# Number Systems (Chapter 1, Exercise 1.3) – Class 9 Maths Solutions) NCERT

This page contains step by step solutions of Exercise 1.3 of Chapter 1 of Class 9 Mathematics under NCERT Syllabus.

## Question 1: Write the following in decimal form and say what kind of decimal expansion each has

i. $\frac{36}{100}$

Answer: Terminating Decimal Expansion.

Explanation:

$\frac{36}{100}=0.36$

Since the value of $\frac{36}{100}$ is $0.36$, therefore it is a Terminating Decimal Expansion.

ii. $\frac{1}{11}$

Explanation:

$\frac{1}{11}=0.090909090909….=0.\overline{09}$

The number $‘09’$ repeats itself after decimal if we keep on dividing $1$ by $11$ and it is never ending. Therefore, the given Decimal Expansion is Recurring and Non-Terminating.

iii. $4\frac{1}{8}$

Explanation:

$4\frac{1}{8}=\frac{32+1}{8}=\frac{33}{8}=4.125$

The value of $4\frac{1}{8}$ is $4.125$ which is a Terminating Number. Therefore, the above decimal expression is a Terminating Decimal Expansion.

iv. $\frac{3}{13}$

Explanation:

$\frac{3}{13}=0.230769230769….=0.\overline{230769}$

If we convert the fraction $\frac{3}{13}$ into decimal, then the number $230769$ keeps on repeating after decimal and it never ends. Therefore, the above Decimal Expansion is called Recurring and Non-Terminating Decimal Expansion.

v. $\frac{2}{11}$

Explanation:

$\frac{2}{11}=0.181818181818….=0.\overline{18}$ If we find the decimal expansion for the fraction $\frac{2}{11}$, the number $18$ will keep on repeating after decimal and never ends. Therefore, the Decimal Expansion is Recurring and Non-Terminating.

vi. $\frac{329}{400}$

Explanation:

$\frac{329}{400}=0.8225$

The decimal expression of the fraction $\frac{329}{400}$ is $0.8225$ which is a Terminating Decimal Expansion.

## Question 2: You know that $\frac{1}{7}=0.\overline{142857}$. Can you what the decimal expansions of $\frac{2}{7},\, \frac{3}{7},\,\frac{4}{7},\,\frac{5}{7},\,\frac{6}{7}$ are, without actually doing the long division? If so, how?

Solution:

Given, $\frac{1}{7}=0.\overline{142857}$

The fraction $\frac{2}{7}$ can be written as, $\frac{2}{7}=2\times \frac{1}{7}$

Therefore, $\frac{2}{7}=2\times 0.\overline{142857}=0.\overline{285714}$

Similarly,

$\frac{3}{7}=3\times \frac{1}{7}=3\times 0.\overline{142857}=0.\overline {428571}$

$\frac{4}{7}=4\times \frac{1}{7}=4\times 0.\overline{142857}=0.\overline {571428}$

$\frac{5}{7}=5\times \frac{1}{7}=5\times 0.\overline{142857}=0.\overline {714285}$

$\frac{6}{7}=6\times \frac{1}{7}=6\times 0.\overline{142857}=0.\overline {857142}$

## Question 3: Express the following in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$?

i. $0.\overline {6}$

Solution:

We know that, $0.\overline {6}=0.666666…$

Now, let us consider

$x=0.666666…$

Multiplying by $10$ in both sides of the equation,

$10\times x=10\times 0.666666…$

$\Rightarrow 10x=6.66666…$

$\Rightarrow 10x=6+0.66666…$

$\Rightarrow 10x=6+x$    [since, $x=0.66666…$]

$\Rightarrow 10x-x=6$

$\Rightarrow 9x=6$

$\Rightarrow x=\frac{6}{9}$

$\Rightarrow x=\frac{2}{3}$

Therefore, the $\frac{p}{q}$ form of $0.\overline{6}$ is $\frac{2}{3}$.

ii. $0.4\overline {7}$

Solution:

We know that, $0.4\overline {7}=0.477777…$

Let us consider,

$x=0.477777…\,\,\,\,\,\,\,\,\,\,\, ………..(i)$ $\therefore\,\,\,10x=4.7777…$

Multiplying by $100$ in both sides of the equation$(i)$,

$100x=47.7777…$

$\Rightarrow\,100x=43+4.7777…$

$\Rightarrow\,100x=43+10x$

$\Rightarrow\,100x-10x=43$

$\Rightarrow\,90x=43$

$\Rightarrow\,x=\frac {43}{90}$

Therefore, the \frac{p}{q} form of 0.4\overline {7} is \frac{43}{90}.

iii. $0.\overline {001}$

Solution:

We know that, $0.\overline {001} =0.001001001001…$

Let us consider,

$x=0.001001…$

Multiplying by 1000 in both sides of the above equation,

$1000x=1.001001…$

$\Rightarrow\,1000x=1+0.001001…$

$\Rightarrow\,1000x=1+x$

$\Rightarrow\,1000x-1x=1$

$\Rightarrow\,999x=1$

$\Rightarrow\,x=\frac{1}{999}$

Therefore, the $\frac{p}{q}$ form of $0.\overline{001}$ is $\frac{1}{999}$.

Question 4: Express $0.99999…$ in the form $\frac{p}{q}$.

Solution:

Let us consider, $x=0.99999…$

Multiplying both sides by $10$,

$10x=9.9999…$

$\Rightarrow\,10x=9+0.9999…$

$\Rightarrow\,10x=9+x$

$\Rightarrow\,10x-x=9$

$\Rightarrow\,9x=9$

$\Rightarrow\,x=\frac{9}{9}=1$

Therefore, the $\frac{p}{q}$ form of $0.99999$ is $1$.

Question 5: What can the maximum number of digits be in one repeating block of digits in the decimal expression of $\frac {1}{17}$?

Solution:

If we divide the number $1$ by $17$, then the Quotient will be,

$\frac{1}{17}=0.05882352941176470588235294117647=0.\overline {0588235294117647}$

From the above, it is clear that the sixteen numbers i.e., $0588235294117647$ is repeating itself.

Therefore, the maximum number of digits in one repeating block is $16$.

Question 6: Write three numbers whose decimal expansions are non-terminating non-recurring?

Solution:

Three numbers whose decimal expansions are non-terminating and non-recurring are:

1. $\pi=3.1415926536…$
2. $\sqrt{2}=1.41421356…$
3. $\sqrt{5}=2.23606798…$

Question 7: Find three different irrational Numbers between the rational numbers $\frac {5}{7}$ and $\frac {9}{11}$.

Solution:

The decimal expansion of $\frac{5}{7}$ and $\frac{9}{11}$ will be:

$\frac{5}{7}=0.714285714285…=0.\overline{714285}$ $\frac{9}{11}=0.818181…=0.\overline{81}$

We know that the Decimal Expansion of an Irrational Number is Non-Terminating and Non-Recurring and three such numbers between $\frac{5}{7}$ and $\frac{9}{11}$ are,

1. $0.730730073000730000…$
2. $0.750750075000750000…$
3. $0.791791179111791111…$

Question 8: Classify the following Numbers as Rational or irrational

i. $\sqrt{23}$

Solution:

$\sqrt{23}=4.795831523312…$ From above, we have seen that the value of $\sqrt{23}$ is Non-Terminating and Non-Recurring. Therefore, $\sqrt{23}$ is an Irrational Number.

ii. $\sqrt{225}$

Solution:

$\sqrt{225}=15$

The value of $\sqrt{225}$ is Terminating. Hence, $\sqrt {225}$ is a Rational Number.

iii. $0.3796$

Solution: The given number $0.3796$ is a Rational Number because it is Terminating.

iv. $7.478478…$

Solution:

$7.478478…=7.\overline{478}$

Here, the number $7.478478…$ is Non-Terminating but Recurring. Therefore, the number is a Rational Number.

v. $1.101001000100001…$

Solution: The given number is Non-Terminating and Non-Recurring. Therefore, it is an Irrational Number.

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