HOTS Questions on Class 10 Maths Chapter 1 ‘Real Numbers’ with Answers

Higher-Order Thinking Skills (HOTS) question set for Chapter 1, 'Real Numbers', Class 10 is crafted to move learners beyond procedural exercises and build rigorous, flexible number sense. Aligned with CBSE and NCERT objectives, the questions focus on deep conceptual understanding and inventive application of real-number ideas such as the Euclidean algorithm and HCF, the fundamental theorem of arithmetic, classification of numbers, decimal expansions, and properties of real numbers on the number line. Each question includes a clear answer and brief reasoning to support classroom discussion and self-study. Download the PDF to access the same HOTS questions in a printable format, making offline learning and exam revision easier.

Solved HOTS Questions on Real Numbers

Question 1: Prove that the square of any positive integer is of the form 4q or 4q + 1 for some integer q.

Solution: 

Step 1: Let n be any positive integer. By Euclid's Lemma with b = 2, every integer is either of the form 2m (even) or 2m + 1 (odd).

Case 1  Even: If n = 2m, then n² = 4m². Setting q = m², we get n² = 4q.

Case 2  Odd: If n = 2m + 1, then n² = 4m² + 4m + 1 = 4(m² + m) + 1. Setting q = m² + m, we get n² = 4q + 1.

Every square is of the form 4q or 4q + 1. This also means a square can never be of the form 4q + 2 or 4q + 3.

Question 2: Show that the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Solution: 

Step 1: Divide any positive integer by 3. The remainder is 0, 1, or 2. So every integer has one of these forms: 3q, 3q + 1, or 3q + 2.

Case 1: n = 3q ⇒ n³ = 27q³ = 9(3q³) = 9m, where m = 3q³.

Case 2: n = 3q + 1 ⇒ n³ = 27q³ + 27q² + 9q + 1 = 9(3q³ + 3q² + q) + 1 = 9m + 1.

Case 3: n = 3q + 2 ⇒ n³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8 = 9m + 8.

∴ Any cube is of the form 9m, 9m + 1, or 9m + 8

Question 3: The prime factorisation of a natural number n is 2³ × 3² × 5 × 7. How many factors does n have? Is n divisible by 45? Justify.

Solution: 

Number of factors: If n = p₁ᵃ × p₂ᵇ × p₃ᶜ..., total factors = (a+1)(b+1)(c+1)...

So factors of n = 2³ × 3² × 5¹ × 7¹ = (3+1)(2+1)(1+1)(1+1) = 4 × 3 × 2 × 2 = 48

Is n divisible by 45? 

45 = 3² × 5. The factorisation of n contains 3² and 5¹. 

Since both prime powers of 45 appear in n, yes, n is divisible by 45.

Question 4: Find the largest number that divides 70 and 125, leaving remainders 5 and 8 respectively.

Solution: 

Step 1: If a number d divides 70 leaving remainder 5, then d exactly divides 70 − 5 = 65. Similarly, d divides 125 − 8 = 117.

Step 2: We need the HCF of 65 and 117.

65 = 5 × 13 and 117 = 9 × 13 = 3² × 13

HCF(65, 117) = 13

Verify: 70 ÷ 13 = 5 remainder 5  and 125 ÷ 13 = 9 remainder 8 

Required number = 13

Question 5: Explain why 7 × 11 × 13 + 13 is a composite number.

Solution: 

Factorise: 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) 

= 13 × (77 + 1) = 13 × 78 

= 13 × 2 × 39 = 13 × 2 × 3 × 13

The number = 2 × 3 × 13², which has more than two factors. By definition, it is a composite number. It is not prime because it has factors other than 1 and itself.

Question 6: Can two numbers have HCF = 18 and LCM = 380? Give reasons.

Solution: 

Key Property: HCF of two numbers always divides their LCM.

Check: Does 18 divide 380?

380 ÷ 18 = 21.11... 

No, 18 does not divide 380.

Two numbers with HCF = 18 and LCM = 380 cannot exist. This is a contradiction of the fundamental property that HCF | LCM.

Question 7: Find the smallest 4-digit number that is exactly divisible by 8, 12, and 20.

Solution: 

Step 1: Find LCM(8, 12, 20).

8 = 2³, 12 = 2² × 3, 20 = 2² × 5

LCM = 2³ × 3 × 5 = 120

Step 2: Find the smallest 4-digit multiple of 120. Smallest 4-digit number = 1000.

1000 ÷ 120 = 8.33... 

The next whole = 9.

120 × 9 = 1080

Smallest 4-digit number = 1080

Question 8: If a and b are rational and p and q are irrational, is a + b√p always irrational? Give a counterexample or prove it.

Solution:

It depends on b: If b = 0, then a + b√p = a, which is rational. So the statement is not always true.

But if b ≠ 0: Assume a + b√p = r (rational). Then √p = (r − a)/b. Since a, b, and r are all rational and b ≠ 0, the RHS is rational. But √p is irrational, which is a contradiction.

a + b√p is irrational whenever b ≠ 0

Question 9: The decimal expansion of a number is 0.142857142857... Is this number rational or irrational? If rational, express it in p/q form.

Solution:

Pattern: ‘142857’ repeats every 6 digits, so this is a non-terminating repeating decimal = rational.

Let x = 0.142857̄. Then 1000000x = 142857.142857̄.

Subtracting: 999999x = 142857, so x = 142857/999999 = 1/7.

Question 10: Three bells ring at intervals of 6, 12, and 18 minutes, respectively. They all ring together at 8:00 AM. At what time will they next ring together? How many times will the first bell have rung by then (after 8 AM)?

Solution:

Step 1: Find LCM(6, 12, 18).

6 = 2 × 3, 12 = 2² × 3, 18 = 2 × 3²

LCM = 2² × 3² = 36 minutes

Step 2: All three bells will ring together again at 8:00 AM + 36 minutes = 8:36 AM.

Step 3: The first bell rings every 6 minutes. In 36 minutes, it rings 36 ÷ 6 = 6 times (at 8:06, 8:12, 8:18, 8:24, 8:30, 8:36).

Questions PDF with worked-out examples for Class 10 Chapter 1: Real Numbers, perfect for last-minute CBSE exam revision.

Class 10 Chapter 1: Real Numbers HOTS PDF

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