complete solutions for NCERT Class 9 Maths Ganita Manjari Chapter 3: The World of Numbers, covering all exercise sets, in-text questions, examples, and end-of-chapter exercises with full explanations.
NCERT Class 9 Maths Ganita Manjari Chapter 3 Solutions
Chapter Overview: The World of Numbers
This chapter traces the historical evolution of numbers—from ancient counting methods (like the Ishango Bone and finger counting) to the formal classification of numbers into Natural, Whole, Integer, Rational, Irrational, and Real numbers. It highlights India’s contribution through Brahmagupta (who formalised zero and negative numbers) and introduces concepts like density of rational numbers, proof by contradiction for irrationals, and decimal expansions.
Exercise Set 3.1 Solutions
1. A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with?
Solution:
Given: 2 bags = 15 ingots
So, 1 bag = 15/2 ingots
For 12 bags = 12 × (15/2) = 6 × 15 = 90 ingots
Explanation: This is a unitary method problem. The key is finding the value of one unit (1 bag) and then multiplying by the required quantity.
2. Look at the sequence of numbers on one column of the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern.
Solution:
These numbers are prime numbers (numbers with exactly two factors: 1 and the number itself).
Next three primes after 19: 23, 29, 31
Explanation: The Ishango bone (c. 20,000 BCE) is one of the oldest known mathematical artefacts. Its columns contain prime numbers, suggesting early understanding of primality.
3. We know that Natural Numbers are closed under addition (the sum of any two natural numbers is always a natural number). Are they closed under subtraction? Provide a couple of examples to justify your answer.
Solution:
Natural numbers are NOT closed under subtraction.
Examples:
5 − 3 = 2 (Natural number) ✓
3 − 5 = −2 (Not a natural number) ✗
Explanation: Closure means performing an operation on two numbers from a set always gives a result inside the same set. Since subtraction can yield negative numbers (which aren't natural numbers), the set fails closure.
4. Ancient Indians used the joints of their fingers to count, a practice still seen today. Each finger has 3 joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to the ancient base-12 counting systems?
Solution:
Fingers used for counting (excluding thumb) = 4 fingers
Joints per finger = 3
Total count on one hand = 4 × 3 = 12
Relation: Counting up to 12 on one hand naturally led to base-12 (duodecimal) systems, which explain why we have terms like "dozen" (12) and "gross" (144 = 12²).
Exercise Set 3.2 Solutions (Integers & Brahmagupta's Laws)
1. The temperature in Ladakh is recorded as 4°C at noon. By midnight, it drops by 15°C. What is the midnight temperature?
Solution:
Initial temperature = 4°C
Drop = 15°C (subtract)
Midnight temperature = 4 − 15 = −11°C
2. A spice trader takes a loan (debt) of ₹850. The next day, he makes a profit (fortune) of ₹1,200. The following week, he incurs a loss of ₹450. Calculate his final financial standing using integers.
Solution:
Debt = −850 (negative)
Profit = +1200 (positive)
Loss = −450 (negative)
Equation: (−850) + 1200 + (−450)
Step 1: −850 + 1200 = 350
Step 2: 350 − 450 = −100
Final standing: −₹100 (a loss of ₹100)
Explanation: This uses Brahmagupta's concept of representing fortunes as positive and debts as negative numbers.
3. Calculate the following using Brahmagupta's laws (Debt = Negative, Fortune = Positive):
| Expression | Law Applied | Answer |
|---|---|---|
| (i) (−12) × 5 | Negative × Positive = Negative | −60 |
| (ii) (−8) × (−7) | Negative × Negative = Positive | 56 |
| (iii) 0 − (−14) | Zero minus debt = Fortune | 14 |
| (iv) (−20) ÷ 4 | Negative ÷ Positive = Negative | −5 |
4. Explain, using a real-world example of debt, why subtracting a negative number equals adding a positive number.
Solution:
Suppose you have ₹10 and you owe ₹5 (i.e., you have −5 debt)
The expression 10 − (−5) means: "remove a debt of ₹5"
If the debt is cancelled/removed, your effective money increases by ₹5
So, 10 − (−5) = 10 + 5 = 15
Thus: Subtracting a negative = Adding a positive.
Exercise Set 3.3 Solutions (Rational Numbers)
1. Prove that the following rational numbers are equal:
| Pair | Simplification | Conclusion |
|---|---|---|
| (i) 2/3 and 4/6 | 4/6 = (4÷2)/(6÷2) = 2/3 | Equal |
| (ii) 5/4 and 10/8 | 10/8 = 5/4 | Equal |
| (iii) −3/5 and −6/10 | −6/10 = −3/5 | Equal |
| (iv) 9/3 and 3 | 9/3 = 3 | Equal |
2. Find the sum:
| Expression | LCM | Solution |
|---|---|---|
| (i) 2/5 + 3/10 | 10 | 4/10 + 3/10 = 7/10 |
| (ii) 7/12 + 5/8 | 24 | 14/24 + 15/24 = 29/24 |
| (iii) −4/7 + 3/14 | 14 | −8/14 + 3/14 = −5/14 |
3. Find the difference:
| Expression | LCM | Solution |
|---|---|---|
| (i) 5/6 − 1/4 | 12 | 10/12 − 3/12 = 7/12 |
| (ii) 11/8 − 3/4 | 8 | 11/8 − 6/8 = 5/8 |
Explanation: For addition/subtraction of rational numbers, always find the LCM of denominators, convert to equivalent fractions, then operate on numerators.
In-Text Questions (Think & Reflect)
Q: Are all integers rational numbers? Justify.
Solution:
Yes, all integers are rational numbers.
Justification: An integer can be written as , which is of the form with . For example:
5 = 5/1
−3 = −3/1
0 = 0/1
Q: Between any two rational numbers, how many rational numbers exist?
Solution:
Infinitely many rational numbers exist between any two distinct rational numbers.
Example: Between 1/2 = 0.5 and 3/4 = 0.75:
5/8 = 0.625
11/16 = 0.6875
9/16 = 0.5625
And countless more. This property is called density of rational numbers.
Q: Is √2 rational or irrational? Prove.
Solution:
√2 is irrational.
Proof by contradiction:
Assume √2 is rational = p/q in simplest form (p, q integers, q ≠ 0, no common factors)
Squaring: 2 = p²/q² → p² = 2q²
So p² is even ⇒ p is even ⇒ p = 2k
Substitute: (2k)² = 2q² → 4k² = 2q² → 2k² = q²
So q² is even ⇒ q is even
This means p and q are both even ⇒ they have common factor 2
This contradicts our assumption that p/q was in simplest form
Therefore, √2 cannot be rational. Hence, it's irrational.
Q: Classify the decimal expansions:
0.375
0.333...
0.1010010001...
Solution:
| Decimal | Type | Reason |
|---|---|---|
| 0.375 | Terminating | Ends after finite digits = 375/1000 |
| 0.333... | Non-terminating repeating | 1/3 in decimal form |
| 0.1010010001... | Non-terminating non-repeating | No pattern repetition → Irrational |
Solved Examples from Chapter
Example: Locate √2 on the number line.
Solution:
Draw a number line. Mark O (0) and A (1).
At A, draw perpendicular AB of length 1 unit.
Join OB. By Pythagoras: OB = √(1² + 1²) = √2
Using compass, draw an arc with centre O and radius OB to intersect the number line at point P.
OP = √2. Point P represents √2.
Example: Simplify
Solution:
End of Chapter Exercises
1. Short Answer Type
Q1: Write three rational numbers between 2 and 3.
Solution: 2.1, 2.5, 2.75 (or as fractions: 21/10, 5/2, 11/4)
Q2: Identify as rational or irrational:
(i) √9 = 3 → Rational
(ii) 2√3 → Irrational (√3 is irrational)
(iii) 0.121221222... → Irrational (non-terminating non-repeating)
Q3: Express 0.999... in p/q form.
Solution: Let x = 0.999...
10x = 9.999...
Subtract: 10x − x = 9.999... − 0.999... → 9x = 9 → x = 1
Thus, 0.999... = 1 (This shows two different decimal representations for the same number!)
2. Long Answer Type
Q4: Prove that is irrational.
Solution:
Assume √3 = p/q (simplest form)
Squaring: 3 = p²/q² → p² = 3q²
So p² is divisible by 3 ⇒ p is divisible by 3 ⇒ p = 3k
Substituting: (3k)² = 3q² → 9k² = 3q² → 3k² = q²
So q² is divisible by 3 ⇒ q is divisible by 3
Both p and q divisible by 3 ⇒ contradiction to simplest form assumption
Hence, √3 is irrational.
Q5: Represent on the number line.
Solution:
Mark O (0) and A (2) on number line
At A, draw perpendicular AB = 1 unit
Join OB. OB = √(2² + 1²) = √4 + 1 = √5
Draw arc with centre O and radius OB to intersect number line at P
OP = √5
3. Multiple Choice Questions (MCQs)
Q1: Which of the following is an irrational number?
(a) √16
(b) 2/3
(c) 0.242242224...
(d) 0.25
Answer: (c) 0.242242224... (non-terminating non-repeating)
Q2: The decimal expansion of 1/7 is:
(a) Terminating
(b) Non-terminating repeating with period 6
(c) Non-terminating non-repeating
(d) Terminating after 2 decimal places
Answer: (b) 1/7 = 0.142857142857... (repeating block of 6 digits) — this is a cyclic number.
Q3: Between any two rational numbers, there exist:
(a) Exactly one rational number
(b) Exactly two rational numbers
(c) Infinitely many rational numbers
(d) No rational number
Answer: (c) Infinitely many rational numbers (density property)
Summary Table: Number Types
| Number Type | Definition | Examples | Closed Under? |
|---|---|---|---|
| Natural (N) | Counting numbers {1, 2, 3, ...} | 1, 7, 23 | Addition: ✓, Subtraction: ✗ |
| Whole (W) | Natural + {0} | 0, 1, 2 | Addition: ✓ |
| Integers (Z) | {..., −2, −1, 0, 1, 2, ...} | −5, 0, 8 | Add, Sub, Mult: ✓; Division: ✗ |
| Rational (Q) | p/q, q ≠ 0 | −3/4, 0.6, 5 | Add, Sub, Mult, Div (≠0): ✓ |
| Irrational | Not p/q | √2, Ο, e | None of the operations are closed |
| Real (R) | Rational ∪ Irrational | All numbers on number line | Add, Sub, Mult, Div (≠0): ✓ |
Key Takeaways
Zero as a number and negative numbers were formalised by Indian mathematician Brahmagupta in 628 CE.
Proof by contradiction is used to prove irrationality of numbers like √2, √3, √5.
Rational numbers have terminating or non-terminating repeating decimal expansions.
Irrational numbers have non-terminating non-repeating decimal expansions.
The set of rational numbers is dense (infinitely many between any two).
A number can be represented in multiple ways (e.g., 0.999... = 1).