Question

Multiply:

viii) $6\frac{2}{3}\text{and}\frac{-3}{8}$

Answer 1

Step 1: Converting the mixed fraction into
an improper fraction
Given: $6\frac{2}{3}=\frac{6\times 3+2}{3}=\frac{20}{3}$

Step: 2 Finding the product of given
numbers
$=\frac{20}{3}\times \left(\frac{-3}{8}\right)$ $\because$ Prduct of two rational numbers
$=\frac{\text{Product of the numerators}}{\text{Product of the denominators}}$
$\therefore \frac{20}{3}\times \left(\frac{-3}{8}\right)=\frac{20\times (-3)}{3\times 8}=-\frac{20\times 3}{3\times 8}=-\frac{20}{8}$
$=-\frac{5}{2}=-2\frac{1}{2}$
Hence, the required product is $-2\frac{1}{2}$.