Two taps fill a tank in 10 minutes and 12 minutes respectively, while a third tap empties it. With the third tap closed, the two fillers together can fill the tank as stated; with all three open, the tank is filled in 15 minutes. If the two fillers are closed, how long will the third tap take to empty a full tank?

Introduction / Context:
We are given individual filling times for two taps, and the net filling time when an emptying tap also runs. From this, we back out the emptying rate and compute the emptying time alone.

Given Data / Assumptions:

• Filler A: 10 min ⇒ 1/10 tank/min.
• Filler B: 12 min ⇒ 1/12 tank/min.
• All three together fill in 15 min ⇒ net = 1/15 tank/min.

Concept / Approach:
Let E be the emptying rate (tank/min). Then 1/10 + 1/12 − E = 1/15 ⇒ E = 1/10 + 1/12 − 1/15.

Step-by-Step Solution:
Compute E: LCM(10,12,15) = 60.1/10 = 6/60; 1/12 = 5/60; 1/15 = 4/60.E = (6 + 5 − 4)/60 = 7/60 tank/min.Emptying time = 1 ÷ (7/60) = 60/7 min = 8 min 34.285… sec ≈ 8 min 34 sec.

Verification / Alternative check:
With all three: 6/60 + 5/60 − 7/60 = 4/60 = 1/15 tank/min, as given.

Why Other Options Are Wrong:
7 min and 6 min are too fast; 9 min 32 sec is too slow compared with 60/7.

Common Pitfalls:
Sign error for the outlet; arithmetic mistakes when converting minutes to seconds.

Final Answer:
8 min. and 34 sec.