Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

10 Jun 2010, 06:23

3

This post received KUDOS

18

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

61% (03:24) correct
39% (01:55) wrong based on 404 sessions

HideShow timer Statictics

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]

Show Tags

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

Show Tags

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]

Show Tags

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]

Show Tags

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

Show Tags

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]

Show Tags

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 563 times ]

Screen Shot 2016-03-31 at 6.26.17 PM.png [ 117.77 KiB | Viewed 562 times ]

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

Show Tags

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\(\frac{(100x - z)}{(x+y)}\)

"Encourage me with kudos, if its worth! " _________________

"Fight the HARDEST battle that anyone can ever imagine"

gmatclubot

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

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...

By Libby Koerbel Engaging a room of more than 100 people for two straight hours is no easy task, but the Women’s Business Association (WBA), Professor Victoria Medvec...