a 6 litre sol is 25 % alcohol . how many litres of pure alcohol must be added to produce a sol that is 50 % alcohol ?
"25 % of 6 = 1.5 50 % of 6 = 3 shortage is 1.5 so we need to have 1.5 / 50 % to get 50 % alcohol content . = 3 d"
subtract(6, multiply(const_2, multiply(divide(25, const_100), 6)))