DCJ wrote: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).
