Solution X contains 20% of material A and 80% of material B. Solution Y contains 30% of material A and 70% of material B. A mixture of both these solutions contains 22% of material A in the final product. how much solution X is present in the mixture?  Choose the most appropriate option
A) 40% B) 60% C) 80% D) 100% E) 110%
We can assume the total weight of the mixture= 100   Conc of A in the final mixture= 22   Let weight of A in the mixture be X. Conc given= 20%=.2   Therefore, weight of B= 100-X. Conc given = 30%= .3   Now, acc to the problem, .2X+ .3(100-X)=22   Solving, we get X=80. Since we assumed the weight of the mixture = 100, Therefore presence of A in the mixture= 80%.   ANSWER:80%
<result>C</result>
Solution X contains 20% of material A and 80% of material B. Solution Y contains 30% of material A and 70% of material B. A mixture of both these solutions contains 22% of material A in the final product. how much solution X is present in the mixture?