Find a rational number between

(i) $\frac{3}{8}$ and $\frac{2}{5}$

Let:

$x=\frac{3}{8}$ and $y=\frac{2}{5}$

Rational number lying between x and y:

$\frac{1}{2}(x+y)=\frac{1}{2}\left(\frac{3}{8}+\frac{2}{5}\right)$

$=\frac{1}{2}\left(\frac{15+16}{40}\right)=\frac{31}{80}$

(ii) 1.3 and 1.4
Let:
x = 1.3 and y = 1.4
Rational number lying between x and y:

$\frac{1}{2}(x+y)=\frac{1}{2}(1.3+1.4)$

$=\frac{1}{2}(2.7)=1.35$

(iii) $-1$ and $\frac{1}{2}$

Let:

$x=-1$ and $y=\frac{1}{2}$

Rational number lying between x and y:

$\frac{1}{2}(x+y)=\frac{1}{2}\left(-1+\frac{1}{2}\right)$

$=-\frac{1}{4}$

(iv) $-\frac{3}{4}$ and $-\frac{2}{5}$

Let:

$x=-\frac{3}{4}$ and $y=-\frac{2}{5}$

Rational number lying between x and y:

$\frac{1}{2}(x+y)=\frac{1}{2}\left(-\frac{3}{4}-\frac{2}{5}\right)$

$=\frac{1}{2}\left(\frac{-15-8}{20}\right)=-\frac{23}{40}$

(v) $\frac{1}{9}$ and $\frac{2}{9}$

A rational number lying between $\frac{1}{9}$ and $\frac{2}{9}$ will be

$\frac{1}{2}\left(\frac{1}{9}+\frac{2}{9}\right)=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$