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:

The chemist will add 400 milliliters of 20% solution to  milliliters of 60% 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: 

60% 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

If the chemist is to use  milliliters of 12% to make the solution, then he will mix it with  milliliters (1,000 milliliters being equal to 1 liter) of the 30% solution. The result will be 1,000 milliliters of 12% 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:

12% solution: 

30% solution: 

15% 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

Using the ratio, we can write the following equation:

There are  pounds of peanuts,  pounds of pistachios, and  pounds of macadamia nuts in the mixture, so that means that in 40 pounds of the mixture, there are

 pounds of peanuts

 pounds of pistachios

 pounds of macadamia nuts

Since the question asked about pistachios, the correct answer is that there are  pounds of pistachios in  pounds of the mixture.

Let  be the amount of water used to dilute the acid. The resulting solution will then have . The amount of acid in the final solution is: . That amount of acid in the final solution is equal to the original  of acid, since there is no acid in water.

The amount of the final solution is , and the amount of water used to dilute the original acid solution is .

Let x equal the number of liters of the 77% solution, the value we need to solve for. If the first solution is 77% glycol, then 0.77x will give us the number of liters of glycol in the first solution. Accordingly, 0.53(42 liters) will give us the number of liters of glycol in the second solution. If we are mixing these two solutions, then the number of liters of glycol in the final mixture will equal the sum of the number of liters of glycol in each solution. The amount of glycol in the final mixture will be its desired concentration times the volume of the total mixture, 0.60(x+42), so we can write the following equation and solve for x:

40% of one liter is 0.4 liters of acid, which is the amount of acid that will be in the solution at the end as well.

After  liters of water is added, there will be  liters of solution. the amount of acid in the solution will be 25% of this, or . 

The amount of acid in the solutions remains constant, so we set up and solve this equation:

The scientist will need to add 0.6 liters of water.

40% of one liter of the original solution is 0.4 liters of acid.

Suppose the scientist adds  liters of acid. Then the scientist will have  liters of solution and  liters of acid, and the concentration will be

 percent. Solve for :

The scientist will need to add 0.2 liters of pure acid to the solution.