Pipes A, B and C can fill a tank in 30 hours, 60 hours and 90 hours,…

To solve this problem efficiently, we can use the LCM (Least Common Multiple) method to determine the total capacity of the tank and the individual efficiencies of each pipe.

Assume Total Capacity:
Find the LCM of the given times (30, 60, and 90 hours).
LCM(30, 60, 90) = 180 units.
Let the total capacity of the tank be 180 units.

Calculate Individual Efficiencies (Units filled per hour):

Efficiency of Pipe A = 180 / 30 = 6 units/hour.

Efficiency of Pipe B = 180 / 60 = 3 units/hour.

Efficiency of Pipe C = 180 / 90 = 2 units/hour.

Determine Cycle Output:
The pipes operate alternatively for one hour each, in the sequence: C -> B -> A.

1st hour (Pipe C): 2 units

2nd hour (Pipe B): 3 units

3rd hour (Pipe A): 6 units

Total work done in one 3-hour cycle = 2 + 3 + 6 = 11 units.

Calculate Total Full Cycles:
Divide the total capacity by the work done in one cycle to find out how many complete cycles can occur without overfilling the tank.
180 / 11 = 16 complete cycles (with a remainder).

Time for 16 cycles = 16 × 3 = 48 hours.

Work completed in 48 hours = 16 × 11 = 176 units.

Calculate Time for Remaining Work:

Remaining capacity = 180 - 176 = 4 units.

49th hour: The cycle restarts with Pipe C. Pipe C fills 2 units.
Remaining capacity = 4 - 2 = 2 units.

50th hour: It is now Pipe B's turn. Pipe B fills at a rate of 3 units/hour.
Time required for Pipe B to fill the remaining 2 units = 2/3 hours.

Find Total Time:
Total time = 48 hours + 1 hour (Pipe C) + 2/3 hour (Pipe B)
Total time = 49 2/3 hours.