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How to solve the problem of chickens and rabbits in the same cage in primary school mathematics?
Draw eight small circles to indicate that there are eight chickens and rabbits * * *, and then draw two legs under each head (indicating that all chickens are assumed), and calculate that a * * * has 16 legs, which is less than the actual situation of the topic 10. Then I asked why 10' s leg was missing. The student answered that rabbits are chickens, and each rabbit is missing two legs. Then we add two legs to some chickens to turn them into rabbits and make up the missing 10 leg. Finally, students can intuitively find five rabbits and three chickens. Students are very interested in drawing without formulas, and most students can understand the thinking method smoothly. So how to solve the above drawing method? (1) suppose all chickens: 8×2= 16 feet; (2) Less than the actual 26: 26- 16= 10 feet; (3) Take the rabbit as a chicken, and each counts its own: 4-2=2 (feet); (4) It is necessary to replace several chickens with rabbits: 10÷2=5 (only), so there are 8-5=3 (only) chickens.