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.

Since  and  are being set to numbers of milliliters, we convert one liter to 1,000 milliliters. The sum of the amounts of the two solutions  and  is 1,000, so

is one equation of the system.

The other equation will add the acid. The amount of acid in an acid solution is the concentration, as a decimal, multiplied by the amount of solution. The amount of acid in the pre-mixture solutions are  and ; add these to get the amount of acid in 1,000 milliliters of a 20% solution, or  milliliters:

The correct system is:

Since a normal batch of candies is one-sixth green candies, then it contains  pounds of green candies; the defective batch contained half this, or 5 pounds. If we let  be the number of pounds of green candies that were added, then, after the addition, there were  pounds of green candies in a batch of  candies. One sixth of the candies are green, so set and solve the equation:

Six pounds of green candies are added.

Initially, there were 48 pounds of candies. In a normal batch, one sixth of them, or  pounds, should be white, but in the defective batch, half this, or 4 pounds, were white. Subsequently, 44 pounds of candies in the defective batch were not white.

Let  be the number of pounds of candy in the batch after the white candies were added. Since  of a normal batch of candies are not white,  pounds of candies are not white; since no candies that weren't white were added, 

The closest choice, and the correct one, is 53 pounds.