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Mathematics — Standard (041) — CBSE Class 10 board question

Q1. [1] § 1.1 Introduction § 1.2 The Fundamental Theorem of Arithmetic
The greatest number which divides 281 and 1249, leaving remainder 5 and 7 respectively, is :
  1. A $23$
  2. B $276$
  3. C $138$
  4. D $69$
Previously asked in CBSE board exam
2024 30/5/1 Q10
Generated by claude-sonnet-4-6 · 2026-06-15 10:29 · grounding rag
Model Answer

Option (B) 276

Required number = HCF(281 − 5, 1249 − 7) = HCF(276, 1242).
276 = 2² × 3 × 23; 1242 = 2 × 3³ × 23. HCF = 2 × 3 × 23 = 138.

Correct answer: C) 138

Explanation

Subtract each remainder from the respective number to get values exactly divisible by the required HCF: 281 − 5 = 276 and 1249 − 7 = 1242. The greatest such divisor is HCF(276, 1242) = 138. A common mistake is stopping at 276 (option B) without computing the actual HCF.

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