Water is filled in a container in such a manner that its volume doubles after every five minutes. If takes 30 minutes for the container to be full, in how much time will it be one-fourth full ?

Since the water doubles every 5 minutes, volume follows V(t) = V_full · 2^{(t-30)/5} if we measure t in minutes and V_full is the volume at t = 30. Set V(t) = V_full/4: 1/4 = 2^{(t-30)/5}. Because 1/4 = 2^{-2}, we have (t-30)/5 = -2, so t - 30 = -10 and t = 20 minutes. Intuitively, one-fourth is two doublings below full; each doubling is 5 minutes, so 2×5 = 10 minutes before full (30 − 10 = 20 minutes).