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Various solutions to the problem of chickens and rabbits in the same cage at the beginning of Xiaoshengchu Olympic Games
20 17 Various solutions to the problem of cooping chickens and rabbits in the early Olympic Games.

Chicken and rabbit in the same cage is one of the famous mathematical problems in ancient China. About 1500 years ago, this interesting question was recorded in Sun Tzu's calculation. It is described in the book that there are pheasant rabbits in the same cage, with fourteen heads on the top and thirty-eight feet on the bottom. What are the geometric shapes of pheasant rabbits? It should be said that it is easy to solve this problem, so how can it be fast and efficient? Can you clarify the relationship and explain the reasons? Let's take a look at the solution to the problem of chickens and rabbits in the same cage!

This means that this is a classic arithmetic problem. It is known that there are several chickens and rabbits in cages, and there are several feet. The problem of how many chickens and rabbits each have is called the first chicken and rabbit cage problem. Given the difference between the total number of chickens and rabbits and the claws of chickens and rabbits, the problem of finding the number of chickens and rabbits is called the second chicken and rabbit cage problem.

Quantitative relationship first: the problem of chickens and rabbits in the same cage;

Suppose all chickens and rabbits = (actual number of feet -2? Total number of chickens and rabbits) (4-2)

Suppose all rabbits and chickens =(4? Total number of chickens and rabbits-actual number of feet)? (4-2)

The second problem of chickens and rabbits in the same cage:

Suppose all chickens and rabbits =(2? Total number of chickens and rabbits-the difference between chicken and rabbit feet)? (4+2)

Suppose all rabbits and chickens =(4? The total number of chickens and rabbits+the difference between chickens and rabbits)? (4+2)

The ideas and methods of solving problems generally adopt the hypothesis method to solve such problems. It can be assumed that they are all chickens or rabbits. If we assume that they are all chickens, then change the rabbits into chickens; If we first assume that they are all rabbits, and then trade chickens for rabbits. This kind of problem is also called substitution problem. Suppose first and then replace, and the problem will be solved.

Example 1 angora rabbit and Lu Hua chicken, put the chicken and rabbit in a cage. There are thirty-five heads to count and ninety-four feet to count. Please count carefully how many rabbits and chickens there are.

Assuming all 35 rabbits, the number of chickens =(4? 35-94)? (4-2)=23 (only)

Number of rabbits =35-23= 12 (only)

It can also be assumed that all 35 chickens are chickens, so the number of rabbits =(94-2? 35)? (4-2)= 12 (only)

Number of chickens =35- 12=23 (only)

A: There are 23 chickens and 0/2 rabbits.

Example 2: 2 mu of spinach should be applied 1 kg, 5 mu of cabbage should be applied with 3 kg, and two kinds of vegetables *** 16 mu should be applied with 9 kg. How many acres of cabbage are there?

Solving this problem is actually a makeover? Chicken and rabbit in the same cage? Question? Fertilize spinach per mu (1? 2) kg? With what? How many feet does a chicken have? Corresponding? Fertilization rate per mu of Chinese cabbage (3? 5) kg? With what? Each rabbit has four feet? Corresponding? 16 mu? With what? The total number of chickens and rabbits? Corresponding? Nine kilograms? With what? What is the total number of feet of chickens and rabbits? Correspondence. Suppose 16 mu is all spinach, with

Chinese cabbage per mu =(9- 1? 2? 16)? (3? 5- 1? 2)= 10 (management unit)

Answer: Chinese cabbage field 10 mu.

Example 3 Teacher Li bought 45 exercise books and diaries for the school with 69 yuan. 3.20 yuan for each exercise book and 0.70 yuan for each diary. How many exercise books and diaries did you buy?

The solution to this problem can be changed to? Chicken and rabbit in the same cage? The question assumes that 45 books are diaries, and there are

Number of exercise books =(69-0.70? 45)? (3.20-0.70)= 15 (Ben)

Number of diaries =45- 15=30 (copies)

Answer: 15 exercise book, 30 diaries.

Example 4 (the second question of chickens and rabbits in the same cage) There are 100 chickens and rabbits, and chickens have 80 more feet than rabbits. How many chickens and rabbits are there?

Suppose 100 chickens are all chickens, and there are

Number of rabbits =(2? 100-80)? (4+2)=20 (only)

Number of chickens = 100-20=80 (only)

There are 80 chickens and 20 rabbits.

Example 5: There are 100 steamed buns for 100 monks, one big monk eats three steamed buns, and three little monks eat 1 steamed buns. How many monks are there?

Assuming that all monks are big monks, then * * * eats steamed stuffed buns (3? 100), eating more than the actual (3? 100- 100), this is because the little monk is also a big monk, so we take it? Small? Change? Big? Replacing a young monk with a big monk can reduce steamed buns (3- 1/3). Therefore, * * * has a little monk (3? 100- 100)? (3- 1/3)=75 (person)

* * * There is a big monk 100-75=25 (person)

A: * * * There are 25 big monks and 75 young monks.

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