NCERT Solutions for Class 9 Maths Ganit Manjari Chapter 3 The World of Numbers

Real numbers are all numbers that can be represented on the number line, including rational and irrational numbers. In Class 9 Maths Chapter 3, students learn how natural numbers, integers, rational numbers, irrational numbers and real numbers developed and how decimals help identify different types of numbers.

NCERT Solutions for Class 9 Maths Ganit Manjari Chapter 3 help students solve textbook questions from The World of Numbers in the new Ganita Manjari Class 9 Maths textbook. This chapter introduces the history of counting, natural numbers, zero or Śhūnya, integers, rational numbers, number line representation, absolute value, density of rational numbers, irrational numbers, proof of irrationality, decimal expansions, cyclic numbers and real numbers.

The chapter connects number systems with historical ideas such as zero, Brahmagupta’s rules, rational number operations and the representation of numbers on a number line. These solutions for CBSE 2026-2027 help students convert decimals into fractions, identify terminating and repeating decimals, prove irrationality and solve number-line questions confidently.

NCERT Solutions for Class 9 Maths Chapter 3 PDF Download

The NCERT Solutions for Class 9 Maths Chapter 3 PDF is useful for revising number system questions that involve calculations, proofs and classifications. Students can use it to practise rational number operations, decimal expansion problems and irrational number proofs in a structured way.

Students can use the PDF to revise:

1. Natural numbers and counting
2. Zero and Brahmagupta’s rules
3. Integers and negative numbers
4. Rational numbers and fractions
5. Operations on rational numbers
6. Number line representation
7. Absolute value and distance
8. Density of rational numbers
9. Irrational numbers
10. Decimal expansions of rational and irrational numbers
11. Terminating and repeating decimals
12. Square root spiral and real numbers

Access Exercise-Wise NCERT Solutions for Class 9 Maths Chapter 3

NCERT Solutions for Class 9 Maths Chapter 3 Exercise Set 3.1

Exercise Set 3.1 focuses on natural numbers, one-to-one correspondence, prime number patterns and closure properties.

Question 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?

Answer:
For every 2 bags of spices, the merchant receives 15 copper ingots.

12 bags = 6 groups of 2 bags

Number of ingots = 6 × 15
= 90

Final answer:
The merchant will leave with 90 copper ingots.

Question 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.

Answer:
The numbers 11, 13, 17 and 19 are all prime numbers.

The next three prime numbers after 19 are:

23, 29, 31

Final answer:
They are prime numbers. The next three numbers are 23, 29 and 31.

Question 3. We know that natural numbers are closed under addition. Are they closed under subtraction? Provide examples to justify your answer.

Answer:
Natural numbers are not closed under subtraction.

Example 1:

5 - 3 = 2

Here, the result is a natural number.

Example 2:

3 - 5 = -2

Here, -2 is not a natural number.

Also:

4 - 4 = 0

If natural numbers are taken as {1, 2, 3, ...}, then 0 is not a natural number.

Final answer:
Natural numbers are not closed under subtraction.

Question 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?

Answer:
There are 4 fingers on one hand, excluding the thumb.

Each finger has 3 joints.

Total joints = 4 × 3
= 12

The thumb can be used to point to each of these 12 joints.

Final answer:
One can count up to 12 on one hand. This connects with base-12 counting systems because 12 becomes a natural counting unit.

NCERT Solutions for Class 9 Maths Chapter 3 Exercise Set 3.2

Exercise Set 3.2 focuses on zero, negative numbers, integers and Brahmagupta’s rules for fortunes and debts.

Question 1. The temperature in the high-altitude desert of Ladakh is recorded as 4°C at noon. By midnight, it drops by 15°C. What is the midnight temperature?

Answer:
Temperature at noon = 4°C
Drop in temperature = 15°C

Midnight temperature = 4 - 15
= -11°C

Final answer:
The midnight temperature is -11°C.

Question 2. A spice trader takes a loan of ₹850. The next day, he makes a profit of ₹1,200. The following week, he incurs a loss of ₹450. Write this sequence as an equation using integers and calculate his final financial standing.

Answer:
Loan of ₹850 = -850
Profit of ₹1,200 = +1200
Loss of ₹450 = -450

Final standing:

-850 + 1200 - 450
= 350 - 450
= -100

Final answer:
Equation: -850 + 1200 - 450 = -100. The trader has a debt of ₹100.

Question 3. Calculate the following using Brahmagupta’s laws.

(i) (-12) × 5

Answer:
A debt multiplied by a fortune gives a debt.

(-12) × 5 = -60

Final answer:
-60

(ii) (-8) × (-7)

Answer:
A debt multiplied by a debt gives a fortune.

(-8) × (-7) = 56

Final answer:
56

(iii) 0 - (-14)

Answer:
Subtracting a negative number is the same as adding the corresponding positive number.

0 - (-14) = 0 + 14
= 14

Final answer:
14

(iv) (-20) ÷ 4

Answer:
A negative number divided by a positive number gives a negative number.

(-20) ÷ 4 = -5

Final answer:
-5

Question 4. Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number.

Answer:
A negative number can represent debt.

Suppose a person has ₹10. If a debt of ₹5 is removed, the person becomes ₹5 richer.

Mathematically:

10 - (-5) = 10 + 5
= 15

Final answer:
Subtracting a negative number means removing a debt, which increases the value. Therefore, 10 - (-5) = 15.

NCERT Solutions for Class 9 Maths Chapter 3 Exercise Set 3.3

Exercise Set 3.3 focuses on rational numbers, equivalent fractions and arithmetic operations on rational numbers.

Question 1. Prove that the following rational numbers are equal.

(i) 2/3 and 4/6

Answer:
Two rational numbers a/b and c/d are equal if ad = bc.

For 2/3 and 4/6:

2 × 6 = 12
3 × 4 = 12

Since both products are equal, the rational numbers are equal.

Final answer:
2/3 = 4/6

(ii) 5/4 and 10/8

Answer:
5 × 8 = 40
4 × 10 = 40

Final answer:
5/4 = 10/8

(iii) -3/5 and -6/10

Answer:
(-3) × 10 = -30
5 × (-6) = -30

Final answer:
-3/5 = -6/10

(iv) 9/3 and 3

Answer:
9/3 = 3

Final answer:
9/3 = 3

Question 2. Find the sum.

(i) 2/5 + 3/10

Answer:
2/5 = 4/10

So:

4/10 + 3/10 = 7/10

Final answer:
7/10

(ii) 7/12 + 5/8

Answer:
LCM of 12 and 8 = 24

7/12 = 14/24
5/8 = 15/24

Sum = 14/24 + 15/24
= 29/24

Final answer:
29/24

(iii) -4/7 + 3/14

Answer:
-4/7 = -8/14

So:

-8/14 + 3/14 = -5/14

Final answer:
-5/14

Question 3. Find the difference.

(i) 5/6 - 1/4

Answer:
LCM of 6 and 4 = 12

5/6 = 10/12
1/4 = 3/12

Difference = 10/12 - 3/12
= 7/12

Final answer:
7/12

(ii) 11/8 - 3/4

Answer:
3/4 = 6/8

11/8 - 6/8 = 5/8

Final answer:
5/8

(iii) -7/9 - (-2/3)

Answer:
-7/9 - (-2/3) = -7/9 + 2/3

2/3 = 6/9

So:

-7/9 + 6/9 = -1/9

Final answer:
-1/9

Question 4. Find the product.

(i) 2/3 × 3/10

Answer:
2/3 × 3/10 = 6/30
= 1/5

Final answer:
1/5

(ii) 7/11 × 5/8

Answer:
7/11 × 5/8 = 35/88

Final answer:
35/88

(iii) -4/7 × 5/14

Answer:
-4/7 × 5/14 = -20/98
= -10/49

Final answer:
-10/49

Question 5. Find the quotient.

(i) 2/3 ÷ 3/10

Answer:
2/3 ÷ 3/10 = 2/3 × 10/3
= 20/9

Final answer:
20/9

(ii) 7/11 ÷ 5/8

Answer:
7/11 ÷ 5/8 = 7/11 × 8/5
= 56/55

Final answer:
56/55

(iii) -4/7 ÷ 5/14

Answer:
-4/7 ÷ 5/14 = -4/7 × 14/5
= -56/35
= -8/5

Final answer:
-8/5

Question 6. Show that (1/2 + 3/4) × 8/3 = 1/2 × 8/3 + 3/4 × 8/3.

Answer:
LHS:

(1/2 + 3/4) × 8/3
= (2/4 + 3/4) × 8/3
= 5/4 × 8/3
= 10/3

RHS:

1/2 × 8/3 + 3/4 × 8/3
= 4/3 + 2
= 4/3 + 6/3
= 10/3

Final answer:
LHS = RHS = 10/3. Hence, the distributive property is verified.

Question 7. Simplify the following using the distributive property: 7/9 × (6/7 - 3/4).

Answer:
Using the distributive property:

7/9 × (6/7 - 3/4)
= 7/9 × 6/7 - 7/9 × 3/4
= 6/9 - 21/36
= 2/3 - 7/12
= 8/12 - 7/12
= 1/12

Final answer:
1/12

Question 8. Find the rational number x such that 5/6 × (x + 3/5) = 5x/6 + 1/2.

Answer:
LHS:

5/6 × (x + 3/5)
= 5x/6 + 5/6 × 3/5
= 5x/6 + 3/6
= 5x/6 + 1/2

This is equal to the RHS for every rational value of x.

Final answer:
Every rational number x satisfies the equation.

NCERT Solutions for Class 9 Maths Chapter 3 Exercise Set 3.4

Exercise Set 3.4 focuses on rational numbers on the number line, absolute value, density of rational numbers and rational numbers between two given rational numbers.

Question 1. Represent the rational numbers 2/3, -5/4 and 11/2 on a single number line.

Answer:
To represent 2/3, divide the interval between 0 and 1 into 3 equal parts and mark the second part to the right of 0.

To represent -5/4, write it as -1¼. It lies between -2 and -1.

To represent 11/2, write it as 5½. It lies between 5 and 6.

Final answer:
2/3 lies between 0 and 1, -5/4 lies between -2 and -1, and 11/2 lies between 5 and 6.

Question 2. Find three distinct rational numbers that lie strictly between -1/2 and 1/4.

Answer:
Some rational numbers between -1/2 and 1/4 are:

-1/4, 0, 1/8

Check:

-1/2 < -1/4 < 0 < 1/8 < 1/4

Final answer:
-1/4, 0 and 1/8

Question 3. Simplify the expression -1/4 + 5/12.

Answer:
LCM of 4 and 12 = 12

-1/4 = -3/12

So:

-3/12 + 5/12 = 2/12
= 1/6

Final answer:
1/6

Question 4. A tailor has 15¾ metres of fine silk. If making one kurta requires 2¼ metres of silk, exactly how many kurtas can he make?

Answer:
Total silk = 15¾ = 63/4 metres
Silk for one kurta = 2¼ = 9/4 metres

Number of kurtas:

63/4 ÷ 9/4
= 63/4 × 4/9
= 63/9
= 7

Final answer:
The tailor can make 7 kurtas.

Question 5. Find three rational numbers between 3.1415 and 3.1416.

Answer:
Three rational numbers between 3.1415 and 3.1416 are:

3.14151, 3.14155, 3.14159

Final answer:
3.14151, 3.14155 and 3.14159

Question 6. Can you think of other ways to find a rational number between any two rational numbers?

Answer:
Yes. One method is to take the average of the two rational numbers.

If a and b are two rational numbers, then:

(a + b)/2

lies between a and b.

Final answer:
The average method can be used to find a rational number between two rational numbers.

NCERT Solutions for Class 9 Maths Chapter 3 Exercise Set 3.5

Exercise Set 3.5 focuses on decimal expansions, rational and irrational numbers, repeating decimals and cyclic numbers.

Question 1. Without performing long division, determine which rational numbers will have terminating decimals and which will be repeating: 7/20, 4/15 and 13/250. Then check your answers by long division.

Answer:
A rational number in lowest form has a terminating decimal if the denominator has only 2 and/or 5 as prime factors.

7/20

20 = 2² × 5

So, 7/20 has a terminating decimal.

7/20 = 0.35

4/15

15 = 3 × 5

Since the denominator has a factor other than 2 and 5, it has a repeating decimal.

4/15 = 0.2666...

13/250

250 = 2 × 5³

So, 13/250 has a terminating decimal.

13/250 = 0.052

Final answer:
7/20 = 0.35 terminating, 4/15 = 0.2666... repeating, 13/250 = 0.052 terminating.

Question 2. Perform the long division for 1/13. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 2/13? Now compute 3/13, 4/13, etc. What do you notice?

Answer:
1/13 = 0.076923076923...

Repeating block = 076923

Now:

2/13 = 0.153846153846...
3/13 = 0.230769230769...
4/13 = 0.307692307692...

The decimal expansions have repeating blocks. Some of these blocks are cyclic shifts of the digits seen in related multiples.

Final answer:
1/13 has repeating block 076923. The multiples of 1/13 show repeating blocks with related cyclic patterns.

Question 3. Classify the following numbers as rational or irrational. Find the explicit fractions in case they are rational.

(i) √81

Answer:
√81 = 9

Final answer:
Rational, 9 = 9/1

(ii) √12

Answer:
√12 = 2√3

Since √3 is irrational, √12 is irrational.

Final answer:
Irrational

(iii) 0.33333...

Answer:
0.33333... = 1/3

Final answer:
Rational, 1/3

(iv) 0.123451234512345...

Answer:
The block 12345 repeats.

So, the number is rational.

Let x = 0.1234512345...

Then:

100000x = 12345.1234512345...

Subtract:

100000x - x = 12345
99999x = 12345
x = 12345/99999
= 4115/33333

Final answer:
Rational, 4115/33333

(v) 1.01001000100001...

Answer:
The digits do not repeat as a fixed block. The number is non-terminating and non-repeating.

Final answer:
Irrational

(vi) 23.560185612239874790120

Answer:
This is a terminating decimal. Therefore, it is rational.

Final answer:
Rational, 23560185612239874790120/1000000000000000000000

Question 4. The number 0.99999... is a rational number. Using algebra, explain why 0.99999... is exactly equal to 1.

Answer:
Let:

x = 0.99999...

Then:

10x = 9.99999...

Subtract the first equation from the second:

10x - x = 9.99999... - 0.99999...
9x = 9
x = 1

Final answer:
0.99999... = 1

Question 5. We have seen that the repeating block of 1/7 is a cyclic number. Try to find more numbers n whose reciprocals 1/n produce decimals with repeating blocks that are cyclic.

Answer:
Some examples are:

1/17 = 0.0588235294117647...
1/19 = 0.052631578947368421...

Their repeating blocks show cyclic behaviour under multiplication by several integers.

Final answer:
Examples include n = 17 and n = 19.

NCERT Solutions for Class 9 Maths Chapter 3 End-of-Chapter Exercises

The end-of-chapter exercises cover decimal expansions, irrationality proofs, conversion of decimals into p/q form, rational numbers between two numbers, equations with rational numbers and the square root spiral.

Question 1. Convert the following rational numbers into terminating or non-terminating repeating decimals by long division.

(i) 3/50

Answer:
3/50 = 6/100
= 0.06

Final answer:
0.06, terminating decimal

(ii) 2/9

Answer:
2 ÷ 9 = 0.2222...

Final answer:
0.2222..., non-terminating repeating decimal

Question 2. Prove that √5 is an irrational number.

Answer:
Assume that √5 is rational.

Then:

√5 = p/q

where p and q are co-prime integers and q ≠ 0.

Squaring both sides:

5 = p²/q²

So:

p² = 5q²

This means p² is divisible by 5. Therefore, p is divisible by 5.

Let p = 5k.

Substitute:

(5k)² = 5q²
25k² = 5q²
q² = 5k²

So, q² is divisible by 5. Therefore, q is also divisible by 5.

This means p and q have a common factor 5, which contradicts the assumption that p and q are co-prime.

Final answer:
√5 is irrational.

Question 3. Convert the following decimal numbers in the form p/q.

Note: The following answers treat the decimals as written in the uploaded text, without assuming any missing repeating bars.

(i) 12.6

Answer:
12.6 = 126/10
= 63/5

Final answer:
63/5

(ii) 0.0120

Answer:
0.0120 = 120/10000
= 12/1000
= 3/250

Final answer:
3/250

(iii) 3.052

Answer:
3.052 = 3052/1000
= 763/250

Final answer:
763/250

(iv) 1.235

Answer:
1.235 = 1235/1000
= 247/200

Final answer:
247/200

(v) 0.23

Answer:
0.23 = 23/100

Final answer:
23/100

(vi) 2.05

Answer:
2.05 = 205/100
= 41/20

Final answer:
41/20

(vii) 2.125

Answer:
2.125 = 2125/1000
= 17/8

Final answer:
17/8

(viii) 3.125

Answer:
3.125 = 3125/1000
= 25/8

Final answer:
25/8

(ix) 2.1625

Answer:
2.1625 = 21625/10000
= 173/80

Final answer:
173/80

Question 4. Locate the following rational numbers on the number line.

(i) 0.532

Answer:
0.532 lies between 0 and 1.

It is slightly more than 0.5 and less than 0.6.

Final answer:
0.532 should be marked between 0.5 and 0.6 on the number line.

(ii) 1.15

Answer:
1.15 lies between 1 and 2.

It is slightly greater than 1.1 and less than 1.2.

Final answer:
1.15 should be marked between 1.1 and 1.2 on the number line.

Question 5. Find 6 rational numbers between 3 and 4.

Answer:
Six rational numbers between 3 and 4 are:

3.1, 3.2, 3.3, 3.4, 3.5, 3.6

Final answer:
3.1, 3.2, 3.3, 3.4, 3.5 and 3.6

Question 6. Find 5 rational numbers between 2/5 and 3/5.

Answer:
Write both fractions with denominator 60:

2/5 = 24/60
3/5 = 36/60

Five rational numbers between them are:

25/60, 26/60, 27/60, 28/60, 29/60

Final answer:
25/60, 26/60, 27/60, 28/60 and 29/60

Question 7. Find 5 rational numbers between 1/6 and 2/5.

Answer:
Use denominator 30:

1/6 = 5/30
2/5 = 12/30

Five rational numbers between them are:

6/30, 7/30, 8/30, 9/30, 10/30

Final answer:
1/5, 7/30, 4/15, 3/10 and 1/3

Question 8. If x/3 + x/5 = 16/15, find the rational number x.

Answer:
x/3 + x/5 = 16/15

LCM of 3 and 5 = 15

5x/15 + 3x/15 = 16/15
8x/15 = 16/15

Multiply both sides by 15:

8x = 16
x = 2

Final answer:
x = 2

Question 9. Let a and b be two non-zero rational numbers such that a/b + 1 = 0. Without assigning numerical values, determine whether ab is positive or negative. Justify your answer.

Answer:
Given:

a/b + 1 = 0

So:

a/b = -1

Therefore:

a = -b

Now:

ab = (-b)b
= -b²

Since b is non-zero, b² is positive.

So, -b² is negative.

Final answer:
ab is negative.

Question 10. A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form p/10⁴, where p is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by 2⁴ or 5⁴? Give reasons.

Answer:
If the last non-zero digit occurs in the 4th decimal place, the number can be written with denominator 10⁴.

So, it can be written as:

p/10⁴

where p is an integer not divisible by 10. If p were divisible by 10, the last non-zero digit would occur before the 4th decimal place.

It is not necessary that the denominator in lowest form is divisible by 2⁴ or 5⁴.

Example:

0.0012 = 12/10000
= 3/250

Here, the denominator 250 = 2 × 5³, which is not divisible by 2⁴ or 5⁴.

Final answer:
The number can be written as p/10⁴, but its denominator in lowest form need not be divisible by 2⁴ or 5⁴.

Question 11. Without performing division, determine whether the decimal expansion of 18/125 is terminating or non-terminating. If it terminates, state the number of decimal places.

Answer:
Denominator = 125 = 5³

Since the denominator has only 5 as its prime factor, the decimal expansion is terminating.

To convert the denominator into a power of 10:

125 × 8 = 1000

So, the decimal has 3 places.

Final answer:
18/125 has a terminating decimal expansion with 3 decimal places.

Question 12. A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your answer.

Answer:
Denominator = 2³ × 5

To make it a power of 10, we need equal powers of 2 and 5.

2³ × 5 = 2³ × 5¹

Multiply by 5² to get:

2³ × 5³ = 10³

Therefore, the decimal expansion will have 3 decimal places.

Final answer:
It will have 3 decimal places.

Question 13. Let a = 7/12 and b = 5/6. Express both a and b in the form k₁/m and k₂/m where k₁, k₂ and m are integers and k₂ - k₁ > 6. Using the same denominator m, write exactly five distinct rational numbers lying between a and b.

Answer:
Given:

a = 7/12
b = 5/6

Choose common denominator 48.

7/12 = 28/48
5/6 = 40/48

Here:

k₁ = 28, k₂ = 40, m = 48

k₂ - k₁ = 40 - 28 = 12 > 6

Five rational numbers between them are:

29/48, 30/48, 31/48, 32/48, 33/48

Final answer:
29/48, 30/48, 31/48, 32/48 and 33/48

To find n rational numbers using the same denominator, there must be at least n integer numerators between k₁ and k₂. That is why the numerator gap must be large enough.

Question 14. Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that all rational numbers x, y, z must be simultaneously zero.

Answer:
Given:

x + y + z = 0

Square both sides:

(x + y + z)² = 0

Expand:

x² + y² + z² + 2xy + 2yz + 2zx = 0

This can be written as:

x² + y² + z² + 2(xy + yz + zx) = 0

Given:

xy + yz + zx = 0

So:

x² + y² + z² = 0

Squares of rational numbers are non-negative. Their sum can be zero only when each square is zero.

So:

x = 0, y = 0, z = 0

Final answer:
x, y and z must all be zero.

Question 15. Show that the rational number (a + b)/2 lies between the rational numbers a and b.

Answer:
Assume a < b.

Now compare a and (a + b)/2:

a < (a + b)/2

Multiplying by 2:

2a < a + b
a < b

This is true.

Now compare (a + b)/2 and b:

(a + b)/2 < b

Multiplying by 2:

a + b < 2b
a < b

This is also true.

Therefore:

a < (a + b)/2 < b

Final answer:
(a + b)/2 lies between a and b.

Question 16. Find the lengths of the hypotenuses of all the right triangles in the square root spiral.

Answer:
In a square root spiral, each new right triangle has one leg of length 1 unit and the previous hypotenuse as the other leg.

First triangle:

Hypotenuse = √(1² + 1²) = √2

Second triangle:

Hypotenuse = √[(√2)² + 1²]
= √3

Third triangle:

Hypotenuse = √[(√3)² + 1²]
= √4
= 2

Continuing this pattern, the hypotenuse lengths are:

√2, √3, √4, √5, √6, ...

Final answer:
The hypotenuse lengths in the square root spiral are √2, √3, √4, √5, √6, and so on.

Topics Covered in NCERT Solutions for Class 9 Maths Chapter 3

1. Natural numbers
2. One-to-one correspondence
3. Prime numbers
4. Zero or Śhūnya
5. Brahmagupta’s rules
6. Integers
7. Positive and negative numbers
8. Rational numbers
9. Equivalent rational numbers
10. Operations on rational numbers
11. Number line representation
12. Absolute value
13. Density of rational numbers
14. Irrational numbers
15. Proof by contradiction
16. √2 and √5 irrationality
17. π and irrational numbers
18. Real numbers
19. Decimal expansions
20. Terminating decimals
21. Repeating decimals
22. Cyclic numbers
23. Square root spiral

Important Formulas and Rules in NCERT Solutions for Class 9 Maths Chapter 3

NCERT Class 9 Maths Ganita Manjari 2026 Chapter Solutions