Example 1: As the weather is getting colder and colder, the grass on the pasture is not growing, but decreasing at a fixed rate. If the grass on a meadow can feed 20 cows for 5 days or 15 cows for 6 days, how many cows can 10 feed?

20 cows graze 20×5= 100 (share) for 5 days, 15 cows graze 15×6=90 (share) for 6 days.

Grass decreased (100-90) ÷ (6-5) = 10 (serving) per day.

Before cattle graze, there was grass in the pasture100+10× 5 = 150 (portions)150 =/kloc-0.

However, due to the reduction of 10 grass per day, which is equivalent to 10 cattle, it can only be used for cattle 15- 10 = 5 (head cattle).

Commentary: The topic grass is decreasing every day. By comparing the two groups of conditions, the daily grass reduction is calculated, and then the cow is regarded as two parts, one is the visible cow, and the other is the invisible cow-the embodiment of cold. Calculate separately and finally get the difference.

Ex. 2: The exhibition started at 9 o'clock, but people lined up to enter. Since the first visitor arrives, if there are the same number of visitors every minute, if there are three entrances, there will be no one waiting in line at 9: 09. If there are five entrances, there will be no queue at 9: 05. Find the time when the first audience arrived.

Let each entrance pass 1 "people every minute.

Then 3 entrances pass through 3×9=27 (people) in 9 minutes and 5 entrances pass through 5×5=25 (people) in 5 minutes.

It shows that the number of arrivals per minute is (27-25) ÷ (9-5) = 0.5 (people). It was 27-0.5× 9 = 22.5 (people) before opening the door.

The time for these people to come to the exhibition is 22.5÷0.5=45 minutes. The time for the first audience to arrive is 9: 00-45 minutes =8: 00, 15 minutes.

A: The first audience will arrive at 8: 00. 15.

Comments: On the surface, this problem is far from the problem of cattle eating grass. It can be said that it is irrelevant, but after careful understanding, there are as many viewers coming every minute in the topic, which is similar to "grass grows and flies"; When the entrance is similar to a cow, the problem becomes a Newton problem. The key to solving problems is to master the essence of methods.

Self-practice:

(1) There is a pasture that can feed 27 cows for 6 weeks or 23 cows for 9 weeks. If the pasture grows at a constant speed every week, how many weeks can it feed 2 1 cow?

(2) There is a well. If the water level drops, the water will continue to spray at a constant speed, and the water level will not rise again to a certain extent. Now, if you hang water in buckets, 15 minutes will dry, and if you hang 8 buckets per minute, it will dry in 7 minutes. Now it takes 5 minutes to dry. How many buckets of water should be suspended per minute?

(3) There is a piece of grass that grows at a constant speed every day. Now send 17 people to mow the grass, which will take 30 days to finish. If you send 19 people to cut it, it will take 24 days to finish it. If it takes six days to finish cutting, how many people will be sent to cut it?

(4) There is a barrel of wine. Because there is a crack in the barrel, the same amount of wine is missed every day. Now, if you give this barrel of wine to six people, you can finish it in four days; If four people drink it, they can finish it in five days. How many people can drink this barrel of wine every day?

(5) A certain amount of water is stored in one water, and the river water is uniformly put into storage. Five pumps can continuously drain water for 20 days; Six identical pumps can continuously drain water 15 days. How many identical pumps do you need in 6 days?

Example 3: The escalator runs at a constant speed from bottom to top. Xiaoming and Xiaohong want to go upstairs by escalator. It is known that Xiaoming takes 20 steps per minute and Xiaohong takes 14 steps per minute. As a result, Xiao Ming went upstairs in 4 minutes and Xiao Hong went upstairs in 5 minutes. How many steps are there in the escalator?

Xiaoming walks in 4 minutes, 20×4=80. Xiaohong walks in 5 minutes 14×5=70.

It shows that the elevator runs 80-70 = 10 per minute (level), so the escalator has (20+ 10) × 4 = 120 (level).

Answer: The escalator has120th floor. Comments: Here, the children who go upstairs are equivalent to "cows" and the steps are equivalent to "grass", so this problem is still a Newton problem, and the speed of the escalator is equivalent to "uniform reduction of grass". We regard the "speed" of going upstairs as two parts, one is the speed of children and the other is the speed of escalators. Comparison problem.

Example 4: Two snails walked from the wellhead to the bottom of the well because they couldn't stand the sunshine, and went down during the day. One snail can walk 20 decimetres in the daytime, and the other can only walk 15 decimetres. In the dark, two snails slide at the same speed. As a result, one snail reached the bottom of the well for 5 days and 5 nights, while the other one just spent 6 days and 6 nights. How deep is this well?

The distance difference between two snails in the daytime is 20× 5- 15× 6 = 10 (decimeter).

Because it finally reached the bottom of the well, the snail's descending speed at night was 10 ÷ (6-5) = 10 (decimeter), and the well depth was (20+ 10) × 5 = 150 (decimeter).

Answer: Well depth 150 decimeter.