Class 9 Maths Ganita Manjari Chapter 3 End of Chapter Exercises: The World of Numbers

Class 9 Maths Ganita Manjari Chapter 3 End of Chapter Exercises revise The World of Numbers through fractions, decimals, rational numbers, irrational numbers, real numbers, number-line questions, proof of √5 irrationality and square root spiral problems.

Chapter 3 ends by bringing together the full journey of numbers: counting numbers, zero, integers, rational numbers, irrational numbers, decimal expansions and real numbers. Class 9 Maths Ganita Manjari Chapter 3 End of Chapter Exercises test whether students can move between fractions, decimals and number-line positions, while also proving important results such as the irrationality of √5.

These questions from The World of Numbers include long division, conversion of repeating decimals into fractions, rational numbers between two numbers, terminating-decimal tests, proof-based questions and the square root spiral Class 9 figure. The solutions below follow the textbook order and keep calculations in a copy-friendly format for school notes and CBSE 2026 revision.

Key Takeaways

Decimal Expansion: Rational numbers have terminating or repeating decimal forms.
Irrationality Proof: √5 is proved irrational using contradiction.
Rational Density: Rational numbers can always be found between two rational numbers.
Square Root Spiral: Right triangles can represent lengths such as √2, √3 and √4.

Class 9 Maths Ganita Manjari Chapter 3 End of Chapter Exercises Structure 2026

Class 9 Maths Ganita Manjari Chapter 3 End of Chapter Exercises

The Ganita Manjari Class 9 Chapter 3 End of Chapter Exercises revise both computation and reasoning. Students need to identify decimal types, convert numbers into p/q form, prove irrationality, locate rational numbers on the number line and understand why rational numbers are dense.

These Class 9 Maths Chapter 3 End of Chapter Exercise Solutions are useful for revising rational numbers Class 9, irrational numbers Class 9, real numbers Class 9, and terminating and repeating decimals Class 9 in one place.

End of Chapter Exercises Question 1

Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal by long division.

(i) 3/50

Solution:

3/50 = 6/100

3/50 = 0.06

Answer: 0.06, terminating decimal.

(ii) 2/9

Solution:

On long division:

2 ÷ 9 = 0.2222…

So,

2/9 = 0.2̅

Answer: 0.2222… = 0.2̅, non-terminating repeating decimal.

End of Chapter Exercises Question 2

Prove that √5 is an irrational number.

Solution:

Assume that √5 is rational.

Then it can be written in lowest form as:

√5 = p/q

where p and q are integers, q ≠ 0, and p and q have no common factor.

Squaring both sides:

5 = p²/q²

p² = 5q²

So, p² is divisible by 5. Therefore, p is also divisible by 5.

Let:

p = 5k

Substitute in p² = 5q².

(5k)² = 5q²

25k² = 5q²

q² = 5k²

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

This means both p and q are divisible by 5, which contradicts the assumption that p/q is in lowest form.

Answer: Therefore, √5 is irrational.

End of Chapter Exercises Question 3

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

(i) 12.6̅

Solution:

Let:

x = 12.6666…

Then:

10x = 126.6666…

Subtract:

10x − x = 126.6666… − 12.6666…

9x = 114

x = 114/9

x = 38/3

Answer: 12.6̅ = 38/3

(ii) 0.0120

Solution:

0.0120 = 120/10000

Simplify:

120/10000 = 3/250

Answer: 0.0120 = 3/250

(iii) 3.0\overline{52}

Solution:

Let:

x = 3.0525252…

There is 1 non-repeating digit after the decimal and 2 repeating digits.

First multiply by 10:

10x = 30.525252…

Then multiply by 1000:

1000x = 3052.525252…

Subtract:

1000x − 10x = 3052.525252… − 30.525252…

990x = 3022

x = 3022/990

x = 1511/495

Answer: 3.0\overline{52} = 1511/495

(iv) 1.\overline{235}

Solution:

Let:

x = 1.235235235…

Since 3 digits repeat, multiply by 1000.

1000x = 1235.235235…

Subtract:

1000x − x = 1235.235235… − 1.235235…

999x = 1234

x = 1234/999

Answer: 1.\overline{235} = 1234/999

(v) 0.\overline{23}

Solution:

Let:

x = 0.232323…

100x = 23.232323…

Subtract:

100x − x = 23.232323… − 0.232323…

99x = 23

x = 23/99

Answer: 0.\overline{23} = 23/99

(vi) 2.0\overline{5}

Solution:

Let:

x = 2.05555…

First multiply by 10:

10x = 20.5555…

Then multiply by 100:

100x = 205.5555…

Subtract:

100x − 10x = 205.5555… − 20.5555…

90x = 185

x = 185/90

x = 37/18

Answer: 2.0\overline{5} = 37/18

(vii) 2.12\overline{5}

Solution:

Let:

x = 2.125555…

There are 2 non-repeating decimal digits and 1 repeating digit.

100x = 212.5555…

1000x = 2125.5555…

Subtract:

1000x − 100x = 2125.5555… − 212.5555…

900x = 1913

x = 1913/900

Answer: 2.12\overline{5} = 1913/900

(viii) 3.\overline{125}

Solution:

Let:

x = 3.125125125…

1000x = 3125.125125…

Subtract:

1000x − x = 3125.125125… − 3.125125…

999x = 3122

x = 3122/999

Answer: 3.\overline{125} = 3122/999

(ix) 2.\overline{1625}

Solution:

Let:

x = 2.162516251625…

Since 4 digits repeat, multiply by 10000.

10000x = 21625.16251625…

Subtract:

10000x − x = 21625.16251625… − 2.16251625…

9999x = 21623

x = 21623/9999

Answer: 2.\overline{1625} = 21623/9999

End of Chapter Exercises Question 4

Locate the following rational numbers on the number line.

(i) 0.532

Solution:

0.532 = 532/1000

So, it lies between 0 and 1, slightly more than 0.5.

Answer: Mark 0.532 between 0 and 1, at 532/1000 of the distance from 0 to 1.

(ii) 1.1\overline{5}

Solution:

1.1\overline{5} = 1.15555…

It lies between 1 and 2.

Convert to fraction:

Let:

x = 1.15555…

10x = 11.5555…

100x = 115.5555…

Subtract:

100x − 10x = 115.5555… − 11.5555…

90x = 104

x = 104/90

x = 52/45

So,

1.1\overline{5} = 52/45 = 1 + 7/45

Answer: Mark 1.1\overline{5} between 1 and 2, at 7/45 of the distance after 1.

End of Chapter Exercises Question 5

Find 6 rational numbers between 3 and 4.

Solution:

Write 3 and 4 with denominator 10.

3 = 30/10

4 = 40/10

Six rational numbers between them are:

31/10, 32/10, 33/10, 34/10, 35/10, 36/10

Answer: 31/10, 32/10, 33/10, 34/10, 35/10 and 36/10

End of Chapter Exercises Question 6

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

Solution:

Convert both numbers to denominator 30.

2/5 = 12/30

3/5 = 18/30

Five rational numbers between them are:

13/30, 14/30, 15/30, 16/30, 17/30

Answer: 13/30, 14/30, 15/30, 16/30 and 17/30

End of Chapter Exercises Question 7

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

Solution:

Use denominator 60.

1/6 = 10/60

2/5 = 24/60

Five rational numbers between them are:

11/60, 12/60, 13/60, 14/60, 15/60

Answer: 11/60, 12/60, 13/60, 14/60 and 15/60

End of Chapter Exercises Question 8

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

Solution:

Given:

x/3 + x/5 = 16/15

Take LCM 15.

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

8x/15 = 16/15

Multiply both sides by 15.

8x = 16

x = 2

Answer: x = 2

End of Chapter Exercises Question 9

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

Solution:

Given:

a + 1/b = 0

So,

a = −1/b

Now multiply both sides by b.

ab = −1

Since −1 is negative:

ab < 0

Answer: ab is negative.

End of Chapter Exercises 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 in lowest form is divisible by 2⁴ or 5⁴?

Solution:

If the last non-zero digit occurs in the 4th decimal place, the number can be written with exactly four decimal places.

So, it can be written as:

p/10000

Since:

10000 = 10⁴

the number is:

p/10⁴

Here, p is not divisible by 10, because the fourth decimal digit is the last non-zero digit. If p were divisible by 10, the decimal would end earlier.

Now:

10⁴ = 2⁴ × 5⁴

When the fraction is reduced to lowest form, factors common with p may cancel. Since p is not divisible by 10, it cannot contain both 2 and 5 as factors together. Therefore, after cancellation, at least one of 2⁴ or 5⁴ remains in the denominator.

Answer: Yes, the number can be written as p/10⁴, where p is not divisible by 10. In lowest form, the denominator is necessarily divisible by 2⁴ or 5⁴.

End of Chapter Exercises 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.

Solution:

Given rational number:

18/125

Denominator:

125 = 5³

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

To make the denominator a power of 10:

125 × 8 = 1000

So, the decimal will have 3 decimal places.

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

End of Chapter Exercises Question 12

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

Solution:

Denominator:

2³ × 5

To make it a power of 10, the powers of 2 and 5 must be equal.

We have:

2³ × 5¹

Multiply by 5².

2³ × 5¹ × 5² = 2³ × 5³

2³ × 5³ = 10³

So, the decimal expansion will have 3 decimal places.

Answer: The decimal expansion will have 3 decimal places.

End of Chapter Exercises Question 13

Let a = 5/6 and b = 7/12. Express both using a common denominator so that k₂ − k₁ > 6. Then write exactly five rational numbers between them.

Solution:

Given:

a = 5/6

b = 7/12

Since:

5/6 = 10/12

we can use a larger common denominator to create enough integer numerators between the two numbers.

Use denominator 72.

7/12 = 42/72

5/6 = 60/72

Here:

k₁ = 42

k₂ = 60

So,

k₂ − k₁ = 60 − 42 = 18

Since 18 > 6, we can choose five integer numerators between 42 and 60.

Five rational numbers are:

43/72, 44/72, 45/72, 46/72, 47/72

These all lie between:

42/72 and 60/72

So, they lie between 7/12 and 5/6.

Why is k₂ − k₁ > n + 1 needed?

If two rational numbers are written as:

k₁/m and k₂/m

then the integer numerators strictly between them are:

k₁ + 1, k₁ + 2, ..., k₂ − 1

The number of available numerators is:

k₂ − k₁ − 1

To find n rational numbers, we need enough integer numerators between the endpoints. So, the gap between k₂ and k₁ must be large enough.

Answer: Five rational numbers are 43/72, 44/72, 45/72, 46/72 and 47/72. The condition ensures enough integer numerators are available between the two endpoints.

End of Chapter Exercises Question 14

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

Solution:

Given:

x + y + z = 0

and

xy + yz + zx = 0

Square the first equation:

(x + y + z)² = 0²

Use identity:

x² + y² + z² + 2xy + 2yz + 2zx = 0

So,

x² + y² + z² + 2(xy + yz + zx) = 0

Since:

xy + yz + zx = 0

we get:

x² + y² + z² = 0

Now, squares of rational numbers are always non-negative. Therefore:

x² = 0, y² = 0, z² = 0

So:

x = 0, y = 0, z = 0

Answer: All three rational numbers must be 0.

End of Chapter Exercises Question 15

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

Solution:

Assume:

a < b

Then:

a + a < a + b < b + b

So:

2a < a + b < 2b

Divide by 2:

a < (a + b)/2 < b

Therefore, (a + b)/2 lies between a and b.

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

End of Chapter Exercises Question 16

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

Solution:

In the square root spiral, each new right triangle has one side of length 1 and the previous hypotenuse as the other side.

For the first triangle:

Hypotenuse² = 1² + 1²

Hypotenuse² = 2

Hypotenuse = √2

For the second triangle:

Hypotenuse² = (√2)² + 1²

Hypotenuse² = 2 + 1 = 3

Hypotenuse = √3

Continuing in the same way, the hypotenuse lengths are:

√2, √3, √4, √5, √6, √7, √8, √9, √10, √11

Answer: The hypotenuse lengths are √2, √3, √4, √5, √6, √7, √8, √9, √10 and √11.

Final Answers for Class 9 Maths Ganita Manjari Chapter 3 End of Chapter Exercises

Concept Used in The World of Numbers End of Chapter Exercises

The end exercises of The World of Numbers combine calculation with proof. Students revise how rational numbers behave as fractions and decimals, and how irrational numbers differ from them.

Important concepts include:

• Rational numbers Class 9: numbers written as p/q, where q ≠ 0.
• Terminating and repeating decimals Class 9: rational numbers either terminate or repeat.
• Irrational numbers Class 9: numbers like √5 cannot be written as p/q.
• Rational numbers on number line Class 9: decimals and fractions can be located as exact positions.
• Density of rational numbers: many rational numbers can be found between any two rational numbers.
• Real numbers Class 9: rational and irrational numbers together form the real number line.
• Square root spiral Class 9: repeated right triangles create lengths such as √2, √3, √4, ...

About Class 9 Maths Ganita Manjari Chapter 3 End of Chapter Exercises

Ganita Manjari Class 9 Chapter 3 End of Chapter Exercises appear after the chapter explains natural numbers, zero, integers, rational numbers, irrational numbers and real numbers. These questions check whether students can connect the whole number-system journey.

These Class 9 Ganita Manjari number system solutions help students revise:

• decimal conversion,
• irrationality proof,
• rational numbers between two values,
• terminating-decimal tests,
• number-line location,
• rational number equations,
• square root spiral lengths.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3