1. Use Euclid's algorithm: HCF(60, 24) = ____.
- A6
- B12
- C24
- D120
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Answer: B) 12
60 = 24×2 + 12; 24 = 12×2 + 0. HCF = 12.
Class 10 Real Numbers — practice questions with answers
Number Systems · 5 sample questions from the school syllabus. Try each one, then open the answer and the worked solution.
2. If a number ends in the digit 0, then for it to be of the form 2^m × 5^n only, it must be divisible by:
- A10
- B3
- C7
- D9
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Answer: A) 10
A number ending in 0 is divisible by 10 = 2 × 5. (Note: 6^n can never end in 0 because it has no factor 5.)
3. If the HCF of 65 and 117 is expressible in the form 65m − 117, then the value of m is:
- A2
- B3
- C4
- D1
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Answer: A) 2
HCF(65, 117): 117 = 65·1 + 52; 65 = 52·1 + 13; 52 = 13·4 + 0. HCF = 13. So 13 = 65m − 117 → 65m = 130 → m = 2.
4. The LCM of two numbers is 14 times their HCF, and the sum of the LCM and the HCF is 600. If one of the numbers is 280, find the other.
- A80
- B20
- C60
- D120
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Answer: A) 80
Let HCF = h, so LCM = 14h and 14h + h = 600 gives h = 40 and LCM = 560. Then HCF × LCM = product: 40 × 560 = 22400 = 280 × (other), so the other number is 22400 ÷ 280 = 80. Verify: HCF(280, 80) = 40 and LCM(280, 80) = 560. Dividing 600 by 14 instead of 15 is the usual slip.
5. Assertion (A): has a terminating decimal expansion. Reason (R): 30 = 2 × 3 × 5. Which option is correct?
- ABoth A and R are true, and R is the correct explanation of A
- BBoth A and R are true, but R is NOT the correct explanation of A
- CA is true but R is false
- DA is false but R is true
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Answer: D) A is false but R is true
R states a correct factorisation: 30 = 2 × 3 × 5. But that factorisation contains a 3, so the denominator is NOT of the form 2ᵐ × 5ⁿ and is non-terminating recurring — A is false. Deciding from the numerator 17, or from "30 ends in 0 so it must divide nicely", is the usual error.
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