In the following question, four number pairs are given: 10 – 121, 12 – 169, 14 – 225 and 16 – 279. In three of these pairs, the second number is the square of an odd integer that is exactly one greater than the first number, while in one pair this square relationship is not satisfied correctly. Which pair is the odd one out?

Introduction / Context:
This is an odd one out question involving number pairs and square relationships. Each pair consists of a smaller first number and a larger second number. In three pairs, the second number is the square of an odd number that is exactly one greater than the first number. In one case, this pattern breaks. Recognising squares and understanding such structured relationships is common in aptitude tests.

Given Data / Assumptions:

• Number pairs: 10 – 121, 12 – 169, 14 – 225, 16 – 279.
• We will compare the first number with the square root of the second number.
• We suspect that the second number is intended to be (first number + 1)^2 for three of the pairs.
• One pair will not fit this formula and will be the odd one out.

Concept / Approach:
We use the idea of perfect squares and simple algebraic relations:

• If a pair is of the form (n, m), we check if m = (n + 1)^2.
• Since the first numbers 10, 12, 14 and 16 are even, the next integers 11, 13, 15 and 17 are odd, and their squares may appear as the second numbers.

We will compute (n + 1)^2 for each first number and see which pair does not match.

Step-by-Step Solution:
Step 1: For the pair 10 – 121: First number n = 10, so n + 1 = 11. Compute (n + 1)^2 = 11^2 = 121. The second number is 121, which matches. Step 2: For the pair 12 – 169: n = 12, so n + 1 = 13. Compute (n + 1)^2 = 13^2 = 169. The second number is 169, which matches. Step 3: For the pair 14 – 225: n = 14, so n + 1 = 15. Compute (n + 1)^2 = 15^2 = 225. The second number is 225, which matches. Step 4: For the pair 16 – 279: n = 16, so n + 1 = 17. Compute (n + 1)^2 = 17^2 = 289. Step 5: The given second number is 279, but the correct square should be 289, so this pair does not satisfy the same relation. Step 6: Therefore, 16 – 279 is the only number pair that fails the pattern m = (n + 1)^2 and is the odd one out.

Verification / Alternative check:
List the intended pattern: 10 → 11^2 = 121 ✔ 12 → 13^2 = 169 ✔ 14 → 15^2 = 225 ✔ 16 → 17^2 = 289 ✘ (but given as 279) The first three pairs follow the formula perfectly, while the last pair uses an incorrect second value, confirming that the pair 16 – 279 breaks the pattern.

Why Other Options Are Wrong:
10 – 121 is not odd because 121 equals 11^2, following m = (n + 1)^2. 12 – 169 is not odd because 169 equals 13^2, again following the pattern. 14 – 225 is not odd because 225 equals 15^2 and fits the rule. Only 16 – 279 violates the rule since 279 is not equal to 17^2.

Common Pitfalls:
Students might miscalculate 17^2 and think it is 279 instead of 289, leading to the wrong conclusion that all pairs follow the pattern. Another pitfall is to attempt more complex relationships, such as mixing multiplication and addition, instead of checking the simple square form first. Carefully computing squares for small integers helps avoid such mistakes.

Final Answer:
The number pair that does not satisfy the square-of-next-odd-number rule and is therefore the odd one out is 16 – 279.