A tank can be filled with water by two pipes A and B together in 36 minutes. If the pipe B was stopped after 30 minutes, the tank is filled in 40 minutes. The pipe B can alone fill the tank in:

Solution(By Examveda Team)

Let the pipes fill the tank in x minutes.
Part of tank filled by pipes A and B in one minute: $$\frac{1}{{36}}$$
Part of the tank filled by pipe A in 1 minute = $$\frac{1}{{36}} - \frac{1}{x}$$

According to the question,

$$\eqalign{ & 30 \times \frac{1}{{\text{x}}} + 40 \left( {\frac{1}{{36}} - \frac{1}{{\text{x}}}} \right) = 1 \cr & \Rightarrow \frac{{30}}{{\text{x}}} + \frac{{40}}{36} - \frac{{40}}{{\text{x}}} = 1 \cr & \Rightarrow \frac{{30}}{{\text{x}}} + \frac{{10}}{9} - \frac{{40}}{{\text{x}}} = 1 \cr & \Rightarrow \frac{{10}}{{\text{x}}} = \frac{1}{9} \cr} $$
Hence, x = 90 minutes.

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Alternatively,
% of tank filled by (A+B) in 1 minute = $$\frac{{100}}{{36}}$$ = 2.78%
% of tank filled in 30 minutes = 2.78 × 30 = 83%
That means 17% of tank can be filled by A in last 10 minutes
Then, % tank filled by A in 1 minute = $$\frac{{17}}{{10}}$$ = 1.7%
% of tank filled by B in 1 minute = 2.78 - 1.7 = 1.1%
So, B alone can fill 100% tank in,
= $$\frac{{100}}{{1.1}}$$ = 90 minutes (approx.)