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    Kristie Jones's Avatar
    Kristie Jones Posts: 1, Reputation: 1
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    #1

    Nov 11, 2013, 09:33 PM
    How many are barefooted
    There are 12 people in a room. 6 people are wearing socks and 4 people are wearing shoes , 3 people are wearing both . How many are in bare feet ?
    mande11's Avatar
    mande11 Posts: 3, Reputation: 1
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    #2

    Nov 11, 2013, 10:41 PM
    None
    joypulv's Avatar
    joypulv Posts: 21,591, Reputation: 2941
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    #3

    Nov 12, 2013, 03:44 AM
    'None' isn't correct. Since the numbers add up to more than 12, clearly the 3 people wearing both are a subset of the other two.
    So that DOES leave a few people barefoot. How many do you think now?
    Reinvented25's Avatar
    Reinvented25 Posts: 7, Reputation: 3
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    #4

    Nov 12, 2013, 03:50 AM
    3 people are only wearing socks
    3 people are wearing both
    1 person is wearing only shoes
    5 people are barefoot
    joypulv's Avatar
    joypulv Posts: 21,591, Reputation: 2941
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    #5

    Nov 12, 2013, 03:56 AM
    Nope - that answer is wrong too.
    Hint: you can't know how many people are in shoes only and socks only. You do know how many are in both, and can know how many are barefoot.
    Reinvented25's Avatar
    Reinvented25 Posts: 7, Reputation: 3
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    #6

    Nov 12, 2013, 04:08 AM
    Assuming all information has been given, and no other foot coverings are present, the answer is five.

    12 - 6 - (4 - 3) = 5

    This word problem is in "Critical and Creative Thinking Activities - Grade 5", though the numbers are slightly different.
    joypulv's Avatar
    joypulv Posts: 21,591, Reputation: 2941
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    #7

    Nov 12, 2013, 04:12 AM
    WHAT?
    How can you subtract those wearing both from those wearing shoes????
    If this is considered correct, our whole educational system is just as horrible as I suspected.
    No wonder most of the rich countries are way ahead of us in math.

    We don't usually give out answers on this site, only guidance.
    But I claim that the correct answer is 2.
    6 wearing socks, 4 wearing shoes. It doesn't matter which of those groups are the ones wearing both. There are 2 who are barefoot!!
    joypulv's Avatar
    joypulv Posts: 21,591, Reputation: 2941
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    #8

    Nov 12, 2013, 10:21 AM
    Does anyone out there agree or not? This is driving me nuts.
    smearcase's Avatar
    smearcase Posts: 2,392, Reputation: 316
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    #9

    Nov 12, 2013, 02:14 PM
    I get 5 joy. I did it by drawing pictures I must admit.
    And I got my wife to do it twice for backup.

    You said: "How can you subtract those wearing both from those wearing shoes????"
    You can, because that tells you how many (one) are wearing ONLY shoes.

    start with 12 x's, put sox beside 6 random x's.
    Add shoes to 3 of those already designated sox (as only 3 have both).
    You will have one of the 4 shoes left that has to go with one of the 6 unused x's. You have used 7 of the 12 x's leaving (12-7) = 5 barefooted.
    Wondergirl's Avatar
    Wondergirl Posts: 39,354, Reputation: 5431
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    #10

    Nov 12, 2013, 02:20 PM
    With a Venn diagram, I came up with 2 who are barefooted.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #11

    Nov 12, 2013, 03:23 PM
    The correct answer is indeed 5.

    Wondergirl - I wonder what your Venn diagran look like. Should be like this:
    Attached Images
     
    joypulv's Avatar
    joypulv Posts: 21,591, Reputation: 2941
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    #12

    Nov 12, 2013, 03:30 PM
    Oh shoot
    smearcase's Avatar
    smearcase Posts: 2,392, Reputation: 316
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    #13

    Nov 12, 2013, 03:55 PM
    WG,
    I think you didn't consider the fact that the 3 folks wearing both shoes and socks, can be 3 of the 6 people counted as wearing socks OR 3 of the 4 people counted as wearing shoes. This allows that 3 more people are barefooted.
    There were no ("only's") in the original problem statement.

    Edited to note comment was to WG. Hadn't seen ebaines diagram.

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