Problem: a 6 litre sol is 40 % alcohol . how many litres of pure alcohol must be added to produce a sol that is 50 % alcohol ?
Rationale: "40 % of 6 = 2.4 50 % of 6 = 3 shortage is 0.6 so we need to have 0.6 / 50 % to get 50 % alcohol content . = 1.2 b"
annotated_formula: subtract(6, multiply(const_2, multiply(divide(40, const_100), 6)))