Question

Question 85

If x = $\frac{1}{10}$ and y = $\frac{-3}{8}$, then evaluate $x+y,x-y,x\times yandx÷y$.

Answer 1

Given, $x=\frac{1}{10}andy=\frac{-3}{8}$
Now, $x+y=\frac{1}{10}+\frac{(-3)}{8}=\frac{1}{10}-\frac{3}{8}=\frac{1\times 4}{10\times 4}-\frac{3\times 5}{8\times 5}=\frac{4}{40}-\frac{15}{40}=\frac{4-15}{40}=-\frac{11}{40}$
And $x-y=\frac{1}{10}-(-\frac{3}{8})=\frac{1}{10}+\frac{3}{8}=\frac{1\times 4}{10\times 4}+\frac{3\times 5}{8\times 5}=\frac{4}{40}+\frac{15}{40}=\frac{4+15}{40}=\frac{19}{40}$
$\therefore$ Product of rational number $=\frac{Productofnumerator}{Productofdenominators}=x\times y=\frac{1}{10}\times \frac{(-3)}{8}=\frac{1\times (-3)}{10\times 8}=\frac{-3}{80}$
And $x÷y=\frac{1}{10}÷\left(\frac{-3}{8}\right)$
The reciprocal of $\left(\frac{-3}{8}\right)=\frac{8}{-3}.$
So, $x÷y=\frac{1}{10}\times \frac{8}{-3}=\frac{1\times 8}{10\times -3}=\frac{-8}{30}=\frac{-8÷2}{30÷2}=\frac{-4}{15}$