A cylindrical overhead tank is filled by two pumps - P1 and P2. P1 can fill the tank in 8 hours while P2 can fill the tank in 12 hours. There is a pipe P3 which can empty the tank in 8 hours. Both the pumps are opened simultaneously. The supervisor of the tank, before going out on a work, sets a timer to open P3 when the tank is half filled so that tank is exactly filled up by the time he is back. Due to technical fault P3 opens when tank is one third filled. If the supervisor comes back as per the plan what percent of the tank is still empty?

P1 can fill the tank in 8 hours while P2 can fill the tank in 12 hours therefore when Both the pumps are opened simultaneously they can fill $$\frac{1}{8}$$ + $$\frac{1}{12}$$ = $$\frac{5}{24}$$ this much of tank in an hour.Therefore Time required by P1 and P2 to fill full tank is $$\frac{24}{}$$ Hence they can fill half the tank in $$\frac{12}{5}$$ = 2.4 hours. After all the 3 pipes are opened the tank will be filled at the rate of $$\frac{1}{8}$$ + $$\frac{1}{12}$$ - $$\frac{1}{8}$$ = $$\frac{1}{12}$$ therefore for filling the remaining tank we will need 6 more hours. If supervisor was planning to come exactly when the tank is fully filled then he must come after 8.4 hours

 Time required by P1 and P2 to fill 1/3rd of the tank is $$\frac{24}{5*3}$$ = $$\frac{8}{5}$$ = 1.6 hours therefore supervisor will come after = 8.4 - 1.6 = 6.8 hours 

In 6.8 hours all the three pipe will fill $$\frac{6.8}{12}$$ = 0.56 of the capacity of tank

therefore remaining tank to be filled when the supervisor comes = 1- 0.56 - 0.33 $$\approx$$ 0.1 i.e. 10%

Therefore option 'C' is our answer