Application problem is the key to determine the primary school students' math scores, and it is also the key to score. Next, I collected the example analysis of solving application problems in primary schools. Welcome to read and check. I hope it will help you.

1, normalization problem

meaning

When solving a problem, first find out how much a copy is (that is, the single quantity), and then find out the required quantity based on the single quantity. This kind of application problem is called standardization problem.

magnitude relation

Total amount/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.

Ideas and methods to solve problems

Find a single quantity first, and then find the required quantity according to the single quantity.

Example 1

It costs 0.6 yuan money to buy five pencils, and how much does it cost to buy the same 16 pencil?

solve

(1) How much is it to buy 1 pencils? 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?

solve

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 be transported in four times 100 tons of steel. If the same 7 vehicles are used to transport 105 tons of steel, how many times do you need to transport it?

solve

(1) 1 How many tons of steel can cars transport 1 time? 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. The problem of induction

meaning

When solving a problem, we often work out the "total amount" first, and then calculate the required problem according to other conditions. This is the so-called 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.

magnitude relation

1 number of copies × number of copies = total amount

Total amount1number of copies = number of copies

Total ÷ another number = another number for each number.

Ideas and methods to solve problems

Find out the total amount first, and then get the required amount according to the meaning of the question.

Example 1

Clothing factory used to make a set of 3.2 meters of clothing cloth, and after improving the cutting method, each set of clothing cloth was 2.8 meters. How many sets of cloth can you make now?

solve

(1) 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 a day, 12 days after reading the book Red Rock. Xiaoming reads 36 pages a day. How many days can he finish Hongyan?

solve

(1) How many pages does Hongyan have? 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.

Example 3

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?

solve

(1) How much are these vegetables? 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 and difference problem

meaning

Given the sum and difference of two quantities, what are these two quantities? This kind of application problem is called sum-difference problem.

magnitude relation

Large number = (sum+difference) ÷2

Decimal = (sum and difference) ÷2

Ideas and methods to solve problems

Simple questions can be directly applied to formulas; Complex topics are modified before using formulas.

Example 1

There 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?

solve

Class A population = (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. ..

Example 2

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

solve

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

Three bags of fertilizer, two bags of fertilizer 32kg, two bags of fertilizer 30kg and two bags of fertilizer 22kg. How many kilograms do you want?

solve

Two bags A and B and two bags B and C contain B, from which we can see 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 the two cars originally contain?

solve

"Take 14 baskets from car A and put them on car B. As a result, car A has 3 more baskets than car B." This means 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 is = (97+/kloc-).

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 the times.

meaning

It is known that the sum of two numbers and the large number are multiples of the decimal (or the decimal is a fraction of the large number), so it is required that these two numbers are different. This kind of application problem is called sum and multiple problem.

magnitude relation

Sum ÷ (several times+1) = smaller number.

Sum-lower number = higher number

Smaller number × several times = larger number

Ideas and methods to solve problems

Simple topics use formulas directly, and complex topics use formulas after adaptation.

Example 1

There are 248 apricot trees and peach trees in the orchard, and the number of peach trees is three times that of apricot trees. How many apricot and peach trees are there?

solve

(1) 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 two warehouses have 480 tons of grain, and the grain quantity in the east is 1.4 times that in the west. How many tons are there in each warehouse?

solve

(1) western grain stocks = 480 ÷ (1.4+ 1) = 200 (tons)

(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.

Example 3

There are 52 cars in Station A 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.

solve

Every day, there are 28 vehicles from Station A to bilibili and 24 vehicles from bilibili to Station A, which is equivalent to 28-24 vehicles from Station A to bilibili ... After a few days, the number of vehicles at Station A is 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, after a few days, the number of vehicles at station A will be reduced to

(52+32) ÷ (2+ 1) = 28 (vehicle)

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 less than A, 4 times less than A, and C is 3 times more than A, 6. What are these three numbers?

solve

Numbers 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.

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5, the problem of differential time

meaning

It is known that the difference between two numbers and a large number is a multiple of a decimal (or a decimal is a fraction of a large number), so these two numbers are required to be different. This kind of application problem is called difference multiple problem.

magnitude relation

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

Smaller number × several times = larger number

Ideas and methods to solve problems

Simple topics use formulas directly, and complex topics use formulas after adaptation.

Example 1

The number of peach trees in the 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?

solve

(1) 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.

Example 2

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

solve

(1) Son's age = 27 ÷ (4- 1) = 9 (years).

(2) Dad's age = 9× 4 = 36 (years old)

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

Example 3

After the reform of management mode, the profit this month is 6.5438+200,000 yuan more than last month, which shows that the profit this month is 300,000 yuan more than last month. What's the profit in these two months?

solve

If last month's profit is taken as 1 times, then (30- 12) ten thousand yuan is equivalent to (2- 1) times of last month's profit, so

Last month's profit = (30-12) ÷ (2-1) =18 (ten thousand yuan)

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.

Example 4

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?

solve

Since 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.

Step 6 double the problem

meaning

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.

magnitude relation

Total ÷ One Quantity = Multiple

Another quantity × multiple = another total quantity

Ideas and methods to solve problems

Find the multiple first, and then use the multiple ratio relationship to find the required number.

Example 1

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

solve

(1) How many times is 3700 kg 100 kg? 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.

Example 2

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?

solve

(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.

Example 3

Fengxiang County has a bumper harvest of apples 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?

solve

(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)

A: The income of 800 mu orchards in the township is 2,222,200 yuan, and the income of10.6 million mu orchards in the county is 44.444 million yuan.

7. Meeting issues

meaning

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

magnitude relation

Meeting time = total distance ÷ (speed A+ speed b)

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

Ideas and methods to solve problems

Simple topics can be directly used in formulas, and complex topics can be modified before use.

Example 1

The waterway from Nanjing to Shanghai is 392 kilometers long. At the same time, ships from each port run relative to each other. Ships from Nanjing run 28 kilometers per hour, while ships from Shanghai run 2 1 kilometer per hour. How many hours passed before the two ships met?

solve

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?

solve

"Meeting for the second time" can be understood as two people running twice.

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

Meanwhile, Party A and Party B ride bicycles in opposite directions from two places, with Party A driving at a speed of15km/h and Party B driving at a speed of13km/h.. They met 3 kilometers from the midpoint to find the distance between the two places.

solve

"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.

8. Catch up with the problem

meaning

Two moving objects start in different places at the same time (or in the same place but not at the same time, or 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.

magnitude relation

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

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

Ideas and methods to solve problems

Simple topics use formulas directly, and complex topics use formulas after adaptation.

Example 1

A good horse walks 120km every day, 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?

solve

(1) How many kilometers can a bad horse 12 walk in a day? 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 200-meter 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?

solve

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

(500-200)÷[40×(500÷200)]

= 300 ÷ 100 = 3 (m)

The speed of Xiao Liang is 3 meters per second.

Example 3

Our People's Liberation Army pursued the fleeing enemy. At 16 pm, the enemy began to flee from A at the speed of 10 km per hour. At 22 o'clock in the evening, the PLA was ordered to pursue from B at a speed of 30 kilometers per hour. As we all know, the distance between A and B is 60 kilometers. How many hours can the PLA catch up with the enemy?

solve

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 = 1 1 (hours)

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

Example 4

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.

solve

This problem can be solved by turning the problem of meeting into the problem of pursuing. 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).

Column into a comprehensive formula (48+40) × [16× 2 ÷ (48-40)]

=88×4

= 352 km

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