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Three printing presses, R, S, and T, working together at the
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Updated on: 17 Sep 2013, 06:25
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Three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours. S and T, working together at their respective constant rates, can do the same job in 5 hours. How many hours would it take R, working alone at its constant rate, to do the same job? A. 8 B. 10 C. 12 D. 15 E. 20
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Originally posted by vivek123 on 04 Mar 2006, 11:41.
Last edited by Bunuel on 17 Sep 2013, 06:25, edited 1 time in total.
Renamed the topic, edited the question and added the OA.




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Re: Three printing presses, R, S, and T, working together at the
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06 Dec 2016, 07:23
vivek123 wrote: Three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours. S and T, working together at their respective constant rates, can do the same job in 5 hours. How many hours would it take R, working alone at its constant rate, to do the same job?
A. 8 B. 10 C. 12 D. 15 E. 20 We are given that three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours. We can let r, s and t be the times, in hours, for printing presses R, S and T to complete the job alone at their respective constant rates. Thus, the rate of printing press R = 1/r, the rate of printing press S = 1/s, and the rate of printing press T = 1/t. Recall that rate = job/time and, since they are completing one printing job, the value for the job is 1. Since they complete the job together in 4 hours, the sum of their rates is 1/4, that is: 1/r + 1/s + 1/t = 1/4 We are also given that printing presses S and T, working together at their respective constant rates, can do the same job in 5 hours. Thus: 1/s + 1/t = 1/5 We can substitute 1/5 for 1/s + 1/t is the equation 1/r + 1/s + 1/t = 1/4 and we have: 1/r + 1/5 = 1/4 1/r = 1/4  1/5 1/r = 5/20  4/20 1/r = 1/20 r = 20 Thus, it takes printing press R 20 hours to complete the job alone. Answer: E
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Re: RTD  Combined Worked Sum
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01 Dec 2012, 04:43
SreeViji wrote: Three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours. S and T, working together at their respective constant rates, can do the same job in 5 hours. How many hours would it take R, working alone at its constant rate, to do the same job? A. 8 B. 10 C. 12 D. 15 E. 20 This is my nightmare . I remember studying some formula for combined work in my school days, but forgot. And so everytime , I come across such sums, my brain looks for the formula and fails . Even after trying to work it in RTD method , I am not able to solve. Somebody please help. Let's assume the work to be something like printing 20 papers. I'm picking 20 as it is the LCM of 4 & 5. Any number in that place will work just as well. Speed of Three machines together = 5 papers per hour Speed of Two machines together = 4 papers per hour So speed of remaining machine = 1 paper per hour So, to print 20 papers, this machine would take 20/1 = 20 hours. Answer is E.
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Re: PS: Work
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04 Mar 2006, 12:16
vivek123 wrote: Three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours. S and T, working together at their respective constant rates, can do the same job in 5 hours. How many hours would it take R, working alone at its constant rate, to do the same job? A. 8 B. 10 C. 12 D. 15 E. 20
E.
R, S and T can do Job J in 4 hours. Which also means the in one hour all three will do J/4 th of the job. Also represented by:
1/r+1/s+1/t = j/4
We know that T and S working together do the same job J in 5 hours. Again in 1 hour of T and S working together they will be done with:
1/s + 1/t = J/5
so 1/r + j/5 = j/4
1/r = j/4j/5=j/20
So in 1 hr working alone R can do 1/20th of J. Therefore R would need 20 hrs.



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Re: PS: Work
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04 Mar 2006, 12:35
=1/41/5=1/20
r can do 1/20 job in 1 hour
r can do the whole job in 20 hours.



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1/R+1/S+ 1/T = 1/4
1/S+ 1/T = 1/5
1/R=1/41/5
R=20



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same logic above 1/r =1/51/4 ... hence r =20 hrs



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RTD  Combined Worked Sum
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01 Dec 2012, 00:42
Three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours. S and T, working together at their respective constant rates, can do the same job in 5 hours. How many hours would it take R, working alone at its constant rate, to do the same job? A. 8 B. 10 C. 12 D. 15 E. 20 This is my nightmare . I remember studying some formula for combined work in my school days, but forgot. And so everytime , I come across such sums, my brain looks for the formula and fails . Even after trying to work it in RTD method , I am not able to solve. Somebody please help.



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Re: RTD  Combined Worked Sum
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02 Dec 2012, 12:13
This is how I solved the problem: 4 hours * (rate of R + rate of S + rate of T) = total job 5 hours * (rate of S + rate of T) = total job
equate the two, reduce them.
4*rate of R = rate of S + rate of T
Plug back into equation 2: 5*(4*rate of R) = total
20* rate of R = total



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Re: Three printing presses, R, S, and T, working together at the
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17 Sep 2013, 06:26



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Re: Three printing presses, R, S, and T, working together at the
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20 Nov 2013, 03:56
I did it like this:
rate R: x rate S: y rate T: z
We know that work(w) = rate (r) * time and we can combine rates.
so I did: R,S,T working together to complete ONE job : 1 = x+y+z *4 S,T working together to complete ONE job: 1 = y+z *5 ==> 1/5 = y+z ==> substitute in first equation I get x = 1/20 which tells me that machine r completes the job in 20h. Hence E.
I took 3 minutes though, because I wasn't 100 % sure that I can solve it like this. Can you please confirm that this is a appropriate way to solve problems like this or explain the answers above a bit more. I think I can follow but not sure.
Thanks!



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Re: Three printing presses, R, S, and T, working together at the
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23 Dec 2015, 07:50
\(\frac{1}{r} + \frac{1}{s} + \frac{1}{t} = \frac{1}{4}\)
\(\frac{1}{r} + \frac{1}{s} = \frac{1}{5}\) \(thus >\) \(\frac{1}{5} + \frac{1}{t} = \frac{1}{4}\)
\(\frac{1}{t} = \frac{1}{4}  \frac{1}{5}\)
\(\frac{1}{t} = \frac{5}{20}  \frac{4}{20}\)
\(t = 20\)



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Re: Three printing presses, R, S, and T, working together at the
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06 Dec 2016, 07:29
vivek123 wrote: Three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours. S and T, working together at their respective constant rates, can do the same job in 5 hours. How many hours would it take R, working alone at its constant rate, to do the same job?
A. 8 B. 10 C. 12 D. 15 E. 20 LCM of ( 4, 5 & T ) = 20T So, let the total work be 20T units... Combined efficiency of R , S & T is 9+T Now, 20T/9 + T = 5 Or, 20T = 45 + 5T So, 15T = 45 Thus, T = 3 So, THe time required by T to do the work will be 20T/T = 20 Hence, correct answer will be (E) 20
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Re: Three printing presses, R, S, and T, working together at the
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26 Jan 2017, 08:34
1/5 + 1/x = 1/4
4/20 + 1/x = 5/20
1/x = 1/20
x = 20



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Re: Three printing presses, R, S, and T, working together at the
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20 Oct 2018, 13:28
Even though this is a relatively easy question, it gives us the opportunity to practice a number of my GMAT timing tips (the links below include growing lists of questions that you can use to practice each tip): Rate problems: Use D = R x T and W = R x TLike most work rate problems, we can start with the equation W = R x T and then plug in the work, rate, and time for each scenario that we are considering. Set the amount of work equal to 1 for a single jobBecause we’re talking about a single printing job, we just set W = 1 for each scenario. Add rates when they are simultaneousLet’s define variables for the rates for printing presses R, S, and T as Rr, Rs, and Rt. Remember that we can add rates when they are simultaneous, so, when all 3 presses are working together, the rate is Rr + Rs + Rt. When just S and R are working together, the rate is Rs + Rt. Rate and time are reciprocals of each other for a single jobSince we are given the amounts of time for each scenario, we can set the rate equal to the reciprocal of the time for each scenario. This means that Rr + Rs + Rt = 1/4 and Rs + Rt = 1/5. In addition, we are solving for the time it takes printing press R to do the job working alone; if we call this time Tr, then Tr = 1/Rr, and we can solve for Tr if we know Rr. Eliminate combinations of variables using substitutionWhile we can’t solve for Rs and Rt separately, we don’t have to. Since we know their sum Rs + Rt = 1/5, we can just plug this value in for (Rs + Rt) in the equation Rr + Rs + Rt = 1/4. This is enough to allow us to solve for Rr, which then allows us to solve for Tr, which is the final answer to this question. Please let me know if you have any questions, or if you want me to post a video solution!
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Re: Three printing presses, R, S, and T, working together at the
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20 Oct 2018, 13:40
Solved in 10 seconds without writing



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Re: Three printing presses, R, S, and T, working together at the
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20 Oct 2018, 13:44
vivek123 wrote: Three printing presses, R, S, and T, working together at their respective constant rates, can do a certain printing job in 4 hours. S and T, working together at their respective constant rates, can do the same job in 5 hours. How many hours would it take R, working alone at its constant rate, to do the same job?
A. 8 B. 10 C. 12 D. 15 E. 20 1/R + 1/S + 1/T = 1/4 We also know 1/S + 1/T = 1/5 So 1/R = 1/4  1/5 = 1/20 Time for R = 20 Answer choice E Posted from my mobile device




Re: Three printing presses, R, S, and T, working together at the
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