Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 months ago

Help me with this question: On school camp a group of children went horse riding. Each child had a horse. Half decided to ride their horse while the other half decided to walk beside it. If there were 70 legs on the ground, how many children were there?

On school camp a group of children went horse riding. Each child had a horse. Half decided to ride their horse while the other half decided to walk beside it. If there were 70 legs on the ground, how many children were there?

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  • 7 months ago

    Let x = number of kids.

    Half are riding a horse so there are 4 legs on the ground, and 4(x/2) or 2x is the number of legs on the ground.

    Half are walking beside a horse so there are 6 legs on the ground, so 6(x/2) or 3x is the number of legs on the ground.

    The total number of legs on the ground is 70, so:

    2x + 3x = 70

    5x = 70

    x = 14

    There are 14 kids.

    Let's check this.

    Each kid has 2 legs, so 14 kids have 28 legs total.

    Half of the kids, which is 7, have 14 legs total.

    Each kid has a horse and each horse has 4 legs.

    For each kid on a horse there are 4 legs total on the ground.

    For each kid walking besides a horse there are 6 legs total on the ground.

    7 kids are on a horse so there are 28 legs total on the ground.

    The other 7 kids are besides a horse so there are 42 legs total on the ground.

    28 legs + 42 legs = 70 legs total.

    There are 14 kids.

  • 7 months ago

    that sounds so weird “70 legs on the ground” lol whoever is writing these word problems really running out of ideas

  • 7 months ago

    INTUITIVE METHOD:

    Half are riding on the horse (4 legs) and half are walking by the horse (6 legs), so the average number of legs per child (with horse) is 5 legs.

    70 / 5 = 14 children

    ALGEBRAIC METHOD 1:

    Let x be the total number of children.

    Half (x/2) are riding --> 4 legs

    4(x/2) --> 2x

    Half (x/2) are walking --> 6 legs

    6(x/2) --> 3x

    Together that's 70 legs:

    2x + 3x = 70

    5x = 70

    x = 70/5

    x = 14

    ALGEBRAIC METHOD 2:

    Let x be the total number of children.

    They each have a horse (4 legs), so that would be 4x legs for the horses.

    Half of them are walking, so that would be (½x)*2 = x legs for the children that are walking.

    Together that's 70 legs:

    4x + 1x = 70

    5x = 70

    x = 70/5

    x = 14

    Answer:

    14 children (with 14 horses, 7 are walking next to them).

  • 7 months ago

    14

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  • 7 months ago

    Continuing Ian's logic, since there were 7 pairs of children, that's 7 * 2 = 14 children.

  • Ian H
    Lv 7
    7 months ago

    The number deciding to ride was equal to the number that decided to walk.

    The horses 4 legs and the 6 legs of horse + walking child means that there are 10 legs for every 2 children, (first observed by Φ² = Φ+1)

    Since were 7 lots of 10 legs on the ground, there were 7 pairs of children

    So that makes .......... ?

  • Amy
    Lv 7
    7 months ago

    number of horses = number of children

    Number of legs on the ground = (4 x number of horses) + (2 x 1/2 x number of children)

  • 7 months ago

    Hint: For every pair of children (one riding their horse, one walking their horse) they are standing on 10 legs.

    If there are 70 legs then ...

    • roderick_young
      Lv 7
      7 months agoReport

      That was exactly my impulse thinking, trying to avoid using paper and pencil. Your answer is good, but I think when someone asks this kind of question, they could benefit from being shown how to set it up in a standard way.

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