Class 10 Maths Chapter 1 Real Numbers Most Important Questions By PW

PW brings you Most Important Questions (MIQs) for this chapter, carefully selected to cover key topics like Euclid’s Division Lemma and the Fundamental Theorem of Arithmetic. With regular practice of these questions, you can strengthen your concepts, reduce mistakes, and approach your Class 10 exams with better speed and confidence.

PW Most Important Questions for Class 10 Maths Chapter 1 Real Numbers 

Below we have provided Important Questions for Class 10 Maths Chapter 1 to help students prepare better for their Class 10 maths exams. Students can prepare these Important Questions for Class 10 Maths Chapter 1 Real Numbers before their exams to understand the concepts better.

Q.1: Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

Solution:

Let x be any positive integer and y = 3. By Euclid’s division algorithm; x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3) Therefore, x = 3q, 3q + 1 and 3q + 2 As per the given question, if we take the square on both the sides, we get; x ² = (3q) ² = 9q ² = 3.3q ² Let 3q ² = m Therefore, x ² = 3m ………………….(1) x ² = (3q + 1) ² = (3q) ² + 1 ² + 2 × 3q × 1 = 9q ² + 1 + 6q = 3(3q ² + 2q) + 1 Substituting 3q ² +2q = m we get, x ² = 3m + 1 ……………………………. (2) x ² = (3q + 2) ² = (3q) ² + 2 ² + 2 × 3q × 2 = 9q ² + 4 + 12q = 3(3q ² + 4q + 1) + 1 Again, substituting 3q ² + 4q + 1 = m, we get, x ² = 3m + 1…………………………… (3) Hence, from eq. 1, 2 and 3, we conclude that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

 

Q.2: Express each number as a product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Solution:

(i) 140 Using the division of a number by prime numbers method, we can get the product of prime factors of 140. Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 2 ² × 5 × 7 (ii) 156 Using the division of a number by prime numbers method, we can get the product of prime factors of 156. Hence, 156 = 2 × 2 × 13 × 3 = 2 ² × 13 × 3 (iii) 3825 Using the division of a number by prime numbers method, we can get the product of prime factors of 3825. Hence, 3825 = 3 × 3 × 5 × 5 × 17 = 3 ² × 5 ² × 17 (iv) 5005 Using the division of a number by prime numbers method, we can get the product of prime factors of 5005. Hence, 5005 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13 (v) 7429 Using the division of a number by prime numbers method, we can get the product of prime factors of 7429. Hence, 7429 = 17 × 19 × 23 = 17 × 19 × 23

 

Q.3: Given that HCF (306, 657) = 9, find LCM (306, 657). 

Solution:

As we know, HCF × LCM = Product of the two given numbers So, 9 × LCM = 306 × 657 LCM = (306 × 657)/9 = 22338 Therefore, LCM(306,657) = 22338

 

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

Solution:

Let 3 + 2√5 be a rational number. Then the co-primes x and y of the given rational number where (y ≠ 0) is such that: 3 + 2√5 = x/y Rearranging, we get, 2√5 = (x/y) – 3 √5 = 1/2[(x/y) – 3] Since x and y are integers, thus, 1/2[(x/y) – 3] is a rational number. Therefore, √5 is also a rational number. But this confronts the fact that √5 is irrational. Thus, our assumption that 3 + 2√5 is a rational number is wrong. Hence, 3 + 2√5 is irrational.

 

Q.5: Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600

Solution:

Note: If the denominator has only factors of 2 and 5 or in the form of 2 ^m × 5 ⁿ then it has a terminating decimal expansion.

If the denominator has factors other than 2 and 5 then it has a non-terminating repeating decimal expansion. (i) 13/3125 Factoring the denominator, we get, 3125 = 5 × 5 × 5 × 5 × 5 = 5 ⁵ Or = 2 ⁰ × 5 ⁵ Since the denominator is of the form 2 ^m × 5 ⁿ then, 13/3125 has a terminating decimal expansion. (ii) 17/8 Factoring the denominator, we get, 8 = 2× 2 × 2 = 2 ³ Or = = 2 ³ × 5 ⁰ Since the denominator is of the form 2 ^m × 5 ⁿ then, 17/8 has a terminating decimal expansion. (iii) 64/455 Factoring the denominator, we get, 455 = 5 × 7 × 13 Since the denominator is not in the form of 2 ^m × 5 ⁿ , therefore 64/455 has a non-terminating repeating decimal expansion. (iv) 15/1600 Factoring the denominator, we get, 1600 = 2 ⁶ × 5 ² Since the denominator is in the form of 2 ^m × 5 ⁿ , 15/1600 has a terminating decimal expansion.

 

Q.6: The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form, p/q what can you say about the prime factors of q?

(i) 43.123456789

(ii) 0.120120012000120000. . .

Solution:

(i) 43.123456789 Since it has a terminating decimal expansion, it is a rational number in the form of p/q and q has factors of 2 and 5 only. (ii) 0.120120012000120000. . . Since it has a non-terminating and non-repeating decimal expansion, it is an irrational number.

Q.7: Check whether 6 ⁿ can end with the digit 0 for any natural number n.

Solution:

If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with a unit place as 0 or 5 is divisible by 5. Prime factorization of 6 ⁿ = (2 × 3) ⁿ Therefore, the prime factorization of 6 ⁿ doesn’t contain the prime number 5. Hence, it is clear that for any natural number n, 6 ⁿ is not divisible by 5 and thus it proves that 6 ⁿ cannot end with the digit 0 for any natural number n.

 

 

Solution: The smallest prime number = 2 The smallest composite number = 4 Prime factorisation of 2 = 2 Prime factorisation of 4 = 2 × 2 HCF(2, 4) = 2 Therefore, the HCF of the smallest prime number and the smallest composite number is 2.

 

 

Solution: 2048 > 960 Using Euclid’s division algorithm, 2048 = 960 × 2 + 128 960 = 128 × 7 + 64 128 = 64 × 2 + 0 Therefore, the HCF of 2048 and 960 is 64.

 

 

Solution:  Prime factorisation of 404 = 2 × 2 × 101 Prime factorisation of 96 = 2 × 2 × 2 × 2 × 2 × 3 = 2 ⁵ × 3 HCF = 2 × 2 = 4 LCM = 2 ⁵ × 3 × 101 = 9696 HCF × LCM = 4 × 9696 = 38784 Product of the given two numbers = 404 × 96 = 38784 Hence, verified that LCM × HCF = Product of the given two numbers.