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It takes machine A 'x' hours to manufacture a deck of cards [#permalink]

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10 Jun 2010, 06:23

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It takes machine A 'x' hours to manufacture a deck of cards that machine B can manufacture in 'y' hours. If machine A operates alone for 'z' hours and is then joined by machine B until 100 decks are finished, for how long will the two machines operate simultaneously?

A. (100xy – z)/(x + y) B. y(100x – z)/(x + y) C. 100y(x – z)/(x + y) D. (x + y)/(100xy – z) E. (x + y – z)/100xy

It takes machine A 'x' hours to manufacture a deck of cards that machine B can manufacture in 'y' hours. If machine A operates alone for 'z' hours and is then joined by machine B until 100 decks are finished, for how long will the two machines operate simultaneously?

A) (100xy – z)/(x + y) B) y(100x – z)/(x + y) C) 100y(x – z)/(x + y) D) (x + y)/(100xy – z) E) (x + y – z)/100xy

Note that we are asked: "for how long will the two machines operate simultaneously?".

In first \(z\) hours machine A alone will manufacture \(\frac{z}{x}\) decks. So there are \(100-\frac{z}{x}=\frac{100x-z}{x}\) decks left to manufacture. Combined rate of machines A and B would be \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\) decks/hour, (remember we can easily sum the rates).

As \(time=\frac{job}{rate}\), then \(time=\frac{100x-z}{x}*\frac{xy}{x+y}=\frac{y(100x-z)}{x+y}\).

It takes machine A 'x' hours to manufacture a deck of cards that machine B can manufacture in 'y' hours. If machine A operates alone for 'z' hours and is then joined by machine B until 100 decks are finished, for how long will the two machines operate simultaneously?

A) (100xy – z)/(x + y) B) y(100x – z)/(x + y) C) 100y(x – z)/(x + y) D) (x + y)/(100xy – z) E) (x + y – z)/100xy

If you're having trouble w/ the above method you can try plugging numbers, but it does take longer. Use values for x, y, and z. Say x = 2, y = 4 and z =20.

We have then 90 (100 - 1/2*20) decks left to complete. So we should have (100 - 10)/(1/x+1/y) hours left. 90/(1/2+1/4) -> 90/.75 = 120hrs.

Now you can eyeball a few of the answer choices and realize that only A/B/C are going to produce anything close to 120hrs.

For B: (100*2*4 - 20*4)/(2+4) -> 720/6 = 120. This is our answer.

Yikes!! I could never do this algebraically like Bunuel did it. Dude's a super human. But I did stay at a holiday in express last night (not really... I just like bad jokes). Here's what I got.

if rt=d then A's rate of work is 1/x and B's rate of work is 1/y. I made x=2 and y=4 so that rate A is 1/2 and rate B is 1/4.

So then we're told that A starts out on 100 decks by itself at 1/2 a deck an hour for z hours. So then I assigned a value for z. I said, "If z>200 then A finishes the 100 decks and B doesn't work at all." So I made z arbitrarily less than 200. For me z=50. So, 100 = (1/2)50 + (1/2+1/4)h, whereas h= the number of hours they worked together that I'll compare all answers to later.

100= 25 + 3h/4 75=3h/4 what do you know? h=100!!

So then I plug it the values I had for x, y and z into answer choices A B C D E to see which one is 100

A) = 550/6 which whatever it is isn't 100 B) = 600/6 which is 100 C) = some large negative number because a positive is multiplied by (x-z) or (2-50) D) = some really small fraction E) = some negative number

We have a winner in B!! _________________

He that is in me > he that is in the world. - source 1 John 4:4

Speed of Machine A = \(\frac{1}{x}\) decks/hour Speed of Machine B = \(\frac{1}{y}\) decks/hour

Combined Speed of both machines = \(\frac{1}{x}+ \frac{1}{y}\) decks/hour

Now, Machine A initially worked for z hours, so the number of decks produced in z hours = \(\frac{1}{x}\) decks/hour * z hours = \(\frac{z}{x}\) decks

Decks remaining to be produced = \(100 - \frac{z}{x}\)

So, the time taken for both to work together and finish this would be = Number of decks left/Combined Speed = \(\frac{100 - \frac{z}{x}}{\frac{1}{x}+ \frac{1}{y}}\) = \(\frac{(100x-z)y}{x+y}\)

Re: It takes machine A 'x' hours to manufacture a deck of cards [#permalink]

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25 Feb 2014, 05:20

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: It takes machine A 'x' hours to manufacture a deck of cards [#permalink]

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18 Mar 2014, 04:23

Folks,

I have seen the replies of experts. However, I have one query on this question.

Solution:

Work to be performed =100 decks

Rate * time = work 100/x * x = 100 Rate 1: 100/x

Similarly Rate 2 : 100/y

Then why posters have taken the rates as 1/x and 1/y.

Rgds, TGC! _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: It takes machine A 'x' hours to manufacture a deck of cards [#permalink]

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18 Mar 2014, 05:39

Expert's post

TGC wrote:

Folks,

I have seen the replies of experts. However, I have one query on this question.

Solution:

Work to be performed =100 decks

Rate * time = work 100/x * x = 100 Rate 1: 100/x

Similarly Rate 2 : 100/y

Then why posters have taken the rates as 1/x and 1/y.

Rgds, TGC!

It takes machine A x hours to manufacture ONE deck --> the rate of A = (job)/(time) = 1/x decks per hour; It takes machine B y hours to manufacture ONE deck --> the rate of B = (job)/(time) = 1/y decks per hour.

Re: It takes machine A 'x' hours to manufacture a deck of cards [#permalink]

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17 May 2015, 10:12

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: It takes machine A 'x' hours to manufacture a deck of cards [#permalink]

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07 Feb 2016, 21:23

Bunuel wrote:

sjayasa wrote:

It takes machine A 'x' hours to manufacture a deck of cards that machine B can manufacture in 'y' hours. If machine A operates alone for 'z' hours and is then joined by machine B until 100 decks are finished, for how long will the two machines operate simultaneously?

A) (100xy – z)/(x + y) B) y(100x – z)/(x + y) C) 100y(x – z)/(x + y) D) (x + y)/(100xy – z) E) (x + y – z)/100xy

Note that we are asked: "for how long will the two machines operate simultaneously?".

In first \(z\) hours machine A alone will manufacture \(\frac{z}{x}\) decks. So there are \(100-\frac{z}{x}=\frac{100x-z}{x}\) decks left to manufacture. Combined rate of machines A and B would be \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\) decks/hour, (remember we can easily sum the rates).

As \(time=\frac{job}{rate}\), then \(time=\frac{100x-z}{x}*\frac{xy}{x+y}=\frac{y(100x-z)}{x+y}\).

Answer: B.

Hope it's clear.

Hi Bunuel, I applied a different approach but failed to get the correct option. Pls. guide.

Working together at x & y rate machines A & B will manufacture 2 decks in x + y hours, so to manufacture 1 deck it will take (x +y)/2 hours. Now to manufacture 100-z/x decks it must take (100-z/x)*2/(x+y).

Re: It takes machine A 'x' hours to manufacture a deck of cards [#permalink]

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31 Mar 2016, 18:26

Attached is a visual that should help. Bundle's solution is the most elegant, but if you can't pull that off (which many test-takers can't), then this illustrates an admittedly more work-intensive second option.

Attachments

Screen Shot 2016-03-31 at 6.25.39 PM.png [ 216.18 KiB | Viewed 783 times ]

Screen Shot 2016-03-31 at 6.26.17 PM.png [ 117.77 KiB | Viewed 781 times ]

It takes machine A 'x' hours to manufacture a deck of cards [#permalink]

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18 Apr 2016, 10:54

It takes machine A 'x' hours to manufacture a deck of cards that machine B can manufacture in 'y' hours. If machine A operates alone for 'z' hours and is then joined by machine B until 100 decks are finished, for how long will the two machines operate simultaneously?

A. (100xy – z)/(x + y) B. y(100x – z)/(x + y) C. 100y(x – z)/(x + y) D. (x + y)/(100xy – z) E. (x + y – z)/100xy

A takes x hours to manufacture a deck of cards, so fraction of work x does is (\(\frac{1}{x}\)) corollary B's fraction of work is (\(\frac{1}{y}\))

1. A operates for 'z' hours = (\(\frac{z}{x}\)) or (\(\frac{1}{x}\))*z 2. On top of z/p, machine B joins with A and works for 'X' hours i.e; (\(\frac{1}{x}\)+\(\frac{1}{y}\))*X => X(\(\frac{x+y}{xy}\)) ; unknown is colored red.

adding 1 and 2 => (\(\frac{z}{x}\))+ \(\frac{(x+y)}{(xy)}\)X = 100

Solve for X => (\(\frac{x+y}{xy})\)X = 100 - (\(\frac{z}{x}\)) =>(\(\frac{x+y}{xy})\)X = \(\frac{(100x - z)}{x}\) ; cancel out x term in the denominator. => X = y\(\frac{(100x - z)}{(x+y)}\)

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It takes machine A 'x' hours to manufacture a deck of cards
[#permalink]
18 Apr 2016, 10:54

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