A tap can fill a tank in 16 minutes while another tap can empty it in 8 minutes. If the tank is already half full and both taps are opened together, what happens and in how much time?

Introduction / Context:
Competing taps—one filling and one emptying—yield a net rate equal to the difference of their individual rates. Starting from half full, we determine whether the tank fills or empties and the time required.

Given Data / Assumptions:

• Fill tap: 16 minutes ⇒ +1/16 tank/min.
• Empty tap: 8 minutes ⇒ −1/8 tank/min.
• Initial content = 1/2 tank.

Concept / Approach:
Net rate = 1/16 − 1/8 = −1/16 tank/min (negative ⇒ net emptying). Time to go from 1/2 to 0 at this rate is (1/2) / (1/16).

Step-by-Step Solution:
Net rate = 1/16 − 1/8 = 1/16 − 2/16 = −1/16 tank/min.Volume to empty = 1/2 tank.Time = (1/2) ÷ (1/16) = 8 minutes.

Verification / Alternative check:
In 8 minutes, the empty tap removes 8 * (1/8) = 1 tank; the fill tap adds 8 * (1/16) = 1/2 tank; net change = −1/2, taking the level from 1/2 to 0.

Why Other Options Are Wrong:
Filled options are impossible since the net rate is negative; 12 minutes overestimates time.

Common Pitfalls:
Adding times rather than rates; forgetting the starting level is half, not empty.

Final Answer:
Emptied in 8 min.