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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]

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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]

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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]

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New post 10 Jun 2010, 06:53
Thanks for the clear explanation Bunuel!
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Re: Deck of Cards [#permalink]

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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]

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

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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]

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New post 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]

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New post 02 Dec 2012, 01:35
\(\frac{1}{x}(z)+\frac{x+y}{yx}(t)=100\)
\(\frac{x+y}{xy}(t)=100-\frac{z}{x}\)
\(t=\frac{100x-z}{x}(\frac{xy}{x+y})\)
\(t=\frac{y(100x-z)}{x+y}\)
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Re: It takes machine A 'x' hours to manufacture a deck of cards [#permalink]

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

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New post 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.

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

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New post 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.

Hope it's clear.
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Re: It takes machine A 'x' hours to manufacture a deck of cards [#permalink]

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

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New post 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

[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.



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

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New post 16 Feb 2016, 14:14
let t=time machines operate simultaneously
z/x+t(1/x+1/y)=100
t=(100-z/x)/[(x+y)/xy]
t=y(100x-z)/(x+y)
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Re: It takes machine A 'x' hours to manufacture a deck of cards [#permalink]

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

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

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New post 02 Apr 2016, 11:32
...and here is a visual version of Bunuel 's explanation.
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Screen Shot 2016-04-02 at 11.31.19 AM.png [ 125.72 KiB | Viewed 711 times ]

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

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New post 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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