Pipe A can fill a tank in 9 minutes, while pipe B can empty the completely filled tank in 10 minutes. Initially, pipe A is opened and after 3 minutes pipe B is also opened. In how  much time (in minutes) will the remaining tank be filled completely?

Correct option is B

Given:

Pipe A fills the tank in 9 min 

rate $$= \frac{1}{9}$$​ tank/min.

Pipe B empties the tank in 10 min

rate $$= \frac{1}{10}$$​ tank/min.

A runs alone for 3 min, then A and B run together.

Formula Used:

Work = Rate × Time

Solution:

In first 3 min, A fills: 3$$\times \frac{1}{9} = \frac{1}{3}$$​ of the tank.
Remaining$$= 1 - \frac{1}{3} = \frac{2}{3}$$​

When both are open, net rate =$$\frac{1}{9} - \frac{1}{10} = \frac{1}{90}$$​ tank/min.

Time to fill the remaining $$\frac{2}{3}$$​ tank
​$$t=\frac{\frac{2}{3}}{\frac{1}{90}}=\frac{2}{3}\times 90=60\ \text{minutes}.$$​​

Alternate Method:
Assume tank capacity = LCM(9,10) = 90 units.
A fills 10 units/min; B empties 9 units/min.
In 3 min, A fills 3$$\times$$​10=30 units → 60 units left.
Together net = 10 - 9 = 1 unit/min → time =60/1=60 min.