Olympiad topic · P2–P6
Chickens and rabbits
One of the best-known olympiad problem types: in the Hua Xia Cup it appears from the P2 first round to the P5 papers.
Try one
★★ MediumIn a cage on a farm there are chickens and rabbits, with 2 more chickens than rabbits.
Each chicken has 2 legs and each rabbit has 4; there are 40 legs in all.
How many rabbits are there?
- Hint 1: Take away the 2 extra chickens first. How do the rest compare?
- Hint 2: Then there are as many of each. A pair of one of each has how many legs?
- 1.First take away the 2 extra chickens. They have 2 × 2 = 4 legs.
- 2.That leaves 40 − 4 = 36 legs, and now there are as many of each.
- 3.Put one chicken and one rabbit together as a pair: 2 + 4 = 6 legs.
- 4.36 ÷ 6 = 6 pairs. Each pair has one rabbit, so there are 6 rabbits.
- ⚠️Common mistake: taking away just 2, forgetting each has 2 legs.
Answer: 6 rabbits
Four levels, step by step
- ★EasyFree2 question types · unlimited questionsSmall numbers you can try one by one.
- ★★MediumFree2 question types · unlimited questionsThe standard suppose-all method.
- ★★★HardFree2 question types · unlimited questionsPoints taken off for wrong answers, or the difference in legs given.
- ★★★★Expert (competition level)VIP2 question types · unlimited questionsThree kinds, or the two amounts mixed up.
Chickens and rabbits: what it is and how to solve it
Two kinds of thing mixed together: you know how many there are and how many legs (wheels, seats, points) in all; find how many of each.
For example: 10 chickens and rabbits have 28 legs. How many rabbits? If all were chickens there'd be 20 legs, 8 short; each rabbit instead of a chicken adds 2, so 8 ÷ 2 = 4 rabbits.
The key idea: Suppose: pretend they're all one kind, see how far off the total is, and how much each swap changes it. The usual method at school (from P5) uses an equation: let there be x rabbits, 4x + 2(10 − x) = 28.
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