izzypop4 wrote:a chemist needs 10 liters of a 25% acid solution. The solution is to be mixed from three solutions whose concentrations are 10% ,20% ,and 50%. how many liters of each solution should the chemist use to satisfy the following?

1. use as little as possible of the 50% solution.

Ten liters of 25% solution will have (0.25)(10) = 2.5 liters of actual acid. Using some logic (and the standard setup for

**mixture-type** word problems), we can see that, if you're to use as little of the 50% solution as possible, then you'll need to use the 20% solution for the rest (since using the 10% solution would only increase your need for a "balancing" amount of 50% solution). So let's label things:

. . . . .liters 20% solution: x

. . . . .liters 50% solution: 10 - x

. . . . .acid in 20% solution: (0.20)(x) = 0.2x

. . . . .acid in 50% solution: (0.50)(10 - x) = 5 - 0.5x

You know that these input amounts of acid have to sum to the known output of 2.5 liters of actual acid. So sum the input amounts, set equal to the output amount, and solve.

izzypop4 wrote:2. use as much as possible of the 50% solution

Use the same sort of reasoning for this part. Of course, since you're wanting to

*maximize* the amount of 50% solution used, this time you'll want to balance it with the 10% solution, since that's the one that is the most dilute. But the set-up and solution are the same.

If you get stuck, please reply showing how far you've gotten in working through the steps. Thank you!