2. Real Numbers : Practice Set 2.1

1. Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type:

(i) \(\displaystyle \frac {13}{5}\)

Solution:

\(\displaystyle \frac {13}{5}\) = 2.6

∴ The decimal representation is terminating.

(ii) \(\displaystyle \frac {2}{11}\)

Solution:

\(\displaystyle \frac {2}{11}\) = 0.1818 ...

∴ The decimal representation is non-terminating but recurring.

(iii) \(\displaystyle \frac {29}{16}\)

Solution:

\(\displaystyle \frac {29}{16}\) = 1.8125

∴ The decimal representation is terminating.

(iv) \(\displaystyle \frac {17}{125}\)

Solution:

\(\displaystyle \frac {17}{125}\) = 0.136

∴ The decimal representation is terminating.

(v) \(\displaystyle \frac {11}{6}\)

Solution:

\(\displaystyle \frac {11}{6}\) = 1.833 ...

∴ The decimal representation is non-terminating but recurring.

2. Write the following rational numbers in decimal form:

(i) \(\displaystyle \frac {127}{200}\)

Solution:

\(\displaystyle \frac {127}{200}\) = 0.635

(ii) \(\displaystyle \frac {25}{99}\)

Solution:

\(\displaystyle \frac {25}{99}\) = 0.2525 ... = \(\displaystyle 0.\overline{ 25 }\)

(iii) \(\displaystyle \frac {23}{7}\)

Solution:

\(\displaystyle \frac {23}{7}\) = 3.285714 ... ... = \(\displaystyle 3.\overline{ 285714 }\)

(iv) \(\displaystyle \frac {4}{5}\)

Solution:

\(\displaystyle \frac {4}{5}\) = 0.8

(v) \(\displaystyle \frac {17}{8}\)

Solution:

\(\displaystyle \frac {17}{8}\) = 2.125

3. Write the following rational numbers in \(\displaystyle \frac{p}{q}\) form:

(i) \(\displaystyle 0.\dot{6}\)

Solution:

Let, x = \(\displaystyle 0.\dot{6}\)

∴ 10x = 6.666 ...

∴ 10x = \(\displaystyle 6.\dot{6}\)

∴ 10x − x = \(\displaystyle 6.\dot{6}\) − \(\displaystyle 0.\dot{6}\)

∴ 9x = 6

∴ \(\displaystyle x = \frac {6}{9}\)

∴ \(\displaystyle x = \frac {2}{3}\)

∴ \(\displaystyle 0.\dot{6}\) = \(\displaystyle \frac {2}{3}\)

(ii) \(\displaystyle 0.\overline{37}\)

Solution:

Let, x = \(\displaystyle 0.\overline{37}\)

∴ 100x = 37.3737 ...

∴ 100x = \(\displaystyle 37.\overline{37}\)

∴ 100x − x = \(\displaystyle 37.\overline{37}\) − \(\displaystyle 0.\overline{37}\)

∴ 99x = 37

∴ \(\displaystyle x = \frac {37}{99}\)

∴ \(\displaystyle 0.\overline{37}\) = \(\displaystyle \frac {37}{99}\)

(iii) \(\displaystyle 3.\overline{17}\)

Solution:

Let, x = \(\displaystyle 3.\overline{17}\)

∴ 100x = 317.1717 ...

∴ 100x = \(\displaystyle 317.\overline{17}\)

∴ 100x − x = \(\displaystyle 317.\overline{17}\) − \(\displaystyle 3.\overline{17}\)

∴ 99x = 314

∴ \(\displaystyle x = \frac {314}{99}\)

∴ \(\displaystyle 3.\overline{17}\) = \(\displaystyle \frac {314}{99}\)

(iv) \(\displaystyle 15.\overline{89}\)

Solution:

Let, x = \(\displaystyle 15.\overline{89}\)

∴ 100x = 1589.8989 ...

∴ 100x = \(\displaystyle 1589.\overline{89}\)

∴ 100x − x = \(\displaystyle 1589.\overline{89}\) − \(\displaystyle 15.\overline{89}\)

∴ 99x = 1574

∴ \(\displaystyle x = \frac {1574}{99}\)

∴ \(\displaystyle 15.\overline{89}\) = \(\displaystyle \frac {1574}{99}\)

(v) \(\displaystyle 2.\overline{514}\)

Solution:

Let, x = \(\displaystyle 2.\overline{514}\)

∴ 1000x = 2514.514514 ...

∴ 1000x = \(\displaystyle 2514.\overline{514}\)

∴ 1000x − x = \(\displaystyle 2514.\overline{514}\) − \(\displaystyle 2.\overline{514}\)

∴ 999x = 2512

∴ \(\displaystyle x = \frac {2512}{999}\)

∴ \(\displaystyle 2.\overline{514}\) = \(\displaystyle \frac {2512}{999}\)