Three taps A, B and C can fill a tank in 20 minutes, 30 minutes and 40 minutes respectively. All three taps are opened together. After 5 minutes tap A is closed, and 6 minutes later tap B is also closed. At that moment a leak develops which, acting alone, can empty a full tank in 70 minutes. How much total time (in minutes) will it take from the start until the tank becomes completely full?

Introduction / Context:
This is a multi stage pipes and cistern problem with changing conditions. Initially three filling taps are open, then one tap is closed, then another tap is closed, and finally a leak develops while the last tap continues to fill. We must carefully track the fraction of the tank filled in each phase and then compute the remaining part under the combined action of the last tap and the leak.

Given Data / Assumptions:
- Tap A fills the tank in 20 minutes
- Tap B fills the tank in 30 minutes
- Tap C fills the tank in 40 minutes
- All taps start together at time zero
- After 5 minutes, tap A is closed
- After a further 6 minutes (that is, at 11 minutes from the start), tap B is closed
- At the moment tap B is closed, a leak starts that can empty a full tank in 70 minutes
- From 11 minutes onward, tap C and the leak act together until the tank is full

Concept / Approach:
We divide the filling into three phases: phase one with A, B and C; phase two with B and C only; and phase three with C and the leak. We compute the filled fraction in each phase using the corresponding net rates and times. The remaining fraction at the start of phase three is then filled by the net rate of C and the leak, from which we find the additional time required.

Step-by-Step Solution:
Step 1: Rates: A = 1/20, B = 1/30, C = 1/40 tank per minute. Step 2: Phase one (0 to 5 minutes): all three taps are open. Combined rate = 1/20 + 1/30 + 1/40. Step 3: LCM of 20, 30 and 40 is 120. So rates are 6/120, 4/120 and 3/120. Combined = 13/120 tank per minute. Step 4: Fraction filled in first 5 minutes = 5 * 13/120 = 65/120 = 13/24. Step 5: Phase two (5 to 11 minutes): A is closed, B and C remain open. Combined rate = 1/30 + 1/40 = (4 + 3) / 120 = 7/120 tank per minute. Step 6: Duration of phase two = 6 minutes, so additional filled fraction = 6 * 7/120 = 42/120 = 7/20. Step 7: Fraction filled by time 11 minutes = 13/24 + 7/20. LCM of 24 and 20 is 120, so 13/24 = 65/120 and 7/20 = 42/120, giving total 107/120. Step 8: Remaining fraction at 11 minutes = 1 - 107/120 = 13/120. Step 9: Phase three: only C and the leak work. Rate of leak = -1/70 tank per minute. Step 10: Net rate in phase three = 1/40 - 1/70. LCM of 40 and 70 is 280, so net rate = (7 - 4) / 280 = 3/280 tank per minute. Step 11: Let t be the time in minutes for phase three. Then t * 3/280 = 13/120. Step 12: Solve for t: t = (13/120) * (280/3) = (13 * 280) / 360 = 364 / 36 = 91/9 ≈ 10.11 minutes. Step 13: Total time from the start = 11 + 91/9 = (99/9 + 91/9) = 190/9 ≈ 21.11 minutes.

Verification / Alternative check:
We can check by approximate reasoning: about 89 percent of the tank is filled in the first 11 minutes, leaving roughly 11 percent. The net rate in phase three is slightly more than 0.01 tank per minute (3/280 ≈ 0.0107). Filling 0.1083 of the tank at this rate requires just over 10 minutes, which matches 91/9 ≈ 10.11 minutes. Adding this to the initial 11 minutes gives about 21.11 minutes overall.

Why Other Options Are Wrong:
24.315, 26.166 and 22.154 minutes: These times are all larger than the correct 21.11 minutes and would imply that the tank is more than full in the third phase when combined with the phases one and two calculations.

Common Pitfalls:
The main challenges are keeping track of the different phases and remembering when the leak starts. Some students mistakenly assume the leak acts from the beginning or confuse which taps are open during each interval. Careful phase separation, clear rate computations and consistent fractions are vital to avoid errors.

Final Answer:
The tank becomes completely full after a total of approximately 21.11 minutes from the start.