I won t 555 on the same cage problem

There are x groups for grade 4 and y for grade 3, and the equation is as follows:

x+y=41 (1)

2x+3y=100 (2)

From equation (1): x=41-y

Bring in equation (2): 2x+3(41-y)=100, and get y=18 (group), so there are 18 groups in grade 3. 3 people per group, 18*3=54 (people).

The answer is that there are 54 third-year students.

The problem of chickens and rabbits in the same cage is very simple to solve by algebra, and this problem should be solved by arithmetic.

1) Assuming that the 41 groups are all 4th grade, 2*41=82 (people) (2) How many groups of 3rd grade students are there?

100-82) divided by (3-2) = 18 divided by 1 = 18 (group) (3) How many 3rd graders are there ? 3*18=54 (person).

A: There are 54 students in the third grade.

The second algorithm (1) assumes that all 41 groups are in grade 3.

3*41=123 (person).

2) How many groups of 4th graders are there?

123-100) divided by (3-2) = 23 (group) (3) How many students are in 4th grade? 2*23=46 (person) (4) How many students are in the third grade?

100-46 = 54 (people).

A: There are 54 students in the third grade.

Assuming that all 41 groups are 4th grade, 2*41=82 (people) (2) How many groups of 3rd grade students are there?

100-82) divided by (3-2) = 18 divided by 1 = 18 (group) (3) How many 3rd graders are there ? 3*18=54 (person).

A: There are 54 students in the third grade.

The second algorithm (1) assumes that all 41 groups are in grade 3.

3*41=123 (person).

2) How many groups of 4th graders are there?

123-100) divided by (3-2) = 23 (group) (3) How many students are in 4th grade? 2*23=46 (person) (4) How many students are in the third grade?

100-46 = 54 (people).

A: There are 54 students in the third grade.

Solution: There are x students in the third grade and y students in the fourth grade.

x+y=100 1 formula.

x 3 + y 2 = 41 2 formula.

Obtained by 2 formula: 2x+3y=246 3 formula.

1 type *2 get:

2x+2y=200 4 formula.

Formula 3-4:

y=46 substitut y=46 into 1 to get:

x+46=100

x=54 so x=54 and y=46 are the solutions of the original system of equations.

There are x people in the third grade and 100-x people in the fourth grade.

100-x) divided by 2+x divided by 3=41

It turns out that there are 54 students in the third grade and 46 students in the fourth grade.

Set up group X for 4th graders and group y for 3rd graders, then x+y=41

2x+3y=100, so x=23, y=18, the third grader equals 54 students, and the fourth grader is 46 students.

Let's say there are x people in 3rd grade.

Then: x 3+(100-x) 2 41

Interpreted, x 54 people.

So 54 in the third grade, 100-54 46 for the fourth grade.

The most primitive method: set up 4-year-old x group, 3-year-old y group, x+y=41, 2x+3y=100, x=23, y=18, so there are 54 people in 3-year-old.

Setting: 4th grade x people, 3rd grade y people.

Solution: x+y=100

x/2+y/3=41

Solution: x=46

y=54Answer: 54 students in the third grade.

4th grader x people, 3rd grader y people.

then there is x 2+y 3=41

x+y=100

The solution is x=46 y=54

Setting: X people in the fourth grade, Y people in the third grade.

x/2+y/3=41

x+y=100

y=54 x=46

Guaranteed right, trust me!

Suppose all are 4th graders: 2*41 82 (person) (100-82) divided by (3-2) 54.

There are 54 students in the third grade.

Setting: X people in the fourth grade, Y people in the third grade.

x/2+y/3=41

x+y=100

y=54 x=46

This kind of problem is unknown, easy to solve, specific, some people have said, I won't repeat it

Such problems can be normalized to the same method.

The sum of 3 5 of A and 2 3 of B is 2136

Multiply 5 3 and get:

The sum of 1 of A and 2 3 5 3 = 10 9 of B is 2136 5 3 = 3560 yuan.

B: (3560-3420) (10 9-1) = 1260 yuan.

A: 3420-1260=2160 yuan.

Let A and B each have money x and y yuan.

x+y=3420

3x/5+2y/3=2136

x=2160

y=1260

A and B each have 2,160 yuan and 1,260 yuan.

The sum of two-fifths of the number of money A and one-third of the number of money B is 3420-2136 = 1284 yuan.

The sum of four-fifths of the number of money A and two-thirds of the number of money B is 1284*2=2568 yuan.

One-fifth of the number of A coins is 2568-2136 = 432 yuan.

The number of A coins = 432 * 5 = 2160 yuan.

B money = 3420-2160 = 1260 yuan.

44 ballpoint pens, 176 pencils, 12 fountain pens.

Solution: Set the ballpoint pen x root. Then the pencil is 4x, and the pen is 232-(x+4x).

100 yuan = 1000 jiao.

2*4x+9*x+21*(232-5x)=1000 to get x=44

1 + 4 = 5 (portions) 100 yuan cavity hail = 1000 jiao.

Solution: If there are x pens, then there are ballpoint pens and (232-x) 5 4 pencils.

The column equation is: 21 x+(232-x) 5 9+(232-x) 5 4 2=1000

x = 12 ballpoint pens have (232-12) 5 = 44 (sock sails) pencils have 44 4 = 176 (sticks).

A: There are 12 fountain pens, 44 ballpoint pens and 176 pencils.

I'm a teacher, thank you.

Solution: Set up a round section of the mountain thick beads only destroy the pen x root. Then the pencil is 4x, and the pen is 232-(x+4x) to grip the town.

100 yuan = 1000 jiao.

2*4x+9*x+21*(232-5x)=1000 to get x=44

1 + 4 = 5 (copies) 100 yuan = 1000 jiao.

Solution: If there are x pens, then there are ballpoint pens and (232-x) 5 4 pencils.

The column equation is: 21 x+(232-x) 5 9+(232-x) 5 4 2=1000

x = 12 ballpoint pens have (232-12) 5 = 44 pencils 44 4 = 176 (pieces).

A: There are 12 fountain pens, 44 ballpoint pens and 176 pencils.

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!

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?

Solution: 2 35 70 (only).

94 70 24 (only).

24 2 12 (only) - Rabbit.

35 12 23 (only) - Chickens.

In ancient China, the "Sun Tzu Sutra" consists of three volumes, which was written in about the 5th century AD. The book is easy to understand and has many interesting arithmetic problems, such as the "chickens and rabbits in the same cage" problem:

Today there are pheasants and rabbits in the same cage, there are thirty-five heads on the top, and there are ninety-four feet under it.

If the two front feet of the rabbit are tied with a rope and regarded as one foot, and the two hind feet are also tied with a rope and regarded as one foot, then the rabbit becomes a chicken with two legs, that is, the rabbit is first regarded as a chicken with two legs. The total number of feet of chickens and rabbits is 35 2 = 70 (only), which is 94-70 = 24 (only) less than the 94 mentioned in the question.

Now, if you untie the rope on a rabbit's foot, the total number of feet will increase by 2, i.e. 70 + 2 = 72 (only), and if you release the rope on a rabbit's foot, the total number of legs will increase by 2, 2, 2 ......and continue until it increases by 24, so the number of rabbits: 24 2 = 12 (only), so that the chickens have 35-12 = 23 (only).

Let's summarize the solution idea of this problem: first assume that they are all chickens, so according to the total number of chickens and rabbits, you can calculate how many feet there are under the assumption, compare the number of feet obtained in this way with the number of feet given in the question, and see how much the difference is, every 2 feet indicates that there is 1 rabbit, and divide the number of legs by 2, you can calculate how many rabbits there are. To sum up, the basic relationship for solving the chicken-rabbit same cage problem is:

Number of rabbits = (actual number of feet - number of chickens per chicken total number of chickens) (number of feet per rabbit - number of chickens and rabbits). Similarly, it can be assumed that it is all rabbits.

We can also use a column equation: let the number of rabbits be x and the number of chickens be y

So: x+y=35 then 4x+2y=94 After solving this equation, we get: there are 12 rabbits and 23 chickens.

Just use the equation method to solve it. If you set the number of known rabbits or chickens to x (as many as there are), then the total number minus x is the total number of rabbits or chickens. Then multiply each of them by their respective foot numbers, the rabbit is 4 and the chicken is 2, which adds up to the total foot number, solve the unknown, and everything is clear.