The problem of chickens and rabbits in the same cage, whatever the number! Help me!

There were several chickens and rabbits in the same cage, counting from above, there were 35 heads; Counting from below, there are 94 feet. How many chickens and rabbits are in the cage?

Hypothetical Method: Solution:

Suppose all rabbits: 4 35 140 (only) less than the total number of feet: 140 94 46 (only) chickens: 46 2 23 (only) rabbits: 35 23 12 (only).

There were several chickens and rabbits in the same cage, counting from above, there were 35 heads; Counting from below, there are 94 feet. How many chickens and rabbits are in the cage? Hypothetical Method:

Solution: Suppose all chickens: 2 35 70 (only) less than the total number of feet:

94 70 24 (pcs) The difference between their legs: 4-2=2 (pcs) Rabbits: 24 2 12 (pcs) Chickens:

35 12 23 (pcs) Equation: Solution: If there are x rabbits, then there are 35-x chickens.

4x+2(35-x)=94 4x+70-2x=94 2x=24 x=24 2x=12 35-x=35-12=23 Answer: There are 12 rabbits and 23 chickens.

Solution 1: (Number of rabbit's feet, total number of feet, total number of feet) (number of rabbit's feet, number of chicken's feet) = number of chickens, total number of chickens, number of chickens = number of rabbits.

Solution 2: ( Total number of feet Number of chicken's feet Total number of chickens) (Number of rabbit feet Number of chicken's feet) = Number of rabbits .

Example: There are 14 chickens and rabbits, with a total of 44 legs.

4 14-44) (4-2) 12 2 6 (chickens), 14-6 8 (rabbits).

or (44-2 14) (4-2) 16 2 8 (rabbits), 14-8 6 (chickens).

The key to solving the problem: The chicken and rabbit problem is one of the famous mathematical problems in ancient China, also known as the "chicken and rabbit in the same cage" problem. To answer the problem of chickens and rabbits in the same cage, the hypothetical method is generally adopted, assuming that all are chickens, calculate the number of feet, compare with the number of feet given in the question, see how much the difference, each difference of one (4 2) feet, it means that there is 1 rabbit, divide the number of feet by (

4 2), you can find the number of rabbits. In the same way, if all of them are rabbits, chickens can be found.

1. There are 80 chickens and rabbits in the same cage, 208 feet, how many chickens and rabbits are there?

Analysis: Assuming that all 80 are chickens, then the feet should be 2 80 160 (only), which is 208 160 48 less than the actual one

Only) feet, this is because 1 rabbit has 4 legs, think of it as a chicken with 2 legs, each rabbit is less than 2 feet, a total of 48 feet are undercounted, 48 in there are several 2s, that is, a few rabbits.

Solution: (208 2 80 ) 4 2 ).

24 (only) - Rabbit.

80 24 56 (only).

A: There are 56 chickens and 24 rabbits.

It can also be assumed that all 80 are rabbits, and the answer is as follows:

Solution: (4 80 208 ) 4 2 ).

56 (pcs) - Chickens.

set, a person who plays a rolling roller coaster, a person b who bumper a car, and a c person who bravely rushes into the rapidsThen:

a+b+c=42

8a+8a+6c=

The left side of the equation is an even number, b is a multiple of 5 to make it an integer, and b must beodd numbers, so that the right side of the equation can be made odd and the right side of the equation can be even

Therefore, b = 5, 15, 25, 35

8a+6c=，134，112，90

4a+3c=78，67，56，45Equation 1.

In addition, a+c = 37,27,17,7

3a+3c=111，81，51，21Equation two.

Equation 1 minus the 2nd lead wheel Ctrip, a = -33, -14, 5, 24

Negative numbers are rounded off, a=5,24, corresponding to b=25,35

and 24+35=59>42, it doesn't fit the topic, give it up.

Therefore, a=5

There are 5 people playing the tumbling roller coaster

The topic is known that the conditions of the mu stool are not enough. Limited to known conditions, only analysis and extrapolation can be made.

Bumper cars are priced per ticket, and the number of people playing bumper cars must be an integer multiple of 5, otherwise the money cannot be an integer yuan. But there can be no more than 25 people, otherwise the total amount of money will be less than 167 yuan. The average ticket price of the other two is 7 yuan, so the number of people cannot exceed 22 people, otherwise the total amount of money will exceed 167 yuan.

Based on the above analysis, Qianpei calculated that there were 25 people playing bumper cars, 5 people playing roller coasters, and 12 people playing rapids. (The latter two can be solved with binary one-dimensional equations).

The most basic solution to the problem of chickens and rabbits in the same cage.

1 Equation method. Set chicken x rabbit y

Number of heads = x+y

Number of legs = 2x + 4y

Solve the equation to find x, y

Comprehension: The most straightforward idea.

The rabbit has a head and four legs.

The chicken has a head and two legs.

So set the number of chickens and the number of rabbits.

Solve by the number of heads and legs.

2. Arithmetic. Number of rabbits = (number of legs - 2 * number of heads) 2

Number of chickens = number of heads - number of rabbits.

Comprehension: Suppose the chicken rabbit is well trained.

Blow a whistle. All chickens and rabbits raise one leg each.

That is, the remaining = the number of legs - the number of heads.

Blow another whistle.

All the chickens and rabbits each lift one more leg.

The chickens are all down. The rabbit stands on both legs.

That is, the number of secondary remaining = the number of legs - the number of heads - the number of heads = the number of legs - 2 * the number of heads, so the number of rabbits = the number of secondary remaining 2 = (the number of legs - 2 * the number of heads) 2 chickens = the number of heads - the number of rabbits.

There are more complex issues.

For example, it involves the exchange of the number of chickens and rabbits.

Using the equation method is the simplest and best understood.

Arithmetic is not easy to think about.

So it is advisable to column equations.

Start chicken x rabbit y

Chicken y rabbit x after change

Just follow the equation of the relationship between the head and the legs.

Let's say it's all one of them, and count the number of legs.

For example, the head has 30 and the legs have 80, for example.

Let's say it's all chickens. The number of legs is 30*2=60

80-60=20 is the number of legs remaining.

The difference in the number of legs between a single rabbit and a chicken is 4-2=2

20 2 = 10, return these 20 legs to the rabbit in pairs, so, there are 10 rabbits, chickens = 30-10 = 20.

Suppose the legs of all rabbits are 30*4=120

In the same way, this is the extra leg of the chicken 120-80 = 40, to subtract so 40 2 = 20 This is the chicken.

30-20=10 This is a rabbit.

Some of the question types are not chickens and rabbits in the same cage, it may be 2 points and 5 points coins or the like, the idea is the same, and the last thing to pay attention to is to divide by their differences.

Suppose the cage is full of chickens.

Total number of legs – number of chickens 2 = number of legs that rabbits have more than chickens.

Rabbits have twice as many feet as chickens.

So the number of feet that a rabbit has more than a chicken 2 = the number of rabbits.

Total number of birds - number of rabbits = number of chickens.

1.Chickens and rabbits have 20 heads and 48 feet in the same cage, how many chickens and rabbits are there?

Solution: Suppose it's all chickens.

20*2=40 (only).

48-48=8 (only).

4-2=2 (only).

8 2 = 4 (only) - rabbit.

20-4 = 16 ——— chickens.

Two times is the number of branches exchanged.

The total number of two strokes of the fountain pen and the sum of the two strokes of the ballpoint pen are the same and equal to the total number of the two strokes each time.

Two times in total: 90 120 210 (yuan).

The total unit price of the two transactions: 5 2 7 (yuan).

The total number of branches of each two transactions is: 210 7 30 (branches) The price difference between the two times: 120 90 30 (yuan).

The difference between the two units: 5 2 3 (yuan).

The difference between the number of branches of the two strokes is: 30 3 10 (branches).

The first time you spend less money than the second.

The number of fountain pens (with a high price) must be less than the number of ballpoint pens (with a lower price) for the first time.

Number of pens: (30 10) 2 10 (branches).

Number of ballpoint pens: (30 10) 2 20 (branches).

(120+90) (5+2)=30 (branches) total.

20 (sticks) ballpoint pens.

30-20 = 10 (branch) fountain pen.

The unit price of the fountain pen - the unit price of the ballpoint pen = 3 yuan.

Therefore, ballpoint pens should be 30 more than fountain pens 3 = 10.

90-2*10=70 yuan, which was used to buy the same number of fountain pens and ballpoint pens.

70 (5+2)=10 pcs.

So. A total of 10 fountain pens and 20 ballpoint pens were bought.

A: Buy back 10 fountain pens and 20 ballpoint pens.

Set up fountain pen x branch, ballpoint pen y branch.

5x+2y=90

5y+2x=120

x=10y=20

There are several chickens and rabbits in the same cage, counting from above, with 35 heads, and counting from below, with 94 legs. Q: How many chickens and rabbits are in each cage?

The hypothetical method assumes that all chickens are: 2 35 = 70 (only).

Chicken feet are less than the total number of feet: 94 70 = 24 (pcs).

The number of feet that rabbits have more than chickens: 4-2=2 (only).

Number of rabbits: 24 2=12 (only).

Number of chickens: 35 12 = 23 (chickens).

Ideas; Assuming that 35 are all chickens, then there should be 35 * 2 = 70 legs, but there are actually 30 more legs than 70, because as long as there is a rabbit, there are 2 legs less.

So the extra legs divided by 2= the number of rabbits.

100-35*2) 2=15 (rabbits) 35-15=20 (chickens).

Suppose there are x chickens and y rabbits.

x+y=35 There are 35 chickens and rabbits in the same cage, this is the formula 2x+4y=100 Each chicken has two legs, each rabbit has 4 legs, a total of 100 legs, this is the formula.

①×2: 2x+4y-(2x+2y)=100-35×22y=30

y=15x=20

So there are 20 chickens and 15 rabbits.

If there are two chickens and four rabbits, then there are 20 chickens and 15 rabbits!

I don't know if you can use a system of binary equations in the sixth grade, so I'll solve it for you from beginning to end.

The chicken rabbit has a total of 35 heads and 94 legs. How many chickens and rabbits are there?

Solution: There are x chickens and y rabbits.

There should be curly braces here: x+y=35

Braces and so on will be more detailed 2x+4y=94 2x is 2x because chickens have two legs

Solution: From x=35-y 4y Since the rabbit has four legs, it is 4y that substituting into .

2(35-y)+4y=94

70-2y+4y=94

70+2y=94

2y=94-70

2y=24y=12

Substitute y=12 into .

x=35-12=23

So the solution of the original system of equations is x=23

There should be curly braces here) y=12

A: There are 23 chickens and 12 rabbits.

Note: Because it is a binary system of equations, it is necessary to enlarge the parentheses "{ The computer is too small, so it is not typed. Be sure to enclose both equations. The final solution should also be enlarged in parentheses, and I have pointed out the arrows!

This is the knowledge of the first year of junior high school, and I think I said it in a very detailed way, and you should be able to understand it. I really don't understand, you can ask me again.

There are 15 chickens and rabbits, a total of 48 feet, how many chickens and rabbits are there?

9 (only) - rabbits.

15-9=6 (only) - chickens.

A: There are 6 chickens and 9 rabbits.

There are 20 chickens and rabbits, with a total of 68 legs, and there are several chickens and rabbits.

2 20 = 40 (only) 68 40 = 28 (only) 4 2=2 (only) Rabbit: 28 2=14 (only) Chicken: 20 14=6 (only).