Pipes M, N, and S can fill a tank in 25, 50, and 100 minutes, respectively. Initially, pipes N and S are kept open for 10 minutes, and then pipe N is shut while pipe M is opened. Pipe S is closed 15 minutes before the tank overflows. How much time (in minutes) will it take to fill the tank if the three pipes work in this pattern?
Show answer & explanation
Correct answer: D
Explanation
The correct answer is Option 4 i.e. 27.
Pipes M, N, and S can fill a tank in 25, 50, and 100 minutes, respectively
L.C.M(25, 50, and 100) = 100
The efficiency of M = 100/25 = 4 units
The efficiency of N = 100/50 = 2 units
The efficiency of S = 100/100 = 1 unit
Initially, pipes N and S are kept open for 10 minutes,
Work done by N and S in 10 minutes = (2 + 1) × 10 = 3 × 10 = 30 units
Remaining work = 100 - 30 = 70 units
M will work for 15 minutes,
Work done by M in 15 minutes = 4 × 15 = 60 units
Remaining work = 70 - 60 = 10 units
Now, M and S work 10 units
Time taken by M and S = 10/(4 + 1) = 10/5 = 2 minutes
Thus,
Total time taken = 10 + 15 + 2 = 27 minutes