Select the distinct pair: Choose the pair in which both numbers are perfect squares (n^2). In the other pairs, at least one of the numbers is not a perfect square.

Difficulty: Easy

Correct Answer: 9, 49

Explanation:


Introduction / Context:
Pair-classification problems often rely on a shared property. Here, we look for the pair in which both numbers are perfect squares. In the remaining pairs, at least one member is not a square.


Given Data / Assumptions:

  • Pairs: (9, 49), (13, 121), (10, 61), (7, 25).
  • Recall: 9 = 3^2, 49 = 7^2, 121 = 11^2, 25 = 5^2.


Concept / Approach:
Check each element of every pair for being a perfect square. A qualifying pair must have both elements equal to n^2 for some integers n.


Step-by-Step Solution:

(9, 49) → 9 = 3^2 and 49 = 7^2 → both squares → qualifies.(13, 121) → 13 is not a square; 121 = 11^2 → at least one non-square → disqualify.(10, 61) → neither 10 nor 61 is a perfect square → disqualify.(7, 25) → 7 is not a square; 25 = 5^2 → at least one non-square → disqualify.


Verification / Alternative check:
Compute integer square roots: sqrt(9) = 3, sqrt(49) = 7 are integers; sqrt(13), sqrt(10), sqrt(61), sqrt(7) are not integers; sqrt(121) = 11 and sqrt(25) = 5 are integers. Only the first pair has both square roots integral.


Why Other Options Are Wrong:
Each contains at least one non-square, so they fail the “both are squares” criterion.


Common Pitfalls:
Assuming that the presence of one square in a pair (e.g., 121 or 25) is sufficient; the requirement is both elements must be squares.


Final Answer:
9, 49 is the only pair where both numbers are perfect squares.

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