HCF of 26 and 91 is:
- A. 13
- B. 7
- C. 26
- D. 1
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Correct option: A — 13
91 = 26 × 3 + 13; 26 = 13 × 2 + 0, so HCF is 13.
Class 10 Mathematics
Important questions
4 Practice Questions — The answers are hidden. Reveal them to check your answers. The Timed Test will be available in Phase 3.
Chapter: Ch. 1
Prove that √3 is irrational (outline the usual contradiction method).
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Assume √3 = p/q in lowest terms. Then p² = 3q² so p is divisible by 3; write p = 3k. Then 9k² = 3q² ⇒ q² = 3k², so q is also divisible by 3. This contradicts lowest terms. Hence √3 is irrational.
The same pattern works for √2, √5, √7.
Use Euclid’s algorithm to find HCF(405, 252) and write 405 and 252 as products of primes.
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405 = 252 × 1 + 153
252 = 153 × 1 + 99
153 = 99 × 1 + 54
99 = 54 × 1 + 45
54 = 45 × 1 + 9
45 = 9 × 5 + 0
HCF = 9.
405 = 5 × 3⁴; 252 = 2² × 3² × 7. Common primes give 3² = 9.
Prime factorisation is a check, not a replacement, for the lemma method in exams.
Next step
These practice sets use original GKSchools wording — they are not a verbatim dump of the NCERT textbook. “Important” means high-weight practice for the chapter, not a claim that these are previous-year exam questions.
