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The speed of a bus during the second hour of the journey is twice that in the first hour. Also, its speed in the third hour is two-third the sum of its speeds in the first two hours. Had the bus travelled for three hours at the speed of the first hour, it would have travelled $120$km less. Find the average speed of the bus for the first three hours.

  1. $100 \text{kmph}$
  2. $80 \text{kmph}$
  3. $70 \text{kmph}$
  4. $60 \text{kmph}$
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Let the speed of the bus during the first hour be $t\,kmph$

So, speed during second hour=$2t\,kmph$ and the speed during third hour works out to be $\frac{2}{3} \times (t+2t)=2t\,kmph$

So, total distance travelled by the bus in 3 hours=$t+2t+2t=5t\,km$

Had the bus travelled for 3 hours at the speed of the first hour it would have travelled 120km less

$5t-3t=120\,km$

$t=60$

So, speed during first,second and third hours are $60kmph,120kmph,120kmph$ respectively.

Average speed=$\frac{60+120+120}{3}=100kmph(A)$
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