A motorist travelled $210 k m$ in $6 h o u r s$ . His average speed for first $150 k i l o m e t r e s$ was $50 k m h^{- 1}$ . What was his average speed for the remaining $60 k i l o m e t r e s$ ?

15 k m h^{- 1}

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20 k m h^{- 1}

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40 k m h^{- 1}

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30 k m h^{- 1}

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Solution

The correct option is B $20 k m h^{- 1}$
Given:
Total distance travelled = $210 k m$ Total time taken = $6 h o u r s$

We know,
$D i s t a n c e = s p e e d \times t i m e,$
$o r, t i m e = \frac{d i s t a n c e}{s p e e d} .$

Time taken to cover the first $150 k m$ = $\frac{150}{50}$ = $3 h o u r s$

So, the next $60 k m$ will be covered in $3 h o u r s$ .

We know,
$a v e r a g e s p e e d = \frac{d i s t a n c e}{t i m e},$
Hence, the average speed for the next $60 k m$ = $\frac{60}{3}$ = $20 k m h^{- 1} .$

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