a cistern is normally filled in 12 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 / 12 - 1 / x = 1 / 14 , solving 1 / x = 1 / 84 so in 84 hours full cistern will be empty . answer : b"
inverse(subtract(divide(const_1, 12), divide(const_1, add(12, 2))))