Some math problems about chickens and rabbits in the same cage .

1.It's all male classmates.

The number of female students is: (3*12-30) (3-2)=6 (people) and the number of male students is: 12-6=6 (people).

Only) 280 14 = 20 (only).

20 (1+3)=5 (copies).

Chicken: 1*5*4=20 (pcs).

Rabbit: 3*5*4=60 (only).

3.Number of seats in the back: (4*2000+1100) (4+) Number of seats in the front row: 2000-1400=600 (sheets).

1.There are x girls and y boys.

x+y=12

2x+3y=30

A: 6 girls and 6 boys.

2.Set chicken x, rabbit y

3x=y 2x+4y=280

A: There are 20 chickens and 60 rabbits.

Sky. Hope it helps

There were 12 students in Class 6 (1) who participated in the tree planting activity, and a total of 30 lessons were planted. Female students planted 2 lessons each, and male students planted 3 lessons each. How many male and female classmates are there?

6 males and 6 females.

Chickens and rabbits are in the same cage, and the number of rabbits is 3 times that of chickens, with a total of 280 feet, chickens and rabbits.

20 chickens and 60 rabbits.

There is a problem with the third question.

1) 12*3=36 36-30=6 Female 6 (3-2)=6 Male 12-6=6

2) 20 chickens and 60 rabbits.

3) There is a problem.

1 If there are x girls and (12-x) boys, then there are 2*x+3*(12-x)=30 and x=6 so there are 6 boys and girls.

If you haven't learned the equation, use the following method: each boy has 1 more tree than the girl, so 30-2*12=6 is the number of boys, and 12-6=6 is the number of girls.

2 A rabbit is equivalent to 2 chickens, so the number of chickens is 280 (2+2*3*2)=20, and the number of rabbits is 20*3=60.

Supplementary question: A completes 1 15 every day, and C completes 1 25 every day, so the time required for two people to complete the project together is 8 15 (1 15 + 1 25) = 3 days.

Set all as: female classmates. 12 2=24 30-24=6 3-2=1 Male:

6 1=6 female: 12-6=6 Question 2: (2 3+1)=7 280 7=40 feet 40 2=20 chickens 20 3=60 rabbits.

1183438942, hello:

There are 46 eyes, and both animals have the following in common:

46 2 23 (only).

Suppose it's all monkeys, with common feet:

23 4 92 (only).

Much more than it actually is :

92 72 20 (only).

Red-crowned cranes: 20 (4 2) 10 (only).

Monkeys have: 23 10 13 (only).

The monkey has 4 legs and the red-crowned crane has 2 legs. ）

Hello, we should consider monkeys as 4 legs.

There are x red-crowned cranes and y monkeys.

2x+2y=46

2x+4y=72

Use 2 to subtract 1 to get:

2y=26y=13.

Substituting y=13 into equation 1 yields:

x = 10.

A: There are 10 red-crowned cranes and 13 monkeys.

Of course, monkeys can walk on four legs. Have you ever seen a monkey walking upright on 2 feet?

Method 1: (Do without equations).

Let's say it's all red-crowned cranes. There are 46 legs, because the eyes of the red-crowned crane are equal to the number of feet.

Whenever a red-crowned crane is replaced with a monkey. That's 2 more feet.

Answer: 13 monkeys, red-crowned cranes = 46 2-13 = 10 Method 2: Equation.

Set monkeys x and red-crowned cranes (23-x).

4x+2(23-x)=72

The solution is x=13 23-13=10Answer: omitted.

The monkey has four legs.

There are x red-crowned cranes and 23-x monkeys.

2x+(23-x)×4=72

2x+92-4x=72

2x=20x=10

10 red-crowned cranes and 13 monkeys.

The monkey has 2 eyes and 4 legs, and the red-crowned crane has 2 eyes and 2 legs. (When the red-crowned crane is standing, it is 1 retreat, but here it is counted with 1, and the result is not an integer).

Red-crowned cranes have (46 2 4-72) (4-2) = 10 monkeys and 46 2-10 = 13 monkeys.

So 10 red-crowned cranes and 13 monkeys.

Solution 46 2=23

A: There are 13 monkeys and 10 red-crowned cranes.

Chickens and rabbits in the same cage math problem is an important type of classical math problem in China, which is commonly solved by arithmetic method. The use of arithmetic method to solve has a positive effect on improving people's logical thinking ability. Although some netizens have used algebra to find the result of this problem and it has been adopted, it is still necessary to use arithmetic to solve this problem.

Chickens and rabbits in the same cage, there are 35 heads, 94 feet, if all 35 are chickens, then there are feet 35 2 = 70 (only), 94 less 70 = 24 feet, this is because there are rabbits not counted, every 1 rabbit is counted less, (4 2) = 2 feet.

So the number of heads of the rabbit is: (94 35 2) (4 2) = 24 2 = 12, and the number of heads of the chicken is: 35 12 = 23, 12 4 + 23 2 = 94;

If all 35 are rabbits, then there are 35 4 = 140 feet, 140 more 94 = 46 feet, this is because there are chickens not counted, for every 1 chicken missing, (4 2) = 2 feet.

So the number of heads of a chicken is: (35 4 94) (4 2) = 46 2 = 23, and the number of heads of a rabbit is: 35 23 = 12, 23 2 + 12 4 = 94.

Question 1, 6 rainy days. From 112 14=8, by 20(8-x)+14x=112, x=6 is obtained

In the second problem, let the frog x, the butterfly y, and the cicada z be solved, and the system of ternary linear equations x+y+z=27, 4x+6y+6z=138, 2y+z=22 is solved to obtain x=12, y=7, z=8

In the third question, set x rabbits, and get x = 19 from 4x+2(12+x)=138, so there are 19 rabbits and 31 chickens.

The number of chickens is:

The number of rabbits is:

It takes 1 50 cubic meters of wood to make a table top and 1 300 cubic meters to make a table leg, so it takes 1 50 + 4 300 = 1 30 cubic meters of wood to make a set of tables. So 10 cubic meters of wood can make 300 sets, that is, 300 table tops and 1,200 table legs.

Chickens and rabbits in the same cage, there are 35 heads, 94 feet, chickens and rabbits each have x chickens, then rabbits have 35-x.

2x+4(35-x)=94

2x+140-4x=94

140-2x=94

2x=46x=23

35-x=12

There are 23 chickens and 12 rabbits.