PT36.S4.G1 - A fruit stand carries at least one kind

Bougie_BreBougie_Bre Member
edited January 2019 in Logic Games 19 karma

I noticed that I have been struggling with In/Out games. Therefore, I’m specifically targeting fool proofing these types of games. The plan is to go from easy, medium, and hard. I’ve done Prep Test 24,29,48,54,63 game 1’s and get all of those in 5 mins or less and it didn't even take me 4 try's. However, I’ve been struggling with Prep Test 36 game 1. I’ve watched JY’s video for it a number of times, and I have attempted the game at least 5 times.I'm still getting only 1-3 right, I know the answers of the others but I'm having a hard time understanding how they are right, and finding why them on my own (4-6). Any tips on understanding specifically 4 and 5?

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Comments

  • blackberryblackberry Alum Member
    218 karma

    Okay, let me give it a shot.

    First, you have your 6 game pieces:
    f k o p t w

    The rules are (I've teased them out, for simplicity for viewing it here):
    1) k --> /p (standard not both rule)

    2) /t --> k (standard or rule)

    3) o --> p and w

    4) w --> f or t or both (inclusive or)

    #4 MBT
    The stimulus is that w is out. This implies that o is out. You know this from rule 3 (apply DeMorgan's rule, /p or /w --> /o; in this case, because w is out, it takes o with it; or in the sufficient condition means either one of the subjects trigger the necessary condition). You still have to remember rule number 1 and 2. 1 is a not both rule. With not both rules, either one of the subjects MUST be out (it can also be that both are out, but at LEAST one MUST be out). So now what we know is out are: w, o, k/p. So a minimum of 3 are out. Remember how many game pieces you had to start with: 6. Therefore, it must be the case that the MAX that are IN are 3, hence (C) The stand carries at MOST three kinds of fruit.

    #5 MBF
    1 MBF
    4 CBTs
    The stimulus tells us w is in. Nothing much triggers except for rule #4, either f or t (or both) are in with w. Let's brute force our way down the answer choices.

    Can (a) be true? Yes! If f is out, it means t is in according to rule #4. Let's randomly assign the remaining 3 game pieces as per below -

    IN OUT
    w f
    t p
    k o

    Can (b) be true? Yup! Goes in the opposite as (a). If t is out, f must be in with w & also randomly assign the remaining -

    IN OUT
    w t
    f o
    k p

    Can (c) be true? Absolutely! p out implies o out, because DeMorgan's law. Let's fill up with the leftovers -

    IN OUT
    w p
    t o
    k
    f

    Can (d) be true? Sure can! With the below chart, it implies k must be OUT because k and p hold a not both rule, and p is in. Next, this means t must be in, because t and k hold an or rule, and k is already out, so t must be in. Now, all you have is f to deal with, can go IN or OUT

    IN OUT
    w o
    p

    IN OUT
    w o
    p k

    IN OUT
    w o
    p k
    t

    Can (e) be true? No :( Since t is out, f must be in. one of k/p must be out, and p is in, so let's put k out. Oh no :( Now, you can't satisfy rule # 2 the or relationship between t and k since both are out. Since MUST BE FALSE, aka the correct answer.

    IN OUT
    w t
    p

    IN OUT
    w t
    p
    f

    IN OUT
    w t
    p k
    f

  • blackberryblackberry Alum Member
    218 karma

    Opps, forgot about

    #6 MBF
    4 CBTs
    1 MBF

    In this stimulus, it tells us to discard #2, so our new comprehensive set of rules are:

    1) k --> /p (standard not both rule)

    3) o --> p and w

    4) w --> f or t or both (inclusive or)

    With this, we need to find which one CANNOT be a complete and accurate list of the in fruits:

    (a) works; must put k out because of the not both rule; if we put w out, #4 becomes irrelevant because sufficient condition not triggered; #3 /p or /w --> /o, so put o out since w is out
    IN OUT
    p k
    w
    f
    t
    o

    (b) works; put k out because p is in, need to satisfy #1; w and o can be out due to DeMorgan's; t can go out
    IN OUT
    f k
    p w
    o
    t

    (c) does NOT work!! If w is, according to rule #4, one of f or t, or both must be in! But it's not :(
    IN OUT
    o
    p
    w

    (d) works! k must be out due to the not both rule with p; rule #4 is satisfied because f is in with w; o can be out
    IN OUT
    f k
    p t
    w o

    (e) works! rule #3 is satisfied; put k out to satisfy #; t can be out because f is already in with w to satisfy #4
    IN OUT
    f k
    o t
    p
    w


    I think it would be good for you to review your lessons on OR and NOT BOTH rules in in/out games. A lot of the questions required an understanding of it, so you know which game piece needs to go in, when the other is out, etc. Hope this helps!

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