Junior Middle School Olympic Mathematics: Analysis of Sum, Difference, Multiple Knowledge Points and Examples

# Junior High School Olympiad # Introduction Learning math and physics well is not afraid to travel all over the world, but there are still many students who are not good at math and need to practice more. !

Sum-difference problem

When it comes to "harmonious but different problems", people in the upper grades of primary school will say, "I will!" The calculation of sum and difference problem is too simple. Yes, knowing the sum and difference of two numbers, there is a calculation formula for finding two numbers:

Large number = (sum+difference) ÷2

Decimal = (sum and difference) ÷2

Can calculate, but also flexible use, turn some application problems into sum and difference problems to calculate.

Let's look at a few simple examples first.

For example 1, in Zhang Ming's final exam, the average score of Chinese and mathematics was 95, and mathematics was 8 points more than Chinese. What is Zhang Ming's score in these two courses?

Solution: 95 times 2 is the sum of the scores of mathematics and Chinese. We know that the difference between the scores of mathematics and Chinese is 8. Therefore,

Math score =(95×2+8)÷2=99.

Chinese score =(95×2-8)÷2=9 1.

Answer: Zhang Ming scored 99 points in mathematics and 9 1 point in Chinese.

Note: Chinese scores can also be calculated from 95×2-99=9 1

Example 2, there are three numbers A, B and C, A plus B equals 252, B plus C equals 197, and C plus A equals 149. Find these three numbers.

Solution: From B+C= 197 and A+C= 149, we know that the difference between B and A is 197- 149, and the title tells us that the sum of B and A is 252. Therefore,

b =(252+ 197- 149)÷2 = 150

A=252- 150= 102，

C= 149- 102=47。

A: The numbers A, B and C are 102, 150 and 47 respectively.

Note: There is an easier way.

(A+B)+(B+C)+(C+A)=2×(A+B+C)。

The above formula shows that adding three numbers and dividing by two is the sum of three numbers.

a+B+C =(252+ 197+ 149)÷2 = 299。 Therefore,

C=299-252=47，

B=299- 149= 150，

A=299- 197= 102。

Example 3: Basket A and Basket B contain 75 kilograms of apples. Take out 5 kilograms of apples from Basket A and put them in Basket B. There are 7 kilograms more apples in Basket A than in Basket B. How many kilograms are there in Basket A and Basket B respectively?

Solution: Draw a simple schematic diagram,

You can see that there are more apples in basket A than in basket B.

5+7+5= 17 (kg)

So the sum of A and B is 75 and the difference is 17.

The number of apples in the basket =(75+ 17)÷2=46 (kg).

The number of apples in basket B =75-46=29 (kg).

A: There are 46 kilograms of apples in basket A and 29 kilograms in basket B. 。

Example 4: Zhang Qiang bought a coat, a hat and a pair of shoes with 270 yuan. Coat is more expensive than shoes 140 yuan, coat and shoes are more expensive than hats 2 10 yuan. How much did Zhang Qiang spend on this pair of shoes?

Solution: Let's treat coats and shoes as one thing. The sum of coat and hat prices is 270 yuan, and the difference is 265,438+00 yuan.

The sum of coat and shoe prices =(270+2 10)÷2=240 (yuan).

The difference between the coat price and the shoe price is 140, so

The price of shoes =(240- 140)÷2=50 (RMB).

Answer: Buy this pair of shoes, 50 yuan.

Give three more complicated examples. If you can calculate like the following solution, you can say that you have been able to use the solution of sum and difference problem flexibly.

Example 5: Uncle Li has to go to work at 3 pm. It's probably time for him to go to work. He went to the house to look at the clock, which had already stopped 12: 08+00. He left home in a hurry and looked at the factory clock. It's still 10 minutes before going to work. At night, it is 165438+.

Solution: When I arrived at the factory, I saw that the clock was 2: 50, and when I left home, it was 12: 00. The difference was 2 hours and 40 minutes, which was caused by the time of stopping the clock and the time of walking on the road.

Time when the clock stops+time spent on the road = 160 (minutes).

After work at night, the clock in the factory is 1 1 o'clock, and it is 9 o'clock when I get home, with a difference of 2 hours. This is because the time when the clock stopped was partly offset by the time spent on the way home.

therefore

Time when the clock stops-time spent on the road = 120 (minutes).

Now the problem has been transformed into a standard sum-difference problem.

Clock stop time = (160+120) ÷ 2 =140 (minutes).

Time spent on the road = 160- 140=20 (minutes).

Uncle Li's clock stopped for 2 hours and 20 minutes.

There is another scheme that can quickly calculate the time spent by Uncle Li on the road:

According to the clock of Uncle Li's house, he went out 12, got home at 9: 00 pm, and spent 8 hours and 50 minutes outside, including going to work for 8 hours, waiting for work 10 minutes, and the rest of the time was the time for him to go back and forth to work, so,

Time spent on the way to work =(8 hours and 50 minutes -8 hours-10 minutes) ÷2=20 minutes.

Clock stop time =2 hours 40 minutes -20 minutes

=2 hours and 20 minutes.

Example 6: Xiaoming bought two kinds of greeting cards with 2 1.4 yuan, one is A card 1.5 yuan, and the other is B card 0.7 yuan, and the money just ran out. However, the salesman counted the number of A as the number of B, and the number of B as the number of A, and asked Xiao Ming to return to 3.2 yuan.

Solution: The price difference between A card and B card is 1.5-0.7=0.8 (yuan), and the salesman returned it to Xiaoming 3.2 yuan by mistake, knowing that A card bought by Xiaoming is 3.2÷0.8=4 (sheets) more than B card.

Now there is a difference between the two cards. As long as we find the sum of the two cards, the problem will be solved. How to find it? attention please

1.5×A +0.7×B =2 1.4.

1.5× B card number +0.7× A card number =2 1.4-3.2.

As can be seen from the above two formulas, the sum of the two card numbers is

[21.4+(21.4-3.2)] ⊙ (1.5+0.7) =18 (sheets)

Therefore, the number of cards is

(18+4) ÷ 2 =11(Zhang).

The quantity of B cards is 18- 1 1=7 (cards).

A: Xiao Ming bought 1 1 A card and 7 B cards.

Note: this problem can also be solved by the method of chicken and rabbit in the same cage. Please look at the next lecture.

Example 7: There are two rectangles with the same size, which are combined into two large rectangles, as shown on the right. The circumference of the big rectangle (a) is 240 cm, and that of the big rectangle (b) is 258 cm. How many centimeters is the length and width of the original rectangle?

Solution: The perimeter of the big rectangle (A) is the original rectangle.

Length ×2+ width × 4.

The perimeter of the big rectangle (B) is the original rectangle.

Length ×4+ width × 2.

So 240+258 is the original rectangle.

Length× 6+width× 6.

The sum of the length and width of the original rectangle is

(240+258)÷6=83 (cm).

The difference between the length and width of the original rectangle is

(258-240)÷2=9 (cm).

Therefore, the length and width of the original rectangle are

Length: (83+9)÷2=46 (cm).

Width: (83-9)÷2=37 (cm).

A: The original rectangle is 46 cm long and 37 cm wide.

Multiple problems

When the sum or difference of two numbers is known and the multiple relation of two numbers is known, these two numbers can be solved immediately. The common "age problem" in elementary school arithmetic is typical of this kind of problem. Let's look at a few basic examples first.

Example 1, there are two piles of chess pieces, the first pile has 87 pieces, and the second pile has 69 pieces. Then how many pieces are taken from the first pile to the second pile can make the number of pieces in the second pile three times that of the first pile.

Solution: Two piles of chess pieces have 87+69= 156 pieces.

In order to make the number of pieces in the second pile three times that in the first pile, it is necessary to divide 156 pieces into 1+3=4 pieces, that is, each piece has a piece.

156( 1+3)= 39 (pieces).

Keep thirty-nine dollars for the first pile and take the rest to the second pile. So the number of pieces from the first pile to the second pile is

87-39=48 (pieces).

Answer: You have to take 48 yuan from the first pile to the second pile.

Example 2: There are two bookshelves with 173 books. After taking 38 books from the first floor, there are still 6 books on the second floor, twice as many as the first floor. How many books are there on the second floor?

Solution: We draw the following schematic diagram:

We count the remaining books on the first floor (after taking 38 books) as 1 book, so there are 2 books on the second floor, 6 more. And then get rid of these six books, that is,

173-38-6= 129 (Ben)

Exactly three copies, each one.

129÷3=43 (Ben).

So, the books on the second floor are all * * *

43×2+6=92 (Ben).

A: There are 92 books on the second floor of the bookshelf.

Note: Let's set "1 copy" first, so that the calculation has a very convenient calculation unit. This is a common method to solve application problems, especially for multiple problems. The number of copies shown on the schematic diagram is more obvious.

Example 3: There are 975 students in a primary school. The number of boys in this school is four times less than that in grade six, with 23 students, and the number of girls in this school is more than three times that in grade six. 1 1 person. How many boys and girls are there in the school?

Solution: Let the number of sixth grade students be "1".

There are 4 -23 boys.

The number of girls is 3+1 1.

There are 7 students in the whole school-(23-11).

Each copy is (975+12) ÷ 7 =141(person).

Number of boys =141× 4-23 = 541(person).

Number of girls =975-54 1=434 (persons).

A: 54 1 for boys and 434 for girls.

Example 2 and Example 3 are the same type of problems, but they are slightly different. Please think about it. Where is the "difference"?

70 pairs of leather shoes. At this time, the number of leather shoes is just twice that of travel shoes. How many pairs of shoes are there?

Solution: In order to facilitate the calculation, the original sneakers were counted as 4 copies, and sold 1 copy, leaving 3 copies. Then the original leather shoes with 70 pairs are 3×2=6 (copies). 400+70 will be 3+ 1+6= 10 (copies). Every copy is

(400+70)÷ 10=47 (double)。

Original sneakers 47×4= 188 (double).

Original single leather shoes 47×6-70=2 12 (double).

Answer: Travel shoes 188 pairs, leather shoes 2 12 pairs.

Set an integer number of copies to make the calculation simple and convenient. Elementary school arithmetic, decimals and fractions should be rounded off as much as possible to make thinking and calculation easier. So "try to round as much as possible" will run through the later chapters.

The following example will be the main content of this section-the age problem.

Age is a common problem in elementary school arithmetic, and this kind of problem often has a "multiple" condition. The key point to solve the age problem is that the age difference between two people remains unchanged.

Example 4: Father is 50 years old and daughter 14 years old. How many years ago, my father was five times as old as my daughter?

Solution: The difference between father and daughter is 36 years old and remains unchanged. A few years ago, I was 36 years old. When the father's age is exactly five times that of his daughter, the father is still 36 years older than his daughter. This 36-year-old is (5- 1) times that of her daughter.

36÷(5- 1)=9.

My daughter was 9 years old at that time, 14-9=5, that is, five years ago.

Five years ago, my father was five times as old as my daughter.

Example 5: There are two pools, a big pool and a small pool. There are 300 cubic meters of water in the big pool and 70 cubic meters in the small pool. Now, after the two pools are injected with the same amount of water, the water in the big pool is three times that in the small pool. Ask how many cubic meters of water is injected into each pool.

Solution: Draw the following schematic diagram:

According to our statistics, the amount of water injected into the small pool is 1, and the amount of water injected into the large pool is 3. As can be seen from the figure, the water quantity injected into the two pools is equal, so the water quantity of the large pool (300-70) is 2.

So every copy is

(300-70)÷2= 1 15 (m3).

The amount of water injected is

1 15-70=45 (m3)?

A: Each pool should be filled with 45 cubic meters of water.

Example 5 is exactly the same as the age problem. "Water injection" is equivalent to "several years later" in the age issue.

Example 6: The ages of the two brothers add up to 55 this year. One year, my brother was the same age as my brother this year. At that time, my brother was just twice as old as my brother. How old is my brother this year?

Solution: When the elder brother's age is exactly twice that of the younger brother, we assume that the younger brother's age is 1 and the elder brother's age is 2, so the age difference between the elder brother and the younger brother is 1. The age difference between them will not change, and this year their age difference is still 1.

The title also tells us that my brother's age at that time was the same as my brother's age this year, so my brother's age this year is also 2 copies, and my brother's age this year should be 2+ 1=3 (copies).

This year, the sum of the ages of the two brothers is

3+2=5 (copies)

Each serving is 55÷5= 1 1 (years old).

My brother's age this year is 1 1×3=33 (years old).

A: My brother is 33 years old.

As the last example in this section, we will change the age slightly.

Example 7: Father is 38 years old, mother is 36 years old, and son is 1 1 year old.

How many years later, the total age of parents is four times that of sons?

Solution: Now the sum of parents' ages is

38+36=74.

Now four times the age of the son is 1 1×4=44. The difference is.

74-44=30.

Considering four times, the annual increase will be 1×4=4, and the sum of parents' ages will be 1+ 1=2.

In order to catch up with the gap of 30, it is necessary.

30÷(4-2)= 15 (year)?

A: After 15, the sum of parents' ages is four times that of their sons.

Please use the problem-solving idea of example 6 to solve problem 7 of exercise 2. Maybe you can master this problem-solving skill completely.

Ask the reader to think about it. Is the solution in Example 7 different from that in Example 5? What are their characteristics?

We can also use the solution of case 15 to solve case 12. The specific method is as follows:

(14×5-50)÷(5- 1)=5 (year).

However, it should be noted that14x5 is more than 50, so it was five years ago.

The problem of surplus and shortage

Nine Chapters Arithmetic is the most colorful book in ancient China. In its seventh chapter, it discusses a kind of surplus and deficiency, the first of which, described in modern language, is the following example.

Example 1, some people buy some things together, and everyone goes out of 8 yuan, which will lead to more 3 yuan; If everyone pays 7 yuan, there will be 4 yuan less. So how many people are there? What is the price?

Solution: There is a difference between "more 3 yuan" and "less 4 yuan"

3+4=7 (yuan).

Everyone needs 8-7= 1 (yuan) more.

So we know that * * * has 7÷ 1=7 (person), and the price is

8×7-3=53 (yuan).

Answer: * * * Seven people bought it together at the price of 53 yuan.

The above 3+4 can be said to be the difference between two sums, while 8-7 is the difference between each copy. The calculation formula is

Total difference ÷ Difference per copy = number of copies.

There are many changes in the content of this kind of problem, forming a kind of problem, which we generally call "surplus shortage" Please look at more examples.

Example 2: Give the child a bag of candy, and each person will give 10 tablets, which is just the end. If everyone gets 16 tablets, three children won't get a candy. How many pills are there in this bag?

Scheme 1: Three children could have been given each 10 tablet, and some * * * had 10×3=30 tablets. If given to other children, each child can increase 16- 10=6 (tablets), so other children have it.

10× 3÷ (16-10) = 5 (person).

Together with these three children, * * * has children 5+3=8 (people). This bag of candy has been.

10×(5+3)=80 (grains).

Scheme 2: If you add 16×3 sweets, everyone can add (1- 10), so * * * will have children.

16× 3÷ (16-10) = 8 (person)?

There are 80 sweets in this bag.

There are 80 sweets in this bag.

Where 16×3 is the total difference, (16- 10) is the difference of each share, and 8 is the number of shares.

Example 3: A class of students went boating. They calculated that if one more boat was added, each boat could only hold six people. If one boat is reduced, each boat can only take nine people. How many students are there in this class?

Solution: If there are six people in each boat, we need to add another boat, that is, there are six people who have no boat to sit on now; If each boat takes nine people, you can reduce one boat, that is, take nine more people. The difference between the number of people who can take a boat is 6+9= 15 (person).

This is because each boat has (9-6) more people, so * * * has a boat.

(6+9)÷(9-6)=5 (articles)?

There are 6×5+6=36 students in this class.

There are 36 students in this class.

Example 4: Xiaoming goes to school from home. If he walks 80 meters every minute, he can get to school six minutes before class. If he walks 50 meters every minute, he will be three minutes late. How far is Xiaoming's home from school?

Scheme 1: based on the time from home to class, compare the distance between two different speeds and the distance from home to school: if you walk 80 meters per minute, you can walk 80×6 meters more; If you walk 50 meters per minute, you will walk 50×3 meters less. Please refer to the following schematic diagram:

Therefore, we can find that Xiao Ming's time from home to the classroom is

(80×6+50×3)÷(80-50)=2 1 (minutes).

The distance from home to school is

800×(2 1-6)= 1200 (m)？

Or 50×(2 1+3)= 1200 (m).

The distance from Xiaoming's home to school is1200m.

Option 2: Take the time required to walk home to school at 80 meters per minute as the starting point of thinking.

At a speed of 50 meters per minute, you need to use 6+3=9 (minutes) more. This 9-minute walk of 50×9 (meters) just makes up for the lack of walking ahead. Therefore, the time required for 80 meters per minute is

50×(6+3)÷(80-50)= 15 (minutes)?

Look at two more complicated examples.

Example 5: Some oranges are distributed to several people, and each person has five oranges, which is too much 10. If the number of people is tripled and there are still five people missing, then everyone is still missing eight oranges. How many oranges are there?

Solution: What makes people feel difficult is that the conditions are "three times as high as five people". We must change this situation first.

Suppose there are 10 oranges, and 10 = 2×5, there can be 5 more people. Leaving aside the condition of "five fewer" for the time being, we only consider three times the number of people, which is equivalent to giving each person 2×3=6 (one) according to the original number.

Five for each person and six for each person. The total number is different.

10+ 10+8=28 (pieces).

So the original number of people is 28÷(6-5)=28 (people).

The total number of oranges is 5×28+ 10= 150.

A: There are 150 oranges.

Example 6: There are some apples and pears. If you pile 2 pears per 1 apple, there will be 5 apples left when you divide the pears. If you pile 5 pears for every 3 apples, there will be 5 pears left after the apples are divided. How many apples and pears are there?

Scheme 1: We assume that there are 10 pears, plus the remaining 5 apples, which are divided according to the previous "1 apple, 2 pears". According to the later "3 apples and 5 pears", the total number of apples can be divisible by 3. So the former can be divided into a large pile, and every three piles can be combined.

There are three apples in each pile, but there are 1 pear (6-5= 1). The total number of pears is different.

Imagine adding 10+ remaining 5 = 15.

( 10+5)÷(6-5)= 15.

It is known that there are 15 piles, and the total number of apples is

15×3=45 (pieces).

The total number of pears is (45-5)×2=80.

There are 45 apples and 80 pears.

Solution 2: Use graphic method.

The former is divided into piles, and 1 means two pears and five apples.

In the latter stacking, as long as three apples are added, a pile can be formed with the remaining five pears. Pears count as five, and apples are exactly three.

Comparing the above and below figures, we can see that 5+3=8 (1) is the "half" in the following figure, that is, 1 is 16. Pear is 5, * * you 16×5=80 (one). Apple has 16.