THIS IS THE AOPS ANSWER SO ITS NOT FROM ME!!!
(a) Let be the number of mL of Solution A that is added. So, the total amount of combined solution x + 500 is mL. Since Solution B is 30% acid, the 500 mL of Solution B has (0.3)(500) = 150 mL of acid. Since Solution A is 80% acid, the x mL of Solution A has 0.8x mL of acid. Therefore, the combined solution has 0.8x + 150 mL of acid. This must be 70% of the whole solution, so we have
Expanding the right-hand side gives 0.8x + 150 = 0.7(x + 500) , so 0.1x = 200 and x-200/0.1 . Therefore, we must add \(\boxed{2000 \text{ mL}}\) of Solution A.
Note that we need four times as much of Solution A as Solution B. Could we have figured that out just by comparing the percentages of acid in the two original solutions to the desired percentage of acid in the final solution?
(b) Let and be the desired amounts of Solution A and B, respectively, in mL. Since we want a total of 100 mL, x+y=100 . Solution A contributes 0.8x mL of acid, and Solution B contributes 0.3y mL of acid, so 0.8x + 0.3y=50.
Multiplying the equation x+y=100 by 0.3, we get 0.3x + 0.3y = 30. Subtracting this equation from the equation 0.8x + 0.3y = 50 , we get 0.5x=20 , so x=40 . Then 0.8x + 0.3y = 50. Therefore, we need to combine \(\boxed{40 \text{ mL}}\) of Solution A and \(\boxed{60 \text{ mL}}\)of Solution B.
(c) To see if we can create a combination of Solution A and Solution B that is 90% acid, we proceed the same way as in part (a). Let V be the volume of solution that is 90% acid, and let and be the desired amounts of Solution A and B, respectively, all in mL.
Then x + y = V , and as in part (a), Solution A contributes 0.8x mL of acid, and Solution B contributes 0.3y mL of acid, so 0.8x + 0.3y = 0.9V. Multiplying the equation x + y = V by 0.3, we get 0.3x + 0.3y = 0.3V . Subtracting this equation from the equation 0.8x + 0.3y = 0.9V , we get 0.5x = 0.6V, so x = 1.2V. Then y = V - x = -0.2V .
Since is y negative, there is\(\boxed{\text{no combination}}\) of Solution A and Solution B that is 90% acid.
+140 