In the first case, the average temperature of the liquid

$=\frac{70+40}{2}=55 °\mathrm{C}$. 
Excess temperature $=55-20=35 °\mathrm{C}$

$\frac{d{\mathrm{\theta }}_1}{dt}=\frac{40-70}{5}=-6 °\mathrm{C} {\mathrm{minutes}}^{-1}$

$-6=K\times 35$ ...$\left(1\right)$

In the second case, the average temperature of liquid,  $=\frac{60+40}{2}={50}^{\mathrm{o}}\mathrm{C}$ .
Excess temperature $=50-20=30 °\mathrm{C}$.
Rate of fall in temperature$\Rightarrow \frac{d{\mathrm{\theta }}_2}{dt}$
$\frac{d{\mathrm{\theta }}_2}{dt}=-\frac{60-40}{t_{\mathrm{m}\mathrm{in}}}$
$-\frac{20}{t_{\mathrm{m}\mathrm{in}}}=K\times 30$ ...$\left(2\right)$
Divide $\left(1\right)$ by $\left(2\right)$, 
$$\begin{gathered} \frac{-6}{{\displaystyle \raisebox{1ex}{$-20$}\!\left/ \!\raisebox{-1ex}{$t_{\min }$}\right.}}=\frac{K\times 35}{K\times 30}\Rightarrow \frac{7}{6}=\frac{3\times t_{min}}{10} \\ \Rightarrow t_{\min }=3.89 \min \end{gathered}$$