Primary School Mathematics Engineering Problem 1, Engineering Problem of Primary School Mathematics

Because it takes 20 days for A to do alone, the efficacy of A is 1 20, and in the same way, the efficacy of B is 1 30.

1 20 5 = 1 4 A did it alone for 5 days and did 1 4.

1-1 4=3 4 3 4 left undone.

1 20 + 1 30 = 1 12 The cooperation efficiency of A and B is 1 12.

3 4 1 12 = 9 (days) Divide the remaining fraction of the total amount by a fraction of the total amount that A and B can do every day, and get that you have to go long.

Fewer days. A: There are still 9 days to complete the task.

If the total project is 1, then the work efficiency of team A is 1 20, and the work efficiency of team B is 1 30

Team A does it for 5 days first, then 5x1 20=1 4 has been completed, and there is still 1-1 4=3 4 work.

After 5 days, Team B and Team A work together, then the work efficiency becomes 1 20 + 1 30 and finally the workload work efficiency = 3 4 (1 20 + 1 30) = 9 (days).

The division does 1 A does 1 20 works every day

B does 1 30 works a day

A and B cooperate to do the project every day 1 20 + 1 30 = 1 122 A does the first 5 days The remaining project 1 - 5 20 = 3 43 A and B also need (3 4) (1 12) = 9 days.

So it will take another 9 days.

Regard this project as a unit "1", then the efficiency of team A is 1 20, and the efficiency of team B is 1 30, it says "team A will do it alone for 5 days" to calculate what team A did at the beginning: 1 20 5 = 1 4, team A completed 1 4, this project is left with 1-1 4 = 3 4, use 3 4 (1 20 + 1 30) = the answer you want.

Brain-dead, give points.

This work is set to unit 1, team A can do 1 20 work a day, B can do 1 30 work a day, A does 5 days first, that is, it does 5x1 20 = 1 4 workload, there is still 3 4 workload, now A and B cooperate, a total of 1 20 + 1 30 = 1 12 workload can be done every day, and the remaining workload 3 4 divided by the speed of two people 1 12 will be completed in 9 days.

1÷20=1/20………Team A does all the fractions every day.

1÷30=1/30………Team B does all the fractions every day.

1/20×5=1/4………Team A does all the fractions alone.

1-1/4=3/4………Team A and Team B do all the fractions together.

1/20+1/30=1/12……Team A and B do all the fractions every day.

3 4 1 12 = 9 (days).

A: There are still 9 days to complete the task.

9 (day) 1-1 20*5 The amount of work left after Team A has been working alone for five days is obtained.

1 20 + 1 30 is the efficiency of team B and team A working together.

Divide it by the number of days.

1-5 20) (1 20+1 30) = 9 (days).

1-5 20): The amount of work left after Team A has been working alone for 5 days.

1 20 + 1 30): Team B participates and team A participates in the same workload every day.

9 days, A did a quarter in 5 days, A and B did 12 a day together, and a quarter plus nine-twelves equals 1, so they had to do it together for 9 days.

Team A Efficiency: 1 20

Team B Efficiency: 1 30

It will take x days to complete the task.

5/20+x（1/20+1/30）=1

Solution x=9

In daily life, doing a certain thing, manufacturing a certain product, completing a certain task, completing a certain project, etc., all involve the workload, work efficiency, working time of these three quantities, the basic quantitative relationship between them is: workload = work efficiency time.

In primary school mathematics, we call the application problem of the relationship between these three quantities "engineering problem".

The essence of an engineering problem is the relationship between workload, working hours and work efficiency. The solution idea of the engineering problem is similar to that of the travel problem, and it is necessary to find out the relationship between the three fundamental quantities, and find out the solution method through the conversion between the three basic quantities.

To solve the engineering problem, first figure out the relationship between these three quantities in the travel problem:

Effort = Time Efficiency (a=t e).

Time = Effort Efficiency (t=a e).

Efficiency = Effort Time (e=a t).

Wherein, workload: the workload in the engineering problem is the overall quantity of the engineering problem, and in the unknown case, the workload is often assumed to be 1; Time: Time in an engineering problem is a factor quantity of an engineering problem; Efficiency:

Like time, efficiency is a factor quantity of an engineering problem, similar in status and form to time.

Engineering problems are an important type of primary school application problems, is a key point in primary school fraction application problems, and is also a difficult point, the quantitative relationship of this type of application problems is relatively hidden, sometimes the usual method is used to solve more complicated, if you use a special method to analyze and think, it can turn the difficult into easy. The following is a list of several types of exercises that are commonly seen, and the ideas are analyzed and briefly commented, aiming to enable students to master the problem-solving rules and problem-solving skills of "engineering problems".

1. Use the unit "1" to answer.

Example 1] How many days does it take for team A to do a project and team B to do it for 20 days?

Analysis] The total amount of this project is regarded as the unit "1". Team A did 1 12 of the 12 to complete the project in one day; Team B did 1 20 of the day to complete the project; The two teams A and B work together to complete the project in one day (1 12 1 20) = 2 15, and the total work "1" contains how many 2 15 is the number of days for the two teams to complete the project. 1 (1 12 1 20 ) = days).

Comments] This is a basic problem of an engineering problem, the total amount of work is regarded as the unit "1", and the total amount of work is divided by the sum of the work efficiency, and the time taken to complete the project can be found.

2. Answer with the number of copies.

Example 2] A project needs 12 days to complete alone, and it takes 15 days for B to do it alone, and now A does it alone for 3 days, and B joins to do it again, how many days does it take to complete?

Analysis] The total amount of this project is divided into (12 15) parts, from the fact that A and B complete the project separately for two days, it is known that A and B complete the project separately every day, and can do (15 12) parts together every day, A first did it for 3 days, that is, (15 3) parts, and the rest is (12 15 15 3) parts, and the time required for B to join the project after joining: (12 15 15 3) (15 12) 5 (days).

Comment] When answering this kind of application problem, the key is to consider the product of the time it takes for A and B to do it separately as the total number of copies.

3. Use the multiple relationship to answer.

Example 3] Processing a batch of parts, the master does it alone for 14 days, if the master and apprentice work together for 10 days, how many days does it take for the apprentice to do it alone? 【Analysis】The master does 10 days, and the apprentice does 10 days to complete all the work;

The master does 14 days (10 days and 4 days) to complete the whole work; From this, we can see that the master's workload of 4 days = the apprentice's workload of 10 days, that is, the master's work efficiency is twice that of the apprentice, so the apprentice needs 14 days to do it alone).

Comments] When answering this question, use the master's work efficiency to be twice that of the apprentice, so as to simply find out the number of days it takes for the apprentice to do it alone.

In the above cases, due to the use of some special methods to analyze and think, it can turn the difficult into easy and the complex into simple, provide new methods for solving engineering problems, open up students' problem-solving ideas, and cultivate students' creative thinking ability.

Assuming that all work is done by A, and A's work efficiency is 1 18 of the total workload completed per day, then 21 18 of the total workload can be completed in 21 days

21 18-1 more than the total amount of work actually done

If a number of these days are replaced by B, the amount of work that can be reduced for each day is 1 18-1 21 of the total workload

So it is necessary to replace (21 18-1) (1 18-1 21) days, i.e. the actual number of days B has worked.

Because the hypothetical total amount of work done by A -1 is actually the difference in the engineering quantity of A and B, and then divided by the efficiency difference of A and B, the number of days of A is obtained.

Subtract it a little more to find the number of days of B.

That is, the master completes 1 4 per day, and the apprentice completes 1 6 per day.

Then the two cooperate to complete it every day: 1 4 + 1 6 = 5 12, so it takes 12 5 to complete it every day.

Variant 1: (1-3 6) (5 12)=

Variant 2: [1-2*(5 12)] 1 6)=1

Assuming that A is not transferred, then complete 6*(1 10+1 15+1 20)=6*12 60=

A less dry day less 1 10

The carapace has been dry for 6-2 = 4 days.

The efficiency of A is 1 10, the efficiency of B is 1 15, and the efficiency of C is 1 20, then within 6 days, B has done 6 15, and C has done 6 20, and if the total amount of work is 1, then the workload of A is 1-6 15-6 20 = 3 10 The time is equal to the total amount of work divided by the work efficiency, and the time is equal to: 3 10 divided by 1 10 is equal to 3 A has done 3 days.

PS: I'm a high school student, I can't be wrong.

The efficiency of A: 1 10, the efficiency of B: 1 15, the efficiency of C: 1 20

In 6 days, B and C were completed: (1 15 1 20) 6 7 10

A did: (1 7 10) (1 10) 3 days.

Solution: The armor was done for x days.

1/10x+1/15*6+1/20*6=11/10x+2/5+3/10=1

1/10x+7/10=1

1/10x=3/10

x=3A: A has been doing it for 3 days.

Suppose the whole project is 1; Then the efficiency of A is 1 10, B 1 15, C 1 20;

1-(1 15+1 20)*6] (1 10) = 3 (days).

A's work efficiency is 1 10, B's work efficiency is 1 15, C's work efficiency is 1 20, and A makes x days of equations.

1/10+1/15+1/20）*x+（1/15+1/20）*（6-x）=1x=3

Use the equation method.

Shejia dried for x days.

1/10+1/15+1/20）x+(1/15+1/20)(6-x)=1

Solve x=3

The difference between 1/15 plus 20/20 multiplied by 6 equals 7/10Then use the unit"1"Minus 7/10 equals 3/10Finally, dividing 3/10 by 1/10 equals 3