NCERT Solutions for Exercise 1.3 Class 9 Maths Chapter 1 - Number Systems

Access Answer of Number Systems Class 9 Chapter 1 Exercise: 1.3

Q1 (i) Write the following in decimal form and say what kind of decimal expansion each has : (i) $\frac{36}{100}$

Answer:

We can write $\frac{36}{100}$ as
$\Rightarrow \frac{36}{100}=0.36$
Since the decimal expansion ends after a finite number of steps. Hence, it is terminating

Q1 (ii) Write the following in decimal form and say what kind of decimal expansion each has : (ii) $\frac{1}{11}$

Answer:

We can rewrite $\frac{1}{11}$ as

$\Rightarrow \frac{1}{11}=0.09090909.....=0.\overline{09}$
Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

Q1 (iii) Write the following in decimal form and say what kind of decimal expansion each has : (iii) $4\frac{1}{8}$

Answer:

We can rewrite $4\frac{1}{8}$ as

$\Rightarrow 4\frac{1}{8}=\frac{33}{8}=4.125$
Since the decimal expansion ends after a finite number. Therefore, it is terminating

Q1 (iv) Write the following in decimal form and say what kind of decimal expansion each has : (iv) $\frac{3}{13}$

Answer:

We can rewrite $\frac{3}{13}$ as

$\Rightarrow \frac{3}{13}=0.230769230769=0.\overline{230769}$
Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

Q1 (v) Write the following in decimal form and say what kind of decimal expansion each has: (v) $\frac{2}{11}$

Answer:

We can rewrite $\frac{2}{11}$ as

$\Rightarrow \frac{2}{11}=0.181818......=0.\overline{18}$
Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion.

Q1 (vi) Write the following in decimal form and say what kind of decimal expansion each has : (vi) $\frac{329}{400}$

Answer:

We can rewrite $\frac{329}{400}$ as

$\Rightarrow \frac{329}{400}=0.8225$
Since decimal expansion ends after finite no. of figures. Hence, it is terminating.

Q2 You know that $\frac{1}{7}=0.\overline{142857}$ Can you predict 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?

Answer:

It is given that $\frac{1}{7}=0.\overline{142857}$

Therefore,

$\Rightarrow \frac{2}{7}=2\times \frac{1}{7}=2\times 0.\overline{142857}=0.\overline{285714}$

Similarly,

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

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

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

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

Q3 (i) Express the following in the form $\frac{p}{q}$ , where p and q are integers and q ≠ 0. (i) $0.\overline{6}$

Answer:

Let $x=0.\bar{6}=0.6666....$ -(i)

Now, multiply by 10 on both sides

$10x=6.6666...$

$\Rightarrow 10x=6+x\text{(}using(i))$

$\Rightarrow 9x=6$

$\Rightarrow x=\frac{6}{9}=\frac{2}{3}$

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

Q3 (ii) Express the following in the form $\frac{p}{q}$ , where p and q are integers and q ≠ 0. (ii) $0.4\overline{7}$

Answer:

We can write $0.4\bar{7}$ as

$\Rightarrow 0.4\bar{7}=\frac{4}{10}+\frac{0.777..}{10}$ -(i)

Now,

Let $x=0.\bar{7}=0.7777....$ -(ii)

Now, multiply by 10 on both sides

$10x=7.7777...$

$\Rightarrow 10x=7+x\text{(}using(ii))$

$\Rightarrow 9x=7$

$\Rightarrow x=\frac{7}{9}$
Now, put the value of x in equation (i). we will get

$\Rightarrow 0.4\bar{7}=\frac{4}{10}+\frac{7}{10\times 9}=\frac{4}{10}+\frac{7}{90}=\frac{36+7}{90}=\frac{43}{90}$

Therefore, $\frac{p}{q}$ form of $0.4\bar{7}$ is $\frac{43}{90}$

Q3 (iii) Express the following in the form $\frac{p}{q}$ , where p and q are integers and q ≠ 0. (iii) $0.\overline{001}$

Answer:

Let $x=0.\overline{001}=0.001001....$ -(i)

Now, multiply by 1000 on both sides

$1000x=1.001001...$

$\Rightarrow 1000x=1+x\text{(}using(i))$

$\Rightarrow 999x=1$

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

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

Q4 Express 0.99999 .... in the form $\frac{p}{q}$ . Are you surprised by your answer?

Answer:

Let $x=0.\bar{9}=0.9999....$ -(i)

Now, multiply by 10 on both sides

$10x=9.999....$

$\Rightarrow 10x=9+x\text{(}using(i))$

$\Rightarrow 9x=9$

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

Therefore, $\frac{p}{q}$ form of $0.999....$ is 1

The difference between 1 and 0.999999 is o.000001 which is almost negligible.

Therefore, 0.999 is too much closer to 1. Hence, we can write 0.999999.... as 1

Q5 What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$ ? Perform the division to check your answer.

Answer:

We can rewrite $\frac{1}{17}$ as

$\Rightarrow \frac{1}{17}=0.05882352941176470588235294117647=0.\overline{0588235294117647}$
Therefore, there are total 16 number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$

Q6 Look at several examples of rational numbers in the form $\frac{p}{q}$ (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Answer:

We can observe that when q is 2, 4, 5, 8, 10… then the decimal expansion is terminating. For example:

$\frac{3}{2}=1.5$ , denominator $q=2^1$

$\frac{8}{5}=1.6$ , denominator $q=5^1$

$\frac{15}{10}=1.5$ , denominator $q=10=2\times 5=2^1,5^1$

Therefore,

It can be observed that the terminating decimal can be obtained in a condition where prime factorization of the denominator of the given fractions has the power of 2 only or 5 only or both.

Q7 Write three numbers whose decimal expansions are non-terminating non-recurring.

Answer:

Three numbers whose decimal expansions are non-terminating non-recurring are
1) 0.02002000200002......
2) 0.15115111511115.......
3) 0.27227222722227.......

Q8 Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$ .

Answer:

We can write $\frac{5}{7}$ as

$\Rightarrow \frac{5}{7}=0.714285714285....=0.\overline{714285}$

And $\frac{9}{11}$ as

$\Rightarrow \frac{9}{11}=0.818181....=0.\overline{81}$
Therefore, three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$ are

1) 0.72737475....
2) 0.750760770780...
3) 0.790780770760....

Q9 (i) Classify the following numbers as rational or irrational : $\sqrt{23}$

Answer:

We can rewrite $\sqrt{23}$ in decimal form as

$\Rightarrow \sqrt{23}=4.7958152....$

Now, as the decimal expansion of this number is non-terminating non-recurring.

Therefore, it is an irrational number.

Q9 (ii) Classify the following numbers as rational or irrational : $\sqrt{225}$

Answer:

We can rewrite $\sqrt{225}$ as

$\Rightarrow \sqrt{225}=15$
We can clearly see that it is a rational number because we can represent it in $\frac{p}{q}$ form

Q9 (iii) Classify the following numbers as rational or irrational : 0.3796

Answer:

We can rewrite 0.3796 as

$\Rightarrow 0.3796=\frac{3796}{10000}$
Now, we can clearly see that it is a rational number as the decimal expansion of this number is terminating and we can also write it in $\frac{p}{q}$ form.

Q9 (iv) Classify the following numbers as rational or irrational : 7.478478....

Answer:

We can rewrite 7.478478.... as

$\Rightarrow 7.478478....=7.\overline{478}$
Now, as the decimal expansion of this number is non-terminating recurring. Therefore, it is a rational number.

Q9 (v) Classify the following numbers as rational or irrational : 1.101001000100001...

Answer:

In the case of number 1.101001000100001...
As the decimal expansion of this number is non-terminating non-repeating. Therefore, it is an irrational number.