Two inlet pipes A and B can fill a cistern in 4 hours and 6 hours, respectively. When full, an outlet pipe C can empty it in 8 hours. If all three are opened together from empty, in how much time will the cistern be filled?

Introduction / Context:
Here, two inlets add water and one outlet removes it. The net rate is the sum of inlet rates minus the outlet rate. Once the net rate is found, invert it to get the total filling time.

Given Data / Assumptions:

• A fills in 4 h ⇒ 1/4 tank/h.
• B fills in 6 h ⇒ 1/6 tank/h.
• C empties in 8 h ⇒ 1/8 tank/h (outflow).

Concept / Approach:
Net rate = 1/4 + 1/6 − 1/8.

Step-by-Step Solution:
LCM(4,6,8) = 24 ⇒ 1/4 = 6/24, 1/6 = 4/24, 1/8 = 3/24.Net rate = (6 + 4 − 3)/24 = 7/24 tank/h.Time = 1 ÷ (7/24) = 24/7 h ≈ 3.4286 h = 3 hours 25.7 minutes ≈ 3 hours 26 minutes.

Verification / Alternative check:
In 24/7 hours, volume added by A and B is (6/24 + 4/24)*(24/7) = 10/7; removed by C is (3/24)*(24/7) = 3/7; net = 1 tank.

Why Other Options Are Wrong:
3:18 and 3:42 are symmetric around the right value but not exact; 3:48 is too slow.

Common Pitfalls:
Adding or subtracting times directly; rounding too early and missing the exact 24/7 h.

Final Answer:
3 hours 26 min.