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logic question driving me insane

EarthenrockEarthenrock Registered User regular
edited September 2009 in Help / Advice Forum
So despite earning all my credits for my AA degree, My degree was withheld because I didn't meet the requirements for computational skill, which is unnerving because I've completed both Calc 1 and 2. But a C in Pre-Calculus somehow did not meet the GPA requirement for mathematics, and now I must take a special portion of the CPT, since the CLAST no longer exists.
Sorry had to vent, now on to the practice question that's bugging me.


Study the information given below. If a logical conclusion is given, select that conclusion. If none of the conclusions given is warranted, select the option expressing this condition.

All beachcombers are swimmers. All swimmers wear swimsuits. Sally is wearing a swimsuit,

A. Sally can swim
B. Sally cannot swim
C. Sally is a beachcomber.
D. None of the above is warranted



The answer is D but I just don't understand that. I thought the answer would be C since sally is wearing a swimsuit, and since all swimmers wear swimsuits, and all swimmers are beachcombers...I think Sally would be a beachcomber.
Anyone have any input?

This is a simple test it seems, but I find it frustrating that I just got done last semester integrating partial fractions and now I have to go back and refresh my memory of basic geometry and other junk.

Earthenrock on

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    DekuStickDekuStick Registered User regular
    edited September 2009
    Lack of information leads to D being the answer.

    All beachcombers are swimmers - Nothing states Sally is at the beach or a beachcomber.

    All swimmers wear swimsuits - Nothing is stopping a non-swimmer from wearing one

    Thats how I saw it.

    DekuStick on
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    OrogogusOrogogus San DiegoRegistered User regular
    edited September 2009
    All swimmers wear swimsuits, but not everyone who wears a swimsuit is a swimmer.

    If Sally wasn't wearing a swimsuit you could deduce that she wasn't a swimmer (since all swimmers wear swimsuits), but you can't go the other way. I forget the terms for these things, geometry class was a long, long time ago.

    Anyway, by the same token, all humans breathe air, but not everything that breathes air is human. If you know that Snarlgax the Destroyer breathes air, you can't automatically assume that it's human.

    Orogogus on
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    Napalm DonkeyNapalm Donkey Registered User regular
    edited September 2009
    I think it is because "all swimmers wear swimsuits" does not mean all people who wear swimsuits are swimmers and "all beachcombers are swimmers" does not mean all swimmers are beachcombers.

    Napalm Donkey on
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    EarthenrockEarthenrock Registered User regular
    edited September 2009
    I think you're right napalm, since using that type of reasoning seems to make sense and is consistent with other questions I have like this.

    Earthenrock on
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    SavantSavant Simply Barbaric Registered User regular
    edited September 2009
    Do you have any experience with mathematical sets or pure logic?

    "All beachcombers are swimmers": Beachcombers are a subset of swimmers. That means that every beachcomber can swim, however there possibly exist swimmers that are not beachcombers. With this information alone you cannot tell if there are non-beachcombers who can swim.

    "All swimmers wear swimsuits": Swimmers are a subset of people who wear swimsuits. Like before, any swimmer is necessarily going to wear a swimsuit, but it may be possible for some people wearing swimsuits to not be swimmers.

    "Sally is wearing a swimsuit": This means that Sally is a member of the set of people wearing swimsuits. So it is possible that she can swim or be a beachcomber, she could be a member of the portion of the swimsuit wearing set that doesn't swim, or if she does swim she could be a swimmer that isn't a beachcomber.

    Note that if she wasn't a swimsuit wearer, then you could be sure that she was neither a swimmer or a beachcomber. This is the sort of thing you may want to draw Venn-diagrams for until you are comfortable with it.

    Savant on
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    musanmanmusanman Registered User regular
    edited September 2009
    if p then q does not mean that if q then p

    That's the fallacy you're not getting. (as everybody is saying, I'm just throwing the logic at ya)

    musanman on
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    MurphysParadoxMurphysParadox Registered User regular
    edited September 2009
    To build on musanman's point:

    Think about the inverse. If swimmer, then swimsuit. if not swimsuit then not swimmer (as indicated, you cannot be a swimmer if you don't have a swimsuit because all swimmers have swimsuits). So if p -> q then the opposite is !q -> !p.

    Furthermore, there's even a truth table!
    If p is true and q is true, good deal (swimmer in swimsuit).
    If q is false and p is false, we are fine (no swimsuit and not a swimmer).
    If p is true and q is false, that is bad (swimmer without a swimsuit)
    If p is false, no one cares about q (if not a swimmer, then it doesn't matter whether they have a swimsuit on).

    My logic professor explained p -> q in this way: If my computer breaks (p = broken computer), and it is under warranty (q = under warranty), then good. If my computer breaks and it isn't under warranty, sadness. If my computer doesn't break, then I could care less about the warranty. If my computer is under warranty doesn't have anything to do with whether or not it is working. Only in cases of p = true do we care to look at q.

    Is Sally a swimmer? Well, it doesn't say she is a beachcomber or a swimmer, so no... thus who cares what she's wearing. Of course, if she was also a beachcomber, then we'd know she's a swimmer and must be wearing a swimsuit.


    OPTION 2 (visual learner):

    Draw a venn diagram. Start with a circle and write beachcombers in it. Then draw another circle around that and write swimmers. then another circle around that and write wears swimsuit. Where does Sally fit into the diagram according to the information you have? All we know is that she has a swimsuit, so she is somewhere in the biggest circle... which contains all swimmers, and thus all beachcombers, but also a lot of other people that aren't swimmers.

    MurphysParadox on
    Murphy's Law: Whatever can go wrong will go wrong.
    Murphy's Paradox: The more you plan, the more that can go wrong. The less you plan, the less likely your plan will succeed.
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    EarthenrockEarthenrock Registered User regular
    edited September 2009
    This is awesome, thanks guys.

    Earthenrock on
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    TasteticleTasteticle Registered User regular
    edited September 2009
    QUICK version:

    'All beachcombers are swimmers' and 'All swimmers wear swimsuits' means that All beachcombers thus wear swimsuits.

    Nothing in the phrase implies a person wearing a swimsuit HAS to be a beachcomber or a swimmer

    Tasteticle on

    Uh-oh I accidentally deleted my signature. Uh-oh!!
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    DarkewolfeDarkewolfe Registered User regular
    edited September 2009
    That type of logic problem is called a syllogism, fyi.

    Edit: And the trick to understanding one of the things they're trying to teach you is to stop thinking about what you know outside the logic problem. The only information you have access to is contained in the first two rules. It throws people for a loop for instance when you do a puzzle like this.

    All frogs are dogs.
    All dogs have 13 legs.
    All frogs have 13 legs.

    That one is correct.

    Darkewolfe on
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    GoodOmensGoodOmens Registered User regular
    edited September 2009
    The fallacy at play here is called "affirming the consequent," if you want to impress your teacher with fancy words.

    GoodOmens on
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    -Phil--Phil- Registered User regular
    edited September 2009
    just like all rectangles are rhombi, however, not all rhombi are rectangles....


    Insert comic strip here....

    -Phil- on
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    DaenrisDaenris Registered User regular
    edited September 2009
    -Phil- wrote: »
    just like all rectangles are rhombi, however, not all rhombi are rectangles....


    Insert comic strip here....

    Well... no. All squares are rhombi, but not all rhombi are squares. A rhombus has 4 equal sides, so not all rectangles are rhombuses

    Daenris on
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    TheMarshalTheMarshal Registered User regular
    edited September 2009
    Daenris wrote: »
    -Phil- wrote: »
    just like all rectangles are rhombi, however, not all rhombi are rectangles....


    Insert comic strip here....

    Well... no. All squares are rhombi, but not all rhombi are squares. A rhombus has 4 equal sides, so not all rectangles are rhombuses

    He was probably thinking of a parallelogram, which is the kind of rhombus... well... you know...

    TheMarshal on
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    shadydentistshadydentist Registered User regular
    edited September 2009
    TheMarshal wrote: »
    Daenris wrote: »
    -Phil- wrote: »
    just like all rectangles are rhombi, however, not all rhombi are rectangles....


    Insert comic strip here....

    Well... no. All squares are rhombi, but not all rhombi are squares. A rhombus has 4 equal sides, so not all rectangles are rhombuses

    He was probably thinking of a parallelogram, which is the kind of rhombus... well... you know...

    If someone tells you that all A is B, then the only other true statement you can draw from that fact is that all NOT b is NOT a.

    So, all rhombi are parallelograms, so therefore all non-parallelograms are non-rhombi,

    shadydentist on
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