Question

(a) How many rational numbers lie between $\frac{2}{5}$ and $\frac{6}{5}$? Mention atleast two.

(b) If $x=\frac{1}{10}$, $y=\frac{-3}{8}$. Evaluate: $(x+y),(x-y),(x\times y),(x÷y)$
[3 MARKS]

Answer 1

a) Each rational number: 0.5 Marks
b) Each operation: 0.5 Marks

(a) There are infinite numbers of rational numbers that lie between two points on a number line.

Consider the rational numbers $\frac{2}{5}$ and $\frac{6}{5}$.
Take the average of two numbers $\frac{2+6}{5\times 2}$ = $\frac{8}{10}$

$\frac{8}{10}$ lies in the middle of $\frac{2}{5}$ and $\frac{6}{5}$.

Now take the average of $\frac{2}{5}$ and $\frac{8}{10}$. It is $(\frac{2}{5}+\frac{8}{10})\times \frac{1}{2}$ = $\frac{4+8}{10}\times \frac{1}{2}$ = $\frac{12}{20}$

So, as we proceed in this, we will get an infinite number of rational number between two points.


(b) (i) $x+y=\frac{1}{10}+\frac{-3}{8}$

$\Rightarrow \frac{1}{10}-\frac{3}{8}=\frac{8-30}{80}$

$\Rightarrow \frac{-22}{80}=\frac{-11}{40}$

(ii) $x-y=\frac{1}{10}-\frac{-3}{8}$

$\Rightarrow \frac{1}{10}+\frac{3}{8}=\frac{8+30}{80}$

$\Rightarrow \frac{38}{80}=\frac{19}{40}$


(iii) $x\times y=\frac{1}{10}\times \frac{-3}{8}=\frac{-3}{80}$


(iv) $x÷y=\frac{1}{10}÷\frac{-3}{8}$

$\Rightarrow \frac{1}{10}\times \frac{8}{-3}=\frac{8}{-30}$

$\Rightarrow \frac{-8}{30}=\frac{-4}{15}$