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Machine A and Machine B are each used to make 660 widgets. [#permalink]

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04 Aug 2013, 11:47

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A

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Difficulty:

85% (hard)

Question Stats:

62% (02:40) correct
38% (03:01) wrong based on 424 sessions

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P.S. I searched but could not find this problem already on gmatclub. This problem took me forever to solve

Machine A and Machine B are each used to make 660 widgets. It take Machine A x hours longer to produce 660 widgets than Machine B. Machine B produces y% more widgets per hour than Machine A. How long does it take Machine A to make 660 widgets?

A. x + 100x/y B. x + x/y C. 100x + 100x/y D. (x/100) + (660/y) E. x + (660/y)

Re: Machine A and Machine B are each used to make 660 widgets. [#permalink]

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04 Aug 2013, 11:50

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summer101 wrote:

P.S. I searched but could not find this problem already on gmatclub. This problem took me forever to solve

Machine A and Machine B are each used to make 660 widgets. It take Machine A x hours longer to produce 660 widgets than Machine B. Machine B produces y% more widgets per hour than Machine A. How long does it take Machine A to make 660 widgets?

A. x + 100x/y B. x + x/y C. 100x + 100x/y D. (x/100) + (660/y) E. x + (660/y)

Re: Machine A and Machine B are each used to make 660 widgets. [#permalink]

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04 Aug 2013, 12:26

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summer101 wrote:

P.S. I searched but could not find this problem already on gmatclub. This problem took me forever to solve

Machine A and Machine B are each used to make 660 widgets. It take Machine A x hours longer to produce 660 widgets than Machine B. Machine B produces y% more widgets per hour than Machine A. How long does it take Machine A to make 660 widgets?

A. x + 100x/y B. x + x/y C. 100x + 100x/y D. (x/100) + (660/y) E. x + (660/y)

\(rate*time=work\) so \(time=\frac{work}{rate}\).

\(\frac{660}{A}-\frac{660}{B}=x\) and \(B=A(\frac{100+y}{100})\).

Re: Machine A and Machine B are each used to make 660 widgets. [#permalink]

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06 Aug 2013, 21:51

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Nice problem. Took me a while to solve too but word problems are my weak area so I need the practice. Anyway I figured I would propose an alternative solution for those (like me ) who aren't as good with the algebra.

You can also pick numbers to solve. (This is my 1st post so please bear with me. )

1) To piggy back off of @Zarrolou we have: x = (660/A) - (660/B) and B = A(1+y/100)

2) From here you can pick numbers for A and B, noting that A should be smaller than B (since A takes more time to complete the job). Let's say A = 10 and B = 66.

3) Solving for x and y you get: x= (660/10)-(660/66) = 66-10 = 56 66=10(1+y/100)......skipping some steps if you solve for y you get y=560. Our target value is A's time: 660/A = 660/10 = 66

4) Now plug x and y into the answer choices and see which one gives you the target value (66):

A) x + 100x/y = 56 + 100(56)/560 = 56 + 10 = 66 **Bingo!!!**

I wont bother to write out the rest but if you plug in x and y into the remaining choices you get different values so (A) is the right answer choice. To me this method is much easier to comprehend. I tried the algebra many times and still couldn't figure it out. To each his own. Regardless there is no way I could do this in 2 minutes!!

Re: Machine A and Machine B are each used to make 660 widgets. [#permalink]

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26 Oct 2014, 21:28

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The best way to solve this problem with in 2 min is to pick numbers : let time taken by B to produce 660 goods = t and take t = 10 hours ; and take x = 10 hours time taken by A to produce 660 goods = t+x = 20 hours then the value of y = 100 % ; bcoz , per hour production of B =660/10 =66 and per hour production of A = 660/20 = 33 there fore y is 66 = 33 +33(y/100) and y = 100. Now we know the time taken by A to produce 660 goods is t+x = 20 hours and Substitute x = 10 and y = 100 in the options above to get 20 hours. Only option A satisfies this.

Re: Machine A and Machine B are each used to make 660 widgets. [#permalink]

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09 Aug 2016, 18:35

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We don't need this crazy number - 660 here. Let's say that A produces 5 widgets in 5 hours, B produces 5 widgets in one hour. So it takes A 4 hours longer to produce 5 widgets than B and B produces 400% more than a in one hour, so it takes A 5 hours to produce 5 widgets. X=4, Y=400, and the answer is A!

Re: Machine A and Machine B are each used to make 660 widgets. [#permalink]

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04 Oct 2016, 22:04

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summer101 wrote:

P.S. I searched but could not find this problem already on gmatclub. This problem took me forever to solve

Machine A and Machine B are each used to make 660 widgets. It take Machine A x hours longer to produce 660 widgets than Machine B. Machine B produces y% more widgets per hour than Machine A. How long does it take Machine A to make 660 widgets?

A. x + 100x/y B. x + x/y C. 100x + 100x/y D. (x/100) + (660/y) E. x + (660/y)

This is a typical VICs (Variable in choices) problem

Assume that x=1, the machine A takes 2 hrs and Machine B takes 1 hr since time taken is 50%, work done would be 100% more, therefore Machine B would produce 100% (=y) more widgets in 1 hr.

Put x=1 and y=100 in answer choices, the option which gives you 2 (Machines A takes 2 hr as assumed), that would be right option.

Re: Machine A and Machine B are each used to make 660 widgets. [#permalink]

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13 May 2017, 00:12

Bunuel wrote:

sidoknowia wrote:

Machine A and Machine B are each used to make 660 widgets. It takes Machine A x hours longer to produce 660 widgets than Machine B. Machine B produces y% more widgets per hour than Machine A. How long does it take Machine A to make 660 widgets?

A) x + 100x/y

B) x + x/y

C) 100x + 100x/y

D) x/100 + 660/y

E) x + 660/y

Got this question in veritas prep exam. The algebraic approach is too complex and time consuming, can someone suggest a different solution.

Although I understand that number picking approach would be the best here, but I had to choose an algebraic approach to do the same, even when I prefer number picking as it saves time. This is because the number picking approach sometimes backfires..in the case when more than one options produce the same result. Please help me and maybe many others here by suggesting how the numbers should be picked and when should this approach be used and not used. It will help me so much.

Re: Machine A and Machine B are each used to make 660 widgets. [#permalink]

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31 Jul 2017, 03:08

The number picking approach may well be easy to apply and understand but has relatively lesser reliability than algebraic approach. It would be really helpful to know the algebraic method application for this question as there isn't a good explanation yet with regards to that.

Machine A and Machine B are each used to make 660 widgets. It take Machine A x hours longer to produce 660 widgets than Machine B. Machine B produces y% more widgets per hour than Machine A. How long does it take Machine A to make 660 widgets?

A. x + 100x/y B. x + x/y C. 100x + 100x/y D. (x/100) + (660/y) E. x + (660/y)

We can let b = the number of hours it takes Machine B to make the 660 widgets. Thus, the time in hours for Machine A to make the 660 widgets = x + b. Furthermore, the rate of Machine B = 660/b and the rate of Machine A = 660/(x + b).

Since Machine B produces y% more widgets per hour than Machine A:

660/(x + b)(1 + y/100) = 660/b

1/(x + b)(1 + y/100) = 1/b

1 + y/100 = (x + b)/b

1 + y/100 = x/b + 1

y/100 = x/b

100/y = b/x

100x/y = b

Thus, the time for Machine A to make the 660 widgets is x + b = x + 100x/y.

Answer: A
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