GRE Math Questions with Solutions: Mixed Bag

Mathematics is an essential component of the GRE, and mastering it is vital for achieving a high score on the exam. However, the GRE Math section is not just about rote memorization of formulas and equations. It requires you to think critically and apply your knowledge to solve complex problems. To succeed in the GRE Math section, you need to be familiar with a wide range of mathematical concepts and techniques. In this blog, we will explore a mixed bag of GRE Math questions that will help you practice and improve your skills for the exam. So, let’s dive into the world of GRE Math and get ready to boost your score!

In this blog:

• GRE Maths questions on Algebra With solutions
• GRE Arithmetic Practice Questions With Solutions
• GRE Data Interpretation Questions With Solutions

GRE Maths Questions on Algebra with solutions

Here are 5 GRE Maths questions on Algebra along with their solutions:

1. If 2x + 3y = 12 and x – 2y = 3, what is the value of x + y?

Solution:

We can solve for x and y using simultaneous equations:
x – 2y = 3
2x + 3y = 12

Multiplying the first equation by 2 and adding it to the second equation, we get:
2x – 4y + 2x + 3y = 6 + 12
4x – y = 18

Now, we can solve for x and y by substituting in the value of y from the first equation:
x – 2y = 3
x – 2(2x/3 – 4) = 3
x = 6

Substituting x = 6 into the first equation:
2x + 3y = 12
2(6) + 3y = 12
3y = 0
y = 0

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

Answer: 6

 

2. If a + b = 4 and a^2 + b^2 = 14, what is the value of ab?

Solution:

We can use the identity (a + b)^2 = a^2 + 2ab + b^2 to solve for ab:
(a + b)^2 = a^2 + 2ab + b^2
(4)^2 = 14 + 2ab
16 – 14 = 2ab
2 = 2ab
ab = 1

Answer: 1

 

3. If x^2 – 3x – 10 = 0, what are the solutions for x?

Solution:

We can solve for x using the quadratic formula:
x = (-b ± sqrt(b^2 – 4ac))/2a
where a = 1, b = -3, and c = -10.

x = (-(-3) ± sqrt((-3)^2 – 4(1)(-10)))/2(1)
x = (3 ± sqrt(9 + 40))/2
x = (3 ± sqrt(49))/2
x = (3 ± 7)/2

Therefore, the solutions for x are x = 5 and x = -2.

Answer: 5 and -2

 

4. If x + y = 5 and x – y = 1, what is the value of x^2 – y^2?

Solution:

We can use the identity a^2 – b^2 = (a + b)(a – b) to solve for x^2 – y^2:
x^2 – y^2 = (x + y)(x – y)
x^2 – y^2 = (5)(1)
x^2 – y^2 = 5

Answer: 5

 

5. If 2x – 3y = 4 and 4x + ky = 10, what is the value of k?

Solution:

We can solve for k by eliminating x from the two equations:
Multiply the first equation by 2 and add it to the second equation:
4x – 6y + 4x + ky = 8 + 10
8x + ky – 6y = 18

We know that 2x – 3y = 4, so we can substitute 2x = 4 + 3y into the above equation:
8(4 + 3y)/2 + ky – 6y = 18
16 + 12y + ky – 6y = 18
ky + 6y = 18 – 16 – 12y
ky + 6y = 2 – 12y
ky + 18y = 2
y(k + 18) = 2

Since y cannot be zero, we can divide both sides by y:
k + 18 = 2/y
k = 2/y – 18

We still need to solve for y. We can do this by substituting 2x = 4 + 3y into the first equation:
2x – 3y = 4
2(4 + 3y)/3 – 3y = 4
8/3 + 2y – 3y = 4
y = -2/3
y = 2/3

Substituting y = 2/3 into k = 2/y – 18:
k = 2/(2/3) – 18
k = 3 – 18
k = -15

Therefore, the value of k is -15.

Answer: -15

GRE Arithmetic Practice Questions With Solutions

Here are five GRE arithmetic practice questions with solutions:

Question 1: If the ratio of the lengths of two sides of a rectangle is 2:3 and the perimeter of the rectangle is 40 cm, what is the length of the longer side?

Solution: Let the shorter side have length 2x and the longer side have length 3x. Then, the perimeter of the rectangle is 2(2x + 3x) = 10x. Setting this equal to 40 cm and solving for x, we get x = 4. Therefore, the longer side has length 3x = 3(4) = 12 cm.

 

Question 2: If 2x – y = 7 and x + 2y = 11, what is the value of x + y?

Solution: Adding the two equations, we get 3x + y = 18. Solving for y in terms of x, we get y = 18 – 3x. Substituting this into the first equation, we get 2x – (18 – 3x) = 7. Simplifying, we get x = 5. Substituting this back into y = 18 – 3x, we get y = 3. Therefore, x + y = 5 + 3 = 8.

 

Question 3: A store sells a certain item for $20, which is a 25% markup over the cost. What is the cost of the item?

Solution: Let the cost of the item be x. Then, the markup is 0.25x, and the selling price is x + 0.25x = 1.25x. Setting this equal to $20, we get 1.25x = 20. Solving for x, we get x = $16.Score GRE 320+ with this 2-month study plan

 

Question 4: A class has 25 students. If the average score on a test is 85 and the average score of the students who passed is 90, what is the average score of the students who failed?

Solution: Let the number of students who passed be x, so the number of students who failed is 25 – x. Then, we have:

(90x + 85(25 – x))/25 = 85

Simplifying and solving for x, we get x = 10. Therefore, there are 10 students who passed and 15 who failed. The average score of the students who failed is: (85*15)/15 = 85

 

Question 5: If 2x + 3y = 14 and 3x – y = 7, what is the value of x + y?

Solution: Multiplying the second equation by 3 and adding the two equations, we get 11x = 35. Solving for x, we get x = 35/11. Substituting this into the first equation, we get 2(35/11) + 3y = 14. Solving for y, we get y = -13/11. Therefore, x + y = (35/11) – (13/11) = 22/11.

GRE Data Interpretation Questions With Solutions

 Here are 5 GRE Data Interpretation questions with solutions:

Question 1: A company produced and sold a total of 10,000 units of a product. The revenue generated by the sales was $50,000. If the price of each unit was increased by 10%, how many units would need to be sold to generate the same revenue?

Solution:

Let the price of each unit be x and the number of units sold be y. Then, we have:

xy = 10,000 (1)

and

1.1xy = 50,000 (2)

Dividing (2) by 1.1, we get:

xy = 45,454.55

Substituting this into (1), we get:

y = 4.5455 ≈ 5 thousand units

Therefore, 5,000 units would need to be sold to generate the same revenue.

 

Question 2: A survey was conducted to determine the favourite type of music among students at a college. The results are shown in the following table:

What percentage of students prefer jazz or classical music?

Solution:

The total number of students surveyed is:

60 + 30 + 20 + 10 = 120

The number of students who prefer jazz or classical music is:

20 + 10 = 30

Therefore, the percentage of students who prefer jazz or classical music is:

(30/120) x 100% = 25%

 

Question 3: The following chart shows the number of students enrolled in a college over a 5-year period:

What is the average annual growth rate of enrollment during this period?

Solution: The initial number of students is 1,000 and the final number of students is 1,800. The growth factor over this period is:

1,800/1,000 = 1.8

The number of years is 5, so the average annual growth rate is:

1.8^(1/5) – 1 ≈ 0.128

Therefore, the average annual growth rate of enrollment during this period is approximately 12.8%.

 

Question 4: The following table shows the number of customers who visited a restaurant during the lunch hour over a 4-day period:

What is the average number of customers who visited the restaurant per day during this period?

Solution:

The total number of customers who visited the restaurant during this period is:

45 + 60 + 75 + 50 = 230

The number of days is 4, so the average number of customers per day is:

230/4 = 57.5

Therefore, the average number of customers who visited the restaurant per day during this period is 57.5.

 

If you find it difficult to solve these questions, don’t worry and don’t be discouraged. It’s essential to remember that your current level of proficiency and learning style do not limit your potential. By practising consistently and dedicating yourself to your studies, you can enhance your skills in GRE Quant. Find out how GRE-ready you are with this free full-length GRE practice test.

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