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

Machine R does a job in 36 Hours. Machine S in 18 Hours.
Together, the machines R & S can finish the job in \(\frac{(36*18)}{(36+18)}= 12 Hours\)
In other words,
1 machine of R + 1 machine of S can finish the job together in = 12 Hours

So, to finish the job in 2 Hours, company needs 6 machines of each type - Option C

\(Let N be the number of machines...\)
\(\frac{N}{36}+\frac{N}{18}(2hours)=1\)
\(\frac{3N}{18}=1==>N=6\)

ANswer: 6 machines

Could someone please clarify why the answer is not 3 each? I understand how to get to 6 machines and do it quickly but why isn't this 6 machines total since it is a combined rate?

x in the solution HERE is the number of type R machines as well as the number of type S machines (we are told that the company used the same number of each type of machine to do the job ). So, when we solve for x, we get what we want.

R finishes job in 36h and S does the same in 18h. So if both work together, the job will be done in 12h:
1/36 + 1/18 will give you 1/12.
Now if one pair of R & S finishes job in 12h, how many such pairs are needed to finish the same job in 2h?
12/2 = 6 pairs of R & S. Hence 6 of R machines are required.

Doubt....

1 machine A and 1 machine B together do 1/12 of the job in 1 hour.

We need to do 6/12 of the job in 1 hour...
So 6 A machines and 6 B machines will be needed for 6/12 job in 1 hour...
and 12 A and B machines will do 12/12 of the job in 2 hours...

Shouldn't the answer be E)12

6 A's and 6 B's do half of the job in 1 hour, so 6 A's and 6 B's will do the whole job in 2 hour and this is what we are asked to find.

We have a combined worker problem for which we can use the following formula:

work (1 machine) + work (2 machine) = total work completed

Since we are completing one job, we can say:

work (1 machine) + work (2 machine) = 1

We are given that 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.

Thus, the rates for the two machines are as follows:

rate of machine R = 1/36

rate of machine S = 1/18

We are also given that the company used the same number of each type of machine to do the job in 2 hours. If we let x = the number of each machine used, we can multiply each rate by x and we have:

rate of x number of R machines = x/36

rate of x number of S machines = x/18

Finally, since we know some number of R and S machines worked for two hours, and since work = rate x time, we can calculate the work done by each type of machine.

work done by x number of R machines = 2x/36 = x/18

work done by x number of S machines = 2x/18 = x/9

Now we can determine x using the combined worker formula:

work (machine R) + work (machine S) = 1

x/18 + x/9 = 1

x/18 + 2x/18 = 1

3x/18 = 1

x/6 = 1

x = 6

Answer: C

A company has two types of machines, Type R and Type S. Operating at a certain 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 uses the same number of each type of machine to do the job in 2 hours, how many machines of Type R were used?

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

Can anybody give me a quickest and simpler way to solve this? I took almost 3.5 min on this question to solve during a mock test. Still ended up guessing as I wasn't sure of the approach.
Thanks.

We are given rates of each machine and that the number of machines of each type being used is the same = x.

\(\frac{x}{36} + \frac{x}{18} = \frac{1}{2}\)

\(\frac{3x}{36} = \frac{1}{2}\)

\(6x = 36\)

\(x = 6\)

Rate(R) = \(\frac{1}{36}\)
Rate(S) = \(\frac{1}{18}\)
Combined rate of both machines ,
Rate(RS)=Rate(R) + Rate(S) = \(\frac{1}{12}\)

we have given time , T = 2hrs , Workdone = 1 (for single job)
Plug in the values :

[Rate] * [Time] * [No. of machines] = Workdone

\(\frac{1}{12}\) * 2 * [No. of machines] = 1
[No. of machines] = 6

Ans : C

Why is the total of Machine R used 6 and not 3?

I understand how to solve using the question Work = Rate x Time x Number.

I solved for each rate and then combined the rates to get Number = 6.

The questions states that the same number of Machine S and R is used. Why is 6 the number of one machine and not the total for both since we solved using combined rates of both machines?

1 = 1/36 + 1/18 x (2) x (N)

Please help!!

Work done by machine R in 1 hr = 1/36
Work done by machine S in 1 hr = 1/18

If both work together, work done in 1 hr = 1/36 + 1/18 = 1/12 (this is the work done by a pair, R and S in 1 hr)

But we need them to do 1/2 work in 1 hr. So we must have 6 pairs (R and S) of them.

So 6 R machines were used and 6 S machines were used.

Answer (C)

myhigheredmba

VeritasKarishma - Oh, I see! It's because the 6 represents pairs!! Thank you so much!

EducationAisle GMATBusters Since machine R does some work in 36hrs while machine S does the same work in 18hrs (i.e. half the time) we can infer that the rate of machine S is twice the rate of machine R.

In other words, we can say that 1R is equivalent to 2S (1R = 2S) Can we leverage this and solve the question without doing much calculation?

You can do as follows :
let the work done by R per hour = x
that work done by S per hour = 2x

Now As R was able to do the work in 36 hrs: total work to be done = 36x

Now, one R and one S can do 2*(x+2x) in 2 hrs = 6x work in 2 hrs
we need to do 36x work done, hence we need 6 Nos of each machine R and S

Let the total work be 36 Units

Efficiency of machine R is 2 Units/Hr ; Efficiency of machine S is 1 Units/Hr

\(2 *n*(2 + 1) = 36\) ; n = no of machines of R and S type....

Thus, \(n = 6\), hence Answer must be (C) 6