a cistern is normally filled in 10 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 / 10 - 1 / x = 1 / 12 , solving 1 / x = 1 / 60 so in 60 hours full cistern will be empty . answer : c"
inverse(subtract(divide(const_1, 10), divide(const_1, add(10, 2))))