Abdul, while driving to school, computes the average speed for his trip to be $20$ km $h^{- 1}$ . On his return trip along the same route, there is less traffic and the average speed is $30$ km $h^{- 1}$ . What is the average speed for Abdul's trip?

Question

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Answer 1

Let us assume that Abdul is travelling $x$ kilometer distance while going to school.
Distance = Speed x Time

1)When it's average speed is 20 km/hr
$x = v_{1} \times t_{1} = 20 \times t_{1}$
2)When it's average speed is 30 km/hr

$x = v_{2} \times t_{2} = 30 \times t_{2}$

$V_{a v e r a g e} = \frac{t o t a l d i s t a n c e t r a v e l l e d}{t o t a l t i m e t a k e n}$
$V_{a v e r a g e} = \frac{x + x}{\frac{x}{v _{1}} + \frac{x}{v _{2}}} = \frac{2 x}{\frac{x}{2 0} + \frac{x}{3 0}} = 24 k m h^{- 1}$

Answer 2

Let us assume that Abdul is travelling

x

kilometer distance while going to school.

Distance = Speed x Time

1)When it's average speed is 20 km/hr

x = v_{1} \times t_{1} = 20 \times t_{1}

2)When it's average speed is 30 km/hr

x = v_{2} \times t_{2} = 30 \times t_{2}

V_{a v e r a g e} = \frac{t o t a l d i s t a n c e t r a v e l l e d}{t o t a l t i m e t a k e n}

V_{a v e r a g e} = \frac{x + x}{\frac{x}{v _{1}} + \frac{x}{v _{2}}} = \frac{2 x}{\frac{x}{2 0} + \frac{x}{3 0}} = 24 k m h^{- 1}