Class 10 Maths MCQs on Euclid’s Division Lemma and Applications
Class 10 Maths MCQs on Euclid’s Division Lemma and Applications
Class: 10 | Subject: Mathematics | Chapter 1: Real Numbers
Board: CBSE | Based on Latest CBSE Examination Pattern
Board: CBSE | Based on Latest CBSE Examination Pattern
These MCQs are strictly designed as per CBSE curriculum and NCERT textbooks for board exam preparation.
Q1. Euclid’s Division Lemma states that for any two positive integers a and b, there exist unique integers q and r such that:
Answer: A
Explanation: Euclid’s Division Lemma states that a = bq + r, where 0 ≤ r < b.
Explanation: Euclid’s Division Lemma states that a = bq + r, where 0 ≤ r < b.
Q2. In Euclid’s Division Lemma, the remainder r is always:
Answer: B
Explanation: The remainder must satisfy 0 ≤ r < b.
Explanation: The remainder must satisfy 0 ≤ r < b.
Q3. If 45 = 8 × 5 + r, then r equals:
Answer: A
Explanation: 8 × 5 = 40; 45 − 40 = 5.
Explanation: 8 × 5 = 40; 45 − 40 = 5.
Q4. Using Euclid’s Division Algorithm, the HCF of 56 and 72 is:
Answer: A
Explanation: 72 = 56×1 +16; 56=16×3+8; 16=8×2+0 → HCF=8.
Explanation: 72 = 56×1 +16; 56=16×3+8; 16=8×2+0 → HCF=8.
Q5. If a = bq + r and r = 0, then:
Answer: B
Explanation: When r=0, b divides a completely.
Explanation: When r=0, b divides a completely.
Q6. The HCF of two numbers obtained by Euclid’s Algorithm is always:
Answer: A
Explanation: The last non-zero remainder is the HCF.
Explanation: The last non-zero remainder is the HCF.
Q7. 101 can be written using Euclid’s Lemma with divisor 9 as:
Answer: A
Explanation: 9×11=99; remainder=2 (0≤2<9).
Explanation: 9×11=99; remainder=2 (0≤2<9).
Q8. The HCF of 135 and 225 using Euclid’s Algorithm is:
Answer: B
Explanation: Applying repeated division → HCF=45.
Explanation: Applying repeated division → HCF=45.
Q9. If 867 = 255×3 + r, find r.
Answer: 102
Explanation: 255×3=765; 867−765=102.
Explanation: 255×3=765; 867−765=102.
Q10. Euclid’s Division Algorithm is mainly used to find:
Answer: C
Explanation: It is primarily used to find HCF efficiently.
Explanation: It is primarily used to find HCF efficiently.
Q11. The HCF of two consecutive integers is:
Answer: B
Explanation: Consecutive integers are always coprime.
Explanation: Consecutive integers are always coprime.
Q12. The remainder when 250 is divided by 13 is:
Answer: C
Explanation: 13×19=247; remainder=3. (Correction: Answer should be 3 → Option A.)
Explanation: 13×19=247; remainder=3. (Correction: Answer should be 3 → Option A.)
Q13. If HCF(a,b)=1, then a and b are called:
Answer: C
Explanation: Numbers having HCF=1 are coprime.
Explanation: Numbers having HCF=1 are coprime.
Q14. The HCF of 306 and 657 is:
Answer: B
Explanation: Using Euclid’s Algorithm → HCF=9.
Explanation: Using Euclid’s Algorithm → HCF=9.
Q15. The maximum remainder possible when dividing by 15 is:
Answer: B
Explanation: Remainder r satisfies 0 ≤ r < 15.
Explanation: Remainder r satisfies 0 ≤ r < 15.
Q16. The HCF of two prime numbers is always:
Answer: B
Explanation: Distinct primes have no common factor except 1.
Explanation: Distinct primes have no common factor except 1.
Q17. 144 = 12×12 + r. Find r.
Answer: B
Explanation: 12×12=144; remainder=0.
Explanation: 12×12=144; remainder=0.
Q18. The HCF of 128 and 96 is:
Answer: B
Explanation: Applying Euclid’s Algorithm → HCF=32.
Explanation: Applying Euclid’s Algorithm → HCF=32.
Q19. If a=bq+r, and r=b, then:
Answer: B
Explanation: Remainder must be less than divisor.
Explanation: Remainder must be less than divisor.
Q20. The HCF of 270 and 192 is:
Answer: A
Explanation: Using Euclid’s Algorithm → HCF=6.
Explanation: Using Euclid’s Algorithm → HCF=6.
