The area under the velocity time graph gives the displacement. The area for $t=0 \mathrm{s} \text{to}\mathit{ }t=25 \mathrm{s}$is 

$\left(Area\right)=\left(\frac{1}{2}\times 5\times 10\right)+\left(5\times 10\right)+\left(\frac{1}{2}\times 30\times 5\right)+\left(\frac{1}{2}\times 20\times 5\right)-\left(\frac{1}{2}\times 20\times 5\right)$

 $\Rightarrow Area=25+50+75+50-50=150 \mathrm{m}$
Hence, the displacement $=150 \mathrm{m}$

The distance is defined as the path taken by the object to travel from initial to final position. Also note that displacement is a vector quantity, but distance is a scalar quantity.
Distance $$\begin{gathered} =\left(\frac{1}{2}\times 5\times 10\right)+\left(5\times 10\right)+\left(\frac{1}{2}\times 30\times 5\right)+\left(\frac{1}{2}\times 20\times 5\right)+\left(\frac{1}{2}\times 20\times 5\right) \\ =(200+50)=250 \mathrm{m} \end{gathered}$$

Therefore, required ratio
$\frac{\text{ distance }}{\text{ displacement }}=\frac{250}{150}=\left(\frac{5}{3}\right)$