You drive on Interstate 10 from San Antonio to Houston, half the time at 72 km/h and the other half at 98 km/h. On the way back you travel half the distance at 72 km/h and the other half at 98 km/h. What is your average speed (a) from San Antonio to Houston, (b) from Houston back to San Antonio, and (c) for the entire trip?

Answer: a. 85km/hr b. 82.3km/hr

c. 84km/hr

Explanation: first let take the total time from San Antonio to Houston to be 2hr.

Half time 1hr was covered with speed of 72km/hr

Distance = speed*time=72km/hr * 1hr

=72km

So too with the second half of 1hr covered with speed of 98km/hr

Distance = 98km

Total distance from Houston to San Antonio is 98+72 = 170km

a. Average speed from San Antonio to Houston is

S1 = 170/2

=85km/hr

b. half distance from Houston to San Antonio which is 170km/2

= 85km was covered with speed of 72km/hr first half, so time

t = dist/speed

t = 85/72 = 1hr 12 mins

Remaining 85 km covered with a speed of 98km/hr

Time = 85/98 = 0.88*60min

= 52 mins

Total time = 1hr + 12mins + 52mins

=2hr4mins = 124/60 hr

So average speed = distance/time

=170/124/60

Using reciprocal law

Average speed S2 = 170*60/124

= 82.3km/hr

C. Average speed to and fro (entire tripe)

= (85+82.3) / 2

=84km/hr