**GRE Quant Practice Questions**

What are the types of questions you could expect on the GRE quants? Take a look at some solved samples we've included below, to get an idea of what you'll be up against!

**Broadly, there are four main types of questions. These are:**

**1. Quantitative Comparison Questions:**

In questions of this type, you will be required to compare two quantities—quantity A and quantity B—and then figure out which of the four answers best describes the comparison:

—Quantity A is greater.

—Quantity B is greater.

—The two quantities are equal.

—The relationship cannot be determined from the information given.

Here’s a classic example of a Quantitative Comparison question:

Solution:

Now, whenever you are given a geometry question without diagrams, try sketching one.

If you draw a square, one side of which is the diameter of a circle, you'll get something like this:

Solution:

Now, whenever you are given a geometry question without diagrams, try sketching one.

If you draw a square, one side of which is the diameter of a circle, you'll get something like this:

If you move the circle over, it fits in the square: The area of the square, which is column B, is larger.

If you move the circle over, it fits in the square: The area of the square, which is column B, is larger.

**2. Multiple-choice Questions — Select One Answer Choice:**

This question type require no introduction, for we’ve encountered these from our school days. Say hello to the classic MCQ style question: here, you’ll have a list of answers from which you have to choose the correct one. Let’s take a look at an example of a multiple-choice question with a single answer:

In the rectangle above, AB = x feet, BC = y feet, and AE = FC = 2 feet. What is the area of triangle DEF, in square feet?

Choices:

xy − 2x − 2y + 4

xy − 2x − 2y − 4

xy/2− x − y + 2

xy/2−x−y-2

xy/2+2

Solution:

Plug in your own numbers for x and y. If x = 4 and y = 5, then the sides of the triangle are 2 and 3.The area of a triangle = 1/2 bh, so the triangle has an area of 3. Circle 3 as your target. Plugging in shows you that only choice (c) matches your target.

In the rectangle above, AB = x feet, BC = y feet, and AE = FC = 2 feet. What is the area of triangle DEF, in square feet?

Choices:

xy − 2x − 2y + 4

xy − 2x − 2y − 4

xy/2− x − y + 2

xy/2−x−y-2

xy/2+2

Solution:

Plug in your own numbers for x and y. If x = 4 and y = 5, then the sides of the triangle are 2 and 3.The area of a triangle = 1/2 bh, so the triangle has an area of 3. Circle 3 as your target. Plugging in shows you that only choice (c) matches your target.

**3. Multiple-choice Questions — Select One or More Answer Choices:**

Now, this is the classic MCQ, but with an interesting twist. Here, instead of choosing just one answer, you will be asked to choose two or more from the given list. The number of answers you are required to compute may or may not be specified in the question itself. Let’s look at what a typical MCQ with multiple answer choices looks like:

A. The average increases by 1 if each number increases by 1

B. The average becomes 3 times, if each number becomes three times

C. If the sum of the numbers increases by 7, the average increases by 1

D. If seven numbers increase by 3 each and three numbers decrease by 7 each, the average remains the same

E. The sum of the numbers is 70

Solution:

Here, because you aren’t told how many statements are actually true, you will have to work out every single claim made to figure out which ones are true.

If each number increases by 1, then the sum increases by 70+10=80

Average = 80/10=8

Average increases by 1

Option A is true.

If each number becomes three times, the sum also becomes three times.

Average = 3*70/10=3*7=21

Option B is true.

If the sum increases by 7, it becomes 70+7=77

Average = 77/10=7.7

Option C is false.

If 7 numbers increase by 3 and 3 decrease by 7, then sum = 70+7*3-3*7 = 70+21-21=70

Average = 70/10=7

Option D is true.

The sum of the numbers is 70

Average = 7 and number of observations = 10

Sum of observations = Average* number of observations = 7*10=70

Option E is also true.

Out of the five claims, then, only one (C) is untrue.

**The average of 10 numbers is 7. Which of the following statements is true? Indicate all true statements.**A. The average increases by 1 if each number increases by 1

B. The average becomes 3 times, if each number becomes three times

C. If the sum of the numbers increases by 7, the average increases by 1

D. If seven numbers increase by 3 each and three numbers decrease by 7 each, the average remains the same

E. The sum of the numbers is 70

Solution:

Here, because you aren’t told how many statements are actually true, you will have to work out every single claim made to figure out which ones are true.

If each number increases by 1, then the sum increases by 70+10=80

Average = 80/10=8

Average increases by 1

Option A is true.

If each number becomes three times, the sum also becomes three times.

Average = 3*70/10=3*7=21

Option B is true.

If the sum increases by 7, it becomes 70+7=77

Average = 77/10=7.7

Option C is false.

If 7 numbers increase by 3 and 3 decrease by 7, then sum = 70+7*3-3*7 = 70+21-21=70

Average = 70/10=7

Option D is true.

The sum of the numbers is 70

Average = 7 and number of observations = 10

Sum of observations = Average* number of observations = 7*10=70

Option E is also true.

Out of the five claims, then, only one (C) is untrue.

**4. Numeric Entry Questions:**

Questions of this type ask you either to enter your answer as an integer or a decimal in a single answer box or to enter it as a fraction in two separate boxes — one for the numerator and one for the denominator. In the computer-delivered test, you will use the computer mouse and keyboard to enter your answer. Here’s what such a question usually looks like:

Solution:

Machine A produces (k/ 10) liters per minute, and machine B produces (k /15) liters per minute. So when the machines work simultaneously, the rate at which the chemical is produced is the sum of these two rates, which is (k/10) +(k /15)= k(1/10 + 1/15) =k(25/150) = (k/6 ). To compute the time required to produce k liters at this rate, divide the amount k by the rate (k/6) to get k / (k/6)Therefore, the correct answer is 6 minutes (or equivalent).

**Working alone at its constant rate, machine A produces k liters of a chemical in 10 minutes. Working alone at its constant rate, machine B produces k liters of the chemical in 15 minutes. How many minutes does it take machines A and B, working simultaneously at their respective constant rates, to produce k liters of the chemical?**Solution:

Machine A produces (k/ 10) liters per minute, and machine B produces (k /15) liters per minute. So when the machines work simultaneously, the rate at which the chemical is produced is the sum of these two rates, which is (k/10) +(k /15)= k(1/10 + 1/15) =k(25/150) = (k/6 ). To compute the time required to produce k liters at this rate, divide the amount k by the rate (k/6) to get k / (k/6)Therefore, the correct answer is 6 minutes (or equivalent).

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