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

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A factory has three types of machines, each of which works a  [#permalink]

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New post Updated on: 11 Jun 2013, 06:09
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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

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Originally posted by emmak on 11 Jun 2013, 03:05.
Last edited by Bunuel on 11 Jun 2013, 06:09, edited 1 time in total.
Edited the question.
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Re: A factory has three types of machines, each of which works a  [#permalink]

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New post 05 Jun 2015, 11:46
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Hi All,

When dealing with complex-"looking" prompts, it's important to remember that they were DESIGNED to be solved, so sometimes you have to 'play around' with what you're given to get to the correct answer.

From this prompt, we can create two equations:

7A + 11B = 250/hour
8A + 22C = 600/hour

We're asked for (A+B+C)/hour over the course of 8 hours.

With the given equations, we have 3 variables but only 2 equations, so this is NOT a typical "system" question. The answers ARE numbers though, so there must be a way to get to A+B+C from the two equations that we have....

Notice how we have 11B in one equation and 22C an another? It's INTERESTING that they're both multiples of 11.....Maybe that's a 'clue' as to how we can proceed....

If we "double" the entire first equation, we get...

14A + 22B = 500/hour

Now I can add this equation to the second equation:

14A + 22B = 500/hour
8A + 22C = 600/hour

22A + 22B + 22C = 1100/hour

Dividing everything by 22, we get....

A+B+C = 50/hour

Over the course of an 8-hour day, that gives us...50(8) = 400 widgets.

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Re: A factory has three types of machines, each of which works a  [#permalink]

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New post 11 Jun 2013, 03:10
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\(rate*time=work\)

\((7A+11B)*1h=250\)

\((8A+22C)*1h=600\) or \(4A+11C=300\). Sum those equations:

\(7A+11B+4A+11C=550\) or \(11A+11B+11C=550\)

\(A+B+C=50\) every hour, in 8 hours \(50*8=400\)
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Re: A factory has three types of machines, each of which works a  [#permalink]

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New post 20 Jan 2014, 03:48
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Let Machine A produce A widgets per hour. B produce B widgets per hour and C produce C widgets per hour.
7A+11B=250 ---(1)
8A+22C=600 ---(2)

(1)+(2)
15A+11B+22C=850 split up
11A+11B+11C + 4A+11C = 850---(3)

From (2) 4A+11C=300
Hence (3) becomes
11(A+B+C) = 550
A+B+C = 50. So working together 1 machine of each type produce 50 widgets an hour. in 8 hours they produce 8*50 = 400 widgets.
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Re: A factory has three types of machines, each of which works a  [#permalink]

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New post 02 Mar 2014, 05:52
gmatprav wrote:
Let Machine A produce A widgets per hour. B produce B widgets per hour and C produce C widgets per hour.
7A+11B=250 ---(1)
8A+22C=600 ---(2)

(1)+(2)
15A+11B+22C=850 split up
11A+11B+11C + 4A+11C = 850---(3)

From (2) 4A+11C=300
Hence (3) becomes
11(A+B+C) = 550
A+B+C = 50. So working together 1 machine of each type produce 50 widgets an hour. in 8 hours they produce 8*50 = 400 widgets.



It as an algebraic manipulation.
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Re: A factory has three types of machines, each of which works a  [#permalink]

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New post 02 Mar 2014, 21:11
2
Let Machine A produce A widgets per hour. B produce B widgets per hour and C produce C widgets per hour.
7A+11B=250 ---(1)
8A+22C=600 ---(2)

Dividing (2) by 2
4A+11C=300.....(3)

Adding (1) & (3)

11A+11B+11C = 550
A+B+C=50 per hour

So for eight hrs = 50*8 = 400 = Answer = A
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Re: A factory has three types of machines, each of which works a  [#permalink]

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New post 27 Jun 2015, 12:36
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The only number divisible by 8 is 400. So choice A
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Re: A factory has three types of machines, each of which works a  [#permalink]

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New post 25 Mar 2016, 09:10
waterstyler wrote:
The only number divisible by 8 is 400. So choice A


I'm bit skeptical of the validity of this approach? Can we apply all this kind of problems?
can someone dive deep into this matter?
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Re: A factory has three types of machines, each of which works a  [#permalink]

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New post 25 Mar 2016, 11:43
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Nevernevergiveup wrote:
waterstyler wrote:
The only number divisible by 8 is 400. So choice A


I'm bit skeptical of the validity of this approach? Can we apply all this kind of problems?
can someone dive deep into this matter?


Hi Nevernevergiveup,

You are correct to be cynical about this 'short-cut.' To start, the prompt did NOT state that each machine produces an integer number of widgets per hour, so the conclusion that an 8-hour shift will produce a 'multiple of 8' widgets is questionable. Second (and assuming that all of the hourly rates were integers), if this question had appeared on the Official GMAT, the writers would have anticipated that type of thinking and would have made at least two of the answers divisible by 8. Even if that thinking was correct, it would likely allow the Test Taker to eliminate a few answers, but still be left with an educated guess.

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Re: A factory has three types of machines, each of which works a  [#permalink]

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New post 27 Mar 2016, 10:02
7A+11B=250-eq 1
8A+22C=600-eq 2
Multiply eq one by 2 we get
14A+22B=500
Add above eq to eq 2
14A+22B=500
8A+22C=600
__________
22A+22B+22C=1100
Divide by 22
A+B+C=50 widgets are produced in one Hr.
In 8 hrs no of widgets produced =50x8=400

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Re: A factory has three types of machines, each of which works a  [#permalink]

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New post 10 Oct 2016, 07:34
emmak wrote:
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


a question that might seem a nightmare..but very easy to solve, if you know how to approach it...
(1) 1/7A + 1/11B = 250 -> multiply by 2 -> (3) 1/14A + 1/22B = 500
(2) 1/8A + 1/22C = 600

add 2 and 3

1/22A + 1/22B + 1/22C = 1100
divide by 22 => 50
since we need to know how much they do in 8 hours, multiply by 8.

result is 400.

A
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Re: A factory has three types of machines, each of which works a  [#permalink]

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New post 13 Nov 2016, 02:40
mvictor wrote:
emmak wrote:
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


a question that might seem a nightmare..but very easy to solve, if you know how to approach it...
(1) 1/7A + 1/11B = 250 -> multiply by 2 -> (3) 1/14A + 1/22B = 500
(2) 1/8A + 1/22C = 600

add 2 and 3

1/22A + 1/22B + 1/22C = 1100
divide by 22 => 50
since we need to know how much they do in 8 hours, multiply by 8.

result is 400.

A


I really need help here. Lets say rate of one machine A is 1/A, then rate of 7 machine As should be 7/A right?? Can you please elaborate 1/7A which you have taken?
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Re: A factory has three types of machines, each of which works a  [#permalink]

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New post 03 Jun 2017, 03:33
do we have more problems such as this to practice ?? I straight away went for calculating the individual rates and screwed myself
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Re: A factory has three types of machines, each of which works a  [#permalink]

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New post 03 Jun 2017, 12:26
Hi anuj11,

Work Formula questions are relatively rare on the Official GMAT - you'll likely see just 1 and it will likely involve 2 entities (re: machines, people, etc.) working on a task. In this prompt, we have 3 entities, which is even rarer. As such, spending a lot of time practicing for this one type of rare prompt probably isn't a good use of your time right now. Make sure that you're nailing all of the BIG categories first before you spend too much energy 'nit-picking' over the rarer question types.

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