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Algebraic equation - machines

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by szDave » Thu Jan 24, 2013 6:49 am
hello,

I can pick numbers to solve this, but can you show me the proper equation?

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?

Answer: 6
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by Brent@GMATPrepNow » Thu Jan 24, 2013 7:17 am
szDave wrote:hello,

I can pick numbers to solve this, but can you show me the proper equation?

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?

Answer: 6
Aside from picking numbers (e.g., letting the entire job be 36 units), we can it as follows:

First, when it comes to questions where we must complete an entire job, I often (not always) like to know what can be accomplished in 1 unit of time (in this case, 1 hour).

Machine R can complete 1/36 of the job in 1 hour.
Machine S can complete 1/18 of the job in 1 hour.
Since 1/36 + 1/18 = 1/12, we know that, combined, machines R and S can complete 1/12 of the job in 1 hour.

From here we can apply some logic.
If 1/12 of the job is completed in 1 hour (with 1 R machine and 1 S machine), then we could complete the entire job in 1 hour if we had 12 of each machine type.
However, the question asks us to find the # of machines required to complete the job in 2 hours. So, we need half as many machines. In other words, we need 6 of each machine.

Answer = 6

Cheers,
Brent
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by Brent@GMATPrepNow » Thu Jan 24, 2013 7:23 am
szDave 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?
Alternatively, we can assign a "nice" value to the job.
Say, the job is to make 36 widgets.

R does a certain job in 36 hours
This means that machine R's rate is 1 widget/hour

S does the job in 18 hours
This means that machine S's rate is 2 widgets/hour

So, their combined rate is 3 widgets/hour.


The question asks us to complete the job in 2 hours.
To make 36 widgets in 2 hours, the combined rate of the two machines must be 18 widgets/per hour.
If the combined rate (of 1 R machine and 1 S machine) is 3 widgets/hour, then we'd need 6 of each machine to reach a rate of 18 widgets/hour.

So, the answer is 6

Cheers,
Brent
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by GMATGuruNY » Thu Jan 24, 2013 7:24 am
szDave wrote:hello,

I can pick numbers to solve this, but can you show me the proper equation?

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?

Answer: 6
GmatKiss 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
Let the job = 36 units.
Rate for machine R = 36/36 = 1 unit per hour.
Rate for machine S = 36/18 = 2 units per hour.
To complete the job in 2 hours, the number of units produced each hour = 36/2 = 18 units.
Since using one of each machine will produce 1+2 = 3 units per hour, and 18 units must be produced, we need 18/3 = 6 of each machine.

The correct answer is C.

Algebraically:
Let the job = 1.
Rate for machine R = w/t = 1/36.
Rate for machine S = w/t = 1/18.
Combined rate for machines R and S = 1/36 + 1/18 = 3/36 = 1/12.

Let x = the number of each machine.
Time for the job = 2 hours.

(number of machines)(rate)(time) = work.
Thus:
x(1/12)(2) = 1.
x(1/6) = 1
x = 6.

Plugging in values seems much easier.
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by Lifetron » Sun Jan 27, 2013 9:41 pm
2*n*((1/18)+(1/36)) = 1
2*n*((2+1)/36) = 1
2*n*(1/12) =1
(n*1)/6 = 1
Hence, n=6 !
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