MeaningWhen solving a problem, you first find out how much one part is (i.e., a single quantity), and then use the single quantity as a criterion to find the required quantity. These types of application problems are called normalization problems.
Quantity RelationshipTotal ÷ number of copies = number of copies
Quantity of 1 x number of copies taken = number of copies requested
Another total ÷ (total ÷ number of copies) = number of copies requested
Solution Ideas and MethodsFirst, a single quantity is found, and then a single quantity is used as a criterion for finding the requested quantity.
Example 1 It costs $0.60 to buy 5 pencils, how much does it cost to buy 16 of the same pencils?
Solution (1) How much does it cost to buy 1 pencil?0.6 ÷ 5 = 0.12 ($)
(2) How much does it cost to buy 16 pencils?0.12 × 16 = 1.92 ($)
To make a composite equation 0.6 ÷ 5 × 16 = 0.12 × 16 = 1.92 ($)
Answer: it costs $1.92.
Example 23 tractors plow 90 hectares of land in 3 days, according to this calculation, how many hectares of land will be plowed by 5 tractors in 6 days?
Solution (1) How many hectares are plowed by 1 tractor in 1 day? 90 ÷ 3 ÷ 3 = 10 (hectares)
(2) How many hectares are plowed by 5 tractors in 6 days? 10 × 5 × 6 = 300 (hectares)
Listed as a composite equation 90 ÷ 3 ÷ 3 × 5 × 6 = 10 × 30 = 300 (hectares)
Answer: 5 tractors in 6 days Plow 300 hectares.
Example 35 cars can carry 100 tons of steel in 4 trips, if the same 7 cars are used to carry 105 tons of steel, how many trips are needed?
Solution (1) How many tons of steel can be transported by 1 car in 1 trip? 100÷5÷4 = 5 (tons)
(2) How many tons of steel can be transported by 7 cars in 1 trip? 5×7 = 35 (tons)
(3) How many times do you need to transport 105 tons of steel in 7 cars? 105÷35 = 3 (times)
Conclude in the combined equation 105 ÷ (100 ÷ 5) ÷ 4 × 7) = 35 (tons). ÷4×7) = 3 (times)
Answer: 3 shipments are needed.
2 summarize the problem
meaning of solving the problem, often first find the "total number", and then according to the other conditions of the problem, called the totalization of the problem. By "total quantity" we mean the total price of the goods, the total amount of work done in a few hours (days), the total amount of production on a few acres of land, the total distance traveled in a few hours, and so on.
Quantity Relationship 1 Quantity x Number of Copies = Total
Total ÷ 1 Quantity = Number of Copies
Total ÷ Another Number of Copies = Another Number of Copies
Solution Ideas and MethodsFirst, find the total number of copies, and then find the number of copies required according to the meaning of the question.
Example 1 A clothing factory originally used 3.2 meters of cloth to make one set of clothes, and after improving the cutting method, it used 2.8 meters of cloth for each set of clothes. How many sets of cloth, which originally made 791 sets of clothes, can now be made?
Solution (1) the total **** how many meters of this batch of cloth? 3.2 × 791 = 2531.2 (m)
(2) how many sets of cloth can be made now? 2531.2 ÷ 2.8 = 904 (sets)
Listed as a composite equation 3.2 × 791 ÷ 2.8 = 904 (sets)
Answer: 904 sets of cloth can be made now.
Example 2 Xiaohua reads 24 pages a day and finishes the book Red Rock in 12 days. How many days can Xiao Ming read 36 pages a day and finish the book "Red Rock"?
Solve (1) How many total **** pages in the book "Red Rocks"? 24 × 12 = 288 (pages)
(2) How many days can Xiao Ming finish reading "Red Rocks"? 288 ÷ 36 = 8 (days)
Let's put it into a comprehensive equation 24 × 12 ÷ 36 = 8 (days)
Answer: Xiao Ming can finish reading "Red Rocks" in 8 days.
Example 3 The cafeteria brought in a batch of vegetables, the original plan was to eat 50 kilograms per day, 30 days slowly consume the vegetables. Later, according to everyone's opinion, 10 kilograms per day more than the original plan to eat, this batch of vegetables can be eaten for how many days?
Solution (1) the batch of vegetables *** how many kilograms? 50 × 30 = 1500 (kg)
(2) the batch of vegetables can be eaten for how many days? 1500 ÷ (50 + 10) = 25 (days)
Listed into the synthesis equation 50 × 30 ÷ (50 + 10) = 1500 ÷ 60 = 25 (days)
Answer: the batch of vegetables can be eaten for 25 days.
3Sum and Difference Problems
The meaning is that if you know the sum and difference of two quantities, and you want to find out how much each of the two quantities is, this type of application problem is called a sum and difference problem.
Quantity Relationship Large number = (sum + difference) ÷ 2
Small number = (sum - difference) ÷ 2
Solution of the ideas and methods of simple topics can be applied directly to the formula; complex topics and then use the formula after the workaround.
Example 1 Class A and B **** there are 98 students, class A is 6 more than class B. How many students are there in each of the two classes?
Solution Number of students in class A = (98 + 6) ÷ 2 = 52 (students)
Number of students in class B = (98 - 6) ÷ 2 = 46 (students)
Answer 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 centimeters and the length is 2 centimeters more than the width, find the area of the rectangle.
Solution Length = (18 + 2) ÷ 2 = 10 (centimeters)
Width = (18 - 2) ÷ 2 = 8 (centimeters)
Area of the rectangle = 10 x 8 = 80 (square centimeters)
Answer.
Example 3 There are three bags of fertilizers A, B and C. Two bags of A and B **** weigh 32 kg, two bags of B and C **** weigh 30 kg, and two bags of A and C **** weigh 22 kg, find how many kg each of the three bags of fertilizers weigh.
Solution Two bags of A and B and two bags of B and C contain B. From this, it can be seen that A is more than C by (32-30) = 2 kg and that A is a large number and C is a small number. From this it can be seen
A bag of fertilizer weight = (22 + 2) ÷ 2 = 12 (kg)
C bag of fertilizer weight = (22-2) ÷ 2 = 10 (kg)
B bag of fertilizer weight = 32-12 = 20 (kg)
Answer: A bag of fertilizer weight 12 kg B bag of fertilizer weighs 20 kg and C bag of fertilizer weighs 10 kg.
Example 4 A and B two cars originally **** loaded 97 baskets of apples, 14 baskets from the A car to the B car, the results of the A car more than the B car also 3 baskets, the two cars were loaded with how many baskets of apples?
Solution "14 baskets from car A to car B, the results of car A than car B also 3 baskets", which means that car A is a large number, car B is a small number, A and B is the difference between (14 × 2 + 3), A and B and 97, so the number of baskets of car A = (97 + 14 × 2 + 3) ÷ 2 = 64 (baskets)
The original car was loaded with 97 baskets of apples. p>The number of baskets in car B = 97 - 64 = 33 (baskets)
Answer: car A was loaded with 64 baskets of apples and car B was loaded with 33 baskets of apples.
4 and times the problem
Meaning the sum of two numbers is known and the larger number is a few times the smaller number (or the smaller number is a few times the larger number), the requirement of how much each of these two numbers, this type of application problem is called and times the problem.
Quantity Relationship Sum ÷ (times + 1) = smaller number
Sum - smaller number = larger number
Smaller number × times = larger number
Solution of ideas and methods for simple topics directly using the formula, and complex topics after the use of the formula.
Example 1 in the orchard there are apricot trees and peach trees **** 248 trees, the number of peach trees is three times the number of apricot trees, how many apricot trees, peach trees each?
Solution (1) How many apricot trees are there? 248 ÷ (3 + 1) = 62 (trees)
(2) How many peach trees are there? 62 × 3 = 186 (trees)
Answer: there are 62 apricot trees and 186 peach trees.
Example 2 Two warehouses in the east and west *** stock 480 tons of grain, the number of grain in the east stock is 1.4 times the number of grain in the west stock, how many tons of grain are stored in each of the two warehouses?
Solution (1) West stock = 480 ÷ (1.4 + 1) = 200 (tons)
(2) East stock = 480-200 = 280 (tons)
Answer: 280 tons of grain in the east stock and 200 tons in the west stock.
Example 3 Station A has 52 vehicles and Station B has 32 vehicles, if 28 vehicles are driven from Station A to Station B and 24 vehicles are driven from Station B to Station A every day, how many days later will the number of vehicles in Station B be twice as many as in Station A?
SolutionThere are 28 vehicles traveling from station A to station B and 24 vehicles traveling from station B to station A every day, which is equivalent to (28-24) vehicles traveling from station A to station B every day. A few days later, the number of vehicles in station A as 1 times the amount, the number of vehicles in station B is 2 times the amount, the total number of vehicles in the two stations (52 + 32) is equivalent to (2 + 1) times,
So, a few days later, the number of vehicles in station A is reduced to
(52 + 32) ÷ (2 + 1) = 28 (vehicles)
The required number of days for (52) -28) ÷ (28-24) = 6 (days)
Answer: after 6 days the number of vehicles at station B is twice as many as at station A.
Example 4 The sum of the three numbers A, B and C is 170, B is 4 less than 2 times A and C is 6 more than 3 times A. What is the sum of each of the three numbers?
SolutionBoth the B and C numbers are directly related to the A number, so take the A number as a 1x quantity.
Because B is 4 less than 2 times the number of A, add 4 to B, and the number B becomes 2 times the number of A.
And because C is 6 more than 3 times the number of A, subtracting 6 from the number of C makes the number of A 3 times the number of A.
This makes (170 + 4 - 6) equal to (1 + 2 + 3) times the number. Then,
Number A = (170 + 4-6) ÷ (1 + 2 + 3) = 28
Number B = 28 × 2 - 4 = 52
Number C = 28 × 3 + 6 = 90
Answer: number A is 28, number B is 52, and number C is 90.
5 Difference and Multiplication Problems
Meaning the difference between two numbers and how many times the greater number is the lesser number (or how many times the lesser number is the greater number) is known, and how much each of the two numbers is required, this type of application problem is called difference and multiplication problem.
Quantity relationship between the difference between two numbers ÷ (times - 1) = smaller number
Smaller number × times = larger number
Solution ideas and methods of simple topics directly using the formula, the complexity of the topic variations on the use of the formula.
Example 1 There are three times as many peach trees as apricot trees in the orchard, and there are 124 more peach trees than apricot trees. How many apricot trees and peach trees are there?
Solution (1) How many apricot trees are there? 124 ÷ (3-1) = 62 (trees)
(2) How many peach trees are there? 62 × 3 = 186 (trees)
Answer: there are 62 apricot trees and 186 peach trees in the orchard.
Example 2 A father is 27 years older than his son, and this year, the father's age is 4 times his son's age, so how old is each of the father and son this year?
Solution (1) Son's age = 27 ÷ (4-1) = 9 (years)
(2) Dad's age = 9 × 4 = 36 (years)
Answer: The ages of father and son this year are 36 and 9 respectively.
Example 3 After the mall reformed its business management practices, this month's profit was 120,000 yuan more than twice last month's profit, and it is known that this month's profit was 300,000 yuan more than last month's profit, so how many millions of yuan is the profit of each of these two months?
Solution If last month's profit is taken as a multiple of 1, then (30-12) million yuan is equivalent to (2-1) times last month's profit, so
Previous month's profit = (30-12) ÷ (2-1) = 18 (million yuan) <
Profit for the month = 18 + 30 = 480,000 yuan
Answer: Last month's profit was 180,000 yuan and this month's profit is 480,000 yuan.
Example 4 There are 94 tons of wheat and 138 tons of corn in the granary, if 9 tons of wheat and 9 tons of corn are shipped out each day, ask how many days later there will be 3 times as much corn left as wheat?
SolutionSince equal amounts of wheat and corn are shipped out each day, the difference in the amounts left is equal to the difference in the original amounts (138-94). Considering the wheat left after a few days as 1x the amount, the corn left after a few days is 3x the amount, so (138-94) is equal to (3-1) times the original amount,
The amount of wheat left = (138-94) ÷ (3-1 ) = 22 (tons)
Number of wheat shipped out = 94-22 = 72 (tons)
Number of days to ship grain = 72 ÷ 9 = 8 (days)
Answer: 3 times as much corn as wheat will be left after 8 days.
6 Times Ratio Problems
Meaning that there are two known quantities of the same kind, one of which is a number of times the other, and when solving the problem you first find the multiple, and then you use the method of times ratio to figure out the required number, this type of application problem is called a times ratio problem.
Quantity RelationshipTotal ÷ One Quantity = Multiple
Another Quantity × Multiple = Another Total
Solution Ideas and MethodsFirst, find out the multiple, and then use the multiplicity relationship to find out the required number.
Example 1100 kg of rapeseed can extract 40 kg of oil, now there are 3700 kg of rapeseed, how much oil can be extracted?
Solution (1) 3700 kg is 100 kg of how many times? 3700 ÷ 100 = 37 (times)
(2) how many kilograms of oil can be extracted? 40 × 37 = 1480 (kg)
Listed into the synthesis equation 40 × (3700 ÷ 100) = 1480 (kg)
Answer: you can extract 1480 kg of oil.
Example 2 this year, the day of Arbor Day, an elementary school 300 teachers and students **** planted 400 trees, according to this calculation, the county 48000 teachers and students **** planted how many trees?
Solution (1) 48000 is 300 how many times? 48000 ÷ 300 = 160 (times)
(2) *** planted how many trees? 400 × 160 = 64000 (trees)
Listed into a comprehensive equation 400 × (48000 ÷ 300) = 64000 (trees)
Answer: countywide 48000 teachers and students*** planted 64000 trees.
Example 3 Fengxiang County this year's bumper crop of apples, Tianjiazhuang a family of 4 acres of orchards income of 11,111 yuan, according to this calculation, the whole township of 800 acres of orchards*** income of how many yuan? The county's 16,000 acres of orchards **** income of how many yuan?
Solution (1) 800 acres is 4 acres of how many times? 800 ÷ 4 = 200 (times)
(2) 800 acres of income how much yuan? 11111 × 200 = 2222200 (yuan)
(3) 16,000 acres of 800 acres of how many times? 16,000 ÷ 800 = 20 (times)
(4) 16,000 mu How many yuan of income?2222200 × 20 = 44444000 (yuan)
Answer: 800 acres of orchards in the township **** income of 2222200 yuan,
16000 acres of orchards in the county **** income of 44444000 yuan.
7 Meet Problems
MeaningTwo moving objects start from two places at the same time traveling in opposite directions and meet each other on the way. These types of application problems are called encounter problems.
Quantity Relationship Encounter Time = Total Distance ÷ (A Speed + B Speed)
Total Distance = (A Speed + B Speed) x Encounter Time
Solution Ideas and MethodsSimple topics can be directly utilized by using the formula, and complicated topics can be adapted and then utilized by using the formula.
Example 1 Nanjing to Shanghai waterway is 392 kilometers long, at the same time from the two ports of a ship relative to each other, the ship from Nanjing 28 kilometers per hour, the ship from Shanghai 21 kilometers per hour, after a few hours of the two ships meet?
Solution 392 ÷ (28 + 21) = 8 (hours)
Answer: After 8 hours the two ships meet.
Example 2 Li and Liu run on a circular track with a circumference of 400 meters, Li runs 5 meters per second, Liu runs 3 meters per second, they start from the same place at the same time, and run in the opposite direction, so how long does it take for the two to meet for the second time from the start?
The "second encounter" can be interpreted as two laps.
So the total distance traveled is 400×2
Meeting time = (400×2)÷(5+3) = 100 (seconds)
Answer: It takes 100 seconds for them to meet each other for the second time.
Example 3 A and B ride bicycles in opposite directions at the same time from two places, A travels 15 kilometers per hour, B travels 13 kilometers per hour, and the two people meet at 3 kilometers from the midpoint, find the distance between the two places.
The solution of "two people meet at 3 km from the midpoint" is the key to the correct understanding of the question. From the question, we can see that A rode fast, B rode slow, A 3 kilometers past the midpoint, B 3 kilometers from the midpoint, that is to say, A more than B traveled (3 × 2) kilometers, therefore,
Meeting time = (3 × 2) ÷ (15-13) = 3 (hours)
Distance between the two places = (15 + 13) × 3 = 84 (kilometers)
Answer: the distance between the two places is 84 km.