GMAT Logic: Level 2

Solution (D): If the list has 15 consecutive integers and starts at -6, the integers are -6, -5, …, 0, …, 8. The largest integer is \(-6 + (15-1) = -6 + 14 = 8\). The range is \(\text{Largest} – \text{Smallest} = 8 – (-6) = 8 + 6 = 14\).

Solution (A): The sum of 5 consecutive terms in an arithmetic sequence is always 5 times the middle term. Let the integers be \(n-4, n-2, n, n+2, n+4\). The sum \(x\) is \(5n\). Since \(n\) is an odd integer, \(x\) is 5 times an odd integer. Therefore, \(x\) must be divisible by 5. (e.g., \(1+3+5+7+9 = 25\), which is divisible by 5).

Solution (D): Let the consecutive odd integers be \(n\), \(n+2\), \(n+4\), and \(n+6\). \(a = n\) \(d = n+6\) \(d – a = (n+6) – n = 6\).

Solution (A): 1. \(xyz > 0\) means the product of the three is positive. 2. \(xy < 0\) means the product of \(x\) and \(y\) is negative. 3. Substitute (2) into (1): We have \( (xy) \times z > 0 \), which means \( (\text{Negative}) \times z > 0 \). 4. For this to be true, \(z\) must be negative. Let’s check the others: If \(z < 0\) and \(xy < 0\), we don't know the sign of \(y\). If \(y > 0\), \(yz < 0\). If \(y < 0\), \(yz > 0\). So II is not “must be true”. We also don’t know the sign of \(x\). So III is not “must be true”. Only statement I must be true.

Solution (D): 1. Find the prime factorization of 120: \(120 = 10 \times 12 = (2 \times 5) \times (2^2 \times 3) = 2^3 \times 3^1 \times 5^1\). 2. We have \( (2^3 \times 3^1 \times 5^1) \times n = \text{Cube}\). 3. For a perfect cube, all exponents must be multiples of 3. 4. \(2^3\) is fine. We need to make \(3^1\) into \(3^3\) and \(5^1\) into \(5^3\). 5. We must multiply by \(3^2\) and \(5^2\). 6. \(n = 3^2 \times 5^2 = 9 \times 25 = 225\).

Solution (A): 1. First, factor the expression: \(x^2 + 2x + 1 = (x+1)^2\). 2. The full expression is \(y \times (x+1)^2\). 3. \((x+1)^2\) is already a perfect square. 4. For the entire product to be a perfect square, \(y\) must also be a perfect square. 5. The question asks for the smallest integer \(y > 1\) that is a perfect square. 6. The perfect squares are 1, 4, 9, 16… The smallest one greater than 1 is 4.

Solution (D): Let \(a=-3, b=-2, c=-1\). A) \(c/a = -1 / -3 = 1/3\). This is positive, but what if they are not integers? Let’s re-check. Oh, they are integers. \(c/a\) is positive. B) \(a – b = -3 – (-2) = -1\). (Negative) C) \(abc = (\text{Neg}) \times (\text{Neg}) \times (\text{Neg}) = \text{Negative}\). D) \((c-a)/b\). \(c-a = -1 – (-3) = 2\) (Positive). \(b = -2\) (Negative). \( \text{Pos} / \text{Neg} = \text{Negative} \). E) \(a+b+c = \text{Neg} + \text{Neg} + \text{Neg} = \text{Negative}\). *Correction:* Let’s re-check my logic for A and D. A) \(c/a\). \(c\) and \(a\) are both negative. \( \text{Neg} / \text{Neg} = \text{Positive} \). This must be positive. D) \((c-a)\): Since \(a < c\), \(c-a\) is positive. \(b\) is negative. \( \text{Pos} / \text{Neg} = \text{Negative} \). This means my logic for A is correct. Let me re-read the question. "must be positive". \(a=-3, b=-2, c=-1\). \(c/a = 1/3\). Not an integer, but positive. What if \(a=-4, b=-2, c=-1\)? \(c/a = 1/4\). Positive. What if \(a=-4, c=-2\)? \(c/a = 2/4 = 1/2\). Positive. It seems A is always positive. Let's re-check my analysis of D. \(c-a\): Since \(c > a\), \(c-a\) is always positive. \(b\): Given as negative. So \((c-a)/b\) is \( \text{Pos} / \text{Neg} \), which is *always negative*. My original answer of D was wrong. The correct answer is A.

Solution (A): A) \(c/a\): \(c\) is negative. \(a\) is negative. \( \text{Negative} / \text{Negative} = \text{Positive} \). This must be true. B) \(a – c\): Since \(a < c\), \(a - c\) must be negative. C) \(abc\): \( (\text{Neg}) \times (\text{Neg}) \times (\text{Neg}) = \text{Negative} \). D) \((c-a)/b\): Since \(c > a\), \(c-a\) is positive. \(b\) is negative. \( \text{Positive} / \text{Negative} = \text{Negative} \). E) \(a+b+c\): Sum of three negatives is negative.

Solution (D): \(a = \text{Negative Odd}\), \(b = \text{Negative Odd}\). \(a+b = (\text{Neg Odd}) + (\text{Neg Odd}) = \text{Negative Even}\). \(c = \text{Negative Integer}\) (could be even or odd). The expression is \( \frac{\text{Negative Even}}{\text{Negative Integer}} \). The result must be \( \text{Negative} / \text{Negative} = \text{Positive} \). This eliminates A, B, and C. Let’s test D and E. Can the result be 2? Let \(a = -3, b = -1\). \(a+b = -4\). Let \(c = -2\). \( \frac{-4}{-2} = 2 \). This is possible.

Solution (B): 1. Find the prime factorization of 210. \(210 = 10 \times 21 = (2 \times 5) \times (3 \times 7)\). 2. List P = {2, 3, 5, 7}. 3. The range is \(\text{Largest} – \text{Smallest} = 7 – 2 = 5\).

Solution (A): 1. Factor the difference of squares: \((x-y)(x+y) = 13\). 2. Since 13 is a prime number, its only positive integer factors are 1 and 13. 3. Since \(x\) and \(y\) are positive integers, \(x+y\) must be the larger factor. 4. \(x – y = 1\) 5. \(x + y = 13\) 6. Add the two equations: \(2x = 14 \implies x = 7\). (Subtracting gives \(2y = 12 \implies y = 6\)).