The velocity-time graph of a body moving in a straight line is shown in figure.

 

The ratio of displacement and distance travelled by the body in time $0$ to $10 \mathrm{s}$ is

Area of upper rectangles, which will give us magnitude of the area under the velocity-time graph,

$A_1=8\times 2+4\times 4 =32$

Area of lower rectangles, which will give us magnitude of the area under the velocity-time graph,

$A_2=4\times 2 +4\times 2 = 16$

Therefore,

$$\begin{gathered} \mathrm{Displacement}={\mathrm{A}}_1-{\mathrm{A}}_2= 32-16 =16 \\ \mathrm{Distance} ={\mathrm{A}}_1+{\mathrm{A}}_2= 32+16=48 \end{gathered}$$

Hence, the required ratio would be

$=\frac{16}{48}=\frac{1}{3}$