Machine A produces pencils at a constant rate of 9000

VJesus12 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 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 2/3
C. 5 1/3
D. 6
E. 6 1/4

To minimize B's time, we must maximize A's time.
Since each machine can work for at most 8 hours, let A's time = 8 hours.
Since A's rate = 9000 pencils per hour, the number of pencils produced by A in 8 hours = rt = 9000*8 = 72,000.
Remaining pencils = 100,000 - 72,000, = 28,000.
Since B's rate = 7000 pencils per hour, the time for B to produce the remaining 28,000 pencils = w/r = 28,000/7000 = 4 hours.

The correct answer is A.

VJesus12 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 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 2/3
C. 5 1/3
D. 6
E. 6 1/4

The OA is the option A .

What is the equation that I should set here in order to get the correct answer? Could anyone help me? Please.

Hello Vjesus12.

I will solve it as follows:

1- Machine A produces 9,000 pencils per hour.
2- Machine B produces 7,000 pencils per hour.
3- Both machines have to produce 100,000 pencils.
4- Each machine can operate for at most 8 hours.

(?) What is the least amount of time, in hours, that machine B must operate?

Since machine A produces more pencils per hour than machine B, then in order to find the least amount of time that machine B must operate, machine A must operate all 8 hours.

In 8 hours, machine A will produce $$9,000\cdot8=72,000.$$ Hence, machine B must produce 100,000-72,000 = 28,000 pencils.

Since the rate of the machine B is 7,000 per hour, then producing 28,000 pencils will take machine B $$\frac{28,000}{7,000}=4\ hours .$$ Hence, the correct answer is the option A .

I hope it is clear enough.