Question

a bus travels the first one third distance at speed of 10kmph.the next one third distance at a speed of 20kmph and the last one third distance at speed of 60kmph the average speed is

Answer 1

I think that average speed will be 1/30 kmph

Answer 2

Answer:

Let total distance be d.

d/3 = 10 × t1

d/3 = 20 × t2

d/3 = 60 × t3

=> 20 × t2 = 60 × t3 => t2 = 3 × t3

AND

=> 10 × t1 = 20 × t2 => t1 = 2 × t2 => t1 = 6 × t3

$average \: speed = \frac{total \: distance}{total \: time} = \frac{ \frac{d}{3} + \frac{d}{3} + \frac{d}{3} }{t1 + t2 + t3} = \frac{(10 \times t1) + (20 \times t2) + (60 \times t3)}{(6\times t3) + (3 \times t3) + t3}$

$= \frac{(10 \times 6 \times t3) + (20 \times 3 \times t3) + (60 \times t3)}{10 \times t3} = \frac{180 \times t3}{10 \times t3} = 18$

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