Algebraic Constraint Testing - GMAT Data Insights

If x is a positive integer, is x divisible by 6? (1) $x^2$ is divisible by 12 (2) x is divisible by 3
Answer: Only (1) is sufficient Because x is an integer, if $x^2$ is divisible by 12 = 4 x 3, then x must contain at least one factor 2 and one factor 3. So x is divisible by 6, and statement (1) alone is sufficient. Statement (2) only tells you x is divisible by 3, not by 2, so it is insufficient. The tempting error is thinking you need statement (2) as well, but the square already guarantees the factor 3.
Is the system of equations $2x + 3y = 7$ and $ax + by = c$ inconsistent? (1) $\frac{a}{2} = \frac{b}{3} ≠ \frac{c}{7}$ (2) $a = 4$ , $b = 6$ , and $c = 15$
Answer: EACH statement ALONE is sufficient to answer the question asked. 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 $f(x) = x^2 + px + q$ where $p$ and $q$ are constants, does the equation $f(x) = 0$ have two distinct real roots? (1) $p^2 - 4q > 0$ (2) $f(0) \cdot f(1) < 0$
Answer: EACH statement ALONE is sufficient to answer the question asked. 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.
If $x$ and $y$ are nonzero integers, is $\frac{x}{y} < 1$ ? (1) $x^2 < y^2$ (2) $x < y$
Answer: (1) alone sufficient From $x^2$ < $y^2$ you get |x| < |y|. If x and y have opposite signs, x/y is negative, hence less than 1; if same sign, the absolute value of x/y is less than 1, so it is also less than 1. Thus (1) alone is sufficient. Statement (2) is not: x < y holds for -3 < -2, but -3/-2 = 1.5, which is not less than 1. The tempting error is to assume x<y always makes x/y <1, ignoring negative denominators.
If $m$ and $n$ are positive integers, is $\frac{m}{n}$ in its simplest form? (1) $\gcd(m+1, n+1) = 1$ (2) $m$ and $n$ are consecutive terms in the Fibonacci sequence
Answer: Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. 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( $F_{k+1}$ , $F_k$ ) = gcd( $F_k$ , $F_{k+1}$ - $F_k$ ) = gcd( $F_k$ , $F_{k-1}$ ) = ... = gcd( $F_2$ , $F_1$ ) = gcd(1,1) = 1. Therefore, statement (2) is sufficient.
For positive integers $a,b$ , is $a$ a multiple of $b$ ? (1) $a^2$ is a multiple of $b$ (2) $a$ is a multiple of $b^2$
Answer: (2) alone sufficient From statement (2), if a is a multiple of $b^2$ , then a = k( $b^2$ ) = (kb)b for some integer k, so a is a multiple of b. Statement (2) alone is sufficient. Statement (1) is not: $a^2$ being a multiple of b does not force a to be. For example, a = 2 and b = 4 gives 4 a multiple of 4, yet 2 is not a multiple of 4. So the answer is (2) alone sufficient.
If x and y are nonzero real numbers, is xy > 0? (1) x/y > 0 (2) x > y
Answer: Only (1) is sufficient From (1), x/y > 0 means x and y have the same sign, so their product xy is positive. That makes (1) sufficient alone. (2) only tells you x is greater than y; for example 2 > 1 gives xy > 0, but 2 > -1 gives xy < 0, so it is not sufficient. The tempting wrong answer is to assume x > y means both are positive, but a larger number can still be negative.
If x is a real number, is x > 1? (1) x + 1 > 0 (2) $x^2$ < x
Answer: Only (2) is sufficient From (2), $x^2$ < x means x(x - 1) < 0, so x is between 0 and 1. That gives a definite no to the question, so (2) alone is sufficient. Statement (1) only says x > -1, which includes values above and below 1, so it is not sufficient. The tempting mistake is thinking $x^2$ < x only means x < 1, while ignoring that x must also be positive.
If $x$ and $y$ are positive integers, what is $x$ ? (1) $x + y = 6$ (2) $x^2 + y^2 = 20$
Answer: Together, not sufficient From x + y = 6, the positive integer pairs are (1,5), (2,4), (3,3), (4,2), and (5,1). Checking squares, only (2,4) and (4,2) give 4 + 16 = 20, so x could be 2 or 4. The tempting error is to think the statements determine x uniquely, but they only fix the unordered pair {2,4}.
If $m$ and $n$ are nonzero integers, is $\frac{m}{n} > 0$ ? (1) $m^2 n > 0$ (2) $m n^2 < 0$
Answer: Both together, not alone From (1), $m^2$ is positive, so n must be positive. From (2), $n^2$ is positive, so m must be negative. Together m/n is negative, so the answer is definitely no, which is sufficient. Neither statement alone works because each leaves the other variable's sign unknown. The tempting mistake is thinking a definite no means insufficient.
If x and y are real numbers, is x > y? (1) $x^2$ > $y^2$ (2) x + y > 0
Answer: Both, neither alone Since $x^2$ > $y^2$ is (x - y)(x + y) > 0, and statement (2) says x + y > 0, dividing by that positive factor gives x - y > 0, so x > y. Statement (1) alone is not enough: |-3| > |1| but -3 > 1 is false. Statement (2) alone is not enough either: 1 + 2 > 0 but 1 > 2 is false. The tempting error is treating $x^2$ > $y^2$ as if it directly implied x > y, which ignores negative numbers.
If x and y are real numbers, is x > 0? (1) x + y > 0 (2) xy < 0
Answer: Together not sufficient From xy < 0, x and y have opposite signs. If x is positive and y is negative, x + y > 0 can hold, as with 3 and -1, so x > 0 is possible. If x is negative and y is positive, x + y > 0 can also hold, as with -1 and 3, so x > 0 fails. Both statements together still don't settle it. The tempting mistake is assuming opposite signs plus a positive sum forces x positive, but y can be the larger positive number.
If $x$ and $y$ are integers, is $x^2 > y^2$ ? (1) $x > y$ (2) $x + y > 0$
Answer: Both together, not alone Together, x > y and x + y > 0 force x > |y|, because x is greater than both y and -y. That makes |x| > |y|, so $x^2$ > $y^2$ . The tempting trap is thinking statement (1) alone is sufficient: x = 1 and y = -2 gives x > y but 1 > 4 is false.
For real numbers $x$ and $y$ , is $x^3 + y^3 > x^2y + xy^2$ ? (1) $x > y > 0$ (2) $x + y > 0$ and $xy > 0$
Answer: Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. 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.

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