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?

Correct option is D

Given:

​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.

Formula Used:

Rate of work = Total Work/Time Taken

Solution:

LCM of 25, 50, and 100 = 100

So, let the capacity of tank be 100 units.

Rate of Pipe M to fill the tank = $$\frac{100}{25}$$​ = 4 units/min

Rate of Pipe N to fill the tank = $$\frac{100}{50}$$​ = 2 units/min

Rate of Pipe S to fill the tank = 1 unit/min

Let the complete time taken by all three pipes together to fill the tank be "t" minutes.

So, applying the condition,

​$$4(t-10)+2\times 10+1\times (t-15)=100$$​​

​$$4t-40+20+t-15=100$$​​

5t = 100 + 35 = 135

t = $$\frac{135}{5}=27$$​ minutes