CBSE Class 10 Mathematics Basic sample paper 2026-27: question 21(A)
From the official CBSE sample question paper for 2026-27, section B, question 21(A), for 2 marks. Try it first; the hints and our step-by-step solution are below.
Q21(A) · 2 marks · Very short answerReal Numbers
Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.
Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.
Hints
Open one hint at a time and try again before the next.
Hint 1
The number plus 17 must be divisible by both 520 and 468. What is the smallest number divisible by both?
Hint 2
Find the LCM of 520 and 468 using prime factorisation, then undo the 'increase by 17'.
Step-by-step solution
Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.
- 1 mark
The required number increased by 17 must be a common multiple of 520 and 468. For the smallest such number, we use the LCM.
Prime factorisation: and
- ½ mark
- ½ mark
Required number
Answer: 4663
Where the marks are
- Correct prime factorisation of 520 and 468 (1 mark)
- LCM = 4680 (½ mark)
- Required number = 4680 − 17 = 4663 (½ mark)
Revise the chapter
- Chapter: Real Numbers (notes, formulas, where marks are lost and every past question)
- Topic: HCF and LCM by Prime Factorisation
More questions from Real Numbers
- 2026 board paper, set 430/2/1 (Basic), Q26(b) · 3 marksState the 'Fundamental Theorem of Arithmetic' and use it to find LCM of 36 and 54.
- 2026 board paper, set 430/5/1 (Basic), Q22 · 2 marksFind the H.C.F. and L.C.M. of 1530 and 2040.
- Sample paper 2025-26 (Basic), Q21(B) · 2 marksFind the smallest number which is divisible by both 306 and 657.
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