First, let's look at a typical cage problem: there are chickens and rabbits in a cage. A cage has 10 head and 28 feet. How many chickens and rabbits are there?

From this point of view, a rabbit has 1 head and 4 feet; A chicken has 1 head and 2 feet. Obviously * * * has 10 head, which means * * * has 10 rabbits and chickens. We can solve it with and without equations.

Use the equation: suppose there are x rabbits, then there are 10-x chickens.

Feet: 4x+2( 10-x)=28, x=4.

Get: 4 rabbits and 6 chickens.

There is no equation: if 10 is a rabbit, it should be 4× 10=40 feet.

Every time a rabbit is reduced and a chicken is added, the foot will be reduced by 4-2=2.

Then, the actual number of chickens: (40-28)÷2=6 (only)

The actual number of rabbits should be 10-6=4.

You can also assume that they are all chickens and then calculate.

Looking back at this topic, the question 10 (actually, it is 10, and it can also be directly thought that * * * has ten chickens and rabbits), a correct question (rabbit) gets 8 points (8 feet), and a wrong question (chicken) gets 5 points (-5 points, -5 feet). Here are rabbits and chickens. We will solve it according to the solution.

Use the equation: Answer question X correctly and answer question 10-x incorrectly.

Score: 8x-5( 10-x)=4 1, x=7.

* * * Answer 7 questions correctly and 3 questions incorrectly.

No equation: Assuming all the answers are correct, the score is 8× 10=80 (points).

For every correct answer to 1 question, there is a wrong answer to 1 question, with a difference of 8+5= 13 (points).

Then, * * * Wrong answer: (80-4 1)÷ 13=3 (question)

* * * Correct answer 10-3=7 (question)

Therefore, we can see that there is a paradigm to solve practical problems, especially in primary and secondary schools. As long as we can recognize the paradigm of the problem, we can find an appropriate solution to the problem.