Challenging P6

On a farm, there are 28 chickens and cows. The cows have 40 more legs than the chickens. How many cows and chickens are there?"


28 x 4 = 112 legs        28 cows

112 - 40 = 72 legs     Difference too much (so needs to be reduced)

4 + 2 = 6 legs     Exchanging 1 cow for 1 chicken (difference in legs increases by 6)

72 / 6 = 12 chickens    12 cows need to be exchanged for 12 chickens

28 - 12 = 16 cows

There are 16 cows and 12 chickens.


Alterrnative method
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40 / 4 = 10 cows   Remove the additonal legs from cows, so now the remaining cows and chickens have the same number of legs

28 - 10 = 18        Remaining cows and chickens

1 + 2 = 3             1 set [1 cow (4 legs) --> 2 chickens (4 legs)]

18 / 3 = 6 sets

6 x 2 = 12 chickens

28 - 12 = 16 cows

There are 16 cows and 12 chickens.


Alternative method
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4 - 2 = 2 legs     1 set [Difference between 1 cow and 1 chicken]

40 / 2 = 20 sets

20 + 20 - 28 = 12     Too many animals, need to reduce the number by 12.

2 + 1 = 3         1 group [To ensure that the difference in number of legs is maintained, whenever 1 cow reduced, 2 chickens would have to be reduced]

12 / 3 = 4 groups    To be reduced

20 - 4 = 16 cows

20 - 4 x 2 = 12 chickens

There are 16 cows and 12 chickens.