Travel problem is one of the four major problems (calculation, number theory, geometry and travel) in senior high school entrance examination and primary school. There are a variety of specific questions, forming 10, all of which have their own relatively unique methods to solve problems.
Junior high school mathematics must take the common questions 1 first, and generally encounter chasing questions.
Including one or two people (at the same time, at different times), in the same place, in different places, in the same direction, in the opposite direction, and with a mixture of time and distance. It appears in a large number in cups, accounting for about 80%. It is suggested to skillfully use the standard solution, that is, s=v×t combined with standard line drawing (basic solution). Because only the basic formulas of encounter and pursuit can be used to solve problems, when solving problems, once the situation changes more, I will analyze the situation with my own drawing.
Second, the complex encounter and pursuit of problems.
(1) Many people encounter and pursue problems. There is one more moving object than the general problem of meeting and chasing, that is, what we can usually encounter is the problem of three people meeting and chasing. The idea of solving problems is exactly the same, but it is relatively complicated. The key is whether the ability of standard drawing can clearly show the motion state of the three.
(2) Meet many times to catch up with the problem. That is, two people meet and catch up repeatedly in the same place or different places at the same time in a certain journey, which is commonly known as the "repeated tossing problem". It can be divided into standard problems (such as knowing the distance and speed of two places, finding the distance or number of encounters between N times of meeting or catching up with a specific place or catching up within a specified time) and pure cycle problems (rare, such as knowing the speed of two places, finding the number of encounters, and catching them after a period of time, that is, when both of them return to the initial point).
The standard solution is fixed, and it will be very complicated if we can't start from the distance. It's much easier to find a meeting with the company at the beginning and hurry up, and then find the distance and number of times. If you use a line chart, you can only have a general perceptual understanding, and you can't get a specific answer unless you carefully draw a standard size chart during non-examination time.
The commonly used time formula is (only the cases where Party A and Party B start from both ends at the same time are listed, and few of them start from the same end, so I won't repeat them):
One-way meeting time: t One-way meeting =s/(v A +v B)
One-way tracking time: t One-way tracking =s/(v A -v B)
The time of the nth meeting: tn= t One-way meeting ×(2n- 1)
Time of M-th tracking: tm= t One-way tracking ×(2m- 1)
Encounters in a limited time: n encounters =[ (tn+ t one-way encounters) /2 t one-way encounters]
Time-limited pursuit: m pursuit =[ (tm+ t one-way pursuit) /2 t one-way pursuit]
Note: [] is an integer symbol.
Then choose a or b to study the relationship between distances, which involves the period problem, so be careful not to make a mistake in the direction of movement.
Simple example: two cars, A and B, start from A at the same time and travel between A and B, which are 300 kilometers apart. It is known that the speed of car A is 30 kilometers per hour and that of car B is 20 kilometers per hour.
Q: (1) How long has it been since the second confrontation? (2) How many kilometers away from the midpoint? (3) Within 50 hours, how many times did the two cars meet head-on?
Third, the train problem.
The characteristic is nothing more than involving command, which is relatively easy. The small question is divided into:
1, train crossing bridge (tunnel): one has length and speed, and the other has length and no speed.
Solution: train length+bridge (tunnel) length (total distance) = train speed × transit time;
2. Train+tree (telegraph pole): one has length and speed, and the other has no length and speed.
Solution: train length (total distance) = train speed × transit time;
3. Train+people: one has length and speed, and the other has no length and speed.
(1), train+oncoming people: equivalent to encountering problems,
Solution: train length (total distance) = (train speed+human speed) × head-on missed time;
(2) Train+people walking in the same direction: equivalent to catching up with the problem,
Solution: train length (total distance) = (train speed-human speed) × catch-up time;
(3) Train+people sitting on the train: the encounter and pursuit of problems between trains and people.
Solution: train length (total distance) = (train speed plus human speed) × head-on miss time (catch-up time);
4. Train+Train: One has length and speed, and the other has length and speed.
(1) Wrong train problem: equivalent to encountering a problem,
Solution: the length of express train+the length of local train (total distance) = (the speed of express train+the speed of local train) × the wrong train time;
(2) Overtaking problem: equivalent to rear-end collision problem,
Solution: the length of express train+the length of local train (total distance) = (the speed of express train-the speed of local train) × the wrong train time;
For such problems as trains crossing bridges, trains meeting people, trains chasing people, and trains meeting and chasing each other, we must combine pictures when analyzing the purpose of the problems.
Fourth, the problem of running water.
Knowing the relative speed, the problem of running water is not difficult. Understand and remember the formula 1:
When the ship speed downstream = the ship speed in still water+the speed of current, other formulas can be easily understood and deduced:
Water velocity = still water velocity-water velocity,
Ship speed in still water = (downstream speed+downstream speed) ÷2,
Current speed = (downstream speed-upstream speed) ÷2.
The technical conclusions are as follows:
(1) Meet to catch up. The speed of the current has no effect on the time of meeting and chasing, that is, the speed difference between the two ships, whether in the same direction or in the opposite direction, does not constitute a "threat", so just use it boldly.
2) Falling into the water. Drift speed = current speed, t 1= t2(t 1: time period from falling to finding, t2: time period from finding to picking up) has nothing to do with ship speed, water speed and back and forth movement. The time equation brought by this conclusion is often very easy to solve the problem of falling water and is also very easy to remember.
There are two piers a and b on a river. Pier A is 50 kilometers upstream of Pier B. A passenger ship and a cargo ship set off from Pier A and Pier B at the same time, heading upstream, with the same still water speed. When the passenger ship set out, an article fell into the water from the ship, and it was 5 kilometers away from the passenger ship after 10 minutes. After 20 kilometers, the passenger ship turned around and caught up with this item, which happened to meet the cargo ship when it caught up. Find the velocity of water flow.
Fifth, the problem of interval departure.
Spatial understanding is a bit difficult, and the proof process is not helpful for solving problems quickly. Once you master the three basic formulas, general problems can be solved easily.
(1) is on the bus. Liuka problem. There is no basic formula, and the quick solution is to draw a time-distance graph directly, then draw many intersecting lines, and count the intersections as required.
Example: A and B are two bus stops, and it is uphill from Station A to Mile Mile ... Every morning from 8: 00 to 165438+ 0: 00, a bus leaves from Station A and bilibili at the same time every 30 minutes. It is known that it takes 105 minutes to get from Station A to Miley Miley, and 80 minutes to get from Miley Miley to Station A. Q: How many cars can drivers who leave Station A at 8: 30 and 9: 00 respectively see from Miley Miley Miley?
(2) Outside the bus. It is convenient to combine the three basic formulas.
Distance between vehicles = (vehicle speed+pedestrian speed) × time interval of meeting events
Distance between vehicles = (vehicle speed-pedestrian speed) × time interval of chasing events
Distance between vehicles = vehicle speed × departure time interval
1 2 combination understanding, namely
Distance between vehicles = relative speed × time interval
Divided into two small problems:
1, general interval departure problem. Answer quickly with three formulas;
2. Find the number of buses that you meet and catch up with when you arrive at your destination. The standard method is: drawing-listing three convenient formulas as far as possible-the whole process of combination S = v× t-the number of combined tree planting problems.
Example: Xiaofeng noticed on her way to a party in Bao Xiao by bike that a bus passed Xiaofeng from behind every 9 minutes. Xiaofeng's bike broke down halfway, so he had to take a taxi to Bao Xiao's house. At this time, Xiaofeng found that taxis also passed a bus every 9 minutes. It is understood that the speed of taxis is five times that of Xiaofeng riding bicycles. How many minutes does it take for a bus stop to send a car if these three vehicles keep constant speed during driving?
Six, the average speed problem
A relatively easy question. Remember that big formula: total distance = average speed × total time. It is easier to understand and standardize the corresponding proportion with s=v×t than to write the proportion formula directly, thus forming a unified solution to the travel problem.
Seven, the circular runway problem
It is a challenging and arduous question, which is divided into such minor questions as "the same way", "the different way", "the real encounter" and "can you see it". It involves the period problem, the geometric position problem (it is easy to miss a variety of position possibilities if the questions are not carefully examined), and the inequality problem (for the problem of "can you see", that is, ask whether A can see B at the corner of the line segment).
Eight, the clock problem
It is a concrete extension of the ring problem. Basic relationship: V minute hand = 12v hour hand.
(1) Summing up memory: Clockwise hands go every minute112, 0.5; Minute hand goes every minute 1 grid, 6 degrees. The hour hand and the minute hand overlap 1 1 times in "half" days, forming a straight line * * *1/times, forming a right angle ***22 times (where you need to draw a summary yourself).
(2) Basic problem-solving ideas: the idea of distance difference. that is
Grid or angle (minute hand) = grid or angle (hour hand)+grid or angle (difference)
Grid: x=x/ 12+ (grid behind the first hour hand+grid outside the last hour hand)
Angle: 6x=x/2+ (angle after starting the hour hand+angle after ending the hour hand)
The questions that can solve most clockwise problems are coincidence, right angle, straight line, arbitrary angle, which two squares are in the middle, and how many angles are formed at a moment.
Example: At 9: 23, what is the angle between the hour hand and the minute hand? How many minutes have passed this moment, and the hour hand and the minute hand are perpendicular for the first time?
(3) the problem that the clock is broken. The solution used is not a trip problem, but a proportional problem, and there is a corresponding proportional formula.
Nine, the escalator problem
Or use the basic relationship s escalator series =(v people v escalator) ×t rise or fall to solve it. The distance units here are all "levels", and the only thing to note is that t should be expressed as actual steps/people's speed.
Example: The escalator in the shopping mall runs at a constant speed from bottom to top. Two children are walking up and down the escalator. Girls go from bottom to top, boys go from top to bottom. As a result, the girl walked 40 steps upstairs and the boy walked 80 steps downstairs. If boys walk twice as many steps per unit time as girls, how many steps can you see when the escalator is stationary?
X. crossroads
That is, traveling in different directions. There is no special problem-solving skill, just align the pictures honestly and then solve them through geometric analysis. Travel problems on square or rectangular roads.
XI。 School bus problem
It is a kind of problem: there are many people in line, few school buses, the school buses shuttle back and forth, the people in line keep walking and riding, and finally reach the destination at the same time (that is, it takes the shortest time to reach the destination without proof). There are four kinds of small questions: according to the school bus speed (different trains), the class speed (different classes at different speeds), whether the class size changes.
(1) Constant speed-Constant speed-Level 2 (most common)
(2) Constant speed-constant speed shift-multiple shifts
(3) The speed is constant-the shift speed changes-and the number of shifts is 2.
(4) Variable speed-constant shift speed -2 shifts.
Standard solution: draw-list three formulas:
1, the total time = the time for a team to take the bus+the time for this team to walk;
2. The total distance traveled by the shuttle bus;
3. The time for a team to leave = the time for the bus to pick up after it leaves at the same time.
Finally, the ratio of several road sections will be obtained, and then according to algebra.
Simple example: Students from Class A and Class B leave school at the same time to play in the park 15km away. The walking speed of Class A and Class B is 4 kilometers per hour. There is a car in the school with a speed of 48 kilometers per hour. This car can just take a class of students. How many kilometers do students in Class A and Class B need to walk in order to get to the park in the shortest time?
Twelve. Guaranteed round-trip shipping space
Simple example: A and B are going to explore the desert. They go deep into the desert for 20 kilometers every day. It is understood that everyone can carry food and water for up to 24 days. If some food is not allowed to be stored on the way, how many kilometers can one of them go deep into the desert (two people are required to return to the starting point)? This kind of problem actually belongs to the category of intelligent application problem. It is suggested to memorize the deduced conclusion so as to answer the questions quickly. Everyone can bring enough food for t days, and the farthest time is t.
(1) returns this class. (Make sure that one person goes farthest and everyone comes back alive)
1, two people: if you don't put food halfway: T = 2/3t；; If you put food halfway: T=3/4t.
2. Many people:
(2) Crossing the desert (ensure that one person does not come back after crossing the desert, and everyone else comes back alive) * * * There are n people (including those crossing the desert), that is, many people help 1 person to cross the desert.
1. There is no food in the middle: t ≤ [2n/(n+ 1) ]× t. T is the number of days crossing the desert.
2. Put the food in half: t = (1+1/3+1/5+1/7+…+1(2n-1)) × t.
The sum and difference of two numbers in junior high school mathematics examination question 2 1 and sum-difference problem are known, so find these two numbers.
Example: It is known that the sum of two numbers is 10, and the difference is 2. Find these two numbers.
Concise memory formula
And the sum and the difference are getting bigger and bigger; Divided by 2, it is big;
And subtract the difference, the smaller the reduction; Divided by 2, it is small.
According to the formula, large number =( 10+2)÷2=6 and decimal number =( 10-2)÷2=4.
2. Difference ratio problem
For example, the number A is greater than the number B 12, and A: B = 7: 4. Find two numbers.
Concise memory formula
I am more than you, and multiples are cause and effect.
Actual difference of numerator, multiple difference of denominator.
Double the quotient and multiply it by their respective multiples to get two numbers.
First, double the amount, 12÷(7-4)=4,
So the number A is 4X7=28 and the number B is 4X4= 16.
3. Age problem
Concise memory formula
The age difference is constant, adding and subtracting at the same time.
With the change of age, the multiple is also changing.
Grasp these three points and everything will be easy.
Example 1: Xiaojun is 8 years old and his father is 34 years old. After a few years, his father is three times older than Xiaojun.
Analysis: precession will not change, this year's age is almost 34-8=26, and it will not change in a few years. Knowing the difference and multiple, it becomes the problem of difference ratio.
26÷(3- 1)= 13. In a few years, dad's age is 13X3=39, and Xiaojun's age is 13X 1= 13, so it should be
Example 2: Sister 13 years old, brother 9 years old. How old should they be when the sum of their ages is 40?
Analysis: precession will not change. The age difference this year is 13-9=4, and it will not change in a few years. After several years, the age sum is 40, and the age difference is 4, which turns into a sum-difference problem.
So a few years later, the elder sister's age is (40+4)÷2=22, and the younger brother's age is (40-4)÷2= 18, so the answer is 9 years later.
4. The sum ratio problem is known as a whole and solved locally.
Example: The sum of three numbers A, B and C is 27, and A: B: C =2:3:4. Find the three numbers A, B and C. ..
Concise memory formula
Family members want everyone to be together, and separation is also principled.
Denominator ratio sum, numerator's own.
And multiply by the ratio, you deserve it.
The denominator ratio sum, that is, the denominator is: 2+3+4 = 9;
If the molecule is its own, the ratios of the three numbers A, B and C to the sum are 2÷9, 3÷9 and 4 ÷ 9, respectively;
And the multiplication ratio, a is 27X2÷9=6, b is 27X3÷9=9, and c is 27X4÷9= 12.
5. The problem that chickens and rabbits are in the same cage
Example: chickens are free in the same cage, with head 36 and feet 120. Find the number of chickens and rabbits.
Concise memory formula
Suppose all chickens, suppose all rabbits.
How many feet are there? How many feet are missing?
Divided by the foot difference, it is the number of chickens and rabbits.
When finding rabbits, it is assumed that they are all chickens, so the number of exemptions =( 120-36X2)÷(4-2)=24.
When looking for chickens, it is assumed that they are all rabbits, so the number of chickens = (4x36-120) ÷ (4-2) =12.
6. Distance problem
(1) encountered a problem.
Example: A and B walk in opposite directions from two places with a distance of 120km. Party A's speed is 40km/h and Party B's speed is 20 km/h. How long did they meet?
Concise memory formula
At the moment we met, the distance disappeared.
Divide by the sum of the speeds and you get the time.
At the moment of meeting, all the distances have passed, that is, the distance between Party A and Party B is exactly 120km.
Divided by the sum of the speeds, the time is obtained, that is, the total speed of Party A and Party B is 40+20=60 (km/h), and the meeting time is 120÷60=2 (h).
(2) Traceability problem
Brother and sister go to town from home. Big sister walks at a speed of 3 kilometers per hour. After walking for 2 hours, my little brother rode at a speed of 6 kilometers per hour. When will he catch up?
Concise memory formula
Slow birds fly first, fast birds chase after them.
Divide the distance you go first by the speed difference, and the time is right.
Distance to go first: 3X2=6 (km)
Speed difference: 6-3=3 (km/h)
Catch-up time: 6÷3=2 (hours)
7. Concentration problem
(1) diluted with water
Example: There is 20kg of sugar water with the concentration of 15%. How many kilograms of water are added, and the concentration becomes 10%.
Concise memory formula
Sugar before adding water, sugar water after adding sugar.
Sugar water minus sugar water is the added water.
Get the sugar before adding water. The original sugar content is 20X 15%=3 (kg).
We've run out of sugar. How much sugar water should we have with a concentration of 10%? 3÷ 10%=30 (kg).
Sugar water MINUS sugar water, the amount of sugar water after subtraction is 30-20= 10 (kg).
(2) Sugar concentration
Example: There is 20kg of sugar water with the concentration of 15%. How many kilograms of sugar are added, and the concentration becomes 20%.
Concise memory formula
Water before adding sugar, syrup after adding water.
If you subtract sugar water from sugar water, you can easily solve the problem.
Water needs to be added before sugar is added. The original water content is 20x (1-15%) =17 (kg).
When the water is used up, we will get sugar water. 17 ÷ (1-20%) = 21.25 (kg) How much sugar water should there be with a concentration of 20%?
Sugar water MINUS sugar water, the amount of sugar water after subtraction is 2 1.25-20= 1.25 (kg).
8, engineering problems
Example: A project will be completed in 4 days by yourself and 6 days by yourself. After Party A and Party B do it at the same time for 2 days, how many days will Party B do it alone?
Concise memory formula
The total project amount is set to 1, and 1 divided by time is the work efficiency.
When you do it alone, your work efficiency is your own. When you do it together, your work efficiency is everyone's efficiency.
1 Subtract what has been done and what has not been done, and divide what has not been done by work efficiency is the result.
[1-(1÷ 6+1÷ 4) x2] ÷1÷ 6) =1(days)
9. Planting trees
Concise memory formula
How many trees to plant and how to ask for directions?
Subtract 1 directly and the circle is the result.
Example 1: Plant trees on a road with a length of 120m, with a spacing of 4m. How many trees have been planted?
If the road is straight, planting trees is 120÷4- 1=29 (tree).
Example 2: Plant trees at the edge of a circular flower bed with a length of 120m and a spacing of 4m. How many trees have been planted?
If the road is round, planting trees is 120÷4=30 (trees).
10, profit and loss problem
Concise memory formula
Full profit and loss, greatly reduced; One profit and one loss, the profit and loss add up.
Divided by the difference in distribution, the result is something or people distributed.
Example 1: Children divide peaches, each peach 10, and 9 peaches are missing; Eight more than seven per person. How many children and peaches do you want?
If there is profit or loss, the formula is: (9+7)÷( 10-8)=8 (person), and the corresponding peach is 8X 10-9=7 1 (person).
Example 2: Soldiers carry bullets. 45 rounds, 680 rounds per person; 50 rounds per person is more than 200 rounds. How many soldiers, how many bullets?
The total residual problem, the big MINUS the small, that is, the formula is: (680-200)÷(50-45)=96 (person), and the corresponding bullet is 96X50+200=5000 (hair).
Example 3: Students distribute books. 10 Each person is missing 90 books; There are eight copies each, and there are still eight left. How many books are suitable for how many students?
For the total loss problem, the big one is subtracted from the small one, that is, the formula is: (90-8)÷( 10-8)=4 1 (person), and the corresponding book is 4 1X 10-90=320 (.
1 1, remainder problem
For example, the clock now shows 18 o'clock. What time is it after the minute hand turns 1990?
Concise memory formula
There are (N- 1) remainders, the smallest is 1 and the largest is (N- 1).
When the cycle changes, don't look at the quotient, just look at the surplus.
Analysis: When the minute hand turns one turn, it is 1 hour, and 24 turns are 1 turn of the hour hand, so that the hand returns to its original position immediately. The remainder of 1980÷24 is 22, so it is equivalent to the minute hand rotating 22 times forward, the clockwise hand moving 22 hours forward, the backward 24-22=2 hours, and the clockwise hand pulling back 2 hours. The instantaneous needle is equivalent to 18-2= 16 (point).
12, cattle grazing problem
Concise memory formula
The amount of grass eaten by each cow per day is assumed to be 1. How much grass did A eat in the first b days? How much grass did M eat in the first n days? Subtract the big one from the small one and divide it by the difference of the corresponding days, and the result is the growth rate of grass. The original amount of grass is correspondingly reversed.
Formula: a MINUS the amount of grass eaten in b days before b days multiplied by the growth rate of grass. Cattle with unknown grazing amount are divided into two parts: a small part eats new grass first, and the quantity is the proportion of grass; Divide some grass by the number of remaining cows to get the required number of days.
The grass grows thick and fast all over the pasture. 27 cows can eat grass for 6 days; 23 cows can eat the grass in 9 days. Q 2 1 How many days will it take to finish the grass?
Assume that the daily grazing amount of each cow is 1, the grazing amount of 27 cows for 6 days is 27×6 = 162, and the grazing amount of 23 cows for 9 days is 23×9 = 207.
Subtract the small from the large, 207-162 = 45; If the difference between two corresponding days is 9-6=3 (days), the growth rate of grass is 45÷3= 15 (cattle/day);
The original amount of grass was reversed from here-
Formula: a MINUS the amount of grass eaten in b days before b days multiplied by the growth rate of grass.
Grass amount =27X6-6X 15=72 (cattle/day).
Cattle with unknown grazing amount are divided into two parts:
A small part eats new grass first, the quantity is the proportion of grass, that is to say, 2 1 cow is divided into two parts, and some 15 cows eat new grass; The remaining 2 1- 15=6 eats the original grass, and the number of days required is:
The original amount of grass-distribution of surplus cattle = 72-6 =12 (days)
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