HOTS Questions on Chapter 3 ‘The World of Numbers’ for Class 9 with Answers

The Higher-Order Thinking Skills (HOTS) question set for Chapter 3: The World of Numbers for Class 9 is designed to move students beyond routine exercises and to build deep, flexible mathematical thinking. Aligned with CBSE and NCERT objectives, these questions focus on analysis, reasoning, pattern recognition, and real-world application of number concepts such as types of numbers, laws of exponents, prime factorization, HCF and LCM strategies, divisibility tests, and number-line reasoning. 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 ‘The World of Numbers’

Question 1: The set of natural numbers is closed under subtraction. True or False? Give two counterexamples.

Solution:

Answer: FALSE

A set is closed under an operation if performing that operation on any two members always produces a member of the same set.

Counter-example 1: 3 − 5 = −2. Here, 3 ∈ ℕ and 5 ∈ ℕ, but −2 ∉ ℕ.

Counter-example 2: 1 − 7 = −6. Both 1 and 7 are natural numbers, but −6 is a negative integer, not a natural number.

Question 2: The temperature in Ladakh at noon is 4°C. By midnight, it drops by 15°C. Write the midnight temperature as an integer and justify using Brahmagupta’s debt-fortune model.

Solution:

Answer: Midnight temperature = 4 + (−15) = 4 − 15 = −11°C

In Brahmagupta's framework, a fortune of 4 (4°C above zero) combined with a debt of 15 (a drop of 15°) gives a net debt of 11 or 11 degrees below zero.

Question 3: Explain why the average of two rational numbers, a and b is always a rational number that lies strictly between them.

Solution:

Let a and b be rational numbers with a < b.

Average = (a + b) / 2 

Step 1: Show it is rational. Since a and b are rational, their sum a+b is rational (closed under addition). 

Dividing by 2 gives (a+b)/2, which is rational (closed under division by non-zero rational). 

Step 2: Show it lies between a and b. We need a < (a + b)/2 < b 

Left inequality: multiply both sides by 2 ⇒ 2a < a + b ⇒ a < b  (true by assumption) 

Right inequality: multiply both sides by 2 ⇒ a + b < 2b ⇒ a < b (true by assumption)

Question 4: Let a and b be two non-zero rational numbers such that a/b + 1 = 0. Without assigning any numerical values, determine whether the product ab is positive or negative. Justify.

Solution: Given: a/b + 1 = 0 ⇒ a/b = −1 ⇒ a = −b

Therefore: ab = (−b) × b = −b²

Since b ≠ 0, b² is always positive (a non-zero number squared is positive). So −b² is always negative.

Hence, ab < 0 for all valid non-zero rational values of a and b satisfying the given condition.

Question 5: Prove that √5 is irrational. Will the same method of proof work for √4? Why or why not?

Solution: 

Proof that √5 is irrational:

Assume √5 = p/q (co-prime, q ≠ 0). 

⇒ 5q² = p² 

⇒ p² is divisible by 5 ⇒ p is divisible by 5. 

Let p = 5k. 

⇒ 5q² = 25k² 

⇒ q² = 5k² 

⇒ q is divisible by 5.

Both p, q divisible by 5 

CONTRADICTION to the assumption that p and q are co-primes.

Hence, √5 is irrational.

Does it work for √4?

Assume √4 = p/q (co-prime, q ≠ 0). 

⇒ 4q² = p² 

Now p² is divisible by 4. 

But p could be even with p = 2k: ⇒ 4q² = 4k² 

⇒ q² = k² ⇒ q = k 

This means p/q = 2k/k = 2. No contradiction arises

The proof by contradiction for √n works when n is NOT a perfect square. For perfect squares like 4, 9, 16, the square root is rational, so the proof correctly ‘fails’.

Question 6: Classify each of the following as rational or irrational:

(i) √81  (ii) √12 iii) 0.333… (iv) 0.12345 12345 12345…

(v) 1.010010001000… (vi) 23.56018561…

Solution:

 (i) √81 = 9 → RATIONAL (perfect square) 

(ii) √12 = √(4×3) = 2√3 → IRRATIONAL (√3 is irrational; multiplying by rational 2 keeps it irrational) 

(iii) 0.333... = 0.3̄ = 1/3 → RATIONAL (repeating decimal) 

(iv) 0.12345 12345 12345... The block ‘12345’ repeats → RATIONAL 

p/q form: let  x=0.12345―; then  100000x=12345.12345―, 99999x = 12345 

x = 12345/99999 = 823/6667 

(v) 1.010010001000... Pattern: 1 zero, then 2 zeros, then 3 zeros: increasing zeros, NO fixed repeating block → IRRATIONAL 

(vi) 23.56018561... (non-terminating, no visible repeating block stated) → IRRATIONAL

Question 7: A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your reasoning.

Solution:

The denominator is 2³ × 5¹. Since it has only factors of 2 and 5, the decimal terminates.

To convert to a power of 10, we need the denominator =  10n=2n×5n. Currently: 2³ × 5¹ 

Need to multiply by 5² = 25 (to balance the powers): 2³ × 5¹ × 5² = 2³ × 5³ = 10³ = 1000 

So the denominator becomes 1000 → 3 decimal places.

Answer: 3 decimal places.

General Rule: If the denominator in lowest form is  2m×5n, the decimal terminates in max(m, n) decimal places.

Question 8: If 2a = 3b = 6c, show that c = ab/(a + b)

Solution: 

Let 2a = 3b = 6c = k (some constant). 

Then: 2=k1/a,3=k1/b,6=k1/c 

Since 6 = 2 × 3:  k1/c=k1/a×k1/b=k1/a+1/b

Therefore: 1/c = 1/a + 1/b 1/c = (a + b) / (ab) 

⇒ c = ab/(a + b)

Question 9: Three rational numbers x, y, and z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that x = y = z

Solution: 

Given: x + y + z = 0 → z = −(x + y) and xy + yz + zx = 0 

Substituting z = −(x+y): xy + y(−x−y) + x(−x−y) = 0 

⇒ xy − xy − y² − x² − xy = 0 

⇒ −x² − xy − y² = 0 x² + xy + y² = 0 

Multiply by 4: 4x² + 4xy + 4y² = 0 

(4x² + 4xy + y²) + 3y² = 0 

⇒ (2x + y)² + 3y² = 0 

Since (2x+y)² ≥ 0 and 3y² ≥ 0, their sum = 0 requires 3y² = 0 

⇒ y = 0 

(2x + y)² = 0 ⇒ 2x + 0 = 0 ⇒ x = 0 

⇒ z = −(0+0) = 0 

Therefore, x = y = z = 0.

 

Questions PDF with worked-out examples for Class 9 Chapter 3: The World of Numbers, perfect for last-minute CBSE exam revision.

Class 9 Chapter 3: The World of Numbers HOTS PDF