a cistern is normally filled in 2 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 / 2 - 1 / x = 1 / 4 , solving 1 / x = 1 / 4 so in 4 hours full cistern will be empty . answer : a"
inverse(subtract(divide(const_1, 2), divide(const_1, add(2, 2))))