Exercise 2.2 - Chapter 2 Real Numbers 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question $1 .$
Express the following rational numbers into decimal and state the kind of decimal expansion.
(i) $\frac{2}{7}$
(ii) $-5 \frac{3}{11}$
(iii) $\frac{22}{3}$
(iv) $\frac{327}{200}$
Solution:
(i) $\frac{2}{7}$

$\frac{2}{7}=0 . \overline{285714}$
Nen-terminating and recurring

Question $2 .$
Express $\frac{1}{13}$ in decimal form. Find the length of the period of decimals.
Solution:

Question $3 .$
Express the rational number $\frac{1}{13}$ in recurring decimal form by using the recurring decimal expansion of $\frac{1}{11}$. Hence write $\frac{71}{33}$ in recurring decimal form.
Solution:
The recurring decimal expansion of $\frac{1}{11}=0.09090909 \ldots . .=0 . \overline{09}$
$\begin{aligned}
\therefore \frac{1}{33}=0.03030303 \ldots . &=0 . \overline{03} \text { Also, } \frac{71}{33}=2 \frac{5}{33}=2+\frac{5}{33} \\
&=2+\left(5 \times \frac{1}{33}\right) \\
&=2+(5 \times 0 . \overline{03}) \\
&=2+(5 \times 0.030303 \cdots) \\
&=2+0.151515 \cdots \cdots \\
&=2.151515 \cdots \cdots=2 . \overline{15}
\end{aligned}$

Question $4 .$
Express the following decimal expression into rational numbers.
(i) $0 . \overline{24}$
(ii) $2 . \overline{327}$
(iii) $-5.132$
(iv) $3.1 \overline{7}$
(v) $17 . \overline{215}$
(vi) $-21.213 \overline{7}$
Solution:
(i) $0 . \overline{24}$
Let $x=0 . \overline{24}=0.24242424 \ldots \ldots . . \ldots .$ (1)
(Here period of decimal is 2 , multiply equation (1) by 100 ) $100 \mathrm{x}=24.242424 \ldots \ldots \ldots \ldots . \ldots(2)$
(2) - (1)
$100 \mathrm{x}-\mathrm{x}=24.242424 \ldots . .0 .242424 \ldots$
$99 \mathrm{x}=24$
$x=\frac{24}{99}$
(ii) $2 . \overline{327}$
Let $x=2.327327327 \ldots \ldots . . \ldots \ldots \ldots .$........
(Here period of decimal is 3, multiply equation (1) by 1000 ) $1000 \mathrm{x}=2327.327 \ldots \ldots \ldots \ldots \ldots . . . .$ (2)
(2) - (1)
$1000 x-x=2327.327327 \ldots-2.327327 \ldots . .$
$999 \mathrm{x}=2325$
$\mathrm{x}=\frac{2325}{999}$

(iii) $-5.132$
$x=-5.132=\frac{-5132}{1000}=\frac{-1283}{250}$
(iv) $3.1 \overline{7}$
Let $\mathrm{x}=3.1777 \ldots \ldots . \ldots \ldots \ldots . .$ (1)
(Here the repeating decimal digit is 7 , which is the second digit after the decimal point, multiply equation (1) by 10 )
$10 \mathrm{x}=31.7777 \ldots \ldots \ldots . \ldots \ldots$ (2)
$100 \mathrm{x}=317.7777 \ldots \ldots . \ldots . \ldots \ldots . \ldots \ldots . .(3)$
$90 \mathrm{x}=286$
Let $x=17.215215 \ldots \ldots . \ldots \ldots \ldots$ (1)
$1000 \mathrm{x}=17215.215215 \ldots \ldots . \ldots \ldots \ldots . \ldots$ (2)
$1000 \mathrm{x}-\mathrm{x}=17215.215215 \ldots-17.215 \ldots$
$x=\frac{286}{90}=\frac{143}{45}$ (v) $17 . \overline{215}$ Let $\mathrm{x}=17.2152$ $1000 \mathrm{x}=17215.2$ $(2)-(1)$ $1000 \mathrm{x}-\mathrm{x}=17215$ $999 \mathrm{x}=17198$ $\mathrm{x}=17198$
$\mathrm{x}=\frac{17198}{999}$
(vi) $-21.213 \overline{7}$
Let $\mathrm{x}=-21.2137777 \ldots \ldots \ldots \ldots . . .$ (1)
$10 \mathrm{x}=-212.137777 \ldots \ldots \ldots \ldots . \ldots . . .(2)$
$100 \mathrm{x}=-2121.37777 \ldots \ldots \ldots \ldots \ldots . . . \ldots$ (3)
$1000 \mathrm{x}=-21213.77777 \ldots \ldots \ldots \ldots . . . \ldots(4)$
$10000 \mathrm{x}=212137.77777 \ldots . . \ldots \ldots . \ldots . .(5)$
(Now period of decimal is 1 , multiply equation (4) it by 10 )
$(5)-(4)$
(vi) $-21.213 \overline{7}$
Let $\mathrm{x}=-21.2137777 \ldots \ldots \ldots \ldots \ldots(1)$
$10 \mathrm{x}=-212.137777 \ldots \ldots \ldots \ldots \ldots(2)$
$100 \mathrm{x}=-2121.37777 \ldots \ldots \ldots \ldots \ldots \ldots(3)$
$1000 \mathrm{x}=-21213.77777 \ldots \ldots \ldots \ldots \ldots(4)$
$10000 \mathrm{x}=212137.77777 \ldots \ldots \ldots \ldots \ldots(5)$
(Now period of decimal is 1, multiply equation (4) it by 10$)$
$(5)-(4)$
$10000 \mathrm{x}-1000 \mathrm{x}=(-212137.7777 \ldots)-(-21213.7777 \ldots)$
$9000 \mathrm{x}=-190924$
$10000 \mathrm{x}-1000 \mathrm{x}=(-212137.7777 \ldots)-(-21213.7777 \ldots)$
$9000 x=-190924$

$x=-\frac{190924}{9000}$

Question $5 .$
Without actual division, find which of the following rational numbers have terminating decimal expansion.
(i) $\frac{7}{128}$
(ii) $\frac{21}{15}$
(iii) $4 \frac{9}{35}$
(iv) $\frac{219}{2200}$
Solution:

So $\frac{7}{128}=\frac{7}{2^{7} 5^{0}}$
This of the form $4 \mathrm{~m}, \mathrm{n} \in \mathrm{W}$
So $\frac{7}{128}$ has a terminating decimal expansion.

$\therefore$ This is not of the form $\frac{p}{5^{1} 7^{1}}$
So $4 \frac{9}{35}$ has a non-terminating recurring decimal expansion.

$\frac{219}{2200}=\frac{219}{2^{3} 5^{2} 11^{1}}$
$\therefore$ This is not of the form $\frac{p}{2^{m} 5^{1}}$
So $\frac{219}{2200}$ has a non-terminating recurring decimal expansion.