SSC CHSL 17 Mar 2020 Shift 3 Paper

Car A and Car are at right angles to each other.
Distance covered by Car A in $20\sec$ $\left(\mathrm{AB}\right)=$ Speed $\times$ time
$=54\times \frac{5}{18}\times 20=300\mathrm{m}$
Distance covered by Car B in $20\sec$
$\left(\mathrm{BC}\right)=72\times \frac{5}{18}\times 20=400\mathrm{m}$
In $\mathrm{\Delta }\mathrm{ABC}$
${\mathrm{AC}}^2={\mathrm{BC}}^2+{\mathrm{AB}}^2$ [by Pythagoras theorem]
⇒ ${\mathrm{AC}}^2={\left(400\right)}^2+{\left(300\right)}^2$
⇒ ${\mathrm{AC}}^2=160000+90000$
⇒ ${\mathrm{AC}}^2=250000$
∴ $\mathrm{AC}=500\mathit{m}$
Hence, distance between Car A and Car $\mathrm{B}=500\mathrm{m}$.