Give a simple example: A work can be completed in 15 days, and B work can be completed in 10 days. How many days can two people cooperate to complete it?

As a whole, an assignment is 1, so the workload can be counted as 1. The so-called work efficiency is the amount of work completed in a unit time. The time unit we use is "day", and 1 day is a unit.

According to the basic quantitative relationship, we get

Workload ÷ work efficiency = working hours

1÷( 1/ 15+ 1/ 10)

=6 (days)

A: It takes six days for two people to cooperate.

This is the most basic problem in engineering problems, and many examples introduced in this lecture are developed from this problem. In order to calculate integers (as far as possible), as in the third example 3 and example 8, the workload is increased. Again, the least common multiple of 10 and 15 is 30. If the total workload is 30 copies, Party A will complete 2 copies every day and Party B will complete 3 copies every day. The number of days required for two people to cooperate is:

30(2+3)= 6 (days)

It is more convenient to calculate with numbers.

3.2. Or "the workload is fixed, and the work efficiency is inversely proportional to the time". The work efficiency ratio of Party A and Party B is 10: 15 = 2: 3.

Engineering problem method summary Edit this paragraph.

1. Basic quantitative relation

1. Work efficiency × working time = total work 2. Work efficiency = total work ÷ working time 3. Working hours = total amount of work ÷ working efficiency

Second, the basic characteristics

Let the total workload be "1" and the work efficiency be = 1/ times.

Three: basic methods

Arithmetic method, proportional method, equation method.

Four: basic ideas

Do it together, do it together.

Five: Types and methods

One: It is divided into two parts: 1. Weather, 2. Hypothetical method, 3. Cleverly grasp the change (proportion), 4. Hypothetical method.

Two: equivalent substitution: solving equations → substitution method, addition and subtraction.

Third, the idea of distribution according to work: the work efficiency per person per day → the workload per person → proportional distribution.

Four: rest and vacation:

Methods: 1. Divide the workload. 2. Hypothetical method: assume no rest.

Five: rest and circulation:

1. The sequence of known conditions: ① work efficiency first, then cycle, ② cycle first, then days.

2. Days: ① approximate days, ② exact days.

3. List the working days.

Six: alternating cycle: estimate the cycle, pay attention to the order!

VII: Water injection and cycle: 1. Sequence, 2. Is there any water in the pool? To fill or overflow.

Eight: changes in work efficiency.

Nine: ratio: 1. Segmentation ratio and connectivity ratio, 2. The idea of standardization, 3. The application of positive and negative proportions, 4. The idea of a false idea (period).

Ten: the problem of cattle eating grass: 1. The amount of new grass, 2. Grass quantity, 3. Solve the problem.

Travel problem travel problem skills

20 1 1-06-30 10:20: 12

Travel problem is to study the relationship between speed, time and distance, and it is the key content of civil servants' examination questions. Tourism questions are often combined with scores, proportions and other knowledge, and their application forms are comprehensive and changeable. Pay attention to several points when answering. Travel problem is to study the relationship between speed, time and distance, and it is the key content of civil servants' examination questions. Tourism questions are often combined with scores, proportions and other knowledge, and their application forms are comprehensive and changeable. Pay attention to the following points when answering: 1. As far as possible, the method of drawing line segments can correctly reflect the changing relationship between quantity and quantity, which is helpful for analysis and thinking. 2. Travel questions are often combined with fractional application questions. When you answer, you should skillfully assume the unit "L" to simplify the question, and sometimes you can contact integer knowledge to understand how many copies the distance is. 3. Complex travel problems are usually applied to proportional knowledge. The speed is certain, and the time is proportional to the distance; Time is constant, and speed is proportional to distance; The distance is fixed, and the speed and. Time is inversely proportional. 4. When encountering a comprehensive problem, you can first decompose the comprehensive problem into several individual problems and then solve them one by one. Example: 1, two cars A and B leave from Station A and bilibili respectively at the same time. The first time we met was 90 kilometers away from Station A. After the encounter, the two cars continued to drive at the original speed and returned immediately after arriving at the destination. The second meeting was 50 kilometers away from Station A ... Find the distances A, 150 km B, 160 km C, 180 km D and 200 km between Station A and bilibili. Analysis: Two cars A and B started in opposite directions from A and bilibili to meet for the second time, and made three whole journeys. Because two cars walk the whole journey together, a car shop is 90 kilometers. In the three whole journeys where the two cars met twice, car A * * * traveled 90×3=270 (km), which is exactly 50 km away from station A. Adding 50 makes the two whole journeys 270+50=320 (km). So the distance between Station A and bilibili is 320÷2= 160 (km). The answer is b practice 1. Two cars leave from the East and West Stations at the same time. After the first meeting 45 kilometers away from the West Railway Station, the two cars continued to drive at the original speed. They all returned immediately after arriving at the station and met again at 15km east of the midpoint. How many kilometers are the two stops apart? A, 80km b,100km c,120km d,140km. Two cars, A and B, leave from A and B respectively at the same time. A travels 42 kilometers per hour and B travels 54 kilometers per hour. After the first encounter, A and B continued to move at the original speed, and immediately returned to the original road after arriving at each other's departure place. It takes five hours for two cars to meet for the second time. How many kilometers is it between A and B? A, 150km B, 160km C, 180km D, 200km analysis: the total distance of two cars driving for 5 hours at the same time is (42+54)×5=480 (km). According to the meaning of the question, the two cars meet for three times from the start to the second time, and the whole journey is 480÷3= 160 (km). The answer is B. Exercise 2. The distance between a and b is 60 kilometers. At 9 o'clock in the morning, the express train and the local train leave from A and B, respectively, in opposite directions. The express train returned immediately after arriving at B, and the local train returned immediately after arriving at A. Their second meeting was at noon 12. At this time, the express train traveled 36 kilometers more than the local train. How many kilometers has the local train traveled? A, 72km b, 68km c, 66km d, 62km-. The main way to solve the travel problem is to set up an equation, which can be set as S = V× T, for example, the fuel carried by an airplane can last up to 6 hours. When leaving with the wind, fly1500km per hour, and when returning with the wind, fly1200km per hour. How many hours does the plane need to fly back at most? A, 8/3 B, 1 1/3 C, 3 D, 5/3 According to the condition that the flight distance is equal to the flight distance, we can list the equations. It is also mentioned in the title that * * * flew for 6 hours, so the equation is listed through these two conditions: if it takes t hours to fly back, the equation is listed as 1500t= 1200(6-t) and the equation is t=8/3 hours. In the travel problem, besides the problem of single object movement, there are also problems of multiple objects movement. The motion of multiple objects involves relative motion. Relative speed is the key in relative motion, and different relative speeds will form different forms of relative motion. There are three main forms of relative movement: meeting, leaving and pursuing. Among them, meeting and leaving can exist as a form of motion, and their characteristics are that the motion directions of two moving objects are opposite, so their relative motion speed is the sum of the speeds of two moving objects, that is to say, the distance of meeting (leaving) = speed × time of meeting (leaving); The problem of catching up is that two moving objects move in the same direction, so their relative speed is the difference between the physical speeds of the two movements, that is to say, the distance difference of catching up = speed difference × catching up time. They are often mixed up when doing practical problems. Example 2: Xiao Ming was sitting on the bus and saw his sister walking in the opposite direction. 1 minute later, Xiao Ming got off and chased her. If his speed is 1 times faster than his sister's and his speed is five times that of Xiao Ming's walking, how many minutes will it take Xiao Ming to catch up with her sister? () A, 5.5 B, 10 C, 1 1 D, 20 should be clear at the beginning. The whole movement process is divided into two parts. The first part is that my sister and the car (Xiao Ming is in the car) made a deviant action, and the second part is that Xiao Ming got off the bus and chased her sister (it is a problem of chasing). In this problem, the speed of my sister, Xiao Ming and the car is uncertain, but they are proportional, so we can set the speed of the three as a special value, which is convenient for us to calculate (the special value method is very key and is often used in our math exam). Let the elder sister's speed be 1, Xiao Ming's speed be 2, and the car's speed be 10, then the sum of the distance = speed and the departure time of the first stage, that is, (10+1) ×1=1. The second action is the pursuit action, and the pursuit time = distance difference ÷ speed difference, that is, t =11÷ (2-1) =11,so C. Example 3: A and B are here. Party A's speed is 3M/S, and Party B's speed is 7m/s. They run in the same direction at the same point, and meet for the first time after 100S. If they run in opposite directions, how many seconds will it take them to meet for the first time () A, 30 B, 40 C, 50 D, 70. The first encounter of running in the same direction means that B catches up with A, the length of the runway is the length of the runway, and the total distance of running in the second encounter is also the length of the runway. Make it clear, then this question is simple. You can try it, and the result is 40 seconds. To do the problem of relative motion, we must grasp the speed of relative motion and determine the relative speed, and the problem of relative motion will be solved. The travel question is a very difficult question. I hope you can do more exercises and be familiar with various types and solutions.

type

1, the problem of running water

2. The problem of multiple encounters on the circular road.

3. Elevator problem

4. Departure problem

5. Pick-up and drop-off problem

Step 6 track down the problem

7. Meeting issues

8 Bridge crossing problem

The main uses of the linear equation can usually be used to solve practical problems, such as engineering problems, travel problems, distribution problems, profit and loss problem, ball game score table problems, telephone (water meter, electricity meter) billing problems, digital problems and so on. The basis for merging similar items (1) is as follows: multiplication and division (2) merging items with the same number of unknowns into one item; Constants are combined into one item after calculation. (3) The number of times is unchanged when merging, but the coefficient is increased or decreased. 6. 1 shift term (1) basis: all the terms with unknown properties in equation (2) are moved to the left of the equation after the sign change, and the terms without unknown numbers are moved to the right. (3) When you move a term from one side of the equation to the other, you must change the sign (for example, change+to-and × to ÷). 6.2 Properties of the equation 1: When a number is added to both sides of the equation or the same number or the same algebraic expression is subtracted, the equation is still valid. Property 2 of the equation: both sides of the equation expand or contract at the same time by the same multiple (except 0), and the equation still holds. Property 3 of the equation: When both sides of the equation are multiplied (or squared) at the same time, the equation still holds. Solving the equation is based on these three properties of the equation. Definition of solution: the value of the unknown quantity that makes the left and right sides of the equation equal is called the solution of the equation. Examples of equation 2a=4a-6 3b=- 1 x= 1 are all one-dimensional linear equations. Deformation formula ax=b(a, b is constant and x is unknown, a≠0) The common solution to the root formula is to remove the denominator → remove brackets → move terms → merge similar terms → the coefficient becomes 1 type. Edit this paragraph 1, running water problem 2, multiple encounters on the roundabout 3, elevator problem 4, departure problem 5, pick-up problem 6, catch-up problem 7, encounter problem 8, bridge crossing problem.