Rational & Irrational Numbers — Exercise 1(A)

Without actual division, find which of the following fractions are terminating decimals :

(i) $\dfrac{9}{25}$

(ii) $\dfrac{7}{12}$

(iii) $\dfrac{13}{16}$

(iv) $\dfrac{25}{128}$

(v) $\dfrac{9}{50}$

(vi) $\dfrac{121}{125}$

(vii) $\dfrac{19}{55}$

(viii) $\dfrac{37}{78}$

(ix) $\dfrac{23}{80}$

(x) $\dfrac{19}{30}$

Answer

In rational numbers, if the denominator of the fraction can be expressed in the form of 2^m. × 5^n., then it is a terminating decimal.

(i) So, 25 can be expressed as 2^0. × 5^2., which is in the form of 2^m. × 5^n..

Hence, $\dfrac{9}{25}$ is a terminating decimal number.

(ii) So, 12 can be expressed as 3 × 2^2. × 5^0., which is not in the form of 2^m. × 5^n..

Hence, $\dfrac{7}{12}$ is not a terminating decimal number.

(iii) So, 16 can be expressed as 2^4. × 5^0., which is in the form of 2^m. × 5^n..

Hence, $\dfrac{13}{16}$ is a terminating decimal number.

(iv) So, 128 can be expressed as 2^7. × 5^0., which is in the form of 2^m. × 5^n..

Hence, $\dfrac{25}{128}$ is a terminating decimal number.

(v) So, 50 can be expressed as 2^1. × 5^2., which is in the form of 2^m. × 5^n..

Hence, $\dfrac{9}{50}$ is a terminating decimal number.

(vi) So, 125 can be expressed as 2^0. × 5^3., which is in the form of 2^m. × 5^n..

Hence, $\dfrac{121}{125}$ is a terminating decimal number.

(vii) So, 55 can be expressed as 11 × 2^0. × 5^1., which is not in the form of 2^m. × 5^n..

Hence, $\dfrac{19}{55}$ is not a terminating decimal number.

(viii) So, 78 can be expressed as 39 × 2^1. × 5^0., which is not in the form of 2^m. × 5^n..

Hence, it is not a terminating decimal number.

(ix) So, 80 can be expressed as 2^4. × 5^1., which is in the form of 2^m. × 5^n..

Hence, $\dfrac{23}{80}$ is a terminating decimal number.

(x) So, 30 can be expressed as 3 × 2^1. × 5^1., which is not in the form of 2^m. × 5^n..

Hence, $\dfrac{19}{30}$ is not a terminating decimal number.

Convert each of the following decimals into vulgar fraction in its lowest terms :

(i) 0.65

(ii) 1.08

(iii) 0.075

(iv) 2.016

(v) 1.732

Answer

(i) Given,

$\Rightarrow 0.65 = \dfrac{65}{100}$

The H.CF. of 65 and 100 is 5, dividing the numerator and denominator by 5,

$\Rightarrow \dfrac{65 ÷ 5}{100 ÷ 5} \\[1em] \Rightarrow \dfrac{13}{20}.$

Hence, 0.65 = $\dfrac{13}{20}$.

(ii) Given,

$\Rightarrow 1.08 = \dfrac{108}{100}$

The H.CF. of 108 and 100 is 4, dividing the numerator and denominator by 4,

$\Rightarrow \dfrac{108 ÷ 4}{100 ÷ 4} \\[1em] \Rightarrow \dfrac{27}{25}$

Hence, 1.08 = $\dfrac{27}{25}$.

(iii) Given,

$\Rightarrow 0.075 = \dfrac{75}{1000}$

The H.CF. of 75 and 1000 is 25, dividing the numerator and denominator by 25,

$\Rightarrow \dfrac{75 ÷ 25}{1000 ÷ 25} \\[1em] \Rightarrow \dfrac{3}{40}$

Hence, 0.075 = $\dfrac{3}{40}$.

(iv) Given,

$\Rightarrow 2.016 = \dfrac{2016}{1000}$

The H.CF. of 2016 and 1000 is 8, dividing the numerator and denominator by 8,

$\Rightarrow \dfrac{2016 ÷ 8}{1000 ÷ 8} \\[1em] \Rightarrow \dfrac{252}{125}$

Hence, 2.016 = $\dfrac{252}{125}$.

(v) Given,

$\Rightarrow 1.732 = \dfrac{1732}{1000}$

The H.CF. of 1732 and 1000 is 4, dividing the numerator and denominator by 4

$\Rightarrow \dfrac{1732 ÷ 4}{1000 ÷ 4} \\[1em] \Rightarrow \dfrac{433}{250}$

Hence, 1.732 = $\dfrac{433}{250}$.

Convert each of the following fractions into a decimal :

(i) $\dfrac{1}{8}$

(ii) $\dfrac{3}{32}$

(iii) $\dfrac{44}{9}$

(iv) $\dfrac{11}{24}$

(v) $\dfrac{12}{13}$

(vi) $\dfrac{27}{44}$

(vii) $2\dfrac{5}{12}$

(viii) $1\dfrac{31}{55}$

Answer

(i) $\dfrac{1}{8}$

On dividing,

$\dfrac{1}{8} = 0.125$

Hence, $\dfrac{1}{8} = 0.125$.

(ii) $\dfrac{3}{32}$

On dividing,

$\dfrac{3}{32} = 0.09375$

Hence, $\dfrac{3}{32} = 0.09375$.

(iii) $\dfrac{44}{9}$

On dividing,

$\dfrac{44}{9} = 4.\overline{8}$

Hence, $\dfrac{44}{9} = 4.\overline{8}$.

(iv) $\dfrac{11}{24}$

On dividing,

$\dfrac{11}{24} = 0.458\overline{3}$

Hence, $\dfrac{11}{24} = 0.458\overline{3}$.

(v) $\dfrac{12}{13}$

On dividing,

$\dfrac{12}{13} = 0.\overline{923076}$

Hence, $\dfrac{12}{13} = 0.\overline{923076}$.

(vi) $\dfrac{27}{44}$

On dividing,

$\dfrac{27}{44} = 0.61\overline{36}$

Hence, $\dfrac{27}{44} = 0.61\overline{36}$.

(vii) $2\dfrac{5}{12}$

On dividing,

$2\dfrac{5}{12} = \dfrac{29}{12} = 2.41\overline{6}$

Hence, $\dfrac{29}{12} = 2.41\overline{6}$.

(viii) $1\dfrac{31}{55}$

On dividing,

$1\dfrac{31}{55} = \dfrac{86}{55} = 1.5\overline{63}$

Hence, $1\dfrac{31}{55} = 1.5\overline{63}.$.

Express $\dfrac{15}{56}$ as a decimal, correct to four decimal places.

Answer

On dividing,

$\dfrac{15}{56} = 0.26785$

Hence, $\dfrac{15}{56} = 0.2679$.

Express $\dfrac{13}{34}$ as a decimal, correct to three decimal places.

Answer

On dividing,

$\dfrac{13}{34} = 0.3823$

Hence, $\dfrac{13}{34} = 0.382$.

By actual division show that :

(i) $\dfrac{11}{9} = 1.\overline{2}$

(ii) $\dfrac{43}{11} = 3.\overline{90}$

(iii) $\dfrac{107}{45} = 2.3\overline{7}$

(iv) $\dfrac{21}{55} = 0.3\overline{81}$

Answer

(i) On dividing,

$\dfrac{11}{9} = 1.222......$

Hence, proved that $\dfrac{11}{9} = 1.\overline{2}$.

(ii) On dividing,

$\dfrac{43}{11} = 3.9090...$

Hence, proved that $\dfrac{43}{11} = 3.\overline{90}$.

(iii) On dividing,

$\dfrac{107}{45} = 2.3777...$

Hence, proved that $\dfrac{107}{45} = 2.3\overline{7}$.

(iv) On dividing,

$\dfrac{21}{55} = 0.38181...$

Hence, proved that $\dfrac{21}{55} = 0.3\overline{81}$.

Express each of following as a vulgar fraction in simplest form :

(i) $0.\overline{5}$

(ii) $0.\overline{43}$

(iii) $0.\overline{158}$

(iv) $1.\overline{3}$

(v) $4.\overline{17}$

(vi) $0.\overline{12}$

(vii) $0.1\overline{36}$

(viii) $1.5\overline{7}$

Answer

(i) $0.\overline{5}$

Let x = $0.\overline{5}$.

⇒ x = 0.555..     ...........(1)

⇒ 10x = 5.555..     ..........(2)

On subtracting (1) from (2), we get :

⇒ 10x - x = 5.555... - 0.555...

⇒ 9x = 5

⇒ x = $\dfrac{5}{9}$

Hence, $0.\overline{5} = \dfrac{5}{9}$.

(ii) $0.\overline{43}$

Let x = $0.\overline{43}$.

⇒ x = 0.434343..     .........(1)

⇒ 100x = 43.4343..     ........(2)

On subtracting (1) from (2), we get :

⇒ 100x - x = 43.4343.... - 0.4343......

⇒ 99x = 43

⇒ x = $\dfrac{43}{99}$

Hence, $0.\overline{43} = \dfrac{43}{99}$.

(iii) $0.\overline{158}$

Let x = $0.\overline{158}$

⇒ x = 0.158158..     ......(1)

⇒ 1000x = 158.158158..     ......(2)

On subtracting (1) from (2), we get :

⇒ 1000x - x = 158.158158.... - 0.158158.....

⇒ 999x = 158

⇒ x = $\dfrac{158}{999}$

Hence, $0.\overline{158} = \dfrac{158}{999}$.

(iv) $1.\overline{3}$

Let x = $1.\overline{3}$

⇒ x = 1.3333..     ......(1)

⇒ 10x = 13.333..     ......(2)

On subtracting (1) from (2), we get :

⇒ 10x - x = 13.333.. - 1.333..

⇒ 9x = 12

⇒ x = $\dfrac{12}{9}$

⇒ x = $\dfrac{4}{3}$

Hence, $1.\overline{3} = \dfrac{4}{3}$.

(v) $4.\overline{17}$

Let x = $4.\overline{17}$

⇒ x = 4.1717..     ........(1)

⇒ 100x = 417.1717..     .......(2)

On subtracting (1) from (2), we get :

⇒ 100x - x = 417.1717..... - 4.1717.....

⇒ 99x = 413

⇒ x = $\dfrac{413}{99}$

Hence, $4.\overline{17} = \dfrac{413}{99}$.

(vi) $0.\overline{12}$

Let x = $0.\overline{12}$

⇒ x = 0.1222..     ...........(1)

⇒ 100x = 12.222...(ii)

On subtracting (i) from (ii), we get

⇒ 99x = 12

⇒ x = $\dfrac{12}{99}$

⇒ x = $\dfrac{12÷3}{99÷3}$

⇒ x = $\dfrac{4}{33}$

Hence, 0.12... = $\dfrac{4}{33}$.

(vii) $0.1\overline{36}$

Let x = $0.1\overline{36}$

⇒ x = 0.13636..     ...........(1)

⇒ 10x = 1.3636..     ...........(2)

⇒ 1000x = 136.3636..     ...........(3)

On subtracting (2) from (3), we get

⇒ 1000x - 10x = 136.3636.. - 1.3636..

⇒ 990x = 135

⇒ x = $\dfrac{135}{990}$

⇒ x = $\dfrac{135 ÷ 45}{990 ÷ 45}$

⇒ x = $\dfrac{3}{22}$

Hence, $0.1\overline{36}\dfrac{3}{22}$.

(viii) $1.5\overline{7}$

Let x = $1.5\overline{7}$.

⇒ x = 1.5777..     ...........(1)

⇒ 10x = 15.777..     ...........(2)

⇒ 100x = 157.7777..     ...........(3)

On subtracting (i) from (iii), we get

⇒ 100x - 10x = 157.7777.. - 15.777......

⇒ 90x = 142

⇒ x = $\dfrac{142}{90}$

⇒ x = $\dfrac{142 ÷ 2}{90 ÷ 2}$

⇒ x = $\dfrac{71}{45}$

Hence, $1.5\overline{7} = \dfrac{71}{45}$.