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A company has two types of machines, type R and type S. Oper

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A company has two types of machines, type R and type S. Oper [#permalink] New post 08 Feb 2007, 21:57
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A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?

A. 3
B. 4
C. 6
D. 9
E. 12

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-company-has-two-types-of-machines-type-r-and-type-s-130108.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 23 Aug 2013, 02:35, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: Work problem [#permalink] New post 08 Feb 2007, 22:16
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above720 wrote:
A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?


R does the job in 36 hours so in 1 hour R finishes 1/36 job
S does the job in 18 hours so in 1 hour S finishes 1/18 job

Both work togther for an hour then they can finish 1/18 (1+1/2) = 1/12 job

So both working together to finish 1 job will take 12 hours, but we need to finish the job in 2 hours that means we will 6 machine of each type.

My answer is 6
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Re: A company has two types of machines, type R and type S. [#permalink] New post 22 Aug 2013, 17:50
general formula :

(number of machine(A)/hours taken by single machine(A)) + (number of machine(B)/hours taken by single machine(B)) = reciprocal of the hours taken by two machines

hours = 1/ reciprocal

To answer the question

Let x be the number of machines

here : company uses same number of machines.

hours - 2
so reciprocal 1/2

x/36 + x/18 = 1/2

after simplification

(2x + x)/36 = 1/2

3x = 36/2


3x = 18

x = 6

so answer is 6
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Re: A company has two types of machines, type R and type S. Oper [#permalink] New post 23 Aug 2013, 02:36
Expert's post
A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?

A. 3
B. 4
C. 6
D. 9
E. 12

Rate of A - \frac{1}{36} job/hour, rate of x machines of A - \frac{1}{36}x job/hour;
Rate of B - \frac{1}{18} job/hour, rate of x machines of B - \frac{1}{18}x job/hour, (same number of each type);

Remember that we can sum the rates, hence combined rate of A and B is \frac{1}{36}x+\frac{1}{18}x=\frac{3}{36}x=\frac{1}{12}x job/hour.

We are told that together equal number (x in our case) of machines A and B can do the job (1 job) in 2 hours --> Time*Rate=2*\frac{1}{12}x=1=Job --> x=6.

Answer: C.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-company-has-two-types-of-machines-type-r-and-type-s-130108.html
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Re: A company has two types of machines, type R and type S. Oper [#permalink] New post 24 Aug 2013, 21:40
The quantity of R and quantity of S is same say Y

and total job is done in 2 hours.

Y/36 + Y/18 = Y(18+36) / (18 x 36) = 3Y/36 = Y/12

12/Y = 2 Hours, Y=6
Re: A company has two types of machines, type R and type S. Oper   [#permalink] 24 Aug 2013, 21:40
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