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Machines A and B always operate independently and at their respective

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Machines A and B always operate independently and at their respective constant rates. When working alone, machine A can fill a production lot in 5 hours, and machine B can fill the same lot in x hours. When the two machines operate simultaneously to fill the production lot, it takes them 2 hours to complete the job. What is the value of x?

(A) \(3 \frac{1}{3}\)

(B) \(3\)

(C) \(2\frac{1}{2}\)

(D) \(2\frac{1}{3}\)

(E) \(1\frac{1}{2}\)


Additional info on problem
Source: Paper Test
Test Code 28
Section 5
Problem #12
[Reveal] Spoiler: OA
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Re: Machines A and B always operate independently and at their respective [#permalink]

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New post 17 Oct 2010, 14:51
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A really nice question, I got confused 2 times before finding the right answer.

1. We need to plug in the size of the lot to have some indication for the tempo of machine A and B. Lets plug in the number 10.
2. The tempo of machine B is 10/5=2. The Tempo of machine B is 10/x.
3. We now that both machines can fill the lot in 2 hours so therefore we can write an equation - 10/x +2 = 10/2 (the tempo that both of the machines can fill the lot together). From the equation it can be found that - X=3.3333.

Thus, the correct answer is A.
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Re: Machines A and B always operate independently and at their respective [#permalink]

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niheil wrote:
Could anyone help me with this problem:

Machines A and B always operate independently and at their respective constant rates. When working alone, machine A can fill a production lot in 5 hours, and machine B can fill the same lot in x hours. When the two machines operate simultaneously to fill the production lot, it takes them 2 hours to complete the job. What is the value of x?

(A) \(3 \frac{1}{3}\)

(B) \(3\)

(C) \(2\frac{1}{2}\)

(D) \(2\frac{1}{3}\)

(E) \(1\frac{1}{2}\)


Additional info on problem
Source: Paper Test
Test Code 28
Section 5
Problem #12


Using the work rate equation :

\(\frac{1}{5} + \frac{1}{x} = \frac{1}{2}\)
\(\frac{1}{x} = \frac{3}{10}\)
\(x = 3 \frac{1}{3}\)

Answer : (a)
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Re: Machines A and B always operate independently and at their respective [#permalink]

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New post 17 Oct 2010, 19:09
Thanks for the help guys. I think shrouded1's explanation is a bit easier.

So shrouded1 (and maybe you could also answer this nov79), I understand that Machine A's production rate is 1 lot every 5 hours and that Machine B's production rate is 1 lot every X hours.

Could you explain how you know that simply adding these two rates will give us the production rate of both machines when they are operating simultaneously and independently to fill 1 production lot?
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That's an assumption you make in work rate problems that combined rate is a simple sum of each individual rate

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Re: Machines A and B always operate independently and at their respective [#permalink]

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New post 17 Oct 2010, 20:21
Awesome. Thanks for the help shrouded1 and nov79!
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New post 18 Oct 2010, 06:06
The speed of A is 1/5
The speed of B is 1/x
2*(1/5+1/x)=1
solve for x
x=10/3 (A)/
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New post 18 Oct 2010, 07:34
Machine A- 5hrs to complete the job
Machine B-X hrs to complete the job

In 1 hr Machine A completes (1/5)th of the job
In 1 hr machine B completes (1/x)th of the job

both together will complete (1/5)th of the job+(1/x)th of the job in 1 hr

(1/5)+(1/x)=((x+5)/5x) of the job in 1 hr
complete job takes (5x/(x+5)) hrs

(5x/(x+5))=2

x=10/3 Ans.A
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Re: Machines A and B always operate independently and at their respective [#permalink]

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New post 18 Oct 2010, 08:18
Work Rate problems are based on the concept that rates are additive. That is to say that if I paint half a wall in an hour and if you paint half a wall in an hour, if we both work together on a wall, we will finish the wall in an hour (Assuming that you are not repainting whatever I am painting to cover up my shoddy work!).
Remember, Rate of work = Work done per unit time
So, the proper way to express rate is 1/2 wall per hour and not 1 wall in 2 hours

If my rate of work is 1/2 wall/hour and yours is 1/2 wall/hour, our total rate of work is 1/2 + 1/2 = 1 wall/hour.

The basic questions of work rate are of the following form:
If A, working independently, completes a job in 10 hours and B, working independently, completes a job in 5 hours, how long will they take to complete the same job if they are working together?

Since A completes a job in 10 hours, his rate of work is 1/10th of the job per hour. B's rate of work is 1/5th of the job per hour.
Their combined rate of work would then be 1/10 + 1/5 = 3/10th of the job per hour.
As we said before, Rate of work = Work done/Time so 3/10 = 1/T (because 1 job has to be done)
or T = 10/3 hours.
This implies that A and B will together take 3.33 hours to do the job.
Note: Time taken when A and B work together will obviously be less than time taken by A or B when they are working independently.

Coming back to your question (finally! I know!), if A takes 5 hours to fill a lot and B takes x hours, and together they fill it in 2 hours, what is x?
Rate of work of A = 1/5th of the lot per hour
Rate of work of B = 1/xth of the lot per hour
Combined rate of work = 1/2 of the lot per hour
1/2 = 1/5 + 1/x
x = 10/3 hours
Note: Without solving, I know that E cannot be the answer since they both together take 2 hours to complete the work so one person alone can definitely not do the work in less than 2 hours.

Time for a Teaser: A and B, working together, can finish a job in 10 days, B and C, working together, can finish the same job in 12 days and A and C, working together, can finish the same job in 15 days. If all three work together, how long will they take to finish the same job?
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Re: Machines A and B always operate independently and at their respective [#permalink]

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New post 15 Jan 2015, 14:34
As most of the times in such problems, I am creating the RTW chart:

_______R_____T___W
A_____1/5____5____1
B____3/10____x____1
Both___1/2____2___1

So, now let me explain:
From the stem we know that A is doing the job (1 job) in 5 hours. For under T we add 5. From R*T=W, we get R=W/T, so in this case R=1/5. So, we add this under R.

From the stem we know that both machines together are doing the job in 2 hours. We add 2 under T and 1/2 under R.

Now, since we have the conbined time of both of the machines and the time of machine A we can find the time for machine B:
1/2 - 1/5 = 3 /10 or even easier 0.5 - 0.2 = 0.3, which is 3/10. We add 3/10 under R for machine B.

Finally, we are asked to find x, which is the time machine B needs to complete the job. Using R*T=W --> (3/10)X=1 -->(3X)/10 = 1 --> 3X = 10 --> X = 10/3 --> X = 3+1/3.

*an easy way to calculate the mixed number (mixed fraction) is like this:
To turn 10/3 to a mixed number you are looking to find a number with which you can multiply the denominator, add sth to it and get the nominator.

So, you will always have the same denominator: in this case 3.
You are looking for a number lower than the nominator. You will multiply your denominator with this number and add sth to get 10 (your nominator).
For example, you have 3 in this case in the denominator, multiply 3 by 3 and you get 9, add 1 and you get 10. You are done. The number you multiplied your denominator with goes to the left of the fraction and what you added goes to the nomintor.
You now have 3 + 1/3.
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Hi All,

This prompt is an example of a "Work Formula" question. Any time a question involves two entities (people, machines, etc.) working on a task together and there are no "twists" to the question (someone stops working, someone shows up late to the job, etc.), you can use the Work Formula:

(A)(B)/(A+B) where A and B are the "times" that it takes for each entity to finish the job on his/her/its own.

Here, we're told:
Machine A can do the job in 5 hours
Machine B can do the job in X hours
Working together, the two machines can do the job in 2 hours.

Using the Work Formula, we have:

(5)(X)/(5 + X) = 2

5X = 10 + 2X
3X = 10
X = 10/3 hours

So, Machine B can do the job on its own in 10/3 = 3 1/3 hours.

Final Answer:
[Reveal] Spoiler:
A


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Re: Machines A and B always operate independently and at their respective [#permalink]

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niheil wrote:
Machines A and B always operate independently and at their respective constant rates. When working alone, machine A can fill a production lot in 5 hours, and machine B can fill the same lot in x hours. When the two machines operate simultaneously to fill the production lot, it takes them 2 hours to complete the job. What is the value of x?

(A) \(3 \frac{1}{3}\)

(B) \(3\)

(C) \(2\frac{1}{2}\)

(D) \(2\frac{1}{3}\)

(E) \(1\frac{1}{2}\)


Additional info on problem
Source: Paper Test
Test Code 28
Section 5
Problem #12


in 2 hours machine A will finish 2/5 of the work i.e. machine B will finish 3/5 of work in same time. hence machine B will take 2hr+80mins = 3hr 20min = \(3\frac{1}{3}\)
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Machines A and B always operate independently and at their respective [#permalink]

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New post 17 May 2016, 06:35
If same work is done by two different individuals at a different hours(suppose a and b hours), then together they will do same work in ab/a+b hours.

In above question,
Machine A and B both work to fill a production lot.
Machine A works for 5 hours and Machine B works for x hours.
By formula,
5*x/5+x = 2 hours.
or, 5x= 2*(5+x)
or,5x=10+2x
or5x-2x=10
therefore, x=10/3 = 3whole 1/3
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Re: Machines A and B always operate independently and at their respective [#permalink]

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New post 07 Sep 2016, 08:59
Quote:
Machines A and B always operate independently and at their respective constant rates. When working alone, machine A can fill a production lot in 5 hours, and machine B can fill the same lot in x hours. When the two machines operate simultaneously to fill the production lot, it takes them 2 hours to complete the job. What is the value of x?


A produces one lot in 5 hours, 1/5 of a lot per hour = 0.20 lots per hour
B produces 1 lot in x hours.

A + B = 1 lot in 2 hours, or 1/2 lot (0.5 lot) in 1 hour.

Production of lots in an hour:
A + B = 0.5 lots
0.2 + x = 0.5 lots
x = 0.3 lots per hour
1 lot / 0.3 lots per hour = 3.333 hours to produce 1 lot (Machine B)

Therefore, B produces 1 lot in 3.333 hours. 3.333 repeating = 3 1/3.

Decimals are our friends. I see many solution approaches get confounded by using fractions, even in the OGs. I come from an electrical engineering background and learned how useful decimals are a long time ago and tens of thousands equations ago. Good luck to all!
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Re: Machines A and B always operate independently and at their respective [#permalink]

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New post 28 Mar 2018, 10:33
niheil wrote:
Machines A and B always operate independently and at their respective constant rates. When working alone, machine A can fill a production lot in 5 hours, and machine B can fill the same lot in x hours. When the two machines operate simultaneously to fill the production lot, it takes them 2 hours to complete the job. What is the value of x?

(A) \(3 \frac{1}{3}\)

(B) \(3\)

(C) \(2\frac{1}{2}\)

(D) \(2\frac{1}{3}\)

(E) \(1\frac{1}{2}\)


The rate of machine A is ⅕, and the rate of machine B is 1/x. Their combined rate is ½. Thus, we can create the equation:

1/5 + 1/x = 1/2

Multiplying by 10x, we have:

2x + 10 = 5x

10 = 3x

x = 10/3 = 3 1/3

Answer: A
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Re: Machines A and B always operate independently and at their respective [#permalink]

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New post 28 Mar 2018, 12:54
What always gets me is that TIME is ALWAYS in the denominator!
Re: Machines A and B always operate independently and at their respective   [#permalink] 28 Mar 2018, 12:54
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