What this quiz covers
This quiz focuses on Algebraic Constraint Testing, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT Data Insights.
For real numbers x and y, is x3+y3>x2y+xy2?
(1) x>y>0 (2) x+y>0 and xy>0
GMAT Data Insights Quiz
Practice Algebraic Constraint Testing in GMAT Data Insights with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
This quiz focuses on Algebraic Constraint Testing, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT Data Insights.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
For real numbers x and y, is x3+y3>x2y+xy2?
(1) x>y>0 (2) x+y>0 and xy>0
Explanation: First, factor the expression: x³ + y³ - x²y - xy² = x³ - x²y + y³ - xy² = x²(x - y) + y²(y - x) = x²(x - y) - y²(x - y) = (x - y)(x² - y²) = (x - y)²(x + y). So we need (x - y)²(x + y) > 0. Since (x - y)² ≥ 0 always, this inequality holds if and only if (x - y)² > 0 and x + y > 0, OR (x - y)² = 0 and x + y > 0. The first condition means x ≠ y and x + y > 0. Statement (1): x > y > 0 implies x ≠ y and x + y > 0, so (x - y)²(x + y) > 0. This is sufficient. Statement (2): x + y > 0 and xy > 0 means both x and y have the same sign and their sum is positive, so both are positive. However, this doesn't tell us whether x ≠ y. If x = y, then (x - y)²(x + y) = 0, making the inequality false. If x ≠ y, the inequality is true. Statement (2) is insufficient.
If m and n are positive integers, is nm in its simplest form?
(1) gcd(m+1,n+1)=1 (2) m and n are consecutive terms in the Fibonacci sequence
Explanation: For m/n to be in simplest form, we need gcd(m,n) = 1. Statement (1): gcd(m+1, n+1) = 1 doesn't guarantee gcd(m,n) = 1. For example, if m = 4 and n = 6, then gcd(5,7) = 1, but gcd(4,6) = 2 ≠ 1. Alternatively, if m = 3 and n = 5, then gcd(4,6) = 2 ≠ 1, so the condition isn't even satisfied. Let me try m = 2, n = 4: gcd(3,5) = 1, but gcd(2,4) = 2. Or m = 1, n = 3: gcd(2,4) = 2. The statement is insufficient. Statement (2): If m and n are consecutive Fibonacci numbers, then gcd(m,n) = 1. This is a well-known property: consecutive Fibonacci numbers are always coprime. This can be proven by the Euclidean algorithm: if F_k and F_{k+1} are consecutive Fibonacci numbers, then gcd(Fk+1, F_k) = gcd(F_k, Fk+1 - F_k) = gcd(F_k, Fk−1) = ... = gcd(F2, F1) = gcd(1,1) = 1. Therefore, statement (2) is sufficient.
Is the system of equations 2x+3y=7 and ax+by=c inconsistent?
(1) 2a=3b=7c (2) a=4, b=6, and c=15
Explanation: A system is inconsistent when the coefficient ratios are equal but the constant ratio is different. For 2x + 3y = 7 and ax + by = c, inconsistency occurs when a/2 = b/3 but c/7 ≠ a/2 (equivalently c/7 ≠ b/3). Statement (1): This directly states the condition for inconsistency, so the system is inconsistent. Sufficient. Statement (2): a = 4, b = 6, c = 15. Check ratios: a/2 = 2, b/3 = 2, c/7 = 15/7 ≈ 2.14. Since a/2 = b/3 = 2 but c/7 ≠ 2, the system is inconsistent. Sufficient.
If f(x)=x2+px+q where p and q are constants, does the equation f(x)=0 have two distinct real roots?
(1) p2−4q>0 (2) f(0)⋅f(1)<0
Explanation: For f(x) = x² + px + q to have two distinct real roots, the discriminant must be positive: Δ = p² - 4q > 0. Statement (1): p² - 4q > 0 directly gives us the discriminant condition. This is sufficient for two distinct real roots. Statement (2): f(0) = q and f(1) = 1 + p + q. If f(0)·f(1) < 0, then q(1 + p + q) < 0, meaning q and (1 + p + q) have opposite signs. By the Intermediate Value Theorem, since f is continuous and f(0) and f(1) have opposite signs, there exists at least one root between 0 and 1. However, we need to verify this guarantees two distinct roots. If f(0) and f(1) have opposite signs, then the parabola crosses the x-axis at least once between x = 0 and x = 1. For a upward-opening parabola (coefficient of x² is 1 > 0), this crossing between 0 and 1, combined with the continuous nature and the fact that the parabola goes to +∞ as x → ±∞, guarantees exactly two distinct real roots. Statement (2) is sufficient.