Some chickens and rabbits in the same cage, similar problems in the same cage of chickens and rabbit

1 Assuming all wheelbarrows, there should be 30 wheels, 40 less.

So there are 40 (3-1)=20 tricycles and 10 unicycles.

Or if it is all tricycles, there should be 3*30=90 wheels, and if there are 20 more, there should be 20 (3-1)=10 wheelbarrows.

2 x unicycles and 30-x tricycles.

x+3(30-x)=70

90-2x=70

x=10

The classic method that I stumbled upon :

If one wheel is removed from each car, the wheelbarrow will have no wheels, and the others will be two-wheelers, and there will be 40 wheels left, so there are 20 tricycles and 10 wheelbarrows.

x is the number of wheelbarrows.

y is the number of tricycles.

x+y=30

x+3y=70

x=10 y=20

70-30) (3-1) = 20 (three).

30-20=10 (single).

Set up tricycle x 3x+30-x=70

x=2030-20=10 (independent).

Hypothetical method: Suppose all tricycles are minus 2 wheels, in the case of all unicycles.

There are 30 wheels left (because there are 30 cars).

Then the tricycle reduced the wheels by 70 30 40 wheels in total.

Each tricycle is reduced by 2 wheels, a total of 40 less, then the number of tricycles.

For 40 2 20, the wheelbarrow is naturally 10.

Equation method: There are x unicycles and y tricycles.

According to the title, it can be obtained: x+y 30

x+3y＝70

Solution x 10, y 20

Understand???

Xiao Ming has a total of 50 2-cent coins and 5-cent coins, a total of 2 yuan and 2 jiao. How many 2 cents and 5 cents does Xiao Ming have?

The advanced version of the chicken and rabbit in the same cage is here, learn it, and you will be able to solve the same type of problem.

The big monk eats 3 for one person, and the small monk eats 1 for three people, and you can divide three small monks and one big monk into a group, so that a group of monks can eat a total of 4 steamed buns.

100 steamed buns need (100 4 25) 25 groups of monks to eat up.

25 groups of monks, i.e. 25 big monks, 75 small monks.

Eq. 100 (3+1) 25

The number of tables and washbasin stands is one hundred, with 4 legs for tables and 3 legs for a total of 340 legs. How many tables and basin stands are there? There are many problems like chickens and rabbits in the same cage.

It seems that it is necessary to calculate the formula, not the equation.

Thought process: 1) There are 20 more chickens than rabbits, and the extra part of the chickens should be 20 * 2 = 40 feet more chickens, and we will get rid of the extra 20 chickens now.

2) The number of chickens and rabbits left is the same, so in the same part of the number, there should be 200 + 40 = 240 more rabbit feet than chicken feet.

3) A rabbit has two more legs than a chicken, so the same number of chickens and rabbits is 240 2 = 120 each.

So rabbits are 120 and chickens are 120 + 20 = 140.

Question: (Classical Chinese) "Today there are pheasants and rabbits in the same cage, with thirty-five heads on the top and ninety-four feet on the bottom. "Modern language: some chickens and rabbits are in the same cage, with a total of 35 heads; 94 feet. How many chickens and rabbits are in the cage?

If you use the equation:

Solution: Suppose there are x rabbits, then chickens have (35-x): 4x+2(35-x)=944x+70-2x=94

2x=24x=12

35-x=35-12=23

A: There are 12 rabbits and 23 chickens.

If you use a simple algorithm: (hypothetical method): 2 35 70 (only) less than the total number of legs: 94 70 24

Only) the difference in their legs: 4 — 2 = 2 (strips).

Only) ......Rabbit 35 12 23 (only) ......Chicken.

There is a similar question that goes like this:

There are several chickens and rabbits in the cage. There are 8 heads from above, and 26 feet from below. How many chickens and rabbits are there?

Think of it this way:

1) If the cage is full of chickens, then there are 8 times 2 = 16 feet, so there are 26-16 = 10 feet more.

2) A rabbit has 2 more legs than a chicken, that is, there are 10 divided by 2 = 5 rabbits (3) so there are 3 chickens and 5 rabbits in the cage.

Think of it this way:

Solution: If there are x rabbits, then there are (8-x) chickens.

4x+2（8-x）=26

2x+16=26

x=58-5=3 (only).

A: There are 5 rabbits and 3 chickens.

Absolutely original, please rest assured, look at the owner of the building plus a few points, thank the owner.

1.100 monks eat 100 steamed buns. The big monk eats 3 for each person, and the small monk eats 1 for 3 people. How many monks are there and how many monks are there?

25 people. Small monk 100-25 = 75 people.

2.Chickens and rabbits are in the same cage, and the number of chickens is 4 times that of rabbits, with a total of 180 legs. How many chickens and rabbits are there?

Rabbit 180 (4+2 4).

15 pcs. Chicken 15 4 = 60.

3.Math chickens and rabbits in the same cage, 8 chickens and 15 rabbits have a total of several legs.

Whistleblowing |14 days and 1 hour until the end of the question Asked by: Anonymous |Bounty Points: 5 |Views: 18.

Whistleblowing |2011-12-8 18:44 Mr. Mad |Level.

76 Feet agrees.

3 .A chicken has 2 legs, a rabbit has 4 legs. Now a total of 8 2 + 15 4 = 76 feet agreed.

76 pcs. 4.(1) There are 30 chickens and rabbits in the cage, and 100 legs. How many chickens and rabbits are there? (All using the hypothetical method).

Chickens (4 30-100) (4-2) = 20 2 = 10.

Rabbit 30-10 = 20.

2) Xiaohong bought a total of 60 stamps with a face value of 1 yuan and stamps with a face value of 5 corners, in short, yuan. How many stamps are there for a $1 stamp and a stamps with a 5-corner denomination?

$1 (sheet.

60-25 = 35 sheets of 5 corners.

3) The school bought 4 basketballs and 5 volleyballs, a total of 185 yuan, 1 basketball is 8 yuan more expensive than 1 volleyball, how much is the unit price of basketball?

Basketball (185 + 8 5) (4 + 5) = 225 9 = 25 yuan.

Volleyball 25-8 = 17

4) Uncle Zhang sells waste paper for 240 yuan, and there are 50 pieces of RMB in three denominations of 2 yuan, 5 yuan and 10 yuan, of which the number of 2 yuan and 5 yuan is the same. How many pieces are there in the $10 denomination?

30 sheets. 5) The school has a total of 26 sets of chess and checkers, 2 people play 1 pair of chess, and 6 people play 1 pair of checkers, which is just enough for 120 students to carry out activities. How many sets of chess and checkers are there? (All of the above are hypothetical).

Checkers (120-2 26) (6-2) = 68 4 = 17 pairs.

Chess 26-17 = 9 pairs.

Let's buy x kg cucumbers, 6-x kg tomatoes, so buy 4 kg cucumbers.

1. First use 4 * 8 = 32 (yuan) to find out how much 4 basketballs are more expensive than volleyball, and then use 185-32 = 153 (yuan) to subtract 32 and then the two balls are just as expensive, and then use 153 (4 + 5) = 17 (yuan) to find the volleyball, and then 17 + 8 = 25 (yuan) to find the basketball.

A: $25 for basketball and $17 for volleyball.

2. Column equations:

Solution: Let there be x rabbits and x+15 chickens.

4x+（x+15）*2=186

4x+2x+30=186

6x+30=186

6x+30-30=186-30

6x=156

x=2626+15=41 (only).

A: There are 26 rabbits and 41 chickens.

1. Solution: Set each volleyball to be X yuan.

4（x+8）+5x=185

x=174(x+8)=100

5x=852, solution: set rabbits have x.

2(x+15)+4x=186

x=26x+15=26+15=41

You made a mistake in your question......Football at the top, volleyball at the bottom......I'll think of it as a football ...... for now

Solution: Set each x yuan for football and 8 + x yuan for basketball.

5x+4(8+x)=185

5x+32+4x=185

9x=153

x=17A: 17 yuan for football, 25 yuan for basketball.

Solution: Set x rabbits and chickens (15+x).

4x+2(15+x)=186

4x+30+2x=186

6x=156

x=26A: 26 rabbits, 41 chickens.

Assuming that each basketball is not 8 yuan more than a football, ** is equivalent to the price of 4 + 5 = 9 footballs each: (185-8 4) 9 = 17 yuan.

Basketball: 17 + 8 = 25 yuan.

Suppose there are 15 more rabbits, there are as many chickens and rabbits There are 186 + 15 4 = 246 legs.

246 (4+2)=41 pcsChickens and rabbits: 41-15=26.

1. I treat the 5 soccer balls you said at the beginning as volleyball, okay......Volleyball is $x, basketball is $+8. Column Equation:

5x+4(x+8)=185, and the solution is x=17

$17 for volleyball and $25 for basketball.

2. There are x rabbits and x+15 chickens. Here's the equation:

4x+2(x+15)=186, the solution is x=26 with 26 rabbits and 41 chickens.

Your first question seems wrong.

Question 2: If there are rabbits x and chickens (x + 15), you will know from the question:

4x+2*(15+x)=186

x=26 so there are 26 rabbits and 41 chickens.