Ah, the GRE quant section. The one section that can make even the most mathematically inclined among us break out in a cold sweat. But fear not! While it may be true that the GRE quantitative reasoning section can be a daunting task, it is also one of the most important sections on the GRE exam. In this blog, we’ll explore some of the most challenging GRE quant questions and provide solutions to help you tackle them head-on.

## What’s on the GRE quantitative reasoning section?

First, let’s break down the GRE quant section. The GRE is a standardized test used for graduate school admissions. The GRE quantitative reasoning section consists of two subsections with 20 questions each, for a total of 40 questions. You’ll have 35 minutes to complete each subsection, giving you a total of 70 minutes to complete the entire quant section.

But what exactly is covered in the GRE quant section? Well, you’ll need to have a strong foundation in a variety of math topics, including arithmetic, algebra, geometry, and data analysis. Some of the specific topics that you’ll need to be familiar with include fractions, percentages, equations and inequalities, coordinate geometry, probability, and statistics. Phew, that’s a lot of ground to cover!

But why is the GRE quant section so important? Well, for starters, it’s often used as a way for graduate schools to assess your ability to handle the rigorous coursework that comes with pursuing an advanced degree. Additionally, your score on the quant section can impact your chances of being accepted into a program and even determine what kind of financial aid you may be eligible for.

## The most challenging GRE quantitative reasoning questions

Now, let’s talk about the hardest GRE quant questions. While there’s no way to predict exactly which questions will be the most difficult, there are certainly some topics that tend to trip people up more than others. Let’s take a look at some of the trickiest arithmetic, algebra and geometry GRE quantitative reasoning questions.

## GRE quant arithmetic questions with solutions and explanations

### Question 1:

If the cost of 4 pens and 5 notebooks is \$25, and the cost of 3 pens and 2 notebooks is \$13, what is the cost of one pen?

Solution:

Let the cost of one pen be p, and the cost of one notebook be n. Then we can set up the following system of equations:

4p + 5n = 25

3p + 2n = 13

We want to solve for p, so let’s eliminate n. Multiplying the second equation by 5 and subtracting it from the first equation multiplied by 2, we get:

8p = 27

Therefore, p = 27/8, or \$3.375. The cost of one pen is \$3.375

### Question 2:

If x – y = 3 and x + y = 9, what is the value of x?

Solution:

Adding the two equations, we get:

2x = 12

Therefore, x = 6. The value of x is 6

### Question 3:

If 4x – 3y = 7 and 5x + y = 11, what is the value of x?

Solution:

Solving the second equation for y, we get:

y = 11 – 5x

Substituting this into the first equation, we get:

4x – 3(11 – 5x) = 7

Expanding and simplifying, we get:

19x = 40

Therefore, x = 40/19. The value of x is approximately 2.11

### Question 4:

A rectangular garden has a perimeter of 30 meters and an area of 36 square meters. What are the dimensions of the garden?

Solution:

Let the length of the garden be L and the width be W. Then we have:

2L + 2W = 30

LW = 36

Solving the first equation for L, we get:

L = 15 – W

Substituting this into the second equation, we get:

W(15 – W) = 36

Expanding and rearranging, we get:

W^2 – 15W + 36 = 0

These quadratic factors are (W – 3)(W – 12), so the possible values of W are 3 and 12. If W = 3, then L = 12, and if W = 12, then L = 3. Therefore, the dimensions of the garden are either 3 meters by 12 meters or 12 meters by 3 meters.

### Question 5:

The sum of three consecutive even integers is 90. What is the smallest of these integers?

Solution:

Let the smallest even integer be x. Then the next two even integers are x + 2 and x + 4. We have:

x + (x + 2) + (x + 4) = 90

Simplifying, we get:

3x + 6 = 90

Therefore, 3x = 84, and x = 28. The smallest even integer is 28.

## GRE quant algebra questions with solutions and explanations

### Question 1:

What is the sum of all values of x that satisfy the equation |x-4| = |x+2|?

Solution:

We’ll solve this equation by considering two cases: when x-4 is positive and when x-4 is negative.

Case 1: x-4 is positive

In this case, |x-4| = x-4. To make |x+2| positive as well, we need x+2 to be greater than or equal to 0, which means x is greater than or equal to -2. So the equation becomes:

x-4 = x+2

-4 = 2

This is a contradiction, so there are no solutions in this case.

Case 2: x-4 is negative

In this case, |x-4| = -(x-4) = -x+4. To make |x+2| positive as well, we need x+2 to be less than or equal to 0, which means x is less than or equal to -2. So the equation becomes:

-x+4 = x+2

2x = 2

x = 1

So the only solution in this case is x=1.

Therefore, the sum of all values of x that satisfy the equation is 1.

### Question 2:

If 2x+3y=12 and 3x+4y=18, what is the value of x+y?

Solution:

We can solve this system of equations by elimination. First, we’ll multiply the first equation by 4 and the second equation by -3:

8x + 12y = 48

-9x – 12y = -54

-x = -6

So x = 6. Substituting this into the first equation gives:

2(6) + 3y = 12

3y = 0

So y = 0. Therefore, x+y = 6+0 = 6.

### Question 3:

If a and b are positive integers such that a+b=7 and ab=12, what is the value of a^2+b^2?

Solution:

We can solve for a and b using the fact that (a+b)^2 = a^2 + 2ab + b^2. We know that a+b=7, so:

(a+b)^2 = 7^2 = 49

Expanding the left side gives:

a^2 + 2ab + b^2 = 49

Substituting ab=12 gives:

a^2 + 2(12) + b^2 = 49

a^2 + b^2 = 25

Therefore, a^2+b^2=25.

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### Question 4:

If 2x-3y=6 and 3x-4y=8, what is the value of x-y?

Solution:

We can solve this system of equations by elimination. First, we’ll multiply the first equation by 4 and the second equation by 3:

8x – 12y = 24

9x – 12y = 24

Subtracting these two equations gives:

-x = 0

So x = 0. Substituting this into the first equation gives:

2(0) – 3y = 6

y = -2

Therefore, x-y = 0 – (-2) = 2.

### Question 5:

If x, y, and z are positive integers such that x+y+z=10 and x^2+y^2+z^2=38, what is the maximum possible value of z?

Solution:

To find the maximum possible value of z, we’ll use the fact that z is a positive integer. Since x, y, and z are positive integers, we know that the minimum possible value of x, y, or z is 1. Therefore, the maximum possible value of z is 5, since if z were 6 or greater, then z^2 would be 36 or greater, which means that x^2+y^2+z^2 would be greater than 38.

To show that it’s possible for z to equal 5, we can use the following values for x and y:

x=1, y=2, z=5

Then we have x+y+z=1+2+5=8 and x^2+y^2+z^2=1+4+25=30, which satisfies the given equations. Therefore, the maximum possible value of z is 5.

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## GRE quant geometry questions with solutions and explanations

### Question 1:

A circle with center O is inscribed in a square ABCD. If AB = 4, find the area of the region outside the circle but inside the square.

Solution:

First, we need to find the radius of the circle. Since the circle is inscribed in the square, its diameter is equal to the side of the square, which is 4. Therefore, the radius of the circle is 2.

Next, we can find the area of the circle using the formula A = πr^2, where r is the radius. So, the area of the circle is A = π(2)^2 = 4π.

Finally, we can find the area of the region outside the circle but inside the square by subtracting the area of the circle from the area of the square. The area of the square is 4^2 = 16. Therefore, the area of the region outside the circle but inside the square is 16 – 4π.

### Question 2:

In the figure below, AB = AC, and CD is perpendicular to AB. If AD = 8 and BD = 6, find the length of the CD. Solution:

Since AB = AC, we know that triangle ACD is an isosceles triangle. Therefore, the angle ACD is equal to the angle ADC. Since CD is perpendicular to AB, we know that angle ACD + ADC = 90 degrees. Therefore, angle ACD = angle ADC = 45 degrees.

We can use the sine function to find the length of CD. In triangle ACD, the sine of angle ACD is equal to CD/AD. Therefore, CD = AD * sin(ACD). Since ACD is 45 degrees, sin(ACD) = √2/2. Therefore, CD = 8 * √2/2 = 4√2.

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### Question 3:

In the figure below, AB = BC = CD, and angle BAC = 120 degrees. Find the measure of angle BCD. Solution:

Since AB = BC = CD, we know that triangle ABC is an equilateral triangle. Therefore, angle BAC = angle BCA = angle CAB = 60 degrees.

We can use the angle sum of triangle BCD to find the measure of angle BCD. Since AB = BC, we know that angle ABC = angle BAC = 60 degrees. Therefore, angle BCD = 360 – 2*60 – 60 = 180 – 120 = 60 degrees.

### Question 4:

In the figure below, ABCD is a rectangle and P is a point on AB. If AP = 5 and PD = 8, find the length of PC. Solution:

Let x be the length of PC. Since ABCD is a rectangle, we know that AD = BC, and since AB = DC, we know that AP + PD = BC. Therefore, BC = 5 + 8 = 13.

Using the Pythagorean theorem in triangle BPC, we have BP^2 + PC^2 = BC^2. Therefore, 5^2 + x^2 = 13^2. Solving for x, we get x = √(169 – 25) = √144 = 12.

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### Question 5:

In the figure below, ABCD is a parallelogram and E is a point on CD such that CE = 3 and ED = 5. The line AE intersects BC at point F. Find the ratio of BF to FC. Solution:

Since ABCD is a parallelogram, we know that AB = DC and BC = AD. Therefore, AB + BC = AD + DC = AC.

Using the same logic as in Question 6, we have BF/FC = BP/PC, where P is the intersection of AE and CD.

Using the Pythagorean theorem in triangle AEC, we have AE^2 + EC^2 = AC^2. Therefore, AE^2 + 3^2 = AC^2. Since AB + BC = AC, we have AB + BC = AE + EC + 3. Substituting AB = DC and BC = AD, we get DC + AD = AE + EC + 3. Since ABCD is a parallelogram, we have DC = AB and AD = BC. Therefore, AB + BC = AE + EC + 3 becomes 2AB = AE + EC + 3. Solving for AE, we get AE = (2AB – 3)/2.

Using the Pythagorean theorem in triangle AEF, we have AF^2 + EF^2 = AE^2. Therefore, AF^2 + EF^2 = ((2AB – 3)/2)^2 + 5^2. Since F is on BC, we have AF = AB – BF and EF = EC – FC. Therefore, (AB – BF)^2 + (EC – FC)^2 = ((2AB – 3)/2)^2 + 5^2. Expanding both sides of this equation, we get AB^2 – 2AB·BF + BF^2 + EC^2 – 2EC·FC + FC^2 = (2AB – 3)^2/4 + 25. Since AB = DC and EC = ED, we have BF + FC = DC – ED = AB – ED. Therefore, AB^2 – 2AB·(AB – ED) + (AB – ED)^2 = (2AB – 3)^2/4 + 25 – EC^2 – FC^2. Simplifying this equation, we get 4ED·AB = 25ED – 4BF·FC. Therefore, BF/FC = (25/4 – ED)/ED = (25/4 – 5)/5 = 5/4.