Question

A vessel which contains 100 litres of salt and sugar solution in the ratio of 22 : 3 . From the vessel 40 litres of mixture is taken out and 4.8 litres of pure salt solution and pure sugar solution , both are added to the mixture . What is the percentage of the quantity of sugar solution in the final mixture less than the quantity of salt solution

Answer 1

Initial amount of salt in 100 litres = $\frac{22}{25}\times 100=88litres$

So , Initial amount of sugar in 100 litres =$100-88$ $=12litres$

Final amount of salt in solution = $88-\frac{22}{25}\times 40+4.8=57.6litres$

Final amount of sugar in solution =$12-\frac{3}{25}\times 40+4.8=12litres$

So, Percentage of sugar solution in final mixture less that quantity of salt solution = $\frac{57.6-12}{57.6}\times 100=79.16$%