Excuse me, what are the types of math application problems in grades one to six of primary school?

Reduction problem: the problem of knowing the result of a number after some changes and asking for the original unknown number is generally called reduction problem.

The reduction problem is the inverse solution of the application problem. Generally speaking, according to the relationship between reciprocal operations of addition, subtraction, multiplication and division. Start with the order of topic description, think in reverse order, start with the last known condition, and get the result in reverse.

Substitution problem: there are two unknowns in the problem, and one of them is often regarded as the other for the time being, and then the hypothetical operation is carried out according to the known conditions. The results are often inconsistent with the conditions, and then appropriate adjustments are made to get the results.

Profit and loss problem (the problem of insufficient income): There are often two distribution schemes in the topic, and the result of each distribution scheme will be more or less (surplus). This kind of problem is usually called profit and loss problem (also known as the problem of insufficient income). To solve this kind of problem, we should first compare the two distribution schemes, find out the change of remainder caused by the change of each share, find out the total number of shares participating in the distribution, and then find out the number of items to be distributed according to the meaning of the question. Its calculation method is:

When there is a surplus at one time and the other is insufficient: each copy = (surplus+shortage) ÷ the difference between the two copies.

When both times have a remainder, the total number of copies = (larger remainder-smaller number) ÷ the difference between each number of copies of the two times.

When two times are insufficient, the total number of copies = (larger shortage-smaller shortage) ÷ the difference between two times per copy.

Age problem: The main feature of the age problem is that the age difference between two people remains unchanged, but the multiple difference changes.

The commonly used calculation formula is:

Age times = age difference ÷ (multiple-1)

Age a few years ago = small gift-multiplied by small age

Age in a few years = times its age-its age when it was very young now.

Chicken and rabbit problem: a kind of application problem of how many chickens and rabbits are known, also called "turtle and crane problem" and "replacement problem"

It is generally assumed that they are all chickens (or rabbits), and then rabbits (or chickens) are used instead of chickens (or rabbits). Commonly used basic formulas are:

(Total number of feet-number of chicken feet × total number of chickens) The difference between the number of feet of each chicken and rabbit = number of rabbits.

(The number of rabbits × the total number of rabbits-the total number of rabbits) ÷ The difference between the number of feet of each chicken and rabbit = the number of chickens.

Common divisor and common multiple problem: solving application problems with the greatest common divisor or the smallest common multiple is called common divisor and common multiple problem.

Score application problem: refers to the application problem solved by score calculation, which is called score application problem, also called score problem.

Fraction application problems are generally divided into three categories:

1. Find the fraction of one number to another.

2. Find the fraction of a number.

3. Know what the score of a number is, and find out this number.

Engineering problem: it is a special case of fractional application problem. It is a problem to find the third quantity from two of the three quantities when the workload, working time and working efficiency are known.

When solving engineering problems, we should generally regard all projects as "1", and then answer them according to the following quantitative relationship:

Work efficiency × working hours = workload

Workload ÷ working time = working efficiency

Workload ÷ work efficiency = working hours?

For the sake of writing so much, adopt me. ..