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Three printing presses, R, S, and T, working together at the
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Updated on: 17 Sep 2013, 07:25
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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, 12:41.
Last edited by Bunuel on 17 Sep 2013, 07:25, edited 1 time in total.
Renamed the topic, edited the question and added the OA.




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Re: RTD  Combined Worked Sum
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01 Dec 2012, 05:43
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, 13:16
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, 13:35
=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, 01:42
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, 13:13
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, 07:26



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Re: Three printing presses, R, S, and T, working together at the
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20 Nov 2013, 04:56
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, 08:50
\(\frac{1}{r} + \frac{1}{s} + \frac{1}{t} = \frac{1}{4}\)
\(\frac{1}{r} + \frac{1}{s} = \frac{1}{5}\) \(thus >\) \(\frac{1}{5} + \frac{1}{t} = \frac{1}{4}\)
\(\frac{1}{t} = \frac{1}{4}  \frac{1}{5}\)
\(\frac{1}{t} = \frac{5}{20}  \frac{4}{20}\)
\(t = 20\)



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Re: Three printing presses, R, S, and T, working together at the
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06 Dec 2016, 08:23
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: Three printing presses, R, S, and T, working together at the
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06 Dec 2016, 08:29
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, 09:34
1/5 + 1/x = 1/4
4/20 + 1/x = 5/20
1/x = 1/20
x = 20



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