a 6 litre sol is 35 % alcohol . how many litres of pure alcohol must be added to produce a sol that is 50 % alcohol ?
"35 % of 6 = 2.1 50 % of 6 = 3 shortage is 0.9 so we need to have 0.9 / 50 % to get 50 % alcohol content . = 1.8 d"
subtract(6, multiply(const_2, multiply(divide(35, const_100), 6)))