​Two pipes, A and B, can fill the tank in 60 hours and 90 hours, respectively. If both thepipes are opened simultaneously, in how many hours will 75% of the tank be filled?

Correct option is D

Given:

Pipe A can fill the tank in 60 hours.

Pipe B can fill the tank in 90 hours.

We need to find the time it will take for both pipes A and B to fill 75% of the tank.

Formula Used:

Time = $$\frac{\text{Portion of the tank}}{\text{Combined rate}}$$​​

Solution:

Rate of Pipe A = $$\frac{1}{60}$$​​

Rate of Pipe B = $$\frac{1}{90}$$​​

Combined rate = $$\frac{1}{60} + \frac{1}{90}$$​​

​$$= \frac{3}{180} + \frac{2}{180} = \frac{5}{180}$$​​

Thus, the combined rate of filling the tank is $$\frac{5}{180} = \frac{1}{36}$$​ of the tank per hour.

To fill 75% of the tank, the portion of the tank to be filled is 0.75

Time required = $$\frac{0.75}{\frac{1}{36}} = 0.75 \times 36 = 27 \, \text{hours}$$​ 

Alternate Solution: ​

Total work = LCM of 60, 90 = 180 

Efficiency of  Pipe A = $$\frac{180}{60} = 3$$

Efficiency of  Pipe B = $$\frac{180}{90} = 2$$​ 

Now to fill 75% of 180 :

= $$\frac{\frac{75}{100} \times 180}{5} =27$$​​