### All GMAT Math Resources

## Example Questions

### Example Question #1 : Understanding Mixture Problems

A scientist needs a 10% saline solution for an experiment. In his closet he finds a 20 ounce bottle of 25% saline solution. How many ounces of pure water should he add to the mixture to produce the correct saline solution?

**Possible Answers:**

30 ounces

25 ounces

20 ounces

35 ounces

15 ounces

**Correct answer:**

30 ounces

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.

### Example Question #1 : Mixture Problems

If Audrey is currently years old, and Matt’s age is 2 years more than of Audrey’s age, what will be Matt’s age in 5 years?

**Possible Answers:**

**Correct answer:**

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:

### Example Question #2 : Mixture Problems

Mike, the barista at Moose Jaw Coffee, has to mix together two kinds of coffee beans - Mocha Madness, which costs $12 a pound, and Sumatra Sweetness, which costs $20 a pound - to produce forty pounds of a coffee that costs $14 a pound. The beans in the mixture sell for the same price as they would separately.

How many pounds of Mocha Madness coffee will Mike put into the mixture?

**Possible Answers:**

**Correct answer:**

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.

### Example Question #3 : Mixture Problems

Some dimes and some quarters are together worth $8.95.

Which of the following is a possible number of dimes in this mixture?

**Possible Answers:**

**Correct answer:**

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.

### Example Question #4 : Mixture Problems

How much of a 20% acid solution would a chemist have to mix with one liter of a 40% acid solution to yield a 36% acid solution?

**Possible Answers:**

**Correct answer:**

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.

### Example Question #5 : Mixture Problems

24 coins, all nickels and dimes, are together worth $1.75.

How many of the coins are nickels?

**Possible Answers:**

Not enough information is given.

**Correct answer:**

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.

### Example Question #1 : Mixture Problems

A chemist has 600 milliliters of a solution of 20% alcohol on hand, and she wants to mix it with enough 50% alcohol solution to turn it into a 30% alcohol solution. How much of the 50% solution will she need?

**Possible Answers:**

The correct answer is not among the other responses.

**Correct answer:**

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.

### Example Question #2 : Mixture Problems

A chemist has 400 milliliters of a solution of 20% alcohol on hand, and he wants to mix it with enough pure alcohol to turn it into a 30% alcohol solution. How much pure alcohol will this require?

Select the response that is closest to the correct answer.

**Possible Answers:**

**Correct answer:**

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.

### Example Question #3 : Mixture Problems

A chemist wants to make one liter of hydrochloric acid solution. He only has two concentrations on hand - and . How much of the solution will the chemist use?

**Possible Answers:**

**Correct answer:**

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.

### Example Question #4 : Mixture Problems

A chemist has milliliters of a solution of alcohol on hand, and he wants to mix it with enough pure alcohol to turn it into a alcohol solution. If is the amount of pure alcohol he needs, which equation can he use to solve for ?

**Possible Answers:**

**Correct answer:**

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:

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