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

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A company has two types of machines, type R and type S.  [#permalink] New post 31 Mar 2008, 05:54
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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 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?

a) 3

b) 4

c) 6

d) 9

e) 12




Can someone show me how to solve this problem? I'm usually good at finding the amount of work, time, and rate. But what was different about this question was that it is asking for the number of a certain type of machine. Would someone please show me how? the OA to this question is C.

Thanks
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Re: PS: Work/Rate problem [#permalink] New post 31 Mar 2008, 06:05
tarek99 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?

a) 3
b) 4
c) 6
d) 9
e) 12


In one hour, R can complete 1/36 of job and S can complete 1/18 of job
So, working together (one R and one S) for one hour, job completed = 1/36+1/18 = 3/36=1/12

Working for 2 hours (one R and one S), job completed = 2/12
So if you wanna complete the whole work in 2 hours, no. of R and S required = 12/2 =6
(This means we need 6Rs and 6Ss)

answer C
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Re: PS: Work/Rate problem [#permalink] New post 31 Mar 2008, 06:14
1 type R machine does the job in 36 hours.
In one hour type R machine will do 1/36th of job.

Similarly in 1 hour type S machine will be 1/18th of job.

Now say x number of type R as well as type S machines are used.

So In one hour x type R machine will do x/36th of job.
So In one hour x type S machine will do x/18th of job.

In 1 hour total job done = x/36 + x/18 = x/12
In two hour total job done = 2*x/12 = x/6
Since total job done in 2 hour should be 100% (which means 1 unit)
=> x/6 = 1 => x = 6

Answer C.
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Re: PS: Work/Rate problem [#permalink] New post 31 Mar 2008, 06:35
Type R does a certain job in 36 hours
Type S does the same job in 18 hours

So, combine Type R and S will do the job in T hrs.

==> (1/T) = 1/36 + 1/18
Therefore T = 12, which means working together R and S will do the job in 12 hrs.

Since the company used the same number of each type of machine to do the job in 2 hours

Therefore he needs (12/2)= 6 times of combined R and S to do the job.

Ans = C
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Re: PS: Work/Rate problem [#permalink] New post 31 Mar 2008, 06:58
abhijit_sen wrote:
1 type R machine does the job in 36 hours.
In one hour type R machine will do 1/36th of job.

Similarly in 1 hour type S machine will be 1/18th of job.

Now say x number of type R as well as type S machines are used.

So In one hour x type R machine will do x/36th of job.
So In one hour x type S machine will do x/18th of job.

In 1 hour total job done = x/36 + x/18 = x/12
In two hour total job done = 2*x/12 = x/6
Since total job done in 2 hour should be 100% (which means 1 unit)
=> x/6 = 1 => x = 6

Answer C.




thanks for the explanation, but I now have a question. By saying 6, you're basically implying that there were no machine S? This is how I looked at this problem:

work = time x rate

rate of machine R = 1/36
rate of machine S = 1/18

total work= (2 hrs * 1/36) + (2 hrs * 1/18) = should result in 1 whole work

therefore: 1/18 + 1/9 = 1 -----> 1/6 = 1

since the job was completed in 2 hours, our 1/6 should equal to 1 when we have 6 machines running in total. since the question says that the same number of each type of machine was used, i assigned the number for each type as x. therefore, 2x = 6 ---------> x = 3. what is logically wrong with this? I don't understand how 6, which seems to be the total, is rather considered the number of machine R alone. I looked at 6 as the total number because we've added everything together. doesn't make sense. help!
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Re: PS: Work/Rate problem [#permalink] New post 31 Mar 2008, 07:28
mnjoosub wrote:
Type R does a certain job in 36 hours
Type S does the same job in 18 hours

So, combine Type R and S will do the job in T hrs.

==> (1/T) = 1/36 + 1/18
Therefore T = 12, which means working together R and S will do the job in 12 hrs.

Since the company used the same number of each type of machine to do the job in 2 hours

Therefore he needs (12/2)= 6 times of combined R and S to do the job.

Ans = C



that helped a lot! thank you man and also thanks to everyone for your great contribution.
really appreciate it!
Re: PS: Work/Rate problem   [#permalink] 31 Mar 2008, 07:28
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