Pipe X can fill a tank in 20 hours and Pipe Y can fill the tank in 35 hours. Both the pipes are opened on alternate hours. Pipe Y is opened first, then in how much time (in hours) the tank will be full?

Let capacity of tank = L.C.M. (20,35) = 140 litres

Pipe X can fill the tank in 20 hours, => X's efficiency = $$\frac{140}{20}=7$$ l/hr

Similarly, Y's efficiency = $$\frac{140}{35}=4$$ l/hr

Pipes Y and X are alternatively opened, => Tank filled in 2 hours = $$4+7=11$$ litres

Multiplying both sides by 12, we get => Tank filled in $$24$$ hours = $$132$$ litres

Now, its Y's turn to be opened, it will 4 more litres, => In $$25$$ hours = $$136$$ litres

Thus, the remaining tank is filled by pipe X in $$\frac{4}{7}$$ hours

=> Total time = $$25+\frac{4}{7}=\frac{179}{7}$$ hours

=> Ans - (C)