For the following question:
Pedro travels by bus to school at an average speed of $40$ km/hr. He is driven home by the same route by a friend's car at an average speed of $50$ km/hr. Which of the following is greatest:
(a) Average speed of both legs of the journey.
(b) $45$.
According to the book, the answer is (b). How did they calculate average speed here? I know that the formula for average speed is $A_v=\frac{Total~ Distance~ Covered}{Total~ Time ~Taken }$. Here we only know the velocity.
Best Answer
You can solve this one without doing any real calculations. It takes longer to do the part of the journey going at $40$ km/h than it does to do the part of the journey going at $50$ km/h, so the majority of the time is spent going at $40$ km/h. Therefore, the average speed is going to be less than $45$ km/h (it would be $45$ km/h if the same time was spent going at both speeds).
Alternatively, you could let the distance for one leg of the journey = $s$. Then, the time taken to do the part of the journey at $40$ km/h $= \frac{s}{40}$, and the time taken to do the part of the journey at $50$ km/h $= \frac{s}{50}$. So the total time taken $= \frac{s}{40} + \frac{s}{50} = \frac{9s}{200}$. As you said, average speed $ = \frac{total ~ distance}{total ~ time} = \frac{2s}{9s/200} = \frac{400}{9} = 44.4\cdots$, which is clearly less than $45$.