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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 27 Sep 2016, 09:53
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I got the values of x=6 and x=4/5 by quadratic equation. We use 6 to solve the question. Why can't we use 4/5
Is there an explanation for this?

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

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New post 08 Jun 2017, 01:09
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, 08:10
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, 18:25
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, 20:56
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, 19:59
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] 04 Oct 2017, 19:59

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