It means that when solving a problem, first find out how much a copy is (that is, the single quantity), and then find out the required quantity according to the single quantity. This kind of application problem is called standardization problem.

Total number of quantity relations ÷ number of copies = 1 number of copies 1 number of copies × number of occupied copies = number of requested copies.

In addition, total amount ÷ (total amount ÷ number of copies) = required number of copies.

The idea and method of solving the problem is to find the single quantity first, and then find the required quantity based on the single quantity.

Example 1 It costs 0.6 yuan money to buy five pencils. How much is it to buy the same pencil 16?

How much is it to buy a 1 pencil? 0.6 ÷ 5 = 0. 12 (yuan)

(2) How much does it cost to buy a 16 pencil? 0.12×16 =1.92 (yuan)

The comprehensive formula is 0.6 ÷ 5×16 = 0.12×16 =1.92 (yuan).

A: 1.92 yuan is required.

Example 2 Three tractors cultivated 90 hectares of land in three days. According to this calculation, how many hectares have been cultivated by five tractors in six days?

How many hectares of arable land is (1) 1 tractor 1 day? 90 ÷ 3 ÷ 3 = 10 (hectare)

(2) How many hectares of farmland are cultivated by five tractors in six days? 10× 5× 6 = 300 (hectare)

It is listed as a comprehensive formula 90 ÷ 3 ÷ 3× 5× 6 = 10× 30 = 300 (hectare).

Five tractors cultivated 300 hectares of land in six days.

Example 3 Five cars can transport 100 tons of steel in four times. If the same 7 cars transport 105 tons of steel, how many times do you need to transport it?

(1) 1 How many tons of steel can a car transport? 100 ÷ 5 ÷ 4 = 5 (ton)

(2) How many tons of steel can be transported by seven cars 1 time? 5× 7 = 35 (ton)

(3) How many times do seven cars 105 tons of steel need to be transported? 105 ÷ 35 = 3 (times)

Column into a comprehensive formula105 ÷ (100 ÷ 5 ÷ 4× 7) = 3 (times).

A: It needs to be shipped three times.

2 Summarize the problem

When solving the meaning problem, we often work out the "total amount" first, and then work out the required problem according to other conditions. This is the inductive problem. The so-called "total amount" refers to the total price of goods, the total workload of several hours (days), the total output of several acres of land, the total distance of several hours of travel, etc.

Quantity relationship 1 number of shares × number of shares = total amount ÷ 1 number of shares = number of shares.

Total ÷ another number = another number for each number.

The idea and method of solving the problem is to get the total amount first, and then get the required amount according to the meaning of the question.

Example 1 The garment factory originally made a set of 3.2-meter clothes cloth, and each set of clothes cloth was 2.8 meters after improving the cutting method. How many sets of cloth can you make now?

How many meters is this batch of cloth? 3.2× 79 1 = 253 1.2 (m)

(2) How many sets can you make now? 253 1.2 ÷ 2.8 = 904 (set)

It is listed as a comprehensive formula of 3.2 × 79 1 ÷ 2.8 = 904 (sets).

A: You can make 904 sets now.

Example 2 Xiaohua reads 24 pages every day, and finishes the book Red Rock in 12 days. Xiaoming reads 36 pages a day. How many days can he finish Hongyan?

How many pages are there in the book (1) Hongyan? 24× 12 = 288 pages

(2) How many days can Xiao Ming finish reading Red Rock? 288 ÷ 36 = 8 (days)

It is listed as a comprehensive formula of 24× 12 ÷ 36 = 8 (days).

Xiao Ming can finish reading Hongyan in eight days.

A batch of vegetables was sent to the canteen. It was originally planned to eat 50 Jin a day and slowly consume the vegetables in 30 days. Later, according to everyone's opinion, I ate 10 kg more than planned every day. How many days can we eat these vegetables?

How many kilograms is this batch of vegetables (1)? 50× 30 = 1500 (kg)

(2) How many days can this batch of vegetables last? 1500 ÷ (50+ 10) = 25 (days)

The comprehensive formula is 50× 30 ÷ (50+10) =1500 ÷ 60 = 25 (days).

A: This batch of vegetables can be eaten for 25 days.

3 sum difference problem

The meaning of the sum and difference of two quantities is known. What are these two quantities? This kind of application problem is called sum and difference problem.

Quantitative relationship Large number = (sum+difference) ÷ 2 Decimal number = (sum-difference) ÷ 2

Formulas can be directly applied to problems with simple problem-solving ideas and methods; Complex topics are modified before using formulas.

Example1There are 98 students in Class A and Class B. Class A has 6 more students than Class B. How many students are there in each class?

Number of disarmament classes = (98+6) ÷ 2 = 52.

Class B population = (98-6) ÷ 2 = 46 people.

A: There are 52 students in Class A and 46 students in Class B. ..

The sum of the length and width of a rectangle is 18 cm, and the length is 2 cm more than the width. Find the area of a rectangle.

Solution length = (18+2) ÷ 2 = 10 (cm) width = (18-2) ÷ 2 = 8 (cm).

Area of rectangle = 10× 8 = 80 (square centimeter)

A: The area of a rectangle is 80 square centimeters.

Example 3 has three bags of fertilizer, two bags of fertilizer weigh 32kg, two bags of fertilizer weigh 30kg and two bags of fertilizer weigh 22kg. How many kilograms do you want to know?

Two bags of solutions A and B contain B, from which it can be seen that A is greater than C (32-30) = 2kg, A is a large number and C is a decimal number. It can be seen that

The weight of fertilizer in bag A = (22+2) ÷ 2 = 12 (kg)

Weight of bagged fertilizer C = (22-2) ÷ 2 = 10 (kg)

Weight of fertilizer in bag B = 32- 12 = 20 (kg)

Answer: The fertilizer in bag A weighs 12kg, the fertilizer in bag B weighs 20kg and the fertilizer in bag C weighs 10kg.

Example 4 Car A and Car B originally contained 97 baskets of apples, and 14 baskets were taken from Car A and put on Car B. As a result, Car A had 3 more baskets than Car B. How many baskets did each car originally contain?

The solution of "take 14 baskets from car A and put them on car B" shows that car A is a large number, car B is a decimal number, the difference between a and b is (14× 2+3), and the sum of a and b is 97, so the number of baskets in car A = (97+ 14× 3).

Number of baskets of car B = 97-64 = 33 (baskets)

A: Car A originally contained 64 baskets of apples, while car B originally contained 33 baskets of apples.

4 and multiple problems

Given the meaning of the sum of two numbers and how many times a large number is a decimal (or how many times a decimal is a large number), this kind of application problem is called the sum and multiple problem.

Sum of quantitative relations ÷ (multiple+1) = sum of smaller numbers-smaller numbers = larger numbers.

Smaller number × several times = larger number

Simple problem-solving ideas and methods directly use formulas, and complex problems are modified with formulas.

There are 248 apricot and peach trees in the orchard of 1. There are three times as many peach trees as apricot trees. How many apricot and peach trees are there?

How many apricot trees are there? 248 ÷ (3+ 1) = 62 (tree)

(2) How many peach trees are there? 62× 3 = 186 (tree)

A: There are 62 apricot trees and 86 peach trees/kloc-0.

Example 2 The east and west warehouses have a grain storage capacity of 480 tons, and the grain storage capacity of the east warehouse is 1.4 times that of the west warehouse. How many tons of grain are stored in each warehouse?

Solution (1) Grain quantity in stock in western China = 480 ÷ (1.4+ 1) = 200 (ton)

(2) Grain in stock in East China = 480-200 = 280 (ton)

A: There are 280 tons of grain in the east and 200 tons in the west.

There are 52 cars in Station 3a and 32 cars in bilibili. If there are 28 cars from Station A to bilibili and 24 cars from bilibili to Station A every day, the number of cars in bilibili will be twice that of Station A in a few days.

There are 28 cars from Station A to bilibili and 24 cars from bilibili to Station A every day, which is equivalent to 28-24 cars from Station A to bilibili every day. After a few days, the number of vehicles at Station A was regarded as 1 time. At this time, the number of vehicles in bilibili is twice, and the total number of vehicles in two stations (52+32) is equivalent to (2+ 1) times. Then, a few days later, the number of vehicles at Station A was reduced to (52+32) ÷ (2+65438).

The required number of days is (52-28) ÷ (28-24) = 6 (days)

A: After 6 days, the number of vehicles in bilibili is twice that of Station A. ..

Example 4 The sum of the three numbers A, B and C is 170, B is 2 times smaller than A, 4 times larger than A, and C is 3 times larger than A. What are these three numbers?

The numbers of solutions B and C are directly related to the number A, so the number A is taken as 1 time.

Because b is 2 times less than a by 4, if b is added by 4, the number of b becomes 2 times that of a;

And because C is three times more than A, the number of C minus 6 becomes three times that of A;

At this time, (170+4-6) is equivalent to (1+2+3) times. So,

A number = (170+4-6) ÷ (1+2+3) = 28.

B Quantity = 28× 2-4 = 52

C = 28× 3+6 = 90

A: The number A is 28, the number B is 52 and the number C is 90.

Five-difference multiple problem

This paper gives the meaning of the difference between two numbers and how many times a large number is a decimal (or how many times a decimal is a large number). This kind of application problem is called the difference multiple problem.

Difference between two numbers ÷ (several times-1) = smaller number.

Smaller number × several times = larger number

Simple problem-solving ideas and methods directly use formulas, and complex problems are modified with formulas.

The number of peach trees in 1 orchard is three times that of apricot trees, and peach trees are more than apricot trees 124. How many apricot and peach trees are there?

How many apricot trees are there? 124 ÷ (3- 1) = 62 (tree)

(2) How many peach trees are there? 62× 3 = 186 (tree)

A: There are 62 apricot trees in the orchard, 186 peach trees.

The father is 27 years older than his son. This year, the father is four times older than his son. How old are the father and son this year?

Solution (1) Subage = 27 ÷ (4- 1) = 9 (year)

(2) Dad's age = 9× 4 = 36 (years old)

A: The father and son are 36 and 9 years old respectively this year.

After the reform of management mode, the profit of this month is 6,543.8+200,000 yuan more than that of last month, which shows that the profit of this month is 300,000 yuan more than that of last month. What's the profit in these two months?

If last month's profit is 1 times, then (30- 12) ten thousand yuan is equivalent to (2- 1) times of last month's profit, then last month's profit = (30-12) ÷ (2-1.

This month's profit = 18+30 = 48 (ten thousand yuan)

A: Last month's profit was 6.5438+0.8 million yuan, and this month's profit was 480,000 yuan.

There are 94 tons of wheat and 0/38 tons of corn in the grain depot. If 9 tons of wheat and 9 tons of corn are shipped out every day, how many days later will the remaining corn be three times that of wheat?

Because the quantity of wheat and corn shipped every day is equal, the remaining quantity difference is equal to the original quantity difference (138-94). If the wheat left after a few days is regarded as 1 times, and the corn left after a few days is three times, then (138-94) is equivalent to (3- 1) times, so

Wheat surplus = (138-94) ÷ (3-1) = 22 (ton)

Quantity of wheat shipped = 94-22 = 72 (tons)

Grain transportation days = 72 ÷ 9 = 8 (days)

A: After eight days, the remaining corn is three times that of wheat.

Six times ratio problem

There are two known quantities of the same species, one of which is several times that of the other. When solving a problem, first find this multiple, and then use the method of multiple ratio to calculate the required number. This kind of application problem is called magnification problem.

Quantity Relationship Total Quantity ÷ One quantity = multiple of another quantity × multiple = another total quantity.

The idea and method of solving the problem is to find the multiple first, and then use the multiple ratio relationship to find the required number.

Example 1 100 kg of rapeseed can extract 40 kg of oil. Now there are 3700 kilograms of rapeseed. How much oil can be squeezed out?

How many times is the solution of (1) 3700kg? 3700 ÷ 100 = 37 (times)

(2) How many kilograms of oil can be squeezed out? 40× 37 = 1480 (kg)

The comprehensive formula is 40× (3700 ÷100) =1480 (kg).

Answer: Oil can be squeezed 1480 kg.

This year's Arbor Day, 300 teachers and students in a primary school planted 400 trees. According to this calculation, how many trees have been planted by 48,000 teachers and students in the county?

(1) How many times is 48000 more than 300? 48000 ÷ 300 = 160 (times)

(2) How many trees have been planted? 400× 160 = 64000 (tree)

Column into a comprehensive formula 400 × (48000 ÷ 300) = 64000 (tree).

A: The county's 48,000 teachers and students planted 64,000 trees.

There is a bumper harvest of apples in Fengxiang County this year. The income of one household in Tianjiazhuang 4 mu orchard111yuan. According to this calculation, what is the income of 800 acres of orchards in the township? What is the income of the county 16000 mu orchard?

(1) How many times is 800 mu more than 4 mu? 800 ÷ 4 = 200 (times)

(2) What is the income of 800 mu? 1111× 200 = 222200 (yuan)

(3) How many times is16000 mu more than 800 mu? 16000 ÷ 800 = 20 (times)

(4) What is the profit of16000 mu? 2222200× 20 = 44444000 (yuan)

Answer: The income of 800 mu orchards in the township is 2,222,200 yuan, and the income of 1.6 million mu orchards in the county is * * *.

44.44 million yuan.

7 encounter problems

It means that two moving objects start from two places at the same time, move in opposite directions and meet on the way. This application problem is called encounter problem.

Quantitative relationship of meeting time = total distance ÷ (speed A+ speed b)

Total distance = (speed A+ speed B) × meeting time

Formulas can be used directly for problems with simple ideas and methods, and formulas can be used after modification for complex problems.

The waterway from Nanjing to Shanghai is 392 kilometers long. At the same time, ships from each port run relative to each other. The speed of ships from Nanjing is 28km/h, while that from Shanghai is 2 1 km. How many hours passed before the two ships met?

Solution 392 ÷ (28+2 1) = 8 (hours)

A: Eight hours later, the two ships met.

Example 2 Xiao Li and Xiao Liu are running on a 400-meter-long circular track. Xiao Li runs 5 meters per second and Xiao Liu runs 3 meters per second. They started from the same place at the same time and ran in the opposite direction. So, how long will it take them to meet for the second time?

The second meeting can be understood as two people running two laps. So the total distance is 400×2.

Meeting time = (400× 2) ÷ (5+3) = 100 (seconds)

A: It takes 100 seconds for them to meet for the second time.

Example 3 Both Party A and Party B ride bicycles from two places at the same time. The cycling speed of Party A is15km, and that of Party B is13km. They met 3 kilometers from the midpoint to find the distance between the two places.

Understanding that "two people meet at a distance of 3 kilometers from the midpoint" is the key to correctly understand the meaning of this question. As can be seen from the title, A rides fast and B rides slowly. A crosses the midpoint 3 kilometers, and B is 3 kilometers away from the midpoint, which means that A has walked (3×2) kilometers more than B. Therefore,

Meeting time = (3× 2) ÷ (15- 13) = 3 (hours)

Distance between the two places = (15+ 13) × 3 = 84 (km)

Attendant: The distance between the two places is 84 kilometers.

Eight follow-up questions

It means that two moving objects start in different places at the same time (or start in the same place but not at the same time, or start in different places but not at the same time) and move in the same direction. The back is fast, the front is slow, and in a certain period of time, the back catches up with the front. This application problem is called tracing problem.

Catch-up time = Catch-up distance ÷ (fast-slow)

Catch-up distance = (fast-slow) × catch-up time

Simple problem-solving ideas and methods directly use formulas, and complex problems are modified with formulas.

Example: 1 A good horse walks every day 120km, and a bad horse walks 75km every day. Bad horses go first 12 days. How many days can a good horse catch up with a bad horse?

How many kilometers can a bad horse walk 1 2 days? 75× 12 = 900 km

(2) How many days does a good horse catch up with a bad horse? 900 ÷ (120-75) = 20 (days)

The comprehensive formula is 75×12 ÷ (120-75) = 900 ÷ 45 = 20 (days).

A: A good horse can catch up with a bad horse in 20 days.

Example 2 Xiaoming and Xiao Liang are running on the 200m circular track. Xiao Ming ran for 40 seconds. They started from the same place and ran in the same direction at the same time. Xiaoming ran 500 meters when he first caught up with Liang Xiao. What is the speed of Xiao Liang per second?

When Xie Xiaoming caught up with Liang Xiao for the first time, he ran one lap more than Liang Xiao, that is, 200 meters. At this time, Xiao Liang ran (500-200m). To know the speed of Xiao Liang, you should know the time, that is, it takes Xiaoming to run 500 meters. We also know that it takes 40 seconds for Xiao Ming to run 200 meters and [40× (500 ÷ 200)] seconds for him to run 500 meters, so the speed of Xiao Liang is (500-200) ÷ [40× (500 ÷ 200)] = 300 ÷ 65438.

The speed of Xiao Liang is 3 meters per second.

Our People's Liberation Army pursued the fleeing enemy. The enemy began to flee from place A at the speed of10km per hour on16km in the afternoon, and the PLA was ordered to pursue from place B at the speed of 30km per hour at 22pm. As we all know, the distance between A and B is 60 kilometers. How many hours can the PLA catch up with the enemy?

The time difference between the enemy's escape time and the PLA's pursuit time is (22- 16) hours. During this period, the enemy's escape distance is [10× (22-6)] km, and the distance between Party A and Party B is 60 km. Infer from this

Catch-up time = [10× (22-6)+60] ⊙ (30-10) = 220 ÷ 20 =11(hour)

A: The PLA can catch up with the enemy after 1 1 hour.

A bus travels from Station A to bilibili at a speed of 48 kilometers per hour. A truck was traveling from bilibili to Station A at the same time, with a speed of 40 kilometers per hour. Two trucks meet at the midpoint of two stations 16 km, and find the distance between the two stations.

Solving this problem can change from meeting a problem to chasing it. As can be seen from the topic, the bus lags behind the truck (16×2) kilometers, and the time when the bus catches up with the truck is the aforementioned meeting time.

This time is 16× 2 ÷ (48-40) = 4 (hours).

So the distance between the two stations is (48+40) × 4 = 352 (km).

The comprehensive formula (48+40) × [16× 2 ÷ (48-40)] = 88× 4 = 352 (km).

A: The distance between Station A and bilibili is 352 kilometers.

Both brother and sister go to school from home at the same time. My brother walks 90 meters per minute and my sister walks 60 meters per minute. When my brother arrived at the school gate, he found that he had forgotten his textbook. He immediately went home to get it along the original road and met his sister at 0/80 meters away from the school/kloc-. How far is their home from school?

The solution requires that the distance and speed are known, so the key is to find the meeting time. As can be seen from the topic, in the same time (from departure to meeting), my brother walks (180×2) meters more than my sister, because my brother walks (90-60) meters more than my sister every minute. So, the time it takes them to walk from home to the meeting place is

180× 2 ÷ (90-60) = 12 (minutes)

The distance from home to school is 90× 12- 180 = 900 (meters).

A: My home is 900 meters from school.

Sun Liang plans to go to school five minutes before class. He walks from home to school at a speed of 4 kilometers per hour. When he walked 1 km, he found that his watch was 10 minutes slow, so he immediately ran forward and arrived at school on time. Later, I calculated that if Sun Liang had run away from home, he would have walked to school nine minutes earlier than before. Seek Sun Liang's running speed.

Taking off the watch is slow 10 minutes, which means it is late 10 minutes. If you continue to walk at the original speed, you will be (10-5) minutes late, and you will arrive at school on time for the second run, which means that running takes (10-5) minutes less than walking. If you run from home, it's 9 minutes less than walking. So running takes [9-( 10-5)] minutes less than walking. therefore

Walking 1 km needs1÷ [9-(10-5)] = 0.25 (hours) = 15 (minutes).

Running time 1 km15-[9-(10-5)] =11(minutes).

Operating speed is1÷11/60 =1× 60/1= 5.5 (km) per hour.

A: The running speed in Sun Liang is 5.5 kilometers per hour.

9 tree planting problem

For equidistant tree planting, of the three quantities of distance, plant spacing and number of plants, two are known and the third quantity is needed. This kind of application problem is called planting trees.

Number of linear trees planted in quantitative relationship = distance ÷ distance+1

Number of trees planted in ring = distance between trees

Number of square trees planted = distance -4.

Number of trees planted in the triangle = distance -3.

Planting area = area ÷ (plant spacing × row spacing)

Ideas and methods of solving problems: first, make clear the types of tree planting problems, and then use formulas.

Example 1 a river bank136m, plant a weeping willow every 2m, end to end. How many weeping willows can a * * * plant?

Solution136 ÷ 2+1= 68+1= 69 (tree)

A: A * * * will plant 69 weeping willows.

A circular pond is 400 meters in circumference, and poplar trees are planted every 4 meters along the shore. How many poplars can a * * * plant?

Solution 400 ÷ 4 = 100 (tree)

A: A * * * can plant 100 poplar trees.

A square sports ground, each side of which is 220 meters long, is equipped with lighting lamps every 8 meters. A * * *, how many lights can you hold?

Solution 220× 4 ÷ 8-4 =110-4 =106 (pieces)

A: A * * * can install 106 lighting lamps.

For a 96-square-meter house, the length and width of the floor tiles used are 60 cm and 40 cm respectively. How many floor tiles do you need at least?

Solution 96 ÷ (0.6× 0.4) = 96 ÷ 0.24 = 400 (block)

A: At least 400 floor tiles are needed.

A bridge is 500 meters long, and street lamps are installed on poles on both sides of the bridge. If there is a telephone pole every 50 meters and two street lamps are installed on each pole, how many street lamps can be installed in a * * *?

(1) How many poles are there on one side of the bridge? 500 ÷ 50+1=11(pieces)

(2) How many telephone poles are there on both sides of the bridge? 1 1× 2 = 22 (pieces)

(3) How many street lamps can be installed on both sides of the bridge? 22× 2 = 44 (lamp)

Answer: 44 street lamps can be installed on both sides of the bridge.

10 age problem

The question of meaning is named after the content of the topic. Its main feature is that the age difference between them is unchanged, but the multiple relationship of their ages changes with age.

The age problem of quantitative relations is often closely related to the sum difference problem, the sum multiple problem and the difference multiple problem. In particular, the idea of solving the difference multiple problem should firmly grasp the characteristic of "the age difference is unchanged"

The ideas and methods of solving the problem can draw lessons from the ideas and methods of "differential multiple problems".

Example 1 Dad is 35 years old and Liang Liang is 5 years old. How old is dad this year? What about next year?

Solution 35 ÷ 5 = 7 (degree) (35+ 1) ÷ (5+ 1) = 6 (degree)

A: This year, my father is seven times as old as Liang Liang, and next year, my father is six times as old as Liang Liang.

Mother is 37 years old and daughter is 7 years old. In a few years, my mother will be four times as old as my daughter.

How old is the mother than her daughter? 37-7 = 30 years old

(2) A few years later, the mother is four times as old as her daughter? 30(4- 1)-7 = 3 (year)

Column into a comprehensive formula (37-7) ÷ (4- 1)-7 = 3 (year).

A: After three years, the mother is four times as old as her daughter.

Three years ago, the age of father and son was 49. This year, the father is four times as old as his son. How old are father and son this year?

It is understood that the total age of father and son this year should be (3×2) years older than that of three years ago, and the total age of the two this year is 49+3× 2 = 55 (years old).

Taking the age of the son this year as 1 time, the sum of the ages of the father and son this year is equivalent to (4+ 1) times. Therefore, the age of my son this year is

55 \ u (4+1) =11(years old)

Father's age this year is 1 1× 4 = 44 (years old).

A: My father is 44 years old and my son is 1 1 year old.

Example 4 A said to B, "My age is your present age, and you are only 4 years old". B said to A, "When my age is your present age in the future, you will be 6 1 year old". What are the ages of Party A and Party B now?

solve

This involves three years: the past year, this year and the next year. List analysis:

Last year, this year, the next year.

A □ Age△ Age 6 1 year

B 4 years old □ years old △ years old

Two □ in the table represent the same number, and two △ represent the same number.

Because the age difference between two people is always equal: □-4 = △-□ = 6 1-△, that is, 4 □, △, 6 1 becomes arithmetic progression, so 6 1 should be three years older than 4, so the age difference between two people is (61.

A The age this year is △ = 6 1- 19 = 42 (years old).

B The age this year is □ = 42- 19 = 23 (years old).

A: A is 42 years old and B is 23 years old.

1 1 navigation problems

The problem of sailing is also related to sailing. To solve this kind of problem, it is necessary to understand the ship speed and water speed, that is, the speed of the ship itself, that is, the speed of the ship sailing in still water; Water speed is the speed of water flow, and the speed of sailing along the water is the sum of ship speed and water speed; The speed of sailing against the current is the difference between the speed of the boat and the speed of the water.

Quantity relationship (downstream speed+upstream speed) ÷ 2 = ship speed

(downstream speed-upstream speed) ÷ 2 = water flow speed

Downstream speed = ship speed ×2- current speed = current speed+current speed ×2

Current speed = ship speed ×2- downstream speed = downstream speed-current speed ×2

In most cases, we can directly use the formula of quantitative relationship.

Example 1 It takes 8 hours for a ship to sail 320 kilometers along the river, and the current speed is 15 kilometers per hour. How many hours does it take for the ship to sail against the current?

According to the conditions, the downstream speed = ship speed+water speed = 320 ÷ 8, and the water speed is 15km/h, then the ship speed is 320÷8- 15 = 25(km/h).

The current speed of the ship is 25- 15 = 10 (km).

The time for the ship to travel against the current is 320 ÷ 10 = 32 (hours).

A: It takes 32 hours for this ship to sail against the current.

Example 2 It takes 65,438+08 hours for a ship to travel 360 kilometers upstream, and 65,438+00 hours to return to its original place. It takes 15 hours for ship b to travel the same distance against the current. How long will it take to get back to the original place?

According to the meaning of the question, ship speed+water speed = 360 ÷ 10 = 36.

Ship speed-water speed = 360 ÷ 18 = 20.

It can be seen that (36-20) is equivalent to twice the speed of water,

So the speed of water is (36-20) ÷ 2 = 8 (km) per hour.

Because, B ship speed-water speed = 360 ÷ 15,

So the speed of ship B is 360 ÷ 15+8 = 32 (km).

The downstream speed of ship B is 32+8 = 40 (km).

So it takes 360 ÷ 40 = 9 (hours) for ship B to sail 360 kilometers along the river.

A: It will take nine hours for the B ship to return to its original place.

An airplane flies between two cities. The speed of the plane is 576 kilometers per hour and the wind speed is 24 kilometers per hour. It takes three hours for the plane to arrive against the wind and several hours to fly back with the wind.

This problem can be solved according to the running water problem.

(1) How many kilometers are the two cities apart? (576-24) × 3 = 1656 (km)

(2) How many hours does it take to fly back with the wind? 1656 ÷ (576+24) = 2.76 (hours)

Column into a comprehensive formula [(576-24)×3]⊙(576+24)= 2.76 (hours).

A: It takes 2.76 hours for the plane to fly back with the wind.