a cistern is normally filled in 4 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 / 4 - 1 / x = 1 / 6 , solving 1 / x = 1 / 12 so in 12 hours full cistern will be empty . answer : e"
inverse(subtract(divide(const_1, 4), divide(const_1, add(4, 2))))