Pipe A can fill an empty tank in 6 hours and pipe B in 8 hours. If both the pipes are opened and after 2 hours pipe A is closed. how much time B will take to fill the remaining tank?

Correct option is B

Given:

Pipe A can fill an empty tank in 6 hours and pipe B in 8 hours.

If both the pipes are opened and after 2 hours pipe A is closed.

Concept use:

Rate of work = x hr

One day rate of work = $$\frac{1}{x}$$ hr

Solution:

Pipe A can fill the tank in 6 hours. Therefore, Pipe A's rate is $$\frac{1}{6}$$​ of the tank per hour.

Pipe B can fill the tank in 8 hours. Therefore, Pipe B's rate is $$\frac{1}{8}$$​ of the tank per hour.
Combined rate of both pipes: ​$$\frac{1}{6} + \frac{1}{8}$$​

= $$\frac{4+3}{24}$$​

= $$\frac{7}{24}$$​ of the tank per hour.

​In 2 hours, both pipes fill: $$2 \times \frac{7}{24} = \frac{14}{24} = \frac{7}{12}$$​​

The tank that remains to be filled: $$1 - \frac{7}{12} = \frac{12-7}{12} = \frac{5}{12}$$

Time required for Pipe B to fill;

$$\left(\frac{5}{12}\right) \div \left(\frac{1}{8}\right)$$​​

= $$\left(\frac{5}{12}\right) \times 8$$​

= $$\frac{40}{12}$$​

= $$\frac{10}{3}$$

= $$3\frac{1}{3}$$ hr

So, $$3\frac{1}{3} hr$$​ time B will take to fill the remaining tank.

Thus, the correct answer is (b).

​Alternative method:​

7 ×2 = 14

24 - 14 = 10

B = $$\frac{10}{3}$$

B = $$3\frac{1}{3}$$ hr​