List of simple equations for solving practical problems
10x+ 1, so there is
3( 105+x)= 10x+ 1，
7x=299999，
x=42857 .
A: This six-digit number is 142857.
Note: This solution has two key points:
The research shows that according to the characteristics of the topic, the method of setting elements "as a whole" is very distinctive.
(1) is good at analyzing the quantitative relationship between known numbers and unknowns in problems; (2) It is the transformation of the relationship between general language and mathematical formal language. Therefore, in order to improve the ability to solve application problems, we must work hard in these two aspects.
Example 2 A team travels at the speed of1.4m/s. At the end, a correspondent has something to inform the platoon leader, so he rushes to the platoon leader from the end at the speed of 2.6m/s and immediately returns to the platoon leader. * * * It took 10 minutes and 50 seconds. Q: How long is the queue?
Analysis: This is a question of "catching up with the past and meeting again". The correspondent chases from the end to the head, and the distance between him and the head is the length of the team; It is a meeting problem for a correspondent to return from the head to the tail. The distance he travels with the tail is the length of the team. If it takes X seconds for the communicator to go from the end to the head, it takes (650-x) seconds for the communicator to go back from the head to the tail, so it is not difficult to make an equation.
Solution: Suppose it takes X seconds for a reporter to walk from the end of the line to the front of the line.
2.6x- 1.4x = 2.6(650-x)+ 1.4(650-x).
The solution is x = 500. Infer that the leader of the team is
(2.6- 1.4)×500=600 (m)。
A: The length of the line is 600 meters.
Note: there are two ways to set the unknown: one is to set the direct unknown, what to seek and what to set; The other is to set indirect unknowns. When it is difficult to list equations directly, set indirect unknowns related to demand. For difficult application problems, it is often easier to establish equations by properly selecting unknowns.
On a path parallel to the railway, a group of pedestrians and cyclists are driving south at the same time. The pedestrian speed is 3.6 km/h and the cyclist speed is 10.8km/h when a train comes from behind them. It took the train 22 seconds to overtake the pedestrian and the cyclist 26 seconds. What is the total length of the train?
Analysis: this topic is called catching up. The pedestrian speed is 3.6 km/h = 1 m/s, the cyclist speed is 10.8 km/h =3 m/s, and the length of the train body is equal to the distance difference between the train tail and pedestrians, as well as the distance difference between the train tail and cyclists. If the speed of the train is assumed to be x m/s, the length of the train body can be expressed as (x- 1)×22 or (x-3)×26, so it is not difficult to list the equations.
Solution: Let the speed of this train be x m/s, and make an equation according to the meaning of the question.
(x- 1)×22=(x-3)×26 .
The solution is x= 14. So the body length of the train is
( 14- 1)×22=286 (m)。
A: This train is 286 meters long.
As shown in the figure, along a square with a side length of 90 meters, counterclockwise, A starts from A and walks 65 meters per minute, while B starts from B and walks 72 meters per minute. When B first caught up with A, which side of the square was he on?
Analysis: This is a circular catch-up problem. This kind of problem can be regarded as a "straight-line" chasing problem first, then the time required for B to catch up with A is calculated, and then the "circular chasing problem" is returned. According to the distance traveled by B during this time, we should calculate which side of the square B is on.
Solution: suppose that B left X when catching up with A ... According to the meaning of the question, A is in front of B.
3×90=270 (meters),
So there is
72x=65x+270 .
Because this square is 90 meters long and has four sides, it is made of
It can be inferred that A and B should be on the edge of the square at this time.
A: When B caught up with A for the first time, it was on the side of DA in the square.
Example 5: A ship is sailing between Port A and Port B, sailing downstream from Port A to Port B, and sailing upstream from Port B to Port B. It is known that when the speed of the ship is 8 km/h in still water, the time ratio of retrograde and forward is 2∶ 1. One day it coincided with a rainstorm, and now the speed is twice as fast as before. It took the ship nine hours to go back and forth. Q: How many kilometers is it between Port A and Port B?
Analysis: This is a tripping problem in flowing water:
Downstream velocity = still water velocity+current velocity,
Velocity = still water velocity-water velocity.
The key to solve this problem is to find out the velocity of water flow first.
Solution: Suppose the distance between port A and port B is X kilometers, and the original current speed is one kilometer per hour. According to the meaning of the question, the ratio of upstream speed to downstream speed is 2: 1, that is,
(8-a)∶8+a = 1∶2，
According to the number of rainstorm days, when the water flow speed becomes 2 km/h, there are
The solution is x=20.
A: The distance between Port A and Port B is 20 kilometers.
Example 6 A school organized 150 teachers and students to travel abroad. These people won't leave until 5 o'clock. In order to catch the train, they must arrive at the railway station at 6: 55. They only have a bus that can take 50 people, with a speed of 36 kilometers per hour. The school is 265,438+0 kilometers away from the railway station. Obviously, they all take the bus all the way and have no time, so they can only walk while taking the bus. If you can walk 4 kilometers per hour, how should you arrange it so that everyone can get to the train station on time?
To get to the railway station, everyone's walking time and riding time should be the same. If everyone takes X, can the bus 1 15 minutes walk?
Solution: Divide 150 people into three groups with 50 people in each group. The walking speed is 4 km/h and the car speed is
The solution is x = 1.5 (hours), that is, everyone walks for 90 minutes and rides for 25 minutes. Three groups of people set out at 5 o'clock at the same time. The first group arrived at point A in 25 minutes by car and got off and walked. The bus immediately returned from A, met the second group of people on foot in B, and took the bus for 25 minutes. The second group got off and walked, and the bus immediately returned, met the third group of people on foot in C, and then directly sent to the railway station.
There is no problem in arranging the first and second groups of people to arrive at the railway station on time. Can the third kind of people just take the bus for 25 minutes? Must be calculated.
The second return time is 20 minutes, so it can be calculated that the second return time of the bus should be 20 minutes, so when the bus meets the third group of people, the bus has used 25× 2+20× 2 = 90 (minutes) and 1 15-90=25 (minutes), which is just right.
So it can be arranged according to the above method.
Note: After solving the equation, it can be arranged by walking for 90 minutes and taking a bus for 25 minutes, but the checking calculation cannot be omitted, because it is related to whether the third group can arrive at the station on time. According to the calculation, the third group of people can just take the bus for 25 minutes and arrive on time. However, if the number of people increases or the speed slows down, although the equation can be listed similarly, there is no guarantee that everyone will arrive at the destination on time.
Secondly, the parametric equation is introduced to solve the application problems.
For application problems with complex quantitative relations or few known conditions, in addition to unknowns, it is necessary to add some parameters of "setting but not seeking" to facilitate the translation of quantitative relations described in natural language into algebraic language, communicate quantitative relations and create conditions for equations.
Someone is walking on the expressway. Every four minutes, a bus will meet him head-on, and every six minutes, a bus will overtake him from behind. If people and cars are driving at a constant speed, how often does the bus stop leave?
Analysis: It seems that it is not easy to find an equal relationship in this issue. Walking on the highway, I noticed that someone met an oncoming car, which was a meeting problem. The sum of the four distances between a person and a car is exactly the distance between two buses driving in the same direction. Every six minutes, a car passes behind the man, which is a catch-up problem. The distance between a car and a person at 6 minutes is exactly the distance between two cars. Then the unknown constant velocity is introduced as a parameter, and the problem is solved.
Solution: We set up a bus stop, and allocate buses every X. The speed of people is v 1, and the speed of cars is v2.
From ① ②, get
Substitute ③ into ① to get.
Note: This question introduces two unknown variables, v 1 and v2, which are eliminated during calculation, that is, the answer to this question has nothing to do with the choice of parameters. There are many solutions to this problem. Please refer to lecture 26 of the fifth grade math activity class in this series.
The grass on the whole pasture grows equally dense and fast. It is known that 70 cows eat grass for 24 days, and 30 cows need 60 days. How many cows are there to finish the pasture grass in 96 days?
Analysis: What is the amount of grass in the pasture in this question? How much grass can you grow every day? How much grass do you eat every day in Niu Yi? If these three quantities are represented by parameters A, B and C, and the number of cattle to be asked is X, three equations can be listed. If A, B and C can be excluded, the problem will be solved.
Solution: Let the original amount of grass in the whole pasture be A, the amount of grass that grows every day be B, and the amount of grass that Niu Yi eats every day be C. If X cows can finish the grass in the pasture within 96 days, there will be.
②-①. Obtain
36b= 120C .④
③-②, yes
96xc= 1800c+36b .⑤
Substitute ④ into ⑤ to get.
96xc= 1800c+ 120c .
The solution is x=20.
A: There are 20 cows.
The expressway from A to B has only uphill and downhill, and there is no smooth road. The speed of the car is 20 kilometers per hour when going uphill and 35 kilometers per hour when going downhill. How many kilometers of uphill does the car have to travel from a to b?
Solution: The uphill road from A to B is the downhill road from B to A; The downhill from A to B is the uphill from B to A. Let the uphill from A to B be X kilometers and the downhill is Y kilometers.
①+②，get
Substitute y = 2 10-x into ① to obtain
The solution is x = 140.
Answer: The expressway between A and B is 2 10/0km, and it is necessary to drive an uphill road of140km from A to B. ..
Third, enumerate indefinite equations to solve application problems
Some application problems are solved by algebraic equations, and sometimes there are more unknowns than the listed equations. In this case, the equation is called indefinite equation. At this time, the equation has multiple solutions, that is, the solutions are not unique. However, we should pay attention to the requirements of the topic for the solution. Sometimes, only some or individual solutions are needed.
Example: Class 6 10 (1) held a math exam and adopted a five-level scoring system (5 points were the highest, 4 points were the second, and so on). The average score of boys is 4, that of girls is 3.25, and that of the whole class is 3.6. If there are more than 30 students and less than 50 students in the class, how many boys and girls have passed the exam?
Solution: Suppose there are X boys and Y girls in this class, then there are.
4x+3.25y=3.6(x+y)，
After simplification, 8x=7y is obtained. So there are students in the class.
Of the natural numbers greater than 30 and less than 50, only 45 can be divisible by 15, so
Infer x = 2 1, y=24.
A: The boys in this class are 2 1 and the girls are 24.
Example 1 1 Xiaoming plays the game of rings. The chicken in the trap gets 9 points at a time, the monkey in the trap gets 5 points and the dog in the trap gets 2 points. Xiao Ming * * * set 10 times, and was trapped every time. Every small toy was trapped at least once. Xiao Ming set 10 times * * and got 6 1 minute. Q: How often does Xiao Ming trap chickens?
Solution: If the chicken is trapped x times, the monkey is trapped y times and the puppy is trapped (10-x-y) times. According to the 6 1 fractional countable equation
9x+5y+2( 10-x-y)=6 1，
The simplified value is 7x = 4 1-3y.
Obviously, the smaller y is, the larger x is. Substitute y= 1 to get 7x=38, and there is no integer solution; If y=2 and 7x=35, the solution is x=5.
A: Xiaoming trapped the chicken five times at most.
Example 12 A sewing company has four groups: A, B, C and D. Group A can sew 8 tops or 10 pairs of pants every day. Group B can sew 9 tops or 12 pairs of pants every day; Group C can sew 7 tops or 1 1 pairs of pants every day; Group D can sew 6 tops or 7 pairs of trousers every day. Now the coat and trousers should be sewn together (each set is a coat and trousers). Q: How many sets of clothes can these four groups sew in seven days?
Analysis: We can't arrange production only according to the number of jackets or trousers. We should consider the efficiency of each group in producing jackets and trousers, and arrange the production with collocation.
First of all, it is necessary to arrange for those who are efficient in making coats to make more coats and those who are efficient in making trousers to make more trousers, so as to make the most sets of clothes.
Generally speaking, it is assumed that Group A can sew shirts of a 1 or pants of b 1 every day, and their proportion is that under the condition that Group A makes as many shirts as possible and Group B makes as many pants as possible, the supporting production will be arranged. this
The efficiency is high, so the two groups are arranged to produce one product in these seven days.
If Group A produces X-day jackets and 7-x-day trousers, and Group B produces Y-day jackets and 7-y-day trousers, then the four groups produce 6× 7+8x+9y (pieces) and 1 1× 7+65438 jackets and trousers respectively. According to the meaning of the question, get
42+8x+9y = 77+70- 10x+84- 12y，
Let u = 42+8x+9y, then
Obviously, the larger X is, the larger U is. Therefore, when x=7, the maximum value of U is 125, and the value of Y is 3.
A: Groups A and D were arranged to make jackets in 7 days, while Group C made trousers in 7 days, while Group B made jackets in 3 days and trousers in 4 days, so the number of sets produced was the largest, totaling 125 sets.
Note: this problem is still two unknowns, one equation, and it is impossible to have a definite solution. In this problem, the number of sets is the largest, which is essentially the maximum value of a "unary function" in a certain range. Pay attention to the reason for getting the maximum value.