GMAT Math : GMAT Quantitative Reasoning

Study concepts, example questions & explanations for GMAT Math

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Example Questions

Example Question #12 : Calculating Profit

The stock we just bought soared  percent, in other words our position increased by . How much did we initially invest?

Possible Answers:

Correct answer:

Explanation:

A 55% increased of the unkown value  resulting as a  increase can be written as follow .

So

, which is the initial value.

Example Question #161 : Word Problems

Mary, a very respected day trader, has the following postions: She invested  in stock A,  in stock B and  in stock C. What is her total return, in dollars on this portfolio considering the fact that since she invested in these stocks, A went up , B went down  and C increased .

Possible Answers:

Correct answer:

Explanation:

To solve this problem, we simply have to multiply each position by their respective rate.

Note that we use a negative rate if the value decreased.

The final answer is then given by following equation 

 and we end up with .

Example Question #162 : Word Problems

We have just lost  percent, or , of our investment, what was the amount invested before the loss?

Possible Answers:

Correct answer:

Explanation:

To find the amount invested prior to the loss set up a proportion that represents what is given mathematically.

To write 60 percent of something is the same as,

If this is equivalent to $6,000 then the statement to solve would read,

.

From here cross multiply and divide to solve for the amount invested prior to the loss.

 

Example Question #1 : 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:

15 ounces

30 ounces

20 ounces

35 ounces

25 ounces

Correct answer:

30 ounces

Explanation:

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 #2 : 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:

Explanation:

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 #1 : 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:

Explanation:

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 #2 : 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:

Explanation:

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 #3 : 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:

Explanation:

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 #4 : 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:

Explanation:

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:

Explanation:

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.

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