Machine A produces pencils at a constant rate of 9,000 pencils per hou

Machine A produces pencils at a constant rate of 9,000 pencils per hour, and machine B produces pencils at a constant rate of 7,000 pencils per hour. If the two machines together must produce 100,000 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?

a) 4
b) 4 and 2/3
c) 5 and 1/3
d) 6
e) 6 and 1/4

To minimize the time that machine B must operate we must maximize the time machine A can operate, so make it operate 8 hours. In 8 hours machine A will produce \(8*9,000 = 72,000\) pencils, so \(100,000 - 72,000 = 28,000\) pencils are left to produce, which can be produced by machine B in \(\frac{28,000}{7,000}=4\) hours.

Answer: A.