The solution he needs has only a 10% salt level. Currently we know that his solution has 20 ounces at 25% salt. We can calculate the amount of salt in the 20 ounce container by utilizing the given information. (20)*(.25) = 5 ounces of salt. Let x be the volume of pure water (in ounces) added. 

Therefore, we know the total volume of our new solution will be 20+x.  We know we want our solution to have 10% salt, so our salt amount in the new solution will have to be (20+x) *(.10). Since we are not adding any salt when we add our "pure water", we know the total salt in the solution will not change. Therefore, we can write the equation. (20+x)*(.10) = 5

Solve for x and you get x = 30 ounces of pure water. 

of Audrey’s age is .

2 years more than  of Audrey’s age is

.

However, be careful because the question is asking for Matt’s age in 5 years, so you need to add 5 years to Audrey’s current age:

Let be the number of pounds of Mocha Madness coffee beans Mike uses. Then he will use  pounds of Sumatra Sweetness coffee beans. 

The Mocha Madness coffee costs $12 a pound times  pounds, or  dollars; similarly, the Sumatra Sweetness coffee costs . The total cost of the coffee will be . Since the beans will sell for the same price as unmixed, we can add the prices to obtain and solve this equation:

30 pounds of the Mocha Madness will go into the mixture.

If the dimes are removed, the amount of money remaining, being only quarters, must be a multiple of $0.25. We can test each choice accordingly.

Of the choices given, only the removal of 17 dimes leaves an amount that coud possibly be made up only of quarters.

Let  be the amount of 20% acid solution used in milliliters. Then the amount of total solution will be  (one liter being 1,000 milliliters). The amount of acid in each solution will be as follows:

In the 20% solution: 

In the 40% solution: 

In the 36% solution: 

Add the acid in the two source solutions to get the acid in the resulting solution; then solve for :

The chemist must add 250 milliliters of the 20% acid solution.

Let  be the number of nickels. Then there are  dimes. 

The amount of money is defined by the expression . Set this equal to 1.75 and solve:

The mix includes 13 nickels.

The chemist will add 600 milliliters of a solution of 20% alcohol to  milliliters of 50% alcohol solution to make  milliliters of 30% solution.

The amount of pure alcohol in each solution will be the amount of solution multiplied by the concentration expressed as a decimal. Therefore, the amounts of alcohol in the three solutions, in milliliters, will be:

20% solution: 

50% solution: 

30% solution (result): 

Since the first two solutions are added to yield the third, the amounts of alcohol are also being added, so the equation to solve is

 milliliters, the correct response.

If the chemist adds 400 milliliters of a solution of 20% alcohol to  milliliters of pure alcohol, he will have  milliliters of the 30% alcohol solution.

The amount of pure alcohol in each solution will be the amount of solution multiplied by the concentration expressed as a decimal. Therefore, the amounts of alcohol in the three solutions, in milliliters, will be:

20% solution:  

30% solution (result): 

Since the alcohol in the first solution is being added to  milliliters of pure alcohol to yield the alcohol in the second, the following equation can be set up:

 milliliters. 

The closest response is 60 milliliters.

The choices are given in milliliters, so we will convert one liter to 1,000 milliliters.

If the chemist is to use  milliliters of 10% to make the solution, then he will mix it with  milliliters of the 50% solution. The result will be 1,000 milliliters of 25% solution.

The amount of pure acid in each solution will be the amount of solution multiplied by the concentration expressed as a decimal. Therefore, the amounts of acid in the three solutions, in milliliters, will be:

10% solution: 

50% solution: 

25% solution (result): 

Since the first two solutions are added to yield the third, the amounts of acid are also being added, so the equation to solve is:

 milliliters of the weaker solution.

If the chemist adds 300 milliliters of a solution of 25% alcohol to  milliliters of pure alcohol, he will have  milliliters of the 40% alcohol solution.

The amount of pure alcohol in each solution will be the amount of solution multiplied by the concentration expressed as a decimal. Therefore, the amounts of alcohol in the three solutions, in milliliters, will be:

25% solution:  

40% solution (result): 

Since the first solution is being added to  milliliters of pure alcohol, the following equation can be set up: