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

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Three printing presses, R, S, and T, working together at the [#permalink]

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New post 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

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: PS: Work [#permalink]

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New post 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/4-j/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 [#permalink]

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New post 04 Mar 2006, 13:35
1
=1/4-1/5=1/20

r can do 1/20 job in 1 hour
r can do the whole job in 20 hours.
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 [#permalink]

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New post 04 Mar 2006, 13:44
2
1
1/R+1/S+ 1/T = 1/4
1/S+ 1/T = 1/5
1/R=1/4-1/5
R=20
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 [#permalink]

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New post 04 Mar 2006, 18:23
same logic above 1/r =1/5-1/4 ... hence r =20 hrs
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RTD - Combined Worked Sum [#permalink]

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New post 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 [#permalink]

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New post 01 Dec 2012, 05:43
7
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: RTD - Combined Worked Sum [#permalink]

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New post 02 Dec 2012, 13:13
2
1
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 [#permalink]

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New post 17 Sep 2013, 07:26
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Re: Three printing presses, R, S, and T, working together at the [#permalink]

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New post 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 [#permalink]

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New post 23 Dec 2015, 08:50
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1
\(\frac{1}{r} + \frac{1}{s} + \frac{1}{t} = \frac{1}{4}\)

\(\frac{1}{r} + \frac{1}{s} = \frac{1}{5}\) \(thus ->\) \(\frac{1}{5} + \frac{1}{t} = \frac{1}{4}\)

\(\frac{1}{t} = \frac{1}{4} - \frac{1}{5}\)

\(\frac{1}{t} = \frac{5}{20} - \frac{4}{20}\)


\(t = 20\)
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Re: Three printing presses, R, S, and T, working together at the [#permalink]

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New post 06 Dec 2016, 08:23
1
1
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 [#permalink]

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New post 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 [#permalink]

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New post 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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Re: Three printing presses, R, S, and T, working together at the [#permalink]

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Re: Three printing presses, R, S, and T, working together at the   [#permalink] 07 Mar 2018, 05:15
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