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SVP  Joined: 14 Dec 2004
Posts: 1542
Three printing presses, R, S, and T, working together at the  [#permalink]

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4
29 00:00

Difficulty:   5% (low)

Question Stats: 90% (01:21) correct 10% (02:16) wrong based on 1672 sessions

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

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6
3
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  [#permalink]

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13
2
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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Intern  Joined: 08 Jan 2006
Posts: 26
Re: PS: Work  [#permalink]

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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

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.
VP  Joined: 29 Dec 2005
Posts: 1240
Re: PS: Work  [#permalink]

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2
=1/4-1/5=1/20

r can do 1/20 job in 1 hour
r can do the whole job in 20 hours.
Senior Manager  Joined: 22 Jun 2005
Posts: 332
Location: London

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2
1
1/R+1/S+ 1/T = 1/4
1/S+ 1/T = 1/5
1/R=1/4-1/5
R=20
Manager  Joined: 13 Dec 2005
Posts: 214
Location: Milwaukee,WI

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same logic above 1/r =1/5-1/4 ... hence r =20 hrs
Intern  Joined: 27 Aug 2012
Posts: 14
RTD - Combined Worked Sum  [#permalink]

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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

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.
Intern  Joined: 27 Nov 2012
Posts: 36
Re: RTD - Combined Worked Sum  [#permalink]

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3
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
Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re: Three printing presses, R, S, and T, working together at the  [#permalink]

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Merging similar topics.
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Re: Three printing presses, R, S, and T, working together at the  [#permalink]

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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!
Manager  Joined: 03 Jan 2015
Posts: 72
Re: Three printing presses, R, S, and T, working together at the  [#permalink]

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4
2
$$\frac{1}{r} + \frac{1}{s} + \frac{1}{t} = \frac{1}{4}$$

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

$$\frac{1}{t} = \frac{1}{4} - \frac{1}{5}$$

$$\frac{1}{t} = \frac{5}{20} - \frac{4}{20}$$

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

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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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1/5 + 1/x = 1/4

4/20 + 1/x = 5/20

1/x = 1/20

x = 20
Intern  B
Joined: 28 Apr 2016
Posts: 42
Location: United States
GMAT 1: 780 Q51 V47 GPA: 3.9
Re: Three printing presses, R, S, and T, working together at the  [#permalink]

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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 T
Like 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 job
Because we’re talking about a single printing job, we just set W = 1 for each scenario.

Add rates when they are simultaneous
Let’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 job
Since 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 substitution
While 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  [#permalink]

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Solved in 10 seconds without writing
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Re: Three printing presses, R, S, and T, working together at the  [#permalink]

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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   [#permalink] 20 Oct 2018, 14:44
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