Exercise 1.2 (Revised) - Chapter 1 - Rational Numbers - Ncert Solutions class 8 - Maths

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NCERT Solutions for Class 8 Maths Chapter 1 - Rational Numbers | Comprehensive Guide

Example 1:

Find $\frac{3}{7}+\left(\frac{-6}{11}\right)+\left(\frac{-8}{21}\right)+\left(\frac{5}{22}\right)$
Solution:

$\frac{3}{7}+\left(\frac{-6}{11}\right)+\left(\frac{-8}{21}\right)+\left(\frac{5}{22}\right)$
$=\frac{198}{462}+\left(\frac{-252}{462}\right)+\left(\frac{-176}{462}\right)+\left(\frac{105}{462}\right)$ (Note that 462 is the LCM of 7, 11, 21 and 22)
$
=\frac{198-252-176+105}{462}=\frac{-125}{462}
$

We can also solve it as.
$
\begin{aligned}
& \frac{3}{7}+\left(\frac{-6}{11}\right)+\left(\frac{-8}{21}\right)+\frac{5}{22} \\
& =\left[\frac{3}{7}+\left(\frac{-8}{21}\right)\right]+\left[\frac{-6}{11}+\frac{5}{22}\right] \quad \text { (by using commutativity and associativity) } \\
& =\left[\frac{9+(-8)}{21}\right]+\left[\frac{-12+5}{22}\right] \quad \text { (LCM of } 7 \text { and } 21 \text { is } 21 ; \text { LCM of } 11 \text { and } 22 \text { is } 22 \text { ) } \\
& =\frac{1}{21}+\left(\frac{-7}{22}\right)=\frac{22-147}{462}=\frac{-125}{462}
\end{aligned}
$
(LCM of 7 and 21 is 21 ; LCM of 11 and 22 is 22 )

Do you think the properties of commutativity and associativity made the calculations easier?
Example 2:

Find $\frac{-4}{5} \times \frac{3}{7} \times \frac{15}{16} \times\left(\frac{-14}{9}\right)$
Solution:

We have

$
\begin{aligned}
\frac{-4}{5} & \times \frac{3}{7} \times \frac{15}{16} \times\left(\frac{-14}{9}\right) \\
& =\left(-\frac{4 \times 3}{5 \times 7}\right) \times\left(\frac{15 \times(-14)}{16 \times 9}\right) \\
& =\frac{-12}{35} \times\left(\frac{-35}{24}\right)=\frac{-12 \times(-35)}{35 \times 24}=\frac{1}{2}
\end{aligned}
$

We can also do it as.
$
\begin{aligned}
\frac{-4}{5} & \times \frac{3}{7} \times \frac{15}{16} \times\left(\frac{-14}{9}\right) \\
& =\left(\frac{-4}{5} \times \frac{15}{16}\right) \times\left[\frac{3}{7} \times\left(\frac{-14}{9}\right)\right] \\
& =\frac{-3}{4} \times\left(\frac{-2}{3}\right)=\frac{1}{2}
\end{aligned}
$
(Using commutativity and associativity)

Example 3:

Find $\frac{2}{5} \times \frac{-3}{7}-\frac{1}{14}-\frac{3}{7} \times \frac{3}{5}$

Solution:
$
\begin{aligned}
\frac{2}{5} \times \frac{-3}{7}-\frac{1}{14}-\frac{3}{7} \times \frac{3}{5} & =\frac{2}{5} \times \frac{-3}{7}-\frac{3}{7} \times \frac{3}{5}-\frac{1}{14} \text { (by commutativity) } \\
& =\frac{2}{5} \times \frac{-3}{7}+\left(\frac{-3}{7}\right) \times \frac{3}{5}-\frac{1}{14} \\
& =\frac{-3}{7}\left(\frac{2}{5}+\frac{3}{5}\right)-\frac{1}{14} \quad \text { (by distributivity) } \\
& =\frac{-3}{7} \times 1-\frac{1}{14}=\frac{-6-1}{14}=\frac{-1}{2}
\end{aligned}
$
(by distributivity)