I. Basic problems

"Chickens and rabbits in the same cage" is a famous arithmetic problem in ancient China. It first appeared in the Art of War by Sun Tzu. Many elementary school arithmetic application problems can be transformed into such problems or solved by a typical solution-hypothesis method. So we should learn its solutions and ideas.

1 How many chickens and rabbits are there? They have 88 heads and 244 feet. How many chickens and rabbits are there?

Solution: We imagine that every chicken is "golden rooster independent" and stands on one foot; And every rabbit uses two hind legs and stands on two feet like a human. Now, half of the total number of feet appear on the ground, that is,

244÷2= 122 (only).

In the number 122, the number of chickens is counted once and the number of rabbits is counted twice. So 122 subtracts the total of 88, and the rest is the number of rabbits.

122-88=34,

There are 34 rabbits. Of course there are 54 chickens.

Answer: There are 34 rabbits and 54 chickens.

The above calculation can be summarized as the following formula:

Total number of feet ÷2- total number of heads = number of rabbits.

The above solution is recorded in Sun Tzu's Art of War. Do a division and subtraction, and you can get the number of rabbits immediately. How simple! This calculation is mainly based on the fact that the number of feet of rabbits and chickens is 4 and 2 respectively, and 4 is twice that of 2. But when other problems are transformed into such problems, the "number of feet" is not necessarily 4 and 2, and the above calculation method will not work. Therefore, we give a general method to solve this kind of problem.

Also say example 1.

If you imagine that 88 rabbits are all rabbits, it is 4×88 feet, which is more than 244 feet.

88×4-244= 108 (only).

Each chicken has (4-2) fewer feet than a rabbit, so * * * has chickens.

(88×4-244)÷(4-2)= 54 (only).

It shows that among the 88 "rabbits" we imagined, 54 are not rabbits, but chickens, so we can list the formulas.

Number of chickens = (number of rabbit feet × total number of heads-total number of feet) ÷ (number of rabbit feet-number of chicken feet).

Of course, we can also imagine that 88 is a "chicken", so * * * has feet 2×88= 176 (only), less than 244 feet.

244- 176=68 (only).

Each chicken has 4-2 feet less than each rabbit.

68÷2=34 (only).

Explain that 34 of the imaginary "chicken" is only a rabbit, and you can also list formulas.

Number of rabbits = (total number of feet-chicken feet × total number of heads) ÷ (number of rabbit feet-chicken feet).

You don't have to use the above two formulas at the same time. Use one to calculate the number of rabbits or chickens, and then subtract the total to get another number.

Assuming that all chickens or rabbits are usually solved in this way, this method is called "hypothesis method".

Now, try the above formula with a specific question.

Example 2 red pencil 0. 19 yuan, blue pencil 0. 1 1 yuan, two kinds of pencils 2.80 yuan. How many red pencils and blue pencils do you buy?

Solution: take "minutes" as the monetary unit. Let's assume that a "chicken" has 1 1 foot and a "rabbit" has 19 foot. They * * * have 16 heads, 280 feet.

Now the problem of buying pencils has turned into a problem of "chickens and rabbits in the same cage". Using the above formula to calculate the number of rabbits, there are

Number of blue pens = (19×16-280) ÷ (19-1)

=24÷8

=3 (branch).

Red pen number = 16-3= 13 (branch).

I bought 13 red pencils and 3 blue pencils.

For the calculation of this kind of problem, we can often make use of the particularity of the known number of feet. The sum of "feet" in Example 2 is 30. We can also imagine that in 16, 8 is just a rabbit and 8 is just a chicken. According to this assumption, the number of feet is 30.

8×( 1 1+ 19)=240.

Forty is less than 280.

40÷( 19- 1 1)=5.

We know that there should be five fewer "chickens" among the eight envisaged, that is, the number of "chickens" (blue pencils) is three.

30×8 is easier to calculate than 19× 16 or1×16. Using the particularity of the known number, the calculation is completed by mental arithmetic.

In fact, you can imagine a convenient number of rabbits or chickens at will. For example, if the number of rabbits is 16 and the number of chickens is 6, there are feet.

19× 10+ 1 1×6=256.

24 is less than 280.

24÷( 19- 1 1)=3,

Imagine six chickens, and you need three less.

Making imaginary numbers easy to calculate often depends on your mental arithmetic ability.

Here are four more difficult examples.

Example 3 For a manuscript, it takes 6 hours for A to type alone, and 65,438+00 hours for B to type alone. Now, A played alone for several hours, and B spent seven hours playing because of something. How many hours of typing?

Solution: We divide this manuscript into 30 copies on average (30 is the least common multiple of 6 and 10), a typing 30÷6= 5 copies per hour, and b typing 30÷ 10= 3 copies per hour.

Now, if the typing time of A is regarded as the number of "rabbits" and the typing time of B is regarded as the number of "chickens", then the total number of heads is 7. The number of feet of the rabbit is 5, the number of feet of the chicken is 3, and the total number of feet is 30, so the problem becomes the problem of "the chicken and the rabbit are in the same cage".

According to the previous formula

"Rabbit" number =(30-3×7)÷(5-3)

=4.5,

Number of chickens =7-4.5.

=2.5,

That is, it took 4.5 hours for A to type and 2.5 hours for B to type.

A: It took 4 hours and 30 minutes to type.

This year is 1998, with parents' age (integer) 78 and brothers' age 17. Four years later (2002), my father was four times as old as my brother, and my mother was three times as old as my brother. So when the father is three times as old as his brother, what year is it?

Solution: After 4 years, the total age of the two will be added by 8. At this time, the total age of brothers is 17+8=25, and the total age of parents is 78+8=86. We can regard the age of brothers as the number of chickens and the age of brothers as the number of rabbits. 25 is the total number of heads and 86 is the total number of feet.

(25×4-86)÷(4-3)= 14 (years old).

1998, brother's age is?

14-4= 10 (years old).

Father's age is

(25- 14)×4-4=40 (year).

Therefore, when the father's age is three times the brother's age, the brother's age is

(40-10) ÷ (3-1) =15 (years old).

This is 2003.

A: In 2003, my father was three times older than my brother.

A spider has 8 legs, a dragonfly has 6 legs and 2 pairs of wings, and a cicada has 6 legs and 1 pairs of wings. Now these three kinds of bugs have 18, 1 18 legs and 20 pairs of wings. How many bugs are there in each species?

Solution: Because both dragonflies and cicadas have six legs, they can be divided into "eight legs" and "six legs" according to the number of legs, and eight legs can be calculated by formula.

Number of spiders = (118-6×18) ÷ (8-6)

=5 (only).

So you know that a six-legged bug * * *

18-5= 13 (only).

In other words, there are 13 dragonflies and cicadas, and they have 20 pairs of wings. Use the formula again.

Cicada number = (13× 2-20) ÷ (2-1) = 6 (only).

So the number of dragonflies is 13-6=7 (only).

There are five spiders, seven dragonflies and six cicadas.

Example 6 In a math exam, 52 students in the class took part in 5 questions, and * * got it right 18 1 question. It is known that everyone answered at least 1 question correctly, 7 people answered 1 question correctly, 6 people answered 5 questions correctly, and the same number of people answered 2 and 3 questions correctly. How many people answered 4 questions correctly?

Solution: Some people have two, three or four problems.

52-7-6=39 (person).

They did the right thing.

181-1× 7-5× 6 =144 (road).

Because there are as many people who answered question 2 as question 3, we can regard them as those who answered question 2.5 ((2+3)÷2=2.5).

Number of rabbit feet =4, number of chicken feet =2.5,

Total number of feet = 144, total number of heads =39.

For these four questions, there are

(144-2.5× 39) ÷ (4-1.5) = 31(person).

Answer: 365,438+0 people answered four questions correctly.

Exercise 1

1. turtle crane * * * 100 with 350 feet. How many turtles and cranes are there?

There are 26 sets of chess and checkers in the school, just enough for students to carry out activities at the same time. There are two sets of chess and six sets of checkers. How many sets of chess and checkers are there?

3. Some 2-cent and 5-cent coins are worth 2.99 yuan, among which the number of 2-cent coins is 4 times that of 5-cent coins. How many nickels are there?

4. Someone got 50 RMB * * in 240 yuan, including 2 yuan, 5 yuan and 10 yuan, of which 2 yuan and 5 yuan had as many. So how many pieces are there in 2 yuan, 5 yuan and 10 yuan?

5. A project was completed in 0/2 days by Party A and 0/8 days by Party B. Now Party A has done it for several days, and then Party B finished the rest alone, so it took 16 days before and after. How many days did Party A do it first?

6. The whole motorcycle race is 28 1km, which is divided into several stages. At each stage, some of them are composed of uphill roads (3km), horizontal roads (4km), downhill roads (2km) and horizontal roads (4km). Some of them are composed of uphill roads (3 kilometers), downhill roads (2 kilometers) and Pingdi roads (4 kilometers). It is known that after the motorcycle completed the journey, * * * ran 25 uphill roads. How many segments does the whole process of these two stages include?

7. Use 1 yuan * * 1 5 stamp to buy 4 stamps, 8 stamps and1stamp. How many stamps can I buy at most?

Second, the problem of "the difference between two numbers"

The total number of chickens and rabbits in the same cage is the sum of two numbers. If you change the condition to "the difference between two numbers", how should you solve it?

Example 7 Buy some stamps with 4 cents and 8 cents, and their price is 80 cents in 6 yuan. As we all know, there are 40 8-cent stamps more than 4-cent stamps, so how many stamps did you buy each?

Option 1: If you take out 40 8-cent stamps, there are as many 8-cent stamps as 4-cent stamps among the remaining stamps.

(680-8×40)÷(8+4)=30 (sheets),

That is to say, among the remaining stamps, there are 30 8-point stamps and 30 4-point stamps.

Therefore, the eight stamps are

40+30=70 (sheets).

I bought 70 8-cent stamps and 30 4-cent stamps.

You can also arbitrarily assume a number.

Solution 2: For example, suppose there are 20 4-point stamps, and according to the condition that "8 points is 40 more than 4 points", there should be 60 8-point stamps. Taking "Minute" as the calculation unit, the total stamp value at this time is

4×20+8×60=560.

It's less than 680, so we need to add stamps. In order to keep the "difference" at 40, 4 points per 1 block, we need to add 1 block 8 points. The quantity of each item to be added is

(680-4×20-8×60)÷(4+8)= 10 (Zhang)

So 4 o'clock is 20+ 10=30 (sheets), and 8 o'clock is 60+ 10=70 (sheets).

If a project is all sunny, 15 days can be completed. If it rains, it will rain all day.

How many days will it take to complete this project?

Solution: Similar to Example 3, the total workload of our project is 65,438+050 copies, with 65,438+00 copies completed every day in sunny days and 8 copies completed every day in rainy days. Using the method to solve the first problem in the above example, there are

(150-8× 3) ÷ (10+8) = 7 (days).

Rainy days are 7+3= 10 days, totaling * * *

7+ 10= 17 (days).

A: The completion time of this project is 17 days.

Please note that if "rainy days are longer than sunny days" is removed, and the project is known to be completed in 17 days, the problem in the previous section will be returned. Difference 3, and 17. Knowing one, you can deduce the other. This illustrates the relationship between Examples 7 and 8 and the basic problems in the previous section.

The total number of feet is the sum of two numbers. If you change the condition to "the difference between two numbers", how should you solve it?

Example 9 There are 100 chickens and rabbits, and the number of feet of chickens is less than that of rabbits. 28. How many chickens and rabbits are there?

Solution 1: If you add 28 chicken feet, that is, 28÷2= 14 chickens, the number of chicken feet and rabbit feet is equal, and the number of rabbit feet is 4÷2=2 times that of chicken feet, then the number of chickens is twice that of rabbits.

(100+28 ÷ 2) ÷ (2+1) = 38 (only).

Chicken is

100-38=62 (only).

Answer: There are 62 chickens and 38 rabbits.

Of course, you can also remove the rabbit 28÷4=7 (only). The number of rabbits is

(100-28 ÷ 4) ÷ (2+1)+7 = 38 (only).

You can also arbitrarily assume a number.

Solution 2: If there are 50 chickens, then there are 100-50=50 rabbits. At this time, the difference between the number of feet is

4×50-2×50= 100,

72 is more than 28. It means that there are more rabbits (fewer chickens). In order to keep the total number of rabbits at 100, one rabbit was replaced by one chicken, four rabbit feet were missing and two chicken feet were added, with a difference of six (note, not two). Therefore, the number of rabbits that need to be reduced is

(100-28)÷(4+2)= 12 (only).

The number of rabbits is

50- 12=38 (only).

There are also the following problems: the total number of heads is changed to the difference between two numbers, and the total number of feet is also changed to the difference between two numbers.

10 ancient poetry, there are four five-character quatrains, each with five words; Seven-character quatrains are four poems with seven words each. There is a book of poetry, with five-character quatrains more than seven-character quatrains 13, but the total number of words is 20 less. How many poems are there in each of the two kinds?

Option 1: If the five-character quatrains of 13 are removed, the number of the first two poems is equal, and the number of words is different at this time.

13×5×4+20=280 (word).

Every song has a different number of words.

7×4-5×4=8 (word).

Therefore, the seven-character quatrain is

28÷(28-20)=35 (first).

The five-character quatrain is

35+ 13=48 (first).

A: There are 48 five-character quatrains and 35 seven-character quatrains.

Solution 2: Suppose there are 23 five-character quatrains, then according to the difference of 13, the seven-character quatrains are 10. The number of words is 20×23=460 (words), 28× 65,438+00 = 280 (words), but there are more words in five-character quatrains.

460-280= 180 (word).

It is different from "20 words less" in the title.

180+20=200.

Explain that the number of hypothetical poems is small. In order to keep the difference between 13 poems, a five-character quatrain and a seven-character quatrain were added, and the word difference increased by 8. So the number of five-character quatrains is more than assumed.

200÷8=25 (first).

The five-character quatrain is

23+25=48 (first).

The seven-character quatrain is

10+25=35 (first).

When writing the formula of "chickens and rabbits in the same cage", we assume that they are all rabbits or chickens. Of course, we can also make the same assumption for the three problems of Example 7, Example 9 and Example 10. Now, let's do it in detail and compare the listed formula with the formula of "chicken and rabbit in the same cage", and we will find something very interesting.

Example 7, suppose that the number of all 8-cent stamps and 4-cent stamps is

(680-8×40)÷(8+4)=30 (sheets).

Example 9, suppose all rabbits, the number of chickens is

(100×4-28)÷(4+2)=62 (only).

10, assuming that they are all five-character quatrains, then the first number of seven-character quatrains is

(20× 13+20)÷(28-20)=35 (first).

First of all, please understand the origin of the above three formulas, and then compare them with the formula of "chicken and rabbit in the same cage" Only one of the three formulas "-"becomes "+". What is the secret?

When you enter junior high school, you have the concept of negative numbers, you will list binary linear equations, and you will understand that mathematically, the examples listed in the first two sections of this class are the same thing.

Example 1 1 There is a truck carrying 2000 glass bottles. The freight is calculated according to the number of intact bottles when they arrive, and each bottle is 20 cents. If there is any damage, the freight is not compensated, and each bottle is compensated 1 yuan. As a result, the freight was 379.6 yuan. How many glass bottles were damaged in this handling?

Solution: If there is no damage, the freight will be 400 yuan. But as long as the damage is reduced 1+0.2= 1.2 (yuan), the number of damage is only.

(400-379.6) ÷ (1+0.2) =17 (only).

Answer: 17 glass bottle was damaged in this handling.

Please think about it, is this the same type of problem as "chickens and rabbits in the same cage"?

The example 12 has two natural tests. The first time, 24 questions, 1 correct answer 5 points, wrong answer (including no answer) 1 point. In the second test, if you answer 15 correctly, 8 points will be scored, and if you answer or don't answer 1, 2 points will be deducted. Xiao Ming answered 30 questions correctly in two tests, but the score of the first test was more than that of the second 10. How many points did Xiaoming get in each of the two tests?

Option 1: If Xiaoming answers all 24 questions correctly in the first test, he will get 5×24= 120 (points). Then he only got 30-24=6 in the second exam. What is the score?

8×6-2×( 15-6)=30 (minutes).

Two phase differences

120-30=90 (minutes).

It is lower than the condition in the question 10, which is 80 points more. Explain that the number of questions answered correctly the first time is too many and should be reduced. The first answer subtracts a question, 5+ 1=6 (points), and the second answer adds 8 points instead of 2 points, so it adds 8+2 = 65438+.

6+ 10= 16 (point).

(90- 10)÷(6+ 10)=5 (problem).

Therefore, the number of questions answered correctly for the first time is 5 less than that assumed (all right), that is, the first time 19 questions, and the second time 30- 19= 1 1 (questions).

Score for the first time

5× 19- 1×(24- 9)=90.

Second score

8× 1 1-2×( 15- 1 1)=80.

Answer: I got 90 points for the first time and 80 points for the second time.

Option 2: Answer 30 questions correctly, that is, answer them wrong twice.

24+ 15-30=9 (topic).

For the first wrong answer, 5+ 1=6 (points) will be deducted from the full score, and for the second wrong answer, 8+2= 10 (points) will be deducted from the full score. If the answer is wrong, exchange it. The difference between the two scores should be 6+ 10= 16.

If it is the first time to answer 9 questions incorrectly, the full score will be deducted by 6×9. However, the perfect score for both times is 120. It is 6× 9+10 less than the condition of "first score10" in the question. Therefore, the number of wrong answers for the second time is

(6×9+ 10)÷(6+ 10)=4 (topic)

Wrong answer for the first time 9-4=5.

The first score is 5× (24-5)-15 = 90 (points).

The second score is 8×( 15-4)-2×4=80 (points).