Question

Evaluate:
iv) $(\frac{-13}{7}\times \frac{-35}{39})-(\frac{-7}{45}\times \frac{9}{14})$

Answer 1

Step 1: Using the concept of multiplication of rational numbers
Given, $(\frac{-13}{7}\times \frac{-35}{39})-(\frac{-7}{45}\times \frac{9}{14})$

$\therefore (\frac{-13}{7}\times \frac{-35}{39})\frac{-13}{7}\times \frac{-35}{39}=\frac{-13\times -35}{7\times 39}=\frac{5}{3}$
Also given,
$(\frac{-7}{45}\times \frac{9}{14})=\frac{-7}{45}\times \frac{9}{14}=\frac{-7\times 9}{45\times 14}=\frac{-1}{10}$

Step 2: Finding the required value
Given,
$(\frac{-13}{7}\times \frac{-35}{39})-(\frac{-7}{45}\times \frac{9}{14})$
$=\frac{5}{3}-\frac{-1}{10}$
$=\frac{5}{3}+\frac{1}{10}$
$=\frac{5\times 10}{3\times 10}+\frac{1\times 3}{10\times 3}$
[LCM of $3,10=3\times 10=30$]
$=\frac{50}{30}+\frac{3}{30}=\frac{50+3}{30}=\frac{53}{30}=1\frac{23}{30}$.

Hence, the required result is $1\frac{23}{30}$.