a cistern is normally filled in 6 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 / 6 - 1 / x = 1 / 8 , solving 1 / x = 1 / 24 so in 24 hours full cistern will be empty . answer : a"
inverse(subtract(divide(const_1, 6), divide(const_1, add(6, 2))))