HELP

majansen0623

If a logic game has a logical and global question combined. How do I go about it with the rules? Let's say we have 6 kinds of food... k,l,m,o,t,z. Each is added once on a scale of 1-6.

IF the M is 3rd then the L is last
IF Z is 1st then L is added before the O
Neither T nor K is 5
M before T or K but not before both.

One question said-- If the lentils are added last, then which one of the following must be true?

but we see the if the L is last then M is 3rd?

idk im confused

Help

valentina.soares-1

Hi there @majansen0623 ,

Great question! I just want to quickly run through the set up/rules on this game to make sure we’re on the same page about how much information you should ideally have on hand before attempting this question.

Rule 1: M3 -> L6
Rule 2: Z1 -> L - O (-> L/6 -> M/3)
If L - O then L cannot be 6, so that activates the contrapositive of Rule 1 meaning if Z is in 1, then L cannot be 6 and M cannot be 3.
Rule 3: T and K slash under 5th spot
Rule 4: M is before either T or K but not both.
Therefore, M is before one of T/K and after one of K/T
Looks something like this T/K - M - K/T
You could also use a switch to represent T and K trading off.

This tells us that M cannot be 1 or 6, because it would be either before both or after both T+K which is not permitted.

Set Up:
You should have 6 slots with M crossed out under 1 and 6 and T and K crossed out under 5. You should also have your list of rules, additionally noting that if Z1, then M cannot be 3.

Question 3 - if L is last, then what MBT?

First thing you want to do is accommodate your additional premise on a new game board. If you’re operating at a level on Logic Games where you can do this in your head just by looking at the rules, then that’s great. However, until then its best to copy over your game board and accommodate that additional premise.

The next thing you want to do is check if any of your rules are implicated by putting L in 6. Our first two rules both talk about L, specifically, L in 6.

Rule 1 has L6 in the necessary condition. Our additional premise affirms that necessary condition, and the rules of conditional logic tell us that affirming a necessary condition does absolutely nothing. It makes the rule fall away. This isn’t very helpful to us in trying to make an inference. Hopefully this answers your question about having L last leading to M in 3rd. It doesn not lead to M in 3 because you can't go backwards on the arrow without negating your ideas. I’ll drop in the rules to remember about conditionals in LG below for your reference:

If the sufficient condition is triggered, the necessary condition is triggered.
If the sufficient condition is failed, the rule falls away (becomes irrelevant/nonexistent).
If the necessary condition is triggered, the rule falls away.
If the necessary condition is failed/negated, the sufficient condition is failed/negated.

However, because we thought about how rules 1 and 2 interact with each other while doing our set up, we know that a certain sufficient condition leads to L being locked out of the 6th spot. In rule 2 we discovered that If Z is in 1, then L cannot be in 6. This is because if Z is in 1 then L is before O and because that makes L a leader of 1 item (O), it cannot go in the last spot.

For more clarity, if we contrapose that chain we created in Rule 2 (Z1 -> L - O (-> L/6 -> M/3)), it looks like this:
L6 -> /(L - O) -> /Z1
You could also represent it this way since its a one-to-one sequencing game
L6 -> O - L -> /Z1

The inference is Z cannot be in 1, and because this is a MBT question, we want to hunt for that inference in the answer choices before proceeding.

Choice A says that at least something is before Z. This is great and must be the case because we know Z cannot be in 1. The AC is essentially just another way of saying that.

Let me know if you have any questions or need further clarification!

Valentina