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It takes machine A 'x' hours to manufacture a deck of cards [#permalink]
10 Jun 2010, 05:23
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Difficulty:
55% (hard)
Question Stats:
62% (03:22) correct
38% (01:52) wrong based on 315 sessions
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
Re: rate : machine A and machine B. [#permalink]
08 Jul 2010, 12:31
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Expert's post
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]
25 Feb 2014, 04:20
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]
18 Mar 2014, 03:23
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]
18 Mar 2014, 04:39
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]
17 May 2015, 09:12
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]
07 Feb 2016, 20:23
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).
gmatclubot
Re: It takes machine A 'x' hours to manufacture a deck of cards
[#permalink]
07 Feb 2016, 20:23
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