Practice Algebra questions for the Graduate Record Examinations. Every question includes a full explanation of why the correct answer is right and why the tempting distractors are wrong.
16 questions available · medium difficulty · GRE · Free, no registration required
10 example questions with full explanations. Use the interactive practice above to work through the complete set.
Question 1medium
The quadratic equation 3x² − 5x + 3 = 0 has two distinct real roots.
Correct answer: B
To determine the nature of the roots, compute the discriminant: b² − 4ac = (−5)² − 4(3)(3) = 25 − 36 = −11. Because the discriminant is negative, the equation has no real roots at all — it has two complex conjugate roots. A common error is to focus only on the coefficients without fully evaluating the discriminant, or to assume a quadratic always has real solutions. The statement is therefore false.
Question 2medium
If f(x) = 2x + 1 and g(x) = (x − 1)/2, then f(g(x)) = g(f(x)) for all real values of x.
Correct answer: A
Compute each composition separately. f(g(x)) = f((x−1)/2) = 2·(x−1)/2 + 1 = (x−1) + 1 = x. Then g(f(x)) = g(2x+1) = (2x+1−1)/2 = 2x/2 = x. Both compositions yield x, the identity function, which means f and g are inverses of each other — and inverse function pairs always satisfy f(g(x)) = g(f(x)) = x. The statement is true. A tempting error is to assume function composition is generally commutative; it is not, but inverse pairs are a specific exception.
Question 3medium
For all real numbers a and b, the equation (a^(1/2))(b^(1/2)) = (ab)^(1/2) holds.
Correct answer: B
The product rule for square roots, √a · √b = √(ab), requires that both a and b be non-negative real numbers. If both are negative — say a = −4 and b = −9 — then in the real number system √(−4) and √(−9) are undefined, yet (ab)^(1/2) = √36 = 6, creating a contradiction. Even when interpreted through complex numbers, (2i)(3i) = −6 ≠ 6. Because the identity fails for negative real numbers, the claim that it holds for ALL real numbers is false. A common error is to apply the rule without checking the domain restriction.
Question 4medium
The complete solution set of the inequality |2x − 3| < 5 is the open interval (−1, 4).
Correct answer: A
Rewrite the absolute value inequality as a compound inequality: −5 < 2x − 3 < 5. Add 3 to all parts: −2 < 2x < 8. Divide by 2: −1 < x < 4. This is exactly the open interval (−1, 4), so the statement is true. A frequent error is to flip the inequality sign incorrectly when handling the negative case, or to write a closed interval [−1, 4] by mistaking a strict inequality for a non-strict one. Both cases here are strict, confirming open endpoints.
Question 5medium
There exists a sequence of real numbers that is simultaneously arithmetic and geometric.
Correct answer: A
A constant sequence — for example, 3, 3, 3, 3, … — satisfies both definitions at once. It is arithmetic with common difference d = 0, and geometric with common ratio r = 1. Because a valid example exists, the statement is true. Many test-takers assume the two sequence types are mutually exclusive because their standard formulas look different, but that intuition fails in the degenerate constant case. It is worth noting this is the only family of sequences (aside from the zero sequence, which some definitions exclude from geometric sequences) for which both properties hold.
Question 6easy
If −2x > 6, then x > −3.
Correct answer: B
Dividing both sides of −2x > 6 by −2 requires flipping the inequality sign, yielding x < −3, not x > −3. A common error is forgetting to reverse the inequality when multiplying or dividing by a negative number. The statement is false.
Question 7easy
For all real numbers a and b, (a + b)² = a² + b².
Correct answer: B
Expanding (a + b)² using the distributive property gives a² + 2ab + b², not a² + b². The middle term 2ab is only zero when a = 0 or b = 0, so the equation does not hold for all real numbers. The statement is false.
Question 8hard
If ax + by = 1 and bx + ay = 0, where a ≠ b and a ≠ −b, what is the value of x in terms of a and b?
- A.b / (b² − a²)
- B.a / (a + b)²
- C.a / (a² − b²)✓
- D.−a / (a² − b²)
Correct answer: C
From the second equation, bx + ay = 0, so y = −bx/a. Substituting into the first equation: ax + b(−bx/a) = 1, which simplifies to x(a − b²/a) = 1, or x(a² − b²)/a = 1, giving x = a/(a² − b²). Choice A is the value of y, not x — a trap for students who confuse which variable they solved for. Choice D has the wrong sign, and B results from incorrectly squaring (a + b) rather than using the difference of squares.
Question 9hard
Pipe A can fill a tank in 6 hours, Pipe B can fill the same tank in 4 hours, and Pipe C can drain the tank in 12 hours. If all three pipes operate simultaneously and the tank is already one-quarter full when they are turned on, how many hours will it take to fill the tank completely?
Correct answer: D
The combined net fill rate is 1/6 + 1/4 − 1/12 = 2/12 + 3/12 − 1/12 = 4/12 = 1/3 tank per hour. Since the tank is already 1/4 full, only 3/4 of the tank remains to be filled. Time = (3/4) ÷ (1/3) = 9/4 hours. Choice B ignores the initial 1/4 already filled, computing time for a full tank. Choice A incorrectly computes 1/4 ÷ (1/3), as if only 1/4 remained. Choice C results from an arithmetic error when combining the three rates.
Question 10hard
A chemist needs to produce 80 liters of a 35% acid solution by mixing a 20% acid solution with a 60% acid solution. After correctly computing the amounts, the chemist accidentally adds 5 extra liters of the 60% solution and compensates by removing 5 liters of the 20% solution from the original plan. What is the acid concentration of the final 80-liter mixture?
Correct answer: A
First, find the original plan: let x = liters of 60% solution. Then 0.6x + 0.2(80 − x) = 0.35(80) → 0.4x + 16 = 28 → x = 30 liters of 60%, and 50 liters of 20%. After the error: 35 liters of 60% and 45 liters of 20% are used (total still 80 liters). New acid amount = 0.6(35) + 0.2(45) = 21 + 9 = 30 liters. Concentration = 30/80 = 37.5%. Choice C incorrectly assumes the swap has no effect on concentration. Choices A and D reflect off-by-one or proportional errors in accounting for the substituted volume.