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.

Please refer to our theory

More on rate time distance https://gre.myprepclub.com/forum/gre-time- ... 24922.html

And work rate problems https://gre.myprepclub.com/forum/gre-work- ... 24921.html

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