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6 May, 10:02

Jordan count 28 legs on 8 animals if the animals are either zebras or pecocks how many zabras are there

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  1. 6 May, 10:21
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    This is obviously a problem for system of two equations.

    Let

    Z=number of zebras (4 legged)

    P=number peacocks (2 legged).

    We know that

    P=8-Z (total of 8 animals)

    and

    4Z+2P=4Z+2 (8-Z) = 28

    expand and solve for Z

    4Z+16-2Z=28

    2Z=12

    Z=6 There are 6 zebras.

    Another way to solve, without pen and paper:

    if all are zebras, there are 4*8=32 legs, or 4 legs too many.

    Each exchange with a peacock reduces 2 legs, so we exchange 2 zebras with 2 peacocks to get 6 zebras.
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