A body cools down from ${75}^{o}Cto{65}^{o}C$ in 10min. It will cool down from ${65}^{o}Cto{55}^{o}C$ in a time

According to Newton’s law of motion 
For the first condition 
 $$\begin{gathered} mc\left(\frac{75-65}{10}\right)=K(\frac{75+65}{2}-T_o)\Rightarrow mc\left(\frac{10}{10}\right)=K(70-T_o) \\ \Rightarrow mc=K(70-T_o) \mathrm{......}\left(i\right) \end{gathered}$$
According to second condition,
 $mc\left(\frac{65-55}{t^{\text{'}}}\right)=K(\frac{65+55}{2}-T_o)\Rightarrow mc\left(\frac{10}{t^{\text{'}}}\right)=K(60-T_o)\mathrm{....}\left(ii\right)$
Dividing eq(i) by eq(ii) we get
 $\frac{1}{\frac{10}{t^{\text{'}}}}=\frac{70-T_o}{60-T_o}\Rightarrow \frac{t^{\text{'}}}{10}=\frac{70-T_o}{60-T_o}\Rightarrow t^{\text{'}}=10\left(\frac{70-T_o}{60-T_o}\right)$
Clearly,  $\Rightarrow \frac{70-T_o}{60-T_o}>1$
$\therefore t^{\text{'}}>10\min utes$