GMAC Work Problem

[DCJ]

Machine A produces pencils at a constant rate of 9000 pencils per hour and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines together must produce 100000 pencils and if each machine can operate for at most 8 hours. What is the least amount of time in hours that machine B must operate?

4
4 2/3
5 1/3
6
6 1/4

Ans: 4

[mpaudena]

Machine A produces pencils at a constant rate of 9000 pencils per hour and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines together must produce 100000 pencils and if each machine can operate for at most 8 hours. What is the least amount of time in hours that machine B must operate?

4
4 2/3
5 1/3
6
6 1/4

Ans: 4

Work problems can be difficult because of the way that you have to place the figures involved. Different people do it differently. I typically put the rate of work as follows: the amount of work done as the numerator and the amount of time that that amount of work is done in as the denominator. In this case the amount of time is 1 hour in both cases so there is no need for a numerator, e.g., 9,000/1 = 9,000 and 7,000/1.

Now that is just the rate at which the work is done. The numerator is just the amount of time that the machine would take to create the amount of pencils. The numerator is not the actual amount of time that the machine will actually be in operation. The amount of time the machines will be in operation, if not given to you, is an unknown and must therefore be represented as a variable. So if we pick x as the variable representing the amount of time that machine B will be in operation at the given rate of 7,000 pencils per hour, it would be like this: (7,000/1)x.

However, in this case you are told the amount of time that the machines are permitted to work for - 8 hours. So if you want to know how much time B will have to work if A can only work for 8 hours then you've got:

(9,000/1)8 + (7,000/1)x = 100,000.

After that it's simply doing the math:

9,000*8 = 72,000;
100,000 - 72,000 = 28,000;
28,000/7,000 = 4.
So x = 4 and x is the number of hours machine B would have to operate to meet it's contribution to the total number of pencils produced (100,000).

[sanjana]

Always remember :
When u have to Maximize one quantity,minimize everything else.
When u have to minimize one quantity,maximize everything else.

Here we want to minimize the hours worked by B so maximize the hourse worked by A.
The max can be 8 Hrs as the question states.

[antondesh]

We basically have to use up all of machine A before we start using machine B.

The most we can use it for is 8 hours, so using machine A we can produce 9,000 x 8 hours = 72,000 pencils. After we produced 72,000 pencils, we now have to move up to machine B.

We still have 100,000 - 72,000 = 28,000 pencils left to produce. Machine B produces them at a rate of 7,000 per hour so it'll take 28,000 divided by 7,000 = 4 hours.

[Scott@TargetTestPrep]

Machine A produces pencils at a constant rate of 9000 pencils per hour and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines together must produce 100000 pencils and if each machine can operate for at most 8 hours. What is the least amount of time in hours that machine B must operate?

4
4 2/3
5 1/3
6
6 1/4

If machine B's operational time is to be minimized, we must maximize the time for machine A to operate. Since each machine can operate for at most 8 hours, we can let machine A operate for 8 hours. Since the rate of machine A is 9000 pencils per hour, machine A produces 8 x 9000 = 72,000 pencils, and thus 28,000 pencils are left to be produced.

Thus, it will take machine B 28,000/7,000 = 4 hours to produce the remaining pencils.

Answer: A