Two pipes A and B can fill a tank in 20 and 30 hours, respectively. Both pipes are opened to fill the tank, but when the tank is one-third full, a leak develops through which one-fourth of the water supplied by both pipes goes out. Find the total time (in hours) taken to fill the tank.

Pipe A fills the tank in 20 hours.

Pipe B fills the tank in 30 hours.

Both pipes are opened together.

When the tank is one-third full, a leak develops that causes one-fourth of the water to be lost.  

Formula Used: 

Total Work = Efficiency $$\times$$ Time

Solution:   

LCM of 20, 30 = 60 

let the total capacity of the tank = 60  

Efficiency of Pipe A = $$\frac{60}{20} = 3 \ unit/h$$ 

Efficiency of Pipe B = $$\frac{60}{30} = 2 \ unit/h$$​​

So,  Time to fill one third of the tank, i.e $$\frac{60}{3} = 20$$ 

Now, Time to fill 20 unit of the tank , 

 = $$\frac{20}{5} = 4 \ hour$$ 

Now,

After the leak develops, the effective water supply is reduced by one-fourth, meaning three-fourths of the water remains in the tank. The new effective rate becomes $$\frac{3}{4}$$ of the combined rate

So, $$\frac{3}{4} \times 5 = \frac{15}{4}$$​

Now, remaining tank to be filled = $$\frac{40}{\frac{15}{4}} = \frac{32}{3} \ hour$$ 

Total Time = $$4 + \frac{32}{3} = \frac{44}{3} =\bf 14\frac{2}{3} \ hours$$​​