---
title: "NCERT Solutions for Class 9 Maths Chapter 3 Exercise 3.5"
url: https://www.swavid.com/maths/class/9/chapter/the-world-of-numbers/ncert-solutions/exercise-3-5
dateModified: 2026-10-07T15:46:03+00:00
---

# NCERT Solutions for Class 9 Maths Chapter 3 Exercise 3.5

Chapter 3: The World of Numbers. Every question from Exercise 3.5, with full working and the final answer.

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## Exercise Set 3.5

### Question 1

*3 marks · Short answer*

Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: $\frac{7}{20}$, $\frac{4}{15}$ and $\frac{13}{250}$. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.

**Solution**

1. Find the prime factorisation of the denominator for each rational number in its lowest form.
2. For $\frac{7}{20}$, the denominator $20 = 2^2 \times 5$, so its decimal expansion terminates ($0.35$).
3. For $\frac{4}{15}$, the denominator $15 = 3 \times 5$, so its decimal expansion is repeating ($0.2\overline{6}$).
4. For $\frac{13}{250}$, the denominator $250 = 2 \times 5^3$, so its decimal expansion terminates ($0.052$).

**Answer:** $\frac{7}{20}$ and $\frac{13}{250}$ are terminating decimals, while $\frac{4}{15}$ is repeating.

> Common mistake: Failing to check if the rational number is in lowest terms before factoring the denominator.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. Perform long division to find $\frac{1}{13} = 0.\overline{076923}$, where the repeating block is $076923$.
2. Evaluate $\frac{2}{13} = 0.\overline{153846}$, which is a cyclic shift of the same digits.
3. Compute other fractions like $\frac{3}{13} = 0.\overline{230769}$ and observe that all reciprocals with denominator $13$ produce cyclic permutations of the same six digits.

**Answer:** The repeating block for $\frac{1}{13}$ is $076923$, and all multiples $\frac{k}{13}$ exhibit cyclic properties.

> Common mistake: Arithmetic errors during long division with 13.

### Question 3

*3 marks · Short answer*

Classify the following numbers as rational or irrational:
(i) $\sqrt{81}$
(ii) $\sqrt{12}$
(iii) $0.33333 \dots$
(iv) $0.123451234512345 \dots$
(v) $1.01001000100001 \dots$ (Notice the pattern: Is it repeating a single block?)
(vi) $23.560185612239874790120$
Find the explicit fractions in case they are rational.

**Part (i)**

1. $\sqrt{81} = 9$, which can be written as $\frac{9}{1}$.

Answer (i): Rational, $\frac{9}{1}$

**Part (ii)**

1. $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$, which cannot be expressed as a ratio of integers.

Answer (ii): Irrational

**Part (iii)**

1. $0.33333 \dots = 0.\overline{3} = \frac{1}{3}$, which is a repeating decimal.

Answer (iii): Rational, $\frac{1}{3}$

**Part (iv)**

1. $0.123451234512345 \dots = 0.\overline{12345} = \frac{12345}{99999}$.

Answer (iv): Rational, $\frac{4115}{33333}$

**Part (v)**

1. The decimal $1.01001000100001 \dots$ is non-terminating and non-repeating because the number of zeros between the ones increases by one each time.

Answer (v): Irrational

**Answer:** Classified all six numbers as rational or irrational with their fractions where applicable.

> Common mistake: Confusing non-repeating patterns with repeating decimal blocks.

### Question 4

*3 marks · Proof*

The number $0.\bar{9}$ (which means $0.99999 \dots$) is a rational number. Using algebra (let $x = 0.\bar{9}$, multiply by $10$, and subtract), explain why $0.\bar{9}$ is exactly equal to $1$.

**Solution**

1. Let $x = 0.\overline{9} = 0.99999 \dots$.
2. Multiply both sides by $10$ to get $10x = 9.99999 \dots$.
3. Subtract the first equation from the second: $10x - x = 9.99999 \dots - 0.99999 \dots$.
4. Simplify to get $9x = 9$, which gives $x = 1$. Hence proved.

**Answer:** $0.\overline{9} = 1$

> Common mistake: Assuming $0.\overline{9}$ is strictly less than 1 due to the infinite series representation.

### Question 5

*3 marks · Short answer*

We have seen that the repeating block of $\frac{1}{7}$ is a cyclic number. Try to find more numbers ($n$) whose reciprocals ($\frac{1}{n}$) produce decimals with repeating blocks that are cyclic.

**Solution**

1. Find the decimal expansions of reciprocals of prime numbers like $7$, $17$, $19$, and $23$.
2. Observe that $\frac{1}{17} = 0.\overline{0588235294117647}$ and $\frac{1}{19} = 0.\overline{052631578947368421}$ produce repeating blocks that are cyclic.

**Answer:** Examples of such numbers whose reciprocals produce cyclic decimals are $7$, $17$, $19$, and $23$.

> Common mistake: Not checking enough decimal places to see the full repeating block.

## Related pages

- [All Chapter 3 NCERT solutions](https://www.swavid.com/maths/class/9/chapter/the-world-of-numbers/ncert-solutions)
- [Exercise 3.1](https://www.swavid.com/maths/class/9/chapter/the-world-of-numbers/ncert-solutions/exercise-3-1)
- [Exercise 3.2](https://www.swavid.com/maths/class/9/chapter/the-world-of-numbers/ncert-solutions/exercise-3-2)
- [Exercise 3.3](https://www.swavid.com/maths/class/9/chapter/the-world-of-numbers/ncert-solutions/exercise-3-3)
- [Exercise 3.4](https://www.swavid.com/maths/class/9/chapter/the-world-of-numbers/ncert-solutions/exercise-3-4)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student