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Chickens and rabbits in the same cage is one of the famous mathematical problems in ancient China. About 1,500 years ago, this interesting question was recorded in the "Sun Tzu's Sutra". Here's how it is narrated in the book:
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. The meaning of these four sentences is: 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 the cage?
There's a simplest algorithm for this.
Total number of feet - total number of heads * number of chicken's feet) (number of rabbit's feet - number of chicken's feet) = number of rabbits.
94 35 2) 2 = 12 (number of rabbits) Total number of heads (35) Number of rabbits (12) = number of chickens (23).
Explanation: Let the rabbit and the chicken lift both feet at the same time, so that the number of feet in the cage is reduced by 2, since the chicken only has 2 feet, so there are only two feet left in the cage of the rabbit, and then divide by 2 to get the number of rabbits. Although in reality, no one is in the same cage.
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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 the cage? (Total number of feet - total number of heads * number of chicken's feet) (number of rabbit feet - number of chicken's feet) = number of rabbits (94 35 2) 2 = 12 (number of rabbits) total number of heads (35) number of rabbits (12) = number of chickens (23) solution:
If there are x rabbits, then there are (35-x) chickens. 4x+2(35-x)=944x+70-2x=942x=94-702x=24x=24 2x=1235-12=23 (only) original text.
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The question and answer of the chicken and rabbit in the same cage are:
There are 100 chickens and rabbits, and the number of legs of chickens is 28 less than that of rabbits.
Solution: 4*100 400,400-0 400 Suppose they are all rabbits, there are a total of 400 rabbit feet, then the chicken's feet are 0, and the chicken's feet are 400 fewer than the rabbit's feet.
400-28 372 The actual number of chickens' feet is 28 fewer than that of rabbits, a difference of 372 feet.
4+2 6 This is because if a rabbit is replaced by a chicken, the total number of rabbits will be reduced by 4 (from 400 to 396), the total number of chickens will increase by 2 (from 0 to 2), and their difference will be 4+2 6 less (i.e. the original difference is 400-0 400, and now the difference is 396-2 394, which is 400-394 6).
372 6 62 indicates the number of chickens, i.e. since there are 62 chickens out of 100 rabbits in the hypothesis, the difference in the number of feet is changed from 400 to 28, a total difference of 372 chickens.
100-62 38 indicates the number of rabbits.
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The problem of chickens and rabbits in the same cage is a widely circulated mathematical problem in the ancient mathematical work "Sun Tzu's Sutra" in China: there are chickens and rabbits in the same cage today, with 35 heads on the top and 94 feet under itTranslated into modern Chinese, it is:
Today, there are chickens and rabbits living in a cage, and it is known that there are 35 chicken heads and rabbit heads, and 94 chicken feet and rabbit feet. Ask how many chickens and rabbits there areThis ancient mathematical problem is common in real life, and there are many ways to solve it, but the general method is hypothetical.
When solving practical problems, sometimes the idea of "hypothesis" is used to analyze in order to find a way to solve the problem. To solve the problem with hypothetical thinking, we must first correctly judge how to assume according to the meaning of the question, pay attention to the changes in the quantitative relationship according to the assumptions made, and make appropriate adjustments from the comparison between the given conditions and the changed quantitative relationship to find the correct answer.
That is: equation problems made by elementary school students.
Put several numbers of chickens and rabbits in the same cage.
Then give a number of heads.
How many feet there are.
So as to find out how many chickens and rabbits are there?
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Chickens and rabbits have a total of 80 heads and 208 feet in the same cage, how many chickens and rabbits are there?
Analysis: Assuming that these 80 heads are all chickens, then the feet should be 2 80 160 (only), which is 208 160 48 (only) less than the actual one
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 have a few 2s in it, that is, a few rabbits.
Solution: (208 2 80) (4 2).
24 (only) - rabbits.
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.
80 56 24 (only).
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There are 4 kinds of reasoning, first of all, there are the same number of heads, and there are not many feet. That's the premise. The question is generally given how many heads and how many feet.
Then here comes the 4 methods:
1. Assuming that these heads are all chickens, count the number of feet in this case. So the number of feet that are more than this number is the number of rabbits multiplied by 2So we get the number of rabbits, and then we calculate the chickens.
2. Suppose these heads are rabbits, the number of hands in this situation. So the number of feet with less than this number is the number of chickens multiplied by 2So you get the number of chickens, and then you can calculate the number of rabbits.
3. Suppose that the feet are all chickens', and count the number of heads. So the number of conditional heads is less than this is the number of rabbits multiplied by 2
4. Suppose these feet are rabbits, count the number of heads. So the number of feet with less than this number is the number of chickens multiplied by 2(In this case, it should be noted that there may be a number of heads calculated at this time, don't panic, and finally multiply by 2).
If you have studied equations, you will find that these two methods are really just a few ways to solve equations.
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