One Machine of type R's one hour work is (1/36)
One Machine of type S's one hour work is (1/18)
If X no.of Machines of both type were used ,i.e X no of R machines and X no of S machines ....then
(1/2) = X [(1/36)+(1/18 )]
Gives X = 6
So 6 machines were used.
Wats the OA ???
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Source: Beat The GMAT — Problem Solving |
Just to restate the question asked:
A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R can complete a 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
The answer is C) 6
I'm not quite sure why you have 1/2 on the left side of the equation.
Do you mind explaining your thought process again please? Are there alternative approaches to this problem?
thx
A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R can complete a 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
The answer is C) 6
I'm not quite sure why you have 1/2 on the left side of the equation.
Do you mind explaining your thought process again please? Are there alternative approaches to this problem?
thx
Yes should be C
1 hr work of machine A = 1/36
1hr wrk of machine B = 1/18
2 hr work of machine A = 2/36
2 hr work of machine B = 2/18
let x number of machine are used
x*2/36 +x*2/18 = 1
so x= 6
1 hr work of machine A = 1/36
1hr wrk of machine B = 1/18
2 hr work of machine A = 2/36
2 hr work of machine B = 2/18
let x number of machine are used
x*2/36 +x*2/18 = 1
so x= 6
Another way of looking at this problem is if you plug in a number for the value of work. (In some previous posts, users have used the number of pizzas..etc)
So lets say that the value of work is 36 units. (The trick to using this method is to use a number that will be evenly divisible by both times. In this case 36 units is divisible by both 36 and 18.) Thus the equation would look like this:
For Machine R:
Work = 36 units
Time = 36 hours
Rate = 1 unit per hour
For Machine S:
Work = 36 units
Time = 18 hours
Rate = 2 units per hour
Together they are producing 3 unitss per hours. If the job (36 units) is to be completed in 2 hours, it means that 18 units will need to be produced per hour (36/2 hours).
If 18 units need to be produced per hour, you will need:
18 units/3 units per hour for both machines = 6.
Thus you will need 6 machines of each.
So lets say that the value of work is 36 units. (The trick to using this method is to use a number that will be evenly divisible by both times. In this case 36 units is divisible by both 36 and 18.) Thus the equation would look like this:
For Machine R:
Work = 36 units
Time = 36 hours
Rate = 1 unit per hour
For Machine S:
Work = 36 units
Time = 18 hours
Rate = 2 units per hour
Together they are producing 3 unitss per hours. If the job (36 units) is to be completed in 2 hours, it means that 18 units will need to be produced per hour (36/2 hours).
If 18 units need to be produced per hour, you will need:
18 units/3 units per hour for both machines = 6.
Thus you will need 6 machines of each.
Oops...i cal exactly the way in the last post...
just made error in last.
i cal 6 machines in 1 hr.So, 12 in 2 hrs...
...
Extra smart ????? Excess of everything is back???/
just made error in last.
i cal 6 machines in 1 hr.So, 12 in 2 hrs...
...Extra smart ????? Excess of everything is back???/
In the question, "How many machines of Type R were used?" do specify "Type R" to throw you off? The problem already says the same number of each type of machine was used so "Type R" and "Type S" should be the same, right? Essentially isn't this question asking "What is x"?
