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Difficulty: MediumNumber Properties and Integer Constraints in Data Sufficiency

If pp and qq are positive real numbers, is pp an integer?

(1) p2+pqp^2 + pq is an integer.
(2) q2+pqq^2 + pq is an integer.

  1. A
    Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
  2. B
    Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
  3. C
    BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
  4. D
    EACH statement ALONE is sufficient.
  5. Statements (1) and (2) TOGETHER are NOT sufficient.Answer

Answer

Statements (1) and (2) TOGETHER are NOT sufficient to determine if pp is an integer.
The correct answer states that statements (1) and (2) together are not sufficient. Statement (1) alone allows p=0.5,q=1.5p = 0.5, q = 1.5 (giving p2+pq=1p^2+pq=1) as well as p=1,q=1p = 1, q = 1 (giving p2+pq=2p^2+pq=2). Statement (2) alone similarly allows p=0.5,q=1.5p = 0.5, q = 1.5 (giving q2+pq=3q^2+pq=3) as well as p=1,q=1p = 1, q = 1 (giving q2+pq=2q^2+pq=2). When both statements are evaluated together, adding the two equations reveals p+q=(p2+pq)+(q2+pq)p+q = \sqrt{(p^2+pq) + (q^2+pq)}. Choosing p2+pq=1p^2+pq = 1 and q2+pq=3q^2+pq = 3 yields p+q=2p+q = 2 and p=0.5p = 0.5, demonstrating that pp does not have to be an integer even when both statements are satisfied.

Step-by-Step Solution

1
Analyze the Question Stem and Constraints
Variables pp and qq are positive real numbers (not constrained to integers). The goal is to determine if pp must be an integer.
Recognizing that variables are real numbers prevents premature integer assumptions.
2
Evaluate Statement (1) independently: p2+pq=p(p+q)p^2 + pq = p(p+q) is an integer
Case 1: If p=1p = 1 and q=1q = 1, then p(p+q)=1(2)=2p(p+q) = 1(2) = 2 (integer), so pp IS an integer.
Case 2: If p=0.5p = 0.5 and q=1.5q = 1.5, then p(p+q)=0.5(2)=1p(p+q) = 0.5(2) = 1 (integer), so pp IS NOT an integer.
Statement (1) is INSUFFICIENT.
Testing non-integer real values tests whether the condition forces pp to be an integer.
3
Evaluate Statement (2) independently: q2+pq=q(p+q)q^2 + pq = q(p+q) is an integer
Case 1: If p=1p = 1 and q=1q = 1, then q(p+q)=1(2)=2q(p+q) = 1(2) = 2 (integer), so pp IS an integer.
Case 2: If p=0.5p = 0.5 and q=1.5q = 1.5, then q(p+q)=1.5(2)=3q(p+q) = 1.5(2) = 3 (integer), so pp IS NOT an integer.
Statement (2) is INSUFFICIENT.
Statement (2) primarily constrains q(p+q)q(p+q), leaving pp unconstrained.
4
Evaluate Statements (1) and (2) Combined
Let p(p+q)=k1p(p+q) = k_1 and q(p+q)=k2q(p+q) = k_2 where k1,k2k_1, k_2 are positive integers.
Adding gives (p+q)2=k1+k2  ⟹  p+q=k1+k2(p+q)^2 = k_1 + k_2 \implies p+q = \sqrt{k_1 + k_2}.
Thus, p=k1k1+k2p = \frac{k_1}{\sqrt{k_1 + k_2}}.
Case 1: Let k1=1k_1 = 1 and k2=3k_2 = 3. Then p+q=4=2  ⟹  p=12=0.5p+q = \sqrt{4} = 2 \implies p = \frac{1}{2} = 0.5 (not an integer).
Case 2: Let k1=2k_1 = 2 and k2=2k_2 = 2. Then p+q=4=2  ⟹  p=22=1p+q = \sqrt{4} = 2 \implies p = \frac{2}{2} = 1 (an integer).
Since pp can still be either an integer or a non-integer, both statements combined are INSUFFICIENT.
Algebraic combination yields p=k1k1+k2p = \frac{k_1}{\sqrt{k_1 + k_2}}, which produces non-integers for appropriate choices of integer constants k1k_1 and k2k_2.

Key Concept

Avoiding Implicit Integer Assumptions in Real-Valued Data Sufficiency
Estimated Time:2m 0s
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Topics

Number Properties and Integer Constraints in Data Sufficiency

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