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It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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10 Jun 2010, 05:23
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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. \(\frac{100xy – z}{x + y}\) B. \(\frac{y(100x – z)}{x + y}\) C. \(\frac{100y(x – z)}{x + y}\) D. \(\frac{x + y}{100xy – z}\) E. \(\frac{x + y – z}{100xy}\)
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Last edited by Bunuel on 11 Feb 2018, 07:24, edited 2 times in total.
Renamed the topic and edited the question.



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It takes machine A 'x' hours to manufacture a deck of cards that [#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. \(\frac{100xy – z}{x + y}\) B. \(\frac{y(100x – z)}{x + y}\) C. \(\frac{100y(x – z)}{x + y}\) D. \(\frac{x + y}{100xy – z}\) E. \(\frac{x + y – z}{100xy}\) 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 \(100\frac{z}{x}=\frac{100xz}{x}\) 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 \(time=\frac{100xz}{x}*\frac{xy}{x+y}=\frac{y(100xz)}{x+y}\). Answer: B. Hope it's clear.
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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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10 Jun 2010, 10:06
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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 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. Next



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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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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 (xz) or (250) D) = some really small fraction E) = some negative number We have a winner in B!!
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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#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 = \(\frac{100  \frac{z}{x}}{\frac{1}{x}+ \frac{1}{y}}\) = \(\frac{(100xz)y}{x+y}\)
So the answer is B.



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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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16 Jul 2010, 01:41
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: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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\(\frac{1}{x}(z)+\frac{x+y}{yx}(t)=100\) \(\frac{x+y}{xy}(t)=100\frac{z}{x}\) \(t=\frac{100xz}{x}(\frac{xy}{x+y})\) \(t=\frac{y(100xz)}{x+y}\)
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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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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!
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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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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 \(100\frac{z}{x}=\frac{100xz}{x}\) 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 \(time=\frac{100xz}{x}*\frac{xy}{x+y}=\frac{y(100xz)}{x+y}\). 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 100z/x decks it must take (100z/x)*2/(x+y).



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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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16 Feb 2016, 13:14
let t=time machines operate simultaneously z/x+t(1/x+1/y)=100 t=(100z/x)/[(x+y)/xy] t=y(100xz)/(x+y)



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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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31 Mar 2016, 17:26
Attached is a visual that should help. Bundle's solution is the most elegant, but if you can't pull that off (which many testtakers can't), then this illustrates an admittedly more workintensive second option.
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Screen Shot 20160331 at 6.26.17 PM.png [ 117.77 KiB  Viewed 6355 times ]
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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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02 Apr 2016, 10:32
...and here is a visual version of Bunuel 's explanation.
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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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18 Apr 2016, 09:54
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! "
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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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30 May 2016, 12:39
Bunuel why is the rate (y+x / yx). Isnt that time? Work rule is 1/r + 1/s = 1/h so doing 1/x + 1/y actually delivers time not rate?



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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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30 May 2016, 12:45
rjivani wrote: Bunuel why is the rate (y+x / yx). Isnt that time? Work rule is 1/r + 1/s = 1/h so doing 1/x + 1/y actually delivers time not rate? Time is a reciprocal of rate: 1/r + 1/s = 1/h (s + r)/(rs) = 1/h h = rs/(r+s). THEORYThere are several important things you should know to solve work problems: 1. Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.\(time*speed=distance\) <> \(time*rate=job \ done\). For example when we are told that a man can do a certain job in 3 hours we can write: \(3*rate=1\) > \(rate=\frac{1}{3}\) job/hour. Or when we are told that 2 printers need 5 hours to complete a certain job then \(5*(2*rate)=1\) > so rate of 1 printer is \(rate=\frac{1}{10}\) job/hour. Another example: if we are told that 2 printers need 3 hours to print 12 pages then \(3*(2*rate)=12\) > so rate of 1 printer is \(rate=2\) pages per hour; So, time to complete one job = reciprocal of rate. For example if 6 hours (time) are needed to complete one job > 1/6 of the job will be done in 1 hour (rate). 2. We can sum the rates.If we are told that A can complete one job in 2 hours and B can complete the same job in 3 hours, then A's rate is \(rate_a=\frac{job}{time}=\frac{1}{2}\) job/hour and B's rate is \(rate_b=\frac{job}{time}=\frac{1}{3}\) job/hour. Combined rate of A and B working simultaneously would be \(rate_{a+b}=rate_a+rate_b=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\) job/hour, which means that they will complete \(\frac{5}{6}\) job in one hour working together. 3. For multiple entities: \(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}\), where \(T\) is time needed for these entities to complete a given job working simultaneously.For example if: Time needed for A to complete the job is A hours; Time needed for B to complete the job is B hours; Time needed for C to complete the job is C hours; ... Time needed for N to complete the job is N hours; Then: \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+...+\frac{1}{N}=\frac{1}{T}\), where T is the time needed for A, B, C, ..., and N to complete the job working simultaneously. For two and three entities (workers, pumps, ...): General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:Given that \(t_1\) and \(t_2\) are the respective individual times needed for \(A\) and \(B\) workers (pumps, ...) to complete the job, then time needed for \(A\) and \(B\) working simultaneously to complete the job equals to \(T_{(A&B)}=\frac{t_1*t_2}{t_1+t_2}\) hours, which is reciprocal of the sum of their respective rates (\(\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{T}\)). General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:\(T_{(A&B&C)}=\frac{t_1*t_2*t_3}{t_1*t_2+t_1*t_3+t_2*t_3}\) hours. Hope this helps
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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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20 Jun 2017, 18:11
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 We can let the rate of machine A = 1/x and the rate of machine B = 1/y. We are given that machine A operates for z hours, so machine A completes (1/x)(z) = z/x decks of cards when operating alone. Thus, there will be 100  z/x decks left to complete when machines A and B work together. We can let n = the number of hours that machines A and B work together to complete (100  z/x) decks and we can create the following equation: (1/x + 1/y)(n) = 100  z/x Multiplying the entire equation by xy, we have: (y + x)(n) = 100xy  zy n = y(100x  z)/(y + x) Answer: B
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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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04 Feb 2018, 11:47
Hi All, This question can be solved by TESTing VALUES. Machine A takes X hours to make a deck of cards. Machine B takes Y hours to make a deck of cards. Since the answers are suitably complexlooking, let's choose really small, easy numbers to work with: X = 1 Y = 2 So… Machine A takes 1 hour to make a deck of cards Machine B takes 2 hours to make a deck of cards When both machines work together for 2 total hours, 3 decks of cards are made. The question goes on to state that Machine A will work alone for Z hours, then be joined by Machine B until 100 decks are made. Z = 1 In that first hour, Machine A will produce 1 deck of cards, leaving 99 decks to go. Since the two machines together can produce 3 decks every 2 hours, the remaining 99 decks will take… 2 hours x 33 sets = 66 hours. We're looking for the answer that equals 66 when we plug in X=1, Y=2 and Z=1 into the answer choices. While it "looks like" there's a lot of math to be done, most of the answers are way too small to be 66 (and it shouldn't take too long to figure that out). Only one answer equals 66 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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04 Feb 2018, 13:55
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 Given, \(ax = by = 1\) \(az + t(a+b) = 100\) \(\frac{x}{z} + t(\frac{x+y}{xy}) = 100\) \(t = y(100xz)/(x+y)\)
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Re: It takes machine A 'x' hours to manufacture a deck of cards that [#permalink]
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05 Feb 2018, 12:06
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 Another easy way Let x = 1 hour..........then rate of A = 1 deck/hour. Let y = 2 hours. If A completes the the whole job..... then the time that A takes to produce all 100 decks = 100 hours. Then z=100. Since A completes the whole job, the number of hours that A and B work together = 0. Let's plug x=1, y=2 and z=100 into the answers choices. y(100x  z)/(x+y) = 2(100*1  100)/(1+2) = 0. Answer: B.




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