Solving practical problems with equations is one of the important contents of junior high school mathematics. Many practical problems boil down to solving an equation or equations, so listing equations or equations to solve practical problems is an important aspect of integrating mathematics with practice and solving practical problems; The following teachers explain common math problems from the following aspects, hoping to help students.
1. sum, difference, multiplication and division:
(1) multiplicity relation: it is reflected by the key words "how many times, how many times, how many times, what percentage, growth rate …".
(2) How much relationship: it is reflected by the key words "more, less, harmony, difference, lack, surplus ……".
Example 1. According to the statistics of the fifth census released by Xinhua News Agency on March 28th, 2000 1, as of 0: 00 on October 28th, 20001,the population with primary school education per 10,000 population in China was 3,5701,while 6,5701.
Analysis: The equivalence relation is:
Solution: Suppose that at the end of June, 1990, about X out of every 654.38+million people have primary school education.
Answer: Omit.
2. Equal product deformation problem:
"Equal area deformation" is based on the premise that the shape changes but the volume remains unchanged. The commonly used equivalence relation is:
(1) The shape area has changed, but the perimeter has not changed;
② Raw material volume = finished product volume.
Example 2. A cylindrical glass (filled with water) with a diameter of 90mm was poured into a cuboid iron box with a bottom area of 81mm. How much mm did the water in the glass drop? (Results are rounded)
Analysis: The equivalent relationship is: the volume of cylindrical glass = the volume of cuboid iron box.
The falling height is the height of the poured water.
Solution: let the water drop in the glass xmm.
Answer: Omit.
3. Labor deployment:
This kind of problem to find out the number of changes, common problems are:
(1) can be transferred in and out;
(2) Only the transfer-in did not turn out, the transfer-in part changed, and the rest remained unchanged;
(3) Only the transfer-out has not been transferred in, some of the transfers have changed, and the rest remain unchanged.
Example 3. There are 85 workers in the processing workshop of the machinery factory, each of whom processes 16 large gears or 10 small gears on average every day. It is known that two large gears and three small gears form a group. How many workers should be arranged to process the big and small gears separately to make the big and small gears processed every day just match?
Analysis: List method.
Number of people per person per day
Gear 16 x person 16x
Pinion 10 person
Equivalence relation: 2 times the number of pinion gears = 3 times the number of large gears.
Solution: We arrange X workers and X workers to process large and small gears respectively.
Answer: Omit.
4. Proportional distribution:
The general idea of this kind of problem is: let one of them be x and write the corresponding algebraic expression by using the known ratio.
Common equivalence relation: sum of parts = total.
Example 4. The ratio of three positive integers is 1: 2: 4, and their sum is 84, so what is the largest of these three numbers?
Solution: Let one part be X, then the three numbers are X, 2x and 4x respectively.
Analysis: equivalence relation: the sum of three numbers is 84.
Answer: Omit.
5. Numbers.
(1) Need to know the representation method of numbers: the hundredth digit of a three-digit number is A, the tenth digit is B, and the first digit is C (where A, B and C are integers, 1≤a≤9, 0≤b≤9, and 0≤c≤9), then this three-digit number is represented as: 65438.
(2) Some representations in the number problem: the relationship between two consecutive integers, the larger one is larger than the smaller one1; Even numbers are represented by 2N, and continuous even numbers are represented by 2n+2 or 2n-2; Odd numbers are represented by 2n+ 1 or 2n- 1.
Example 5. For two-digit numbers, one digit is twice as much as ten digits. If the tenth digit is reversed, the two digits obtained are 36 larger than the original two digits. Find the original two digits.
Equivalence relation: original two digits +36= new two digits after switching.
Solution: Let the number x on the ten digits, then the number on the one digit is 2x.
The solution of10× 2x+x = (10x+2x)+36 is x=4, 2x=8.
Answer: Omit.
6. Engineering problems:
Three quantities in engineering problems and their relationship are: total work = working efficiency × working time.
Often when the total workload is not given in the title, the total workload is set to 1.
Example 6. For a project, it takes 15 days for Party A to do it alone and 12 days for Party B to do it alone. Now that Party A and Party B have cooperated for three days, Party A has other tasks, and the remaining projects will be completed by Party B alone. How many days does it take Party B to complete all the projects?
The analysis assumes that the total project amount is 1, and the equivalent relationship is: a completed workload +B completed workload = total workload.
Solution: Assume that it takes X days for B to complete all projects, and the total workload is 1. From the meaning of the question, (115+12) × 3+x12 =1,solve this equation, 65438.
12+ 15+5x = 60 5x = 33∴x = 335 = 635
Answer: Omit.
7. Travel problems:
(1) Three basic quantities in the travel problem and their relationships: distance = speed × time.
(2) The basic types are
(1) meeting problems; (2) follow up the problem; Common ones are: running for opponents; Navigation problems; Circular runway problem.
(3) The key to solve this kind of problem is to grasp the time relationship or distance relationship between two objects, so that the problem can be solved as a whole. And often sketch to analyze and understand the trip problem.
Example 7. The distance between Station A and bilibili is 480 kilometers. The local train departs from Station A at a speed of 90km per hour, and the express train departs from bilibili at a speed of140km per hour.
(1) The local train starts first 1 hour, and the express train starts again. The two cars are driving in opposite directions. How many hours after the express train leaves, will the two cars meet?
(2) After two cars started at the same time and walked in opposite directions for several hours, the two cars were 600 kilometers apart?
(3) Two cars start at the same time, and the local train runs in the same direction behind the express train. How many hours later, the distance between the express train and the local train is 600 kilometers?
(4) Two cars leave in the same direction at the same time, and the express train is behind the local train. How many hours will the express catch up with the local train?
(5) After the local train 1 hour, the two cars are driving in the same direction, and the express train is behind the local train. How many hours after the express train leaves, will it catch up with the local train?
The key to this problem is to understand the meaning of opposite direction, opposite direction and same direction, and to understand the driving process. So it can be combined with graphic analysis.
(1) parsing: When encountering a problem, draw it as:
The equivalent relationship is: distance traveled by local train+distance traveled by express train = 480km.
Solution: Suppose that two cars meet after the express train leaves X hours. From the meaning of the title, 140x+90(x+ 1)=480.
To solve this equation, 230x=390.
∴ x= 1 1623
Answer: Omit.
Analysis: run in reverse, and the drawing is expressed as:
The equivalent relationship is: distance traveled by two cars +480 km =600 km.
Solution: suppose that after x hours, the two cars are 600 kilometers apart.
Judging from the meaning of the question, (140+90)x+480=600 solves this equation, and 230x= 120.
∴ x= 1223
Answer: Omit.
(3) Analysis: The equivalent relationship is: the distance traveled by the express train-the distance traveled by the local train +480km = 600km.
Solution: Let the distance between two cars be 600 kilometers after X hours. From the meaning of the question, (140-90) x+480 = 600 50x =120.
∴ x=2.4
Answer: Omit.
Analysis: Trace back the problem and draw it as follows:
The equivalent relationship is: express distance = local distance+480km.
Solution: Set the express train to catch up with the local train after X hours.
From the meaning of the title, 140x=90x+480.
To solve this equation, 50x=480 ∴ x=9.6.
Answer: Omit.
Analysis: Tracing back to the source, the equivalent relationship is: express distance = local distance +480 km.
Solution: Set the express train to catch up with the local train after X hours. From the meaning of the title, 140x=90(x+ 1)+480.
50x=570，x= 1 1.4。
Answer: Omit.
8. profit and loss problem
(1) The quantities that often appear in sales problems are: purchase price, sale price, bid price, profit, etc.
(2) Relationship:
Commodity profit = commodity price-commodity purchase price = commodity price × discount rate-commodity purchase price
Commodity profit rate = commodity profit/commodity purchase price
Commodity price = commodity price × discount rate
Example 8. A store will increase the purchase price of a certain clothing by 40%, then price it and sell it at a 20% discount. As a result, each piece of clothing still earned 15 yuan. What is the purchase price of each piece of clothing?
Analysis: It is the key to explore the implied conditions in the topic, and the cost can be directly set to X yuan.
Purchase discount rate, bid price, preferential price and profit
Twenty percent off (1+40%)x twenty percent off (1+40%)x 15.
Equivalence relationship: (profit = discount price-purchase price) discount price-purchase price = 15.
Solution: Let the purchase price be X yuan, 80% x (1+40%)-x = 15, X= 125.
Answer: Omit.
9. Savings problem
(1) The money deposited by the customer in the bank is called the principal, and the reward paid by the bank to the customer is called interest. Principal and interest are collectively referred to as the sum of principal and interest, the time of deposit in the bank is called the number of periods, and the ratio of interest to principal is called the interest rate. Interest tax is paid at 20% of interest.
(2) Interest = principal × interest rate × number of periods
Sum of principal and interest = principal+interest
Interest tax = interest × tax rate (20%)
Example 9. A classmate deposited 250 yuan in the bank for half a year. After half a year, * * * got the principal and interest and 252.7 yuan. What is the annual interest rate of the bank for half a year? (excluding interest tax)
Analysis: equivalence relation: sum of principal and interest = principal ×( 1+ interest rate)
Solution: let the real interest rate for half a year be x,
250( 1+x)=252.7，
x=0.0 108
So the annual interest rate is 0.0 108×2=0.02 16.
Answer: Omit.