A city bus travels the distance between two stops, moving on schedule, with an average speed of 30 km/h. One day he got into a jam, so the average speed, with which it travelled, was 5 km/h less than at usual. By what average speed should the bus move to catch the delay?
Answer: 37,5 km/h
I would be very grateful if you give me a hint. When I try it, I will share what I've got. With this problem, I can't even start.
On
Average speed is all about distance travelled and time taken. We don't know the exact distance here, but that's irrelevant as 1) all distances are the same, and 2) we don't care about the actual time taken. So you can just make a guess (60 kilometers, for instance, is often a choice that makes for rather easy calculation, or 30km, making for 60 km round trip). Or be general and say "The distance is $d$ kilometers". The answer you find in the end will be the same.
So, with such a guess in mind, how long does the trip take according to the schedule? How far behind schedule is the bus due to traffic? How much time does it have to travel back? How fast does that mean that it's going?
The bus travels a distance $d$. If it travels at $25$ km/h, the time it takes is $\dfrac d{25}$. It order to catch the delay, it will have to move at a speed $v$, and the time it will take will be $\dfrac dv$. So, the total time will be $\dfrac d{25}+\dfrac dv$. And you want it to take the same time that it would it to make two journeys at $30$ km/h , which is $2\times\dfrac d{30}$. So…