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It takes machine A 'x' hours to manufacture a deck of cards [#permalink]
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10 Jun 2010, 06:23
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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. (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
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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 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: Deck of Cards [#permalink]
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10 Jun 2010, 06:53
Thanks for the clear explanation Bunuel!



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Re: Deck of Cards [#permalink]
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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 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: Deck of Cards [#permalink]
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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 (xz) or (250) 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 = \(\frac{100  \frac{z}{x}}{\frac{1}{x}+ \frac{1}{y}}\) = \(\frac{(100xz)y}{x+y}\)
So the answer is B.



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Re: Deck of Cards [#permalink]
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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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02 Dec 2012, 01:35
\(\frac{1}{x}(z)+\frac{x+y}{yx}(t)=100\) \(\frac{x+y}{xy}(t)=100\frac{z}{x}\) \(t=\frac{100xz}{x}(\frac{xy}{x+y})\) \(t=\frac{y(100xz)}{x+y}\)
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Re: It takes machine A 'x' hours to manufacture a deck of cards [#permalink]
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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!
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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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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 \(100\frac{z}{x}=\frac{100xz}{x}\) 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 [#permalink]
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16 Feb 2016, 14:14
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 [#permalink]
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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 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 4197 times ]
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It takes machine A 'x' hours to manufacture a deck of cards [#permalink]
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02 Apr 2016, 11:32
...and here is a visual version of Bunuel 's explanation.
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It takes machine A 'x' hours to manufacture a deck of cards [#permalink]
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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! "
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Re: It takes machine A 'x' hours to manufacture a deck of cards [#permalink]
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30 May 2016, 13:39
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 [#permalink]
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30 May 2016, 13:45
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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