It takes machine A 'x' hours to manufacture a deck of cards

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

10 Jun 2010, 06:23

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60% (03:07) correct

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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

[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Dec 2012, 04:16, edited 1 time in total.

Renamed the topic and edited the question.

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Re: Deck of Cards [#permalink]

10 Jun 2010, 06:51

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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

[Reveal] Spoiler:

OA B

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.
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Re: Deck of Cards [#permalink]

10 Jun 2010, 06:53

Thanks for the clear explanation Bunuel!

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Re: Deck of Cards [#permalink]

10 Jun 2010, 11:06

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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

[Reveal] Spoiler:

OA B

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.

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Re: Deck of Cards [#permalink]

10 Jun 2010, 17:17

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!!
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rate : machine A and machine B. [#permalink]

08 Jul 2010, 13:23

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?

1. (100xy – z)/(x + y)

2. y(100x – z)/(x + y)

3. 100y(x – z)/(x + y)

4. (x + y)/(100xy – z)

5. (x + y – z)/100xy

[Reveal] Spoiler:

OA : B

src : manhattan CAT.

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Re: rate : machine A and machine B. [#permalink]

08 Jul 2010, 13:31

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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}

So the answer is B.

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Re: Deck of Cards [#permalink]

16 Jul 2010, 02:41

whats i did:

lets x= 10hrs
y= 20hrs

they both can do 20/3 Deck in 1 hrs

now lets say both A and B work together and made 90 decks, while 10decks made by A alone

A & B both time will be 600 hrs
A alone time will be 100 hrs which is the value of Z

now put all the values in the answer choices.

Correct answer is B

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Re: Deck of Cards [#permalink]

02 Dec 2012, 01:35

\frac{1}{x}(z)+\frac{x+y}{yx}(t)=100
\frac{x+y}{xy}(t)=100-\frac{z}{x}
t=\frac{100x-z}{x}(\frac{xy}{x+y})
t=\frac{y(100x-z)}{x+y}

Re: Deck of Cards [#permalink] 02 Dec 2012, 01:35

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It takes machine A 'x' hours to manufacture a deck of cards

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