

(w/x)+(w/y)=5w/12----2 eqn

solve 1 and 2 we get.
2/x=11/12
x=24/11

I think iam making mistake in eqn 1.Can someone please elaborate the mistake,which I made.
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Running at their respective constant rates, machine X takes  [#permalink]

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New post 12 Nov 2017, 08:49
balagevr wrote:
I solved this way.
What am i making mistake?
Let x,y be the time for X and Y machines.
(w/x)-(w/y)=w/2 ---------1 eqn.

(w/x)+(w/y)=5w/12----2 eqn

solve 1 and 2 we get.
2/x=11/12
x=24/11

I think iam making mistake in eqn 1.Can someone please elaborate the mistake,which I made.


Don't get bogged down in algebra. As others above have already shown suppose numbers for variables like w. For instance we can suppose w=4 here, therefore 5w/4=5 and 2w=8. Now when the two machines run together, they produce 5 widgets in 3 days. Therefore 4 widgets will require 12/5 days. This is our target OR goal. Start plugging in answer choices always starting with C and see which choice matches our target or goal. Choice E says Machine X alone takes 12 days to produce 2w OR 8 widgets. Therefore Machine X will take 6 days to produce w OR 4 widgets. Machine Y takes 2 days short to produce 4 widgets i.e. it takes 4 days to produce 4 widgets. Both will take 6X4/(6+4) days or 12/5 days to produce 4 widgets. This matches our target or goal we calculated earlier and is therefore the right answer.
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Running at their respective constant rates, machine X takes  [#permalink]

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New post 16 Jan 2018, 12:01
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



My Method to solve - let w = 12 widgets
Given - x takes two more days than y
assume x takes 6 days
Therefore y will take 4 days

Both together produce 5w/4 = (5 * 12)/4
Total 15 widgets.

x takes 6 days to produce w(12) widgets so x alone will take 12 days to produce (2w) 24 widgets.
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Running at their respective constant rates, machine X takes  [#permalink]

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New post 22 Jan 2018, 01:18
I solved by applying substitution:
1)
X=Y+2
X+Y=2Y+2=3 <=> Y=0.5 X=2.5=5/2

2)
(5/4)w=3
5w=12
w=12/5

To answer question: 2*w*X=2*(12/5)*(5/2)=120/10=12
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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New post 20 Feb 2018, 23:12
yyz5028 wrote:
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.

I approached it in exactly the same manner and am stuck at exactly the same place :?
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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New post 27 Feb 2018, 04:29
You have three variables and two equations. Your method would work if you think of w widgets as 1 job. Then you would have two equations in two variables.

Job per day (days)= job

X (1/x)= 1

Y. (1/y)=. 1

3(x+y)=5/12
(1/x)=(1/y)+2

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

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New post 27 Feb 2018, 04:47
E. 12
Let Y does in y days
X does in y+2 days
1/y + 1/y+2 = 5/12
Putting y=4 solve this equation, hence X will take (4+2)2 days to make 2w widgets

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

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New post 05 Mar 2018, 09:14
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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
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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New post 06 Mar 2018, 04:41
The equation from first two conditions would be -
3/(P+2) + 3/(P) = 5/4
Now we need to solve for P to get 2*(P+2)?
The trial and error method and a manipulation of P+2=x saves the time.
Answer comes out to be E.
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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New post 17 Jul 2018, 07:30
Can we not take days for x machine as t+2 and y machine as t. I am.getting a different answer

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

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New post 04 Aug 2018, 23:11
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


Both machines together take 3 days to produce 5/4*w, hence to produce w units, they'll take 12/5 days.
Let x be the # of days that machine X takes to produce w units.
Hence, Y will take x-2 days to produce w units.

We get the equation:
1/x + 1/(x-2) = 5/12

(2x-2)/x(x-2) = 5/12

(x-1)/x(x-2) = 5/24

(x-1)/x(x-2) = (6-1)/6(6-2)

equating both the sides, we get x=6
Hence machine X will take 2*6=12 days to produce 2w units.
Answer is E.
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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New post 10 Sep 2018, 01:21
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


plugging the number is the only solution for this problem in the test room.

we get an equation
1/x +1/(x-2)= 5/12

now plug the number, remember we require 2w, and the above equation is for 1w,

so, we plug half of the answer choice, which are 2,3,4,5,6

only x=6 is ok

answer is E.

PLUGGIN NUMBER
this is tricky.
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Re: Running at their respective constant rates, machine X takes  [#permalink]

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New post 10 Sep 2018, 03:23
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 &nbs [#permalink] 10 Sep 2018, 03:23

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