Chicken and rabbit in the same cage

Reprint , Make a note of ：
Why is this question so classic , It was included in 《 Sun Tzu's Sutra 》 Inside , Because there are two kinds of animals involved , The two animals have different numbers of feet , You need to make assumptions 、 Use imagination to solve , It is very helpful to train students' thinking ability .

actually , There are many ways to solve this problem , I myself have summed up more than a dozen , Of course , You don't have to remember every practice , in fact , Some practices are not very suitable , It is only used to expand our thinking , Provide alternative solutions , For your reference .

I will list these solutions one by one , Go and find the right solution for you ！

Classical solution I ： First easy, then difficult

First, I will introduce the most basic method , This approach seems to be “ stupid ”, But it has its advantages , The concrete solution is to count one by one , Try it one by one .

First make a table , You can see in the table , Chicken and rabbit have different feet , therefore , For different numbers of chickens and rabbits , Although their total number is equal （ All equal to 35 only ）, But the number of legs varies , We will list the number of feet of chicken and rabbit under different combinations , It is shown in the form of a table as follows ：

When listing , Let's assume that the chicken has 35 only , Then the rabbit can only be 0 only , In this way, the total number of feet is calculated , Then decrease the number of chickens , The number of rabbits is increasing , Calculate the total number of legs in turn , Finally, it can be calculated that when a chicken has 23 only , The rabbit has 12 Only when , The total number of legs equals 94 only , Meet the conditions in the title , And then get the final result .

actually , When filling out the form , It is not necessary to calculate all the results completely , Just fill in a few blanks , Careful students observe the changing rules of numbers , You can easily determine the number of chickens and rabbits .

Many students and parents are dismissive of this method , I think this method is clumsy and troublesome , I don't think so , Actually , For students in the lower grades of primary school , I think this method is the most recommended method , Because in the process of making tables , Students need to explore chicken on their own 、 When the number of rabbits changes , The change characteristics of the total number of feet , Draw by hand and observe with eyes , Through analysis and comparison, it is concluded that , Because the rabbit has two more feet than the chicken , So when the number of chickens is small 1 only , There are many rabbits 1 Only when , The total number of legs will increase 2 Only the regular knowledge of . This is to cultivate students' spirit of exploration , An important way to improve students' mathematical thinking .

Classical solution II ： The idea Hypothesis

In the solution of chicken and rabbit in the same cage , My personal favorite and most recommended method for students is the hypothetical method , Because the hypothetical method can solve almost all types of chicken and rabbit cage problems , Even if the title has been greatly adapted and changed , Hypothesis method can effectively solve the problem , Of course , For some deformed chicken and rabbit in the same cage , Using the hypothetical method will burn your brain . in application , The hypothetical method is almost the most classical , The most efficient way , Students use assumptions , Different situations （ Chicken and rabbit have different feet ） Into the same situation , It helps to simplify the problem , ideas .

The difficulty of chicken and rabbit in the same cage is that the number of feet of each chicken and each rabbit is different , This is the difficulty of the problem , But it is also the key point or breakthrough to solve the problem , Suppose the chicken and the rabbit have the same number of feet , that , The topic will be greatly simplified , Simplify complex problems , It is a common way to solve mathematical problems .

Suppose one ： All the rabbits stood up , Hiding 2 One foot . In this case , Each chicken and each rabbit have the same number of feet , All are 2 only , under these circumstances , Altogether 35 Head , In other words, there are 35 Animals , Every animal has 2 One foot , So the total number of legs =35×2=70 only , This is more than the title gives 94 One foot is missing 24 only , Think about why there is less ？ Because every rabbit stood up , Put it away 2 One foot , One rabbit is few 2 One foot , It's less than 24 One foot , So there is 24÷2=12 A rabbit , Reuse 35-12=23 Is the number of chickens .

Hypothesis two ： We can also assume that a chicken is a rabbit , here , All chickens increase 2 One foot . In this case , Each chicken and each rabbit have the same number of feet ,（ All are 4 only ）, under these circumstances , Altogether 35 Head , In other words, there are 35 Animals , Every animal has 4 One foot , So the total number of legs =35×4=140 only , This is more than the title gives 94 There are too many feet 46 only , Think about why there are so many feet this time ？ Because there are more chickens 2 One foot , More than one chicken 2 One foot , That's more 46 One foot , So there is 46÷2=23 chicken , Reuse 35-23=12 It's the number of rabbits .

actually , The hypothetical method can not only solve the classical chicken and rabbit cage problem , For some chicken and rabbit in the same cage , Including the chicken and rabbit cage problem solved by the grouping method , Can be solved very well , We should constantly try to solve such problems with the method of hypothesis .

Classical solution three ： Fair exchange substitution

actually , We can also use the method of variable substitution learned in grade one and two to solve the chicken rabbit cage problem . Use the red circle to represent the chicken , Use the blue circle to represent the rabbit . According to the meaning , We can list the following expressions ：

This method is also highly recommended by me , Because this method only uses the knowledge of grade one and grade two , But in essence, it is the preliminary application of equation thought , It is the rudiment of the problem of setting unknown numbers , In this question , We use red circles and blue circles to represent chicken and rabbit respectively , It is essentially a mathematical abstraction , It has a very good effect on improving students' ability to analyze and summarize problems .

Of course , For students under the fourth grade , Listening is understandable , But it may be difficult for them to do it again , Because the essence of this method is equation solution , It's just a symbol instead of x、y, It is difficult for the junior students .

I suggest that parents use this method to try to tell their children , Look at the child's reaction , Let's also take a look at the children's problems in the future , Which method will be used , Based on this, we can make a preliminary judgment on the child's understanding , If he is still willing and able to use this method to solve the problem of chicken and rabbit in the same cage , I think we can tell him about the equation in advance .

Classical solution IV ： Graphic method at a glance

Of course, the problem of chicken and rabbit in the same cage can also be solved by graphic methods , such as , Now I will use the line segment to represent the chicken and the rabbit , The blue line represents the number of chickens , The red line represents the number of rabbits .

We know , A chicken 2 One foot , A rabbit 4 One foot , We are based on the above figure , Expand out , Form the following figure .

so , The area of the blue area is equal to that of the chicken ×2, The number of feet of a chicken , The area of the red area is equal to that of the rabbit ×4, The number of feet of a rabbit .

There is a condition here that we should not ignore , That's the number of chickens and rabbits 35 only . So we can build the following graph .

From the above figure, we can calculate 35×4=140, Is the area of the whole figure , From the above analysis, we can know , The additional shadow area should be equal to 140-94=46, And the width of this rectangle =4-2=2, That length should be equal to 46÷2=23, That is, the number of chickens , Then we can conclude that the number of rabbits is 35-23=12 only .

I also recommend this method , It's not how fast it calculates , How convenient , It is said that this method broadens our vision of solving the chicken rabbit cage problem , Let us from the rigid single number operation , Happily transition to the world of graphics , For inspiring students' thought of the combination of number and form , It is very helpful to stimulate students' creativity .

At this point, I would like to remind parents , I use the area method , Parents should be able to understand , But actually , Third graders , According to the progress of the knowledge points in the textbook , This method may be beyond his comprehension , Of course , Some students can understand .

The performance of this method is the combination of numbers and shapes , The essence is this in the concept of area “ product ” Application , The so-called product is the multiplication of two numbers , Judging from the formula , Namely 35×4, From the graph , It's long 35, wide 4 The graphic area of a rectangle , obviously , We can use the area of a graph to represent the product of two numbers , Will this kind of thinking inspire students to solve the travel problem , Concentration problem , Engineering problems ？

Classical solution five ： Simple and crude x

Set chicken x only , Because the chicken and the rabbit have 35 only , Then the rabbit has 35-x only , According to the meaning , A chicken 2 One foot , So the number of feet of a chicken is 2x only , A rabbit 4 One foot , So the number of feet of a rabbit is 4×(35-x), We know that the total number of legs is 94 only , So you can list the following formula ：

2x+(35-x)×4=94, figure out x=23, Namely chicken has 23 only , So the rabbit is 12 only .

For seniors , I highly recommend using this method to solve problems , You can say that , For seniors , The method of setting unknowns is the preferred method . Because the method of setting the equation of unknown sequence is the fastest and the most concise , Long term use of equation thinking to solve practical problems , It is of great help to improve students' problem abstraction ability . But for junior students , I still think we should carefully tell them about this method , Because of the premature learning of equation solution , On the one hand, for lower grade students , Their cognitive level is limited , Will cause their cognitive problems , Just like the equivalent substitution method introduced above , Children can understand , But you can't do it yourself . On the other hand , If they can understand and master this method , Will give up other methods , This is very important for developing their exploration ability 、 The ability to combine numbers with shapes 、 For the ability of analysis and induction , It was a disaster ！

Classic connection 6 ： A system of bivariate first-order equations

ok , I admit that it's a bit of a fuss to solve the chicken rabbit cage problem with a system of binary linear equations , But it is also a good method .

Set chicken x only , The rabbit has y only , So according to the question , We can list the following equations ：

x+y=35 ①

2x+4y=94 ②

Multiply the first equation by 2, obtain 2x+2y=70 ③

Reuse ②-③, obtain 2y=24, And then get y=12, That is, rabbits have 12 only , Chicken has 35-12=23 only .

This method is really simple and crude , But the problem is that many students can't master , It does have its limitations , Students who can spare no effort in learning , Students with strong understanding ability , Try this method .