Are you ready to tackle the toughest of the tough GRE math questions? Buckle up and get ready to put your thinking cap on – we’re about to dive into some GRE math problems!
The Quantitative Reasoning or the GRE math section measures a test-taker’s ability to solve mathematical problems and interpret data. Preparing for the GRE math test can be challenging, but practice tests for the GRE can help you get familiar with the types of questions you’ll encounter. In this blog post, we’ll take a look at 15 of the hardest GRE math questions with solutions to help you prepare for the exam.
First up, we have a classic GRE math question that’ll have you feeling like a math whiz in no time.
1. What’s the sum of the first 50 positive integers?
Solution: The formula for the sum of the first n positive integers is (n)(n+1)/2. Plugging in 50, we get (50)(51)/2 = 1275. So the answer is 1275.
Next, we’ve got a GRE math question that’ll have you scratching your head and wondering if you’ve been transported to an alternate dimension where integers behave differently.
2. If a, b, and c are consecutive integers such that a < b < c, which of the following must be true?
- A) b is even
- B) c is odd
- C) a + b + c is even
- D) b – a is odd
- E) b + c is even
Solution: Let’s say that a = x, b = x + 1, and c = x + 2. Now we can use the answer choices to narrow down our options. A is true because x + 1 is always even. B is false because x + 2 can be even or odd. C is false because the sum of three consecutive integers is always odd. D is true because x + 1 – x = 1, which is odd. E is true because x + 1 + x + 2 = 2x + 3, which is odd.
So the answer is A, D, and E.
Now, let’s take things up a notch with another GRE math question that’ll have you feeling like a math wizard.
3. If a = (2x+1)/(x-1) and b = (x+1)/(2x-1), what is the value of (a+b)/(a-b)?
Solution: We can start by finding the value of a+b and a-b separately:
a+b = (2x+1)/(x-1) + (x+1)/(2x-1)
a-b = (2x+1)/(x-1) – (x+1)/(2x-1)
To simplify these expressions, we can first find a common denominator:
a+b = [(2x+1)(2x-1) + (x+1)(x-1)]/[(x-1)(2x-1)]
a-b = [(2x+1)(2x-1) – (x+1)(x-1)]/[(x-1)(2x-1)]
a+b = (5x^2-3)/(2x^2-3x+1)
a-b = (3x^2+3x-1)/(2x^2-3x+1)
Therefore, (a+b)/(a-b) = [(5x^2-3)/(2x^2-3x+1)] / [(3x^2+3x-1)/(2x^2-3x+1)] = (5x^2-3)/(3x^2+3x-1).
Now, we’ve got a problem that’ll have you feeling like a detective on the hunt for clues.
4. If f(x) = x^3 – 3x^2 – 4x + 12, what is the sum of the roots of the equation f(x) = 0?
Solution: To find the sum of the roots, we can use Vieta’s formulas, which state that the sum of the roots of a cubic equation ax^3 + bx^2 + cx + d = 0 is -b/a. In this case, a = 1, b = -3, c = -4, and d = 12, so the sum of the roots is -(-3)/1 = 3.
Another GRE math question to tingle on Math nerves.
5. If x is a positive integer and 2^x + 2^(x+1) + 2^(x+2) = 336, what is the value of x?
Solution: We can start by factoring out 2^x from the left-hand side of the equation: 2^x(1 + 2 + 4) = 2^x(7) = 336. Dividing both sides by 7, we get 2^x = 48, which means x = 5.
For the next question…ok, let’s cut to the chase. Here are 10 more hardest GRE Math questions–nonstop!
6. A circle with center O has radius 1. A square with side length 2 is inscribed in the circle, and a smaller circle is inscribed in the square. What is the radius of the smaller circle?
Solution: The diameter of the smaller circle is equal to the diagonal of the square, which is 2√2. Therefore, the radius of the smaller circle is √2 – 1.
7. If x + 1/x = 3, what is the value of x^5 + 1/x^5?
Solution: We can start by squaring x + 1/x = 3 to get x^2 + 2 + 1/x^2 = 9, which means x^2 + 1/x^2 = Squaring again, we get x^4 + 2 + 1/x^4 = 49, which means x^4 + 1/x^4 = 47. Multiplying both sides by x + 1/x, we get x^5 + 1/x^5 + x^3 + 1/x^3 = 141. Now we can use the identity x^3 + 1/x^3 = (x + 1/x)(x^2 – x + 1) to get x^5 + 1/x^5 = 141 – (3)(7) = 120.
8. In a certain year, January 1 was a Monday. What day of the week was April 1 of the same year?
Solution: There are 31 days in January, 28 (or 29 in a leap year) days in February, and 31 days in March. Therefore, there are a total of 31 + 28 + 31 = 90 days between January 1 and April 1. 90 divided by 7 leaves a remainder of 6, which means April 1 is 6 days after Monday, or Sunday.
9. A bag contains 5 red balls and 7 blue balls. If two balls are drawn at random without replacement, what is the probability that both balls are red?
Solution: The probability of drawing a red ball on the first draw is 5/12. Since one red ball has been removed, there are now 4 red balls and 11 total balls left in the bag. Therefore, the probability of drawing another red ball on the second draw is 4/11. The probability of both events happening is the product of the probabilities, or (5/12) * (4/11) = 5/33.
10. A cube has a volume of 64 cubic meters. What is the length of its diagonal?
Solution: Let s be the length of each edge of the cube. Since the volume is 64 cubic meters, we have s^3 = 64, which means s = 4. The diagonal of the cube can be found using the Pythagorean theorem: d^2 = s^2 + s^2 + s^2 = 3s^2. Plugging in s = 4, we get d^2 = 3(4^2) = 48, which means d = 4√3.
11. The sum of the first 10 terms of an arithmetic sequence is 225. The sum of the next 10 terms is 525. What is the first term of the sequence?
Solution: Let a be the first term of the sequence, and let d be the common difference. Then the sum of the first 10 terms is (10/2)(2a + (10-1)d) = 5(2a + 9d), and the sum of the next 10 terms is (10/2)(2a + (10+9)d) = 10a + 95d. We can set up two equations using these formulas and solve for a and d:
5(2a + 9d) = 225
10a + 95d = 525
Solving for a, we get a = 5.
12. The expression (2^x – 3)(2^x + 7) is equal to 0. What is the value of x?
Solution: The expression is equal to 0 when either 2^x – 3 = 0 or 2^x + 7 = 0. Solving the first equation, we get 2^x = 3, which means x = log_2(3). Since log_2(3) is not an integer, it does not satisfy the equation. Solving the second equation, we get 2^x = -7, which has no real solutions. Therefore, there is no solution to the equation.
13. If a and b are positive integers such that (a+b)^2 + (a-b)^2 = 80, what is the value of a^2 + b^2?
Solution: Expanding the left-hand side of the equation, we get 2a^2 + 2b^2 = 80, which means a^2 + b^2 = 40.
14. In a rectangular coordinate system, the point (-3, 5) lies on a line with slope 4. What is the y-coordinate of the point where the line intersects the x-axis?
Solution: Since the slope of the line is 4, we know that the rise over run is 4/1. Therefore, for every increase of 1 in x, there is an increase of 4 in y. Starting from (-3, 5), we can move 2 units to the left and 8 units down to get to the x-axis. Therefore, the y-coordinate of the point where the line intersects the x-axis is 5 – 8 = -3.
15. A group of 10 friends are ordering pizzas for a party. Each pizza has 8 slices, and each friend wants to have at least 2 slices of pizza. What is the minimum number of pizzas that must be ordered?
Solution: We need to find the minimum number of pizzas that can satisfy the condition that each of the 10 friends gets at least 2 slices. This means that we need to distribute at least 20 slices among the 10 friends. However, each pizza has 8 slices, so the number of pizzas must be rounded up to the nearest integer that satisfies the condition. Therefore, the minimum number of pizzas that must be ordered is ceil(20/8) = 3.
Aaaand…a bonus GRE Math question
16. A triangle has side lengths 6, 8, and 10. What is the radius of the circle inscribed in the triangle?
Solution: To find the radius of the circle inscribed in a triangle, we can use the formula r = A/s, where A is the area of the triangle and s is the semiperimeter (half the perimeter) of the triangle. The semiperimeter of the triangle is (6+8+10)/2 = 12, and the area of the triangle can be found using Heron’s formula: A = sqrt(s(s-a)(s-b)(s-c)), where a, b, and c are the side lengths of the triangle. Plugging in the values, we get A = sqrt(12(12-6)(12-8)(12-10)) = 24. Therefore, the radius of the circle inscribed in the triangle is r = A/s = 24/12 = 2.