Question

Расстояние между пунктами А и Б равно 280 км. Автобус прошёл это расстояние с постоянной скоростью. Через 3 ч после выезда из А он находился более, чем в 100 км от Б. Через 5 ч езды ему оставалось проехать до Б менее 10 км. В каких пределах находилась скорость автобуса?

Answer 1

1)280-100=180(км)проехал через 3 ч
2)180/3=60(км/ч)скорость автоб
3)280-10=270(км)проехал через 5 ч
4)270/5=54(км/ч)скорость автоб
автобус ехал в пределах 60км/м

Answer 2

1)280-10=270(км)-проехал автобус за 3 ч
2)270/5=54(км/ч)-скорость
3)54×3=162(км)-за 3 ч проехал
4)54×2=108(км)-ещё за 2ч проехал 
5)162+108=270(км)-за 5 ч ост 10 км 
я думаю так 

Answer 3

Problem Analysis

We are given that the distance between points A and B is 280 km. An bus travels this distance at a constant speed. After 3 hours of leaving point A, the bus is more than 100 km away from point B. After 5 hours, the bus has less than 10 km left to travel to reach point B. We need to determine the range within which the speed of the bus falls.

Solution

Let's assume the speed of the bus is v km/h.

After 3 hours, the bus has traveled a distance of 3v km. According to the given information, it is more than 100 km away from point B. Therefore, we can write the following inequality:

280 - 3v > 100 Simplifying the inequality, we get:

3v < 180

Dividing both sides by 3, we find:

v < 60

So, the speed of the bus must be less than 60 km/h.

After 5 hours, the bus has traveled a distance of 5v km. According to the given information, it has less than 10 km left to travel to reach point B. Therefore, we can write the following inequality:

280 - 5v < 10 Simplifying the inequality, we get:

5v > 270

Dividing both sides by 5, we find:

v > 54

So, the speed of the bus must be greater than 54 km/h.

Therefore, the speed of the bus must be within the range of 54 km/h < v < 60 km/h.