Drill Press M can bore 3000 holes an hour and Drill Press...

LUANDATO wrote:Drill Press M can bore 3000 holes an hour and Drill Press N can bore 2500 holes an hour. If the two machines together must bore 45,000 holes, and if each machine can operate for at most 9 hours, what is the least amount of time, in hours, Drill Press N must operate?

A. 6
B. 7
C. 36/7
D. 8
E. 90/11

The OA is C.

I'm really confused by this PS question. Experts, any suggestion about how to solve it? Thanks in advance.

Hello LUANDATO.

There is a mistake with the option C. The correct answer should be 36/5. Let's see why.

We know that Drill Press M has to operate for 9 hours (in this way, the number of hours that has to operate Drill Press N will be minimum). Hence, Drill Press M will bore 3000*9 = 27.000 holes.

Now, Drill Press N has to bore 45.000-27.000 = 18.000 holes.

The time that Drill Press N has to operate is 18.000 / 2.500 = 180 / 25 = 36/5 hours.

This is why the correct answer is 36/5 hours.

I hope this explanation may help you.

Feel free to ask me again if you have a doubt.

Regards.