a cistern is normally filled in 8 hrs , but takes 2 hrs longer to fill because of a leak on its bottom , if cistern is full , how much time citern would empty ?
if leakage / hour = 1 / x , then 1 / 8 - 1 / x = 1 / 10 , solving 1 / x = 1 / 40 so in 40 hours full cistern will be empty . answer : d
inverse(subtract(divide(const_1, 8), divide(const_1, add(8, 2))))