Could please somebody explain solution of the problem?

A company has two types of machines, type R and type S. Operating as a constant rate, a machine 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 machines 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

The correct answer is C. 6.

Thanks!

Machine R can do 1 job / 36hrs. S does 1 job / 18 hrs. We want the total to be 1 job / 2hrs

So this is the equation where x represents the number of machines you need:

rate of R + rate of S = rate of total
x/36 + x/18 = 1/2
(multiply all by 36)
x + 2x = 18
3x = 18
x = 6 machines

That's it.

There are some ways to streamline this problem - such as using a 'rate chart' with (work rate) x (time) = (total work done) - but the essence of the problem will still be exactly what's here.

I have it is saying how many type of R machine used. I presume 6 means ( same no. of S + same no. of R)


so total number of R machine should be 3. Please correct if I am wrong

Let me correct I post wrong 3 Answer is 6

It is like this

In 12 hours R + S can finish a job

so in 1 hours the set finish 1/12 of the work

in 2hours the set (R + S) finish 1/6 of the work

so inorder to finish full work we need 6 sets of (R+S).
Hence R machine needed R=6

Yep - if you look at the setup of the algebra done earlier in the post, you'll notice that the equation includes 'x' of machine R *AND* 'x' of machine S.

When I set this problem up I had it like this: x/36 + y/18 = 1/2. Where x and y where the number of each type of machine.

If the company used the same number of machines to do the job in 2 hours, how many machines of type R were used?

Where it says same number of machines, can that be interpreted as the same number of machine r as the number of machine s?

Kind of crucial wording that totally screwed me up.[/b][/i]

I'm assuming that was a typo - a 'streamlined' version of the original question that obfuscates its original meaning. Interestingly, I didn't even notice that when I responded to this thread earlier.

Moreover, that's what the problem has to mean; otherwise, there are many combinations of machines that will solve the problem, as machine S is exactly twice as fast as machine R: we could just take away one machine S and replace it with two R's (or vice versa), and the new combination would also work. Since this isn't data sufficiency (i.e., we need a UNIQUE solution), that won't do.

------------1 machines-----1 machines
Rate-----------Rr---------------Sr----------

Time --------36----------------18---------(hrs)

Work--------1 job-----------1 job

W=rt

1*36*Rr=1*18*Sr

Sr=2*Rr

------------x machines-----x machines -----(Same machines)
Rate-----------Rr---------------Sr----------

Time --------2----------------2---------(hrs)

Work--------1 job-----Total work done by both machines

Work=2*Rr*x+2*Sr*x
=2x(Rr+Sr)
=2x(3Rr) ---Since Sr=2Rr
=6xRr

6xRr=36Rr ---job done by Rr from step 1

 x=6

assume the task is to lay bricks so R - 1 brick per hour =36 hours : s 2 bricks per hour = 18 hours the ratio is 2:1

hence inorder to do the job in 2 hours need to lay 18 bvricks per hour in the ratio 1:2 hence the answer is 6

nice wealth of solutions to this problem.
sweetness

m seein complex solutions here :)
in one hour machine a does 1/36 B does 1/ 18
one hour combined 1/36 + 1/18 = 1/12 ----> 1 hour
2 hours amount of wok done = 2/12 == 1/ 6

since num of machines is the same it shud be 6 .....

this is a nice compact solution.

next time, please write out words in full, so that your posts are easier to read. thank you.