For a train that travels from one station to another at a uniform speed of $40 k m h^{- 1}$ and returns to same station at speed of $60 k m h^{- 1}$ then its average speed is

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$98 k m / h r$

$0 k m / h r$

$50 k m / h r$

$48 k m / h r$

answer is D.

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Detailed Solution

Train travels from one station to another with a speed $40 k m h^{- 1}$ in time $t_{1}$
Train returns to the same station with a speed $60 k m h^{- 1}$ in time $t_{2}$
$A v e r a g e s p e e d = \frac{t o t a l d i s \tan c e}{t o t a l t i m e} - - - (1)$
Let total distance is ‘2S’
$s p e e d = \frac{d i s \tan c e}{t i m e} \Rightarrow t i m e = \frac{d i s \tan c e}{s p e e d} s u b s t i t u t e, t_{1} = \frac{S}{40}; t_{2} = \frac{S}{60}; t_{1} + t_{2} = t = t o t a l t i m e$
$A v e r a g e s p e e d = \frac{2 S}{t_{1} + t_{2}} s u b s t i t u t e t_{1}, t_{2} v a l u e s A v e r a g e s p e e d = \frac{2 S}{\frac{S}{40} + \frac{S}{60}} A v e r a g e s p e e d = \frac{2}{(\frac{40 + 60}{60 \times 40})}$
$A v e r a g e s p e e d = \frac{2 \times 60 \times 40}{40 + 60} A v e r a g e s p e e d = \frac{48 \times 100}{100} A v e r a g e s p e e d = 48 k m h^{- 1}$