Answer this simple yet incredibly hard math question?

I bet you can't get it!

kids on a field trip are gathered on in the barn with cows. If there are 11 heads and 28 legs, how many kids and how many cows are in the barn?


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Most Helpful Guy

  • 8 children
    8x2=16
    3 cows
    3x4=12

    16+12=28

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Most Helpful Girl

  • None cuz I made dinner Ayyy

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What Guys Said 8

  • 8 kids
    3 cows

    Or possibly
    6 kids
    4 cows

    Or if some kids have one leg or more then two or some cows are deformed than it could be many answers!

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  • If we don't consider gene mutation, as in, every kid has one head and 2 legs and every cow has got one head and 4 legs, we can kill this question like:
    Assume we have X kids and Y cows
    Then we got these equationz:
    x + y =11
    2x + 4y = 28

    ∴x=8, y=3

    There u go!!

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  • 8 kids and 3 cows. Hope we helped with your homework.

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  • Do your own home work, cheater :-P

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  • Its a problem of least common denominator. I'm not doing the math.

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  • 3 cows
    8 kids.

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  • 8 kids and 3 cows.. nothing difficult 😑

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  • A + B = 11
    2A + 4B = 28

    A = kids
    B = cows

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    • great. you turned it into an equation. but you didn't solve it. what was the point?

    • @ksoma didn't want to bother since it's a simple two equation with two unknowns.

What Girls Said 1

  • 8 people, 3 cows

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