A factory has three types of machines, each of which works

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A factory has three types of machines, each of which works at its own constant rate. If 7 Machine As and 11 Machine Bs can produce 250 widgets per hour, and if 8 Machine As and 22 Machine Cs can produce 600 widgets per hour, how many widgets could one machine A, one Machine B, and one Machine C produce in one 8-hour day?

A. 400
B. 475
C. 550
D. 625
E. 700

OA A

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by ceilidh.erickson » Tue Oct 23, 2018 7:45 am
At first, it might seem as if we don't have enough information to solve. We have 3 unknowns (the rates of Machines A, B, and C), but only 2 equations. Since this is Problem Solving and not Data Sufficiency, though, (and there is no "cannot be determined" answer) there must be some way to solve.

Let's let A = the rate of Machine A, and so on. Construct equations for (number of machines)(hourly rate of machines) = (output per hour per machine)

7 Machine As and 11 Machine Bs can produce 250 widgets per hour --> 7A + 11B = 250
8 Machine As and 22 Machine Cs can produce 600 widgets per hour --> 8A + 22C = 600

As stated before, we have 2 equations for 3 variables, and it looks like we're stuck. But since it's PS, there must be a way to combine them! Simplify the 2nd equation, then combine it with the 1st:

4A + 11C = 300
7A + 11B = 250
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11A + 11B + 11C = 550

--> A + B + C = 50

If we had one of each machine, the 3 together would produce 50 widgets per hour. We don't have to know what each machine does individually.

Since the question asks how many widgets can be produced by these 3 in 8 hrs, (50)(8) = 400.

The answer is A.
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by GMATGuruNY » Tue Oct 23, 2018 7:45 am
AAPL wrote:Veritas Prep

A factory has three types of machines, each of which works at its own constant rate. If 7 Machine As and 11 Machine Bs can produce 250 widgets per hour, and if 8 Machine As and 22 Machine Cs can produce 600 widgets per hour, how many widgets could one machine A, one Machine B, and one Machine C produce in one 8-hour day?

A. 400
B. 475
C. 550
D. 625
E. 700

OA A
Rates can be ADDED TOGETHER.

7 Machine As and 11 Machine Bs can produce 250 widgets per hour.
7A + 11B = 250

8 Machine As and 22 Machine Cs can produce 600 widgets per hour.
8A + 22C = 600
4A + 11C = 300

Adding together 7A + 11B = 250 and 4A + 11C = 300, we get:
11A + 11B + 11C = 550
A + B + C = 50

Since A+B+C together produce 50 widgets per hour, the number of widgets produced by A+B+C in an 8-hour day = 50*8 = 400.

The correct answer is A.
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by fskilnik@GMATH » Tue Oct 23, 2018 11:07 am
Hi there!

I would like to suggest that we all avoid posting solutions that are IDENTICAL to solutions that were posted in the very same topic.

This is not only in respect for our colleagues but also for the benefit of the whole BTG community.

On this matter, if I violate this "rule" I kindly ask anyone who felt uncomfortable to complain. I apologise in advance, by the way.

Thank you for your understanding.

Regards,
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similar or identical solutions

by GMATGuruNY » Tue Oct 23, 2018 12:04 pm
fskilnik@GMATH wrote:Hi there!

I would like to suggest that we all avoid posting solutions that are IDENTICAL to solutions that were posted in the very same topic.

This is not only in respect for our colleagues but also for the benefit of the whole BTG community.

On this matter, if I violate this "rule" I kindly ask anyone who felt uncomfortable to complain. I apologise in advance, by the way.

Thank you for your understanding.

Regards,
Fabio.
In many cases, similar or identical solutions are posted when two experts respond at virtually the same time, with the result that the second poster is not aware of the first.
The time stamps for the two solutions here indicate that both solutions were posted at 11:45am.
Last edited by GMATGuruNY on Tue Oct 23, 2018 12:09 pm, edited 1 time in total.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

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by ceilidh.erickson » Tue Oct 23, 2018 12:09 pm
Fabio, I definitely agree with your sentiment generally! But Mitch & I have worked side by side for years, with much mutual respect (I'm assuming!). I've never known him to knowingly step on anyone's toes. I'm sure we've all had the experience of someone else posting while we were in the middle of writing, being none the wiser.
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by fskilnik@GMATH » Tue Oct 23, 2018 12:18 pm
First of all, I owe you my apologies, Mitch. (I simply did not see the time stamps. My fault.)

Very elegant observations, Ceilidh! I have already shown my respect to Mitch´s competence in the near past, by the way:

https://www.beatthegmat.com/agatha-chri ... 03990.html

Anyway, you all have my compromise on this matter and I hope (and believe) you all agree to adopt the same posture.

Regards,
Fabio.
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by Scott@TargetTestPrep » Thu Oct 25, 2018 8:25 am
AAPL wrote:Veritas Prep

A factory has three types of machines, each of which works at its own constant rate. If 7 Machine As and 11 Machine Bs can produce 250 widgets per hour, and if 8 Machine As and 22 Machine Cs can produce 600 widgets per hour, how many widgets could one machine A, one Machine B, and one Machine C produce in one 8-hour day?

A. 400
B. 475
C. 550
D. 625
E. 700
Letting a, b, and c be the hourly output of Machines a, b, and c, respectively, we can create the equations:

7a + 11b = 250

And

8a + 22c = 600

Multiplying the first equation by 2, we have:

14a + 22b = 500

Adding the two equations together we have:

22a + 22b + 22c = 1100

a + b + c = 50

So in 8 hours the 3 machines can produce 400 widgets.

Answer: A

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