From a to B, it took 6 hours to go and 111 hours to return A. 211B. 50%C. 12D. 111

6-1 ^ [16 × (1 + 111)], = 6-1 ^ [16 × 1211], = 6-1 ^ 211, = 6-5.5, = 0.5 (hours). Answer: it takes 0.5 hours less than the last time

Zhang Ming goes back and forth between a and B by bike. It takes 4 o'clock to get there and 5 o'clock to get back. How much faster is the speed when he returns than when he goes back?

(4-5)/5=20%

A man walked from a to B. There were regular bus trips between a and B, and the headways between the two places were equal. He found that a bus to a came every 6 minutes, and a bus to B came every 12 minutes. Q: how many minutes did the bus leave from their respective departure stations?

Suppose that the person's speed is B and the car's speed is a, we can get 12 × (a − b) = ax, 1 6 × (a + b) = ax, 2 (a − b) a + B = 12a-2b = a + B, the solution is a = 3b, and we can get x = 8 by substituting equation 6 (a + b) = ax

When someone goes from a to B, there are regular buses between a and B, and the time interval between the two places is equal. He finds that a bus to a comes every 10 minutes, and a bus to B comes every 15 minutes. How many minutes does the bus leave from their respective departure station?

We set the speed as V car and the speed of people as v people,
Because each car has the same distance,
There are (V car - V person) * 15 = (V car + V person) * 10,
The deformed v-car = 5V people,
In this way, the interval = (V vehicle - V person) * 15 / V vehicle = 12 (min),
Or (V vehicle + V person) * 10 / V vehicle = 12 (points)
A: buses leave from their respective departure stations every 12 minutes

From a to B, it took 6 hours to go and 111 hours to return A. 211B. 50%C. 12D. 111