Three machines operating independently, simultaneously, and at the same constant rate can fill a certain production order in 36
hours. If one additional machine were used under the same operating conditions, in how many fewer hours of simultaneous operation could the production order be fulfilled? A. 6
B. 9
C. 12
D. 27
E. 4
Let the total production order be X. The combined rate is thus x/36 orders per hour.
Now, we know that the three machines are working at the same constant rate. This means that individual rate for each of the machines will be x/36 divided by 3 and that gives x/108 per machine.
Now, we are having another machine coming at the same constant rate. This means we are adding an x/108 rate to the preexisting x/36.
The new total rate thus becomes x/108 + x/36 = 4x/108
Now we know that the total new rate is 4x/108. Since the total work doesn’t change and it is still x, the time taken to complete a work of x orders at a rate of 4x/108 order per hour would be x divided by 4x/108 and this is x * 108/4x = 108/4 = 27 hours
