Question

A man walks at a speed of 6km/hr for 1 km and 8 km/hr for next 1 km.What is the average speed of a walk of 2 km

Answer 1

Here's your answer!!

__________________________

In first case,
Speed =6km/h
Distance=1km

So,
Time = D/S
$= > \frac{1}{6} h$
In Second case,
Speed = 8km/h
Distance=1km

So,
Time =D/S
$= > \frac{1}{8} h$
Now,
Total distance travelled=2km
Total time taken =
$= > \frac{1}{6} + \frac{1}{8} \\ = > \frac{4 + 3}{24}$
$= > \frac{7}{24}$
We know that,
$average \: speed = \frac{total \: distance \: travel}{total \: time}$
$= > \frac{2 \times 24}{7} \\ = > \frac{48}{7}$
$= > 6.85$
Hence,
Average speed is 6.85 (approx)

________________________________


Hope it helps you !! :)

Answer 2

Hope u like my process
=====================
=> A man walks for 1 km with speed 6km/hr

So,

$= > d_{1} = 1 \: \: km \\ \\ = > s_{1} = 6 \: \: km \: \: {hr}^{ - 1} \\ \\ = > t_{1} = \frac{ d_{1} }{ s_{1} } = \frac{1}{6} \: \: hr$

Now,... Then he covers next 1km with a speed of 8 km/hr

So,.

$= > d_{2} = 1 \: \: km \\ \\ = > s_{2} = 8 \: \: km \: \: {hr}^{ - 1} \\ \\ = > t_{2} = \frac{ d_{2}}{ s_{2}} = \frac{1}{8} \: \: hr$

Thus average speed for 2 km is

$= > \frac{ d_{1} + d_{2} }{ t_{1} + t_{2} } = \frac{1 + 1}{ \frac{1}{6} + \frac{1}{8} } \\ \\ = \frac{2}{ \frac{8 + 6}{8 \times 6} } = 2 \times \frac{8 \times 6}{14} = 6.85 \: km \: \: {hr}^{ - 1}$

Thus the required average speed = 6.85 km/hr

_____________________________

Hope this is ur required answer

Proud to help you