a 6 litre sol is 30 % alcohol . how many litres of pure alcohol must be added to produce a sol that is 50 % alcohol ?
30 % of 6 = 1.8 50 % of 6 = 3 shortage is 1.2 so we need to have 1.2 / 50 % to get 50 % alcohol content . = 2.4 e
subtract(6, multiply(const_2, multiply(divide(30, const_100), 6)))