Explanation

(d) [image:3_image160.png]
Let $t_{AB},\ t_{BC},\ t_{CD}$ and $t_{DA}$ be the time taken by the aeroplane to go from $A$ to $B$, $B$ to $C$, $C$ to $D$ and $D$ to $A$ respectively
$\therefore t_{AB} = \frac{1000\ km}{250\ km\ h^{- 1}} = 4h$
$\therefore t_{BC} = \frac{1000\ km}{500\ km\ h^{- 1}} = 2h$
$\therefore t_{CD} = \frac{1000\ km}{200\ km\ h^{- 1}} = 5h$
$\therefore t_{DA} = \frac{1000\ km}{100\ km\ h^{- 1}} = 10h$
Average speed$= \frac{Total\ distance\ covered}{Total\ time\ taken}$
$= \frac{1000km + 1000km + 1000km + 1000km}{t_{AB} + t_{BC} + t_{CD} + t_{DA}}$
$= 190.5\ km\ h^{- 1}$