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

Question

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 hrs 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

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

Bunuel · Accepted Answer

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.

arnabghosh · Answer

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

harshavmrg · Answer

Take total work as 360

R:
work : 360
time: 36hrs
rate: 10(calculaed)

S:
Work:360
time 18hrs
rate: 20( calculated)

together

given work:360
time2 hrs
calculated: 180
rate: 180/(10+20) = 180/30 = 6 each

mbaiseasy · Answer

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

ANswer: 6 machines

mns18051985 · Answer

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?

Bunuel · Answer

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.

gauravshakya · Answer

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.

gmathopeful90 · Answer

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

Bunuel · Answer

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.

JeffTargetTestPrep · Answer

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

vitaliyGMAT · Answer

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

prajaktt · Answer

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 :

*   *   = Workdone

\frac{1}{12}  * 2 *   = 1
  = 6

Ans : C

KarishmaB · Answer

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)

totaltestprepNick · Answer

C https://www.youtube.com/watch?v=tTF9_4Q1B1A Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience tutoring@totaltestprep.net

Usernamevisible · Answer

R:S speed = 1:2 → combined = 3 units
Total time = 2 hrs → need 1 job
One pair does 3/36 = 1/12 per hour → in 2 hrs = 1/6
Need 6 such pairs → 6 R machines

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

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