Problem: a 6 litre sol is 20 % alcohol . how many litres of pure alcohol must be added to produce a sol that is 50 % alcohol ?
Rationale: "20 % of 6 = 1.2 50 % of 6 = 3 shortage is 1.8 so we need to have 1.8 / 50 % to get 50 % alcohol content . = 3.6 d"
annotated_formula: subtract(6, multiply(const_2, multiply(divide(20, const_100), 6)))