Class 10 Mathematics: Real Numbers important questions

Real Numbers is a chapter of Class 10 Mathematics for the Andhra Pradesh State Board. These 12 questions are marked Very Important, Important or Practice, and each answer has been checked by our team. Use them to revise, not as a prediction of what will come in the exam.

1. Important · 2 marks

   1. Find the HCF of 220 and 40.

   By prime factorisation: 220 = 2² × 5 × 11 and 40 = 2³ × 5, so HCF = 2² × 5 = 20. (By Euclid's division: 220 = 40 × 5 + 20 and 40 = 20 × 2 + 0, so HCF = 20.)

2. Very Important · 8 marks

   2. Prove that √2 + √5 is irrational.

   Suppose √2 + √5 = r, a rational number. Then √5 = r − √2. Squaring: 5 = r² − 2r√2 + 2, so 2r√2 = r² − 3 and √2 = (r² − 3)/(2r). The right side is rational (r ≠ 0), so √2 would be rational. This contradicts the fact that √2 is irrational. So √2 + √5 is irrational.

3. Practice · 1 marks

   3. Find the prime factorisation of 30.

   30 = 2 × 3 × 5.

4. Very Important · 8 marks

   4. Prove that 2 + 5√3 is irrational.

   Suppose 2 + 5√3 is rational: 2 + 5√3 = a/b, where a, b are integers and b ≠ 0. Then 5√3 = (a − 2b)/b, so √3 = (a − 2b)/(5b), which is rational. This contradicts the fact that √3 is irrational. So 2 + 5√3 is irrational.

5. Practice · 1 marks

   5. Two positive integers are A = p³q² and B = pq³, where p and q are prime numbers. Find HCF(A, B). A) p³q² B) pq² C) p³q D) pq

   B) pq² HCF takes the smallest power of each common prime: p¹ and q². So HCF = pq².

6. Very Important · 8 marks

   6. Is 3 + 2√5 irrational? Justify your answer.

   Yes. Suppose 3 + 2√5 is rational, say a/b with integers a, b and b ≠ 0. Then 2√5 = a/b − 3 = (a − 3b)/b, so √5 = (a − 3b)/(2b), which is rational. But √5 is irrational, which is a contradiction. So 3 + 2√5 is irrational.

7. Practice · 1 marks

   7. If HCF(20, 35) = 5, then LCM(20, 35) = ? A) 140 B) 175 C) 250 D) 350

   A) 140 HCF × LCM = product of the numbers, so LCM = (20 × 35)/5 = 140.

8. Very Important · 8 marks

   8. Is √5 irrational? Justify your answer.

   Yes. Suppose √5 is rational: √5 = a/b, where a and b are co-prime integers and b ≠ 0. Squaring: 5b² = a², so 5 divides a², and hence 5 divides a. Write a = 5c. Then 5b² = 25c², so b² = 5c². So 5 divides b², and hence 5 divides b. Now 5 divides both a and b, which contradicts that they are co-prime. So √5 is irrational.

9. Practice · 1 marks

   9. In a factor tree, the number y splits into 4 and x, and x splits into 3 and 7. Find the values of x and y. A) 10, 14 B) 21, 25 C) 21, 84 D) 84, 21

   C) 21, 84 x = 3 × 7 = 21, and y = 4 × x = 4 × 21 = 84.

10. Very Important · 8 marks

    10. Given that √5 is irrational, prove that 2 + 3√5 is also irrational.

    Suppose 2 + 3√5 is rational. Then 2 + 3√5 = a/b, where a and b are integers and b ≠ 0. So 3√5 = a/b − 2 = (a − 2b)/b, and √5 = (a − 2b)/(3b). The right side is a ratio of integers, so it is rational. That means √5 is rational, which contradicts the fact that √5 is irrational. So our assumption is wrong, and 2 + 3√5 is irrational.

11. Practice · 1 marks

    11. The prime factorisation of 156 is a² × b × c. Find a + b + c.

    156 = 2 × 2 × 3 × 13 = 2² × 3 × 13, so a = 2, b = 3, c = 13. a + b + c = 2 + 3 + 13 = 18.

12. Very Important · 8 marks

    12. Given that √3 is irrational, prove that (5 + 2√3)/7 is also irrational, using the method of contradiction.

    Suppose (5 + 2√3)/7 is rational. Then (5 + 2√3)/7 = a/b, where a, b are integers and b ≠ 0. 5 + 2√3 = 7a/b, so 2√3 = (7a − 5b)/b and √3 = (7a − 5b)/(2b). The right side is a ratio of integers, so √3 would be rational. This contradicts the given fact that √3 is irrational. So (5 + 2√3)/7 is irrational.