How to solve math problems Itinerary, math itinerary problems How to solve them

a) Encounter Issues.

Two moving objects moving in opposite directions or in opposite motion on a circular runway, with the development of time, will inevitably meet face to face, this kind of problem is called the encounter problem. It is characterized by two moving objects traveling together for the entire distance.

The itinerary problem in primary school mathematics textbooks generally refers to the encounter problem.

Encounter problems can be divided into three types according to the quantitative relationship: finding the distance, finding the meeting time, and finding the speed.

Their basic relationship is as follows:

Total distance = (speed A + speed B) Time of encounter.

Encounter time = total distance (speed A + speed B).

Another velocity = A and B velocity and - one known velocity.

2) Catch up on the problem.

The location of the chase problem can be the same (e.g. the catch up problem on a circular runway) or it can be different, but the direction is generally the same. Due to the difference in speed, the problem of fast catching up and slow happens.

According to the relationship between the velocity difference, the distance difference and the catch-up time, the following formula is commonly used:

Distance difference = speed difference catch-up time.

Catch-up time = distance difference speed difference.

Speed difference = distance difference catch-up time.

Speed difference = fast - slow.

The key to solving the problem is to find out the two of the three that are related and correspond to each other, such as distance difference, speed difference, and chasing time, and then use the formula to find the third party to achieve the solution.

3) 2. The problem of separation.

When two moving objects move apart due to their opposite motion, they are separated from each other. The key to solving the distance problem is to find the distance (velocity sum) of the common tendency of two moving objects.

The basic formulas are:

Distance between two places = speed and time apart.

Separation time = distance between two places, speed sum.

Speed sum = distance between two places and time apart.

Running water problems. The problem of going down the river and going up the river is often called the flow problem, which is a travel problem, and it is still solved by using the relationship between speed, time, and distance. When answering, pay attention to the meaning of each speed and the relationship between them.

When a boat travels in still water, the distance traveled per unit of time is called rowing speed or stroked force; The speed of the boat traveling along the river is called the speed of the river; The speed of the boat against the current is called the countercurrent velocity; The boat releases the middle stream, and does not rely on power to travel along the water, and the distance traveled per unit time is called the current velocity. The relationship between the various speeds is as follows:

1) Paddle speed + water velocity = downstream speed.

2) Paddle speed - water velocity = countercurrent speed.

3) (Downstream velocity + Countercurrent speed) 2 = Paddle speed.

4) (Downstream velocity - Countercurrent velocity) 2 = Water velocity.

The quantitative relationship of the flow problem is still the relationship between speed, time and distance. i.e.: velocity time = distance; Distance: Speed = Time; Distance Time = Velocity.

However, the river water flows, so there is a difference between going with and against the current. When calculating, it is necessary to clarify the relationship between the various velocities.

First, grasp the basic quantitative relationship in the itinerary problem: distance = speed * time.

Understand the sum or difference of the three in the journey problem, such as the encounter problem, the catch up problem, and the train problem (or marching, crossing a tunnel, etc.).

The solution is intuitive with the equation, and the velocities are set to be A and B.

A + B = 15

3A-3B = 15

Solve the equation to get B = kilometers of hours.

Solution: Set the distance between the two places of AB X kilometers, and after T hours, the two cars A and B meet.

Because the two cars met at a distance of 40 kilometers, it means that car A traveled 80 kilometers longer than car B.

Lianjiede; t = 4 hours.

x = km.

If you walk to and from home, it will take 70 minutes--- then it will take 70 2 = 35 minutes to walk home, 50-35 = 15 minutes to ride to school, and 15 * 2 = 30 minutes to ride back and forth. 2. Question 2: The train takes 120 seconds from the beginning of the bridge to the complete disembarkation--- the distance of the train running at a point (tail point or head point) is Bridge length + body length, the time for the whole train to be completely on the bridge is 80 seconds--- the distance traveled by the train at a point (tail point or head point) is Bridge length - body length 120 + 80 = 200 seconds time--- distance traveled by the train at a point (tail point or head point) is Bridge length + body length + bridge length - body length = 2 bridge lengths. Then the train speed = 1000 * 2 (120 + 80) = 10 meters per second, body length = 10 * 120-1000 = 200 meters.

1. Let the speed of the gray wolf be x meters per minute, and the speed of the red wolf be 3x-6500+5x=(3x-6)*5

x = 53 red wolf speed is 3x-6 = 153 meters per minute.

2. Set the time I t

153t-53t=200*2

t=4 min.

Velocity ratio: A:B = 4:5

Full process: 4 + 5 = 9 portions.

The first encounter A walks 4 parts, and the distance from point A is 4 parts.

Velocity ratio after encounter: A: B = 4x (1 + 1 4):

5x(1+1 3)=3:4B to point A, A goes: (4 4)x3=3 (copies) A to point B, B goes:

5-3) 3x4=8 3 (parts) left: 9-8 3=19 3 (parts) A and B go together, need: (19 3) (3+4) = 19 21 (time).

19 21 time B goes: 19 21x4 = 76 21 copies.

B distance point a: 76 21 + 8 3 = 132 21 parts.

The second meeting point is from the first meeting point: 132 21-4 = 48 21 parts.

The second meeting point is 48 km from the first meeting point.

Per serving: 48 (48 21) = 21 (km).

Journey: 21x9=189 (km).

At the first encounter, car A traveled the whole journey 4 (4+5)=4 9, that is, the first meeting point is 4 9 away from place A

The velocity ratio of A to B for the second time is (4 4 1 4):(5+5 1 3) =3:4

Since from the first encounter to the second encounter, the two of them traveled together for 2 whole journeys, so.

The time required for the two to meet from the first meeting to the second meeting is 2 (3+4)=2 7 B, and after the first encounter, 2 7 4=8 7 from the second meeting point is 8 7-4 9=44 63ab, and the distance between the two places is 48 (44 63-4 9)=189 km.

When they meet for the first time, car A travels the whole way 4 (4+5)=4 9 After the meeting, the speed ratio of car A and car B is 4 (1+1 4): 5 (1+1 4)=3:4

At the second encounter, car B traveled another 4 2 (3+4)=8 7, the two meeting points were 8 7-4 9 2=16 63, and the two places were 48 16 63=189 kilometers apart.

Set the speed of A and B to 4x, 5xm of the time of the first encounter, and the time of the first encounter and the second encounter is nA system of column equations.

4x+5x）m=1

1+1 4)*4x +(1+1 3)*5x)*n=21+1 3)*5x*n-4x*m=48, a system of ternary equations, which can be solved.

No way. Wait.

I won't do it, please master.

Question 1: 25 60 + (360-72 * 25 60 + 100) (72 + 48) = 240 60 = 4 (hours).

Question 2: Set up x people in advance.

x+2（x+15）=60

x=10 Question 3: If A buys x tail, then B buys (6000-x) tail.

x=4000

6000-x=2000 (survival rate is a distractor, meaningless.) 3600 yuan is the buying price, not the selling price).

Question 4: Let the distance be x kilometers.

x/（26-2）-3=x/（26+2）

x=504

You can solve it with equations.

Question 1 is actually how far B has more lines than A in 4 minutes.

So it's (96-80) 4=64 meters.

Topic 2: 126 3 = 42 km per hour.

126 kilometers per hour against the current.

When the hydrostatic velocity is (42+36) 2=39 km, the current velocity is 42-39=3 kmh.

Question 3: 30 5 4 = 24 kilometers per hour against the current.

Still water travels per hour (30+24) 2=27 km.

The current is 30-27 = 3 kilometers per hour.

Question 4: Downwind per hour 6 17

The headwind travels 1 to 3 per hour

The wind travels per hour (6 17-1 3) 2 = 1 102, so the distance between the two cities is 24 1 102 = 2448 km, Question 5 (I feel that the unit of the question is a bit wrong).

The length of the train is the length of the bridge minus the train line.

The train traveled 2160 meters.

The length of the train is 5680-5400=280 meters.

Question 6: Explain that line B is 5 hours away, and A takes 4 hours.

Then for a 5-hour journey for line A, B needs 5 5 4 = 25 4 hours, that is, B needs 25 4 hours to arrive.

Question 7: B runs 360 meters, 360-(400-50) = 10 meters more than A, B runs 400 meters, 10 400 360 = 100 9 meters more than A, so when B runs to the finish line, A is still 100 9 meters away from the finish line.

The formula for the sum and difference problem.

Sum difference) 2 large numbers.

and difference) 2 decimals.

and times the problem. and (multiples of 1) decimals.

Decimals, multiples, and large numbers.

or with decimal large numbers).

The problem of difference times. Difference (multiple of Zao Kai 1) decimal.

Decimals, multiples, and large numbers.

or a decimal difference for a large number).

Tree planting issues. 1. The problem of tree planting on non-closed lines can be mainly divided into the following three situations:

If trees are to be planted at both ends of the unenclosed line, then:

Number of plants Number of stages 1 Full length Plant spacing 1

Full length plant spacing (number of plants 1).

Plant spacing: Full length (number of plants: 1).

If trees are to be planted at one end of the unenclosed line and not at the other end, then:

Number of plants, number of stages, full length, plant spacing.

Full length, plant spacing, number of plants.

Plant spacing, full length, number of plants.

If trees are not planted at both ends of the unenclosed line, then:

Number of plants Number of stages 1 Full length Plant spacing 1

Full length plant spacing (number of plants 1).

Plant spacing: Full length (number of plants: 1).

2. The relationship between the number of tree planting problems on the closed line is as follows.

Number of plants, number of stages, full length, plant spacing.

Full length, plant spacing, number of plants.

Plant spacing is the total length of the number of plants.

Profit and loss issues. Profit and loss) The difference between the two distributions The number of shares participating in the distribution.

Big Profit Small Profit) The difference between the two allocations The number of shares participating in the allocation.

Big loss, small loss) The difference between the two distributions The number of shares participating in the distribution.

Encounter Problems. The distance traveled by the encounter speed and the time of the encounter.

Encounter time, encounter distance, speed and.

Speed and distance traveled to meet time.

Catch up on the problem. Chase distance, speed difference, chase time.

Chase time, chase distance, speed difference.

Speed difference, chase distance, chase time.

Running water problems. Downstream velocity Hydrostatic velocity Water velocity velocity.

Countercurrent velocity Hydrostatic velocity Water velocity velocity.

Hydrostatic velocity (downstream velocity countercurrent velocity) 2

Water velocity (downstream velocity counter-current velocity) 2

Concentration issues. The weight of the solute The weight of the solvent The weight of the solution.

The weight of the solute is 100% concentrated by the weight of the solution.

The weight of the solution The concentration of the solute is the weight of the solute.

The weight of the solute concentration of the solution.

Profit vs. discount issues.

Profit, Sell Price, Cost.

Profit Margin Profit Cost 100% (Sell Price Cost 1) 100%.

1. Let ab be separated by x, then = 1-3 8, and the solution is x=417

2. x*(3 7) (4* = (x*4 7-25) (3*4 3), and the solution is x=700

1. When A arrives at place B, the ratio of B's distance from B to A's journey is 3:8, that is, B's distance is 5:8 compared to A's

When 40% of the whole journey of line A, line B is 1 4 of the whole journey, B is still 3 4 of the whole journey from place B, and B is still 150 kilometers away from place B, so the two places of AB are 200 kilometers together.

2, when the passenger car and the truck meet, the passenger car goes the whole 4 7, the truck goes the whole 3 7, and the speed ratio of the passenger car and the freight car after the encounter is (4 * 4 5) :(3 * 4 3) = 4:5

When the passenger car went to the remaining 3 7, the truck went (3 7) * (5 4) = 15 28At this time, the truck went 3 7 + 15 28 = 27 28 of the whole journey, the truck still had 1 28 left in the whole journey, and the truck was still 25 kilometers away from place A, so the distance between the two places in AB was: 25 * 28 = 700 kilometers.