If the goblet is in case 6, which one of the following artifacts must be next to the javelin?

zia305 on May 16, 2020

September 2019 logic games explanation

Can someone please explain the all the logic games for September 2019 LSAT. Thanks!

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Skylar on May 18, 2020

@zia305, happy to help! For the sake of clarity, I will post walkthroughs to each of the four games in separate messages. Here is Game #1:

Kittens: FGHJ
Puppies: RSTW
So, we have a total of 8 animals (4 kittens and 4 puppies), but we can only select 5 animals to put in the pens. This means we start out with any of the following possible combinations:
4 kittens and 1 puppy
3 kittens and 2 puppies
2 kittens and 3 puppies
1 kitten and 4 puppies

__ __ __ __ __ | __ __ __
1 2 3 4 5 OUT

Rule #1: Spots 1 and 5 = kittens (so F/G/H/J)
Rule #2: T cannot be next to any kitten (F/G/H/J). From Rule #1, we know that the 1st and 5th pens hold kittens. Therefore, T cannot be in pens 1 or 5. This also means that T cannot be in pens 2 or 4. So, if T is in, T must be 3rd and the 2nd and 4th spots must be puppies.
Rule #3: One of G/H is in, and the other is out.
Rule #4: W -> G(2). The contrapositive of this is: NOT G(2) -> NOT W.

__ __ __ __ __ | (G/H) __ __
1 2 3 4 5 OUT
K K

Now, let's revisit the possible combinations we wrote out before we looked at our rules. We can eliminate the "1 kitten and 4 puppies" option because Rule #1 tells us that we always have at least 2 kittens in. We can also eliminate the "4 kittens and 1 puppy" option because Rule #3 tells us that we cannot have all 4 puppies in. This leaves our options as:
3 kittens and 2 puppies
2 kittens and 3 puppies

QUESTION 1:
With these types of questions, we should go through rule-by-rule and eliminate any answer choices that break the rules.
- (E) breaks Rule #1 because it places T (a puppy) in the 1st spot.
- (D) breaks Rule #2 because it places T next to G (a kitten).
- (B) breaks Rule #3 because it places both G and H in.
- (C) breaks Rule #4 because W is in, but G is not in the 2nd spot.
- This leaves (A) as the correct answer. We should also write this diagram to the side quickly to use as reference for later questions if needed.

QUESTION 2:
This question places Wags in, so let's start at Rule #4, where we have a Sufficient condition of "W."
- Rule #4 tells us: W -> G(2). So, we know that G is in and placed in the 2nd spot.
- Rule #3 tells us that if G is in, H cannot be in. So H is out.
- Rule #2 tells us that if T is in, it would be placed 3rd and the 2nd and 4th spots would have to be dogs. However, we established that G (a kitten) is in the 2nd spot in this scenario, so T must be out.
- Since we can conclude that if W is in, both H and T must be out, (B) is correct.

QUESTION 3:
This refers to the second possible combination we identified - 2 kittens and 3 puppies.
- We know from Rule #1 that the 2 kittens must be in spots 1 and 5.
- Therefore, the 3 puppies fill spots 2, 3, and 4.
- So, we can first eliminate (A) because it places a kitten in spot 2.
- We know from Question 2 that if W is in, then G will be second, which again places a kitten in spot 2. So, (E) can be eliminated and W must be out.
- This leaves R, S, and T as the 3 puppies that must be in.
- Rule #2 tells us that if T is in, it must be placed 3rd. So, (D) is correct.

QUESTION 4:
This places F next to J.
- We know that we can only have a maximum of 3 kittens and 2 of these are assigned to spots 1 and 5. So, F and J must either be in spots 1 and 2 or in spots 4 and 5.
- Since a kitten must be in a spot 2-4, T cannot be in.
- W also cannot be in, because it would require that G is placed 2nd. This is impossible because if F and J are in spots 1 and 2, spot 2 would already be occupied. If F and J are in spots 4 and 5, there would only be one spot left open for a kitten and it would have to be spot 1. So, G could not go 2nd.
- Therefore, (E) Wags is correct.

QUESTION 5:
This tells us that J is out.
- We know that either G or H is always out. In addition to J now being out, this means that only 2 kittens are in. Following Rule #1, we know they are in the 1st and 5th spots. So (A) is out.
- As in our last question, this means that W must also be out because it would require G (a kitten) to be placed 2nd. So (E) is out.
- This leaves R, S, and T to be the three puppies in. Again, Rule #3 says that if T is in, T must be third. This eliminates (C) and (D).
- Therefore, (B) is the correct answer.

QUESTION 6:
This tells us that S is out.
- Let's see what happens if we try to put the remaining 3 puppies in. We notice that we run into a problem trying to place both T and W in. W requires that G be placed 2nd, which means that T cannot go 3rd.
- However, this same issue would happen if we tried to put the remaining 3 kittens in instead, because T cannot be in unless there are a total of three puppies in. So, T has to be out.
- This leaves (F/J) to go 1st and 5th, G to go 2nd, and (W/R) to go 3rd and 4th.
- So, (C) is correct.

Hope this helps!

Skylar on May 18, 2020

@zia305, here is Game #2.

GHJMNPS

<- __ __ __ __ __ __ __ ->
1 2 3 4 5 6 7

We are told that the display cases are arranged in a circle. Don't let this throw you off! We can still draw a linear diagram for clarity, but I recommend putting arrows on the end to remind yourself of the circular arrangement. The key is that we know case 1 and case 7 are next to each other.

Rule #1: spot 7 = H/J
Rule #2: N < M
Rule #3: HM or MH
Rule #4: NOT SP and NOT PS
Rule #5: NOT PJ and NOT JP and NOT SJ and NOT JS

<- __ __ __ __ __ __ (H/J) ->
1 2 3 4 5 6 7

QUESTION 1:
The best approach to this type of question is to go through rule by rule and eliminate answer choices with violations.
- (C) violates Rule #1.
- (A) violates Rule #2.
- (D) violates Rule #3.
- (E) violates Rule #5. Remember, 7 and 1 are next to each other.
- This leaves (B) as the correct answer. We should write out the arrangement to the side as a free hypothetical to be used for later questions.

QUESTION 2:
We know that G is 6th. Let's write this out with our two scenarios for the 7th spot.
<- __ __ __ __ __ G H ->
1 2 3 4 5 6 7
- Under this scenario (when H is 7th), M must be 1st. This is because Rule #3 states that H and M must be next to each other and G is in the other spot adjacent to H.
- However, this poses a problem because if M is in the first spot, it is impossible for N to be in a lower-numbered spot. Therefore, this setup violates Rule #2.
- So, H cannot be 7th. Let's look at the alternate scenario.
<- __ __ __ __ __ G J ->
1. 2 3 4 5 6 7
- Rule #5 says that neither P nor S can be next to J. Remember, spots 1 and 7 are next to each other since the cases are arranged in a circle. Therefore, neither P nor S can be 1st.
- Rule #3 says that M and H must be next to each other and Rule #2 says that M must be in a higher-numbered case than N. Therefore, neither M nor H can be 1st.
- This leaves only N to be 1st, which is answer choice (C).

QUESTION 3:
We know that S is 3rd. Let's write this out with our two scenarios for the 7th spot.
<- __ __ S __ __ __ H ->
1 2 3 4 5 6 7
- Rule #3 says that M and H must be next to each other and Rule #2 says that M must be in a higher-numbered case than N. Therefore, M must be 6th.
- Rules #4 and #5 say that S, P, and J must all be separated. This means that either P or J needs to go 1st and the other needs to go 5th.
- This leaves either N or G to fill spot 2 and the other to fill spot 4.
<- (P/J) (N/G) S (G/N) (J/P) M H ->
1 2 3 4 5 6 7
- Before we try to write out the scenario where J is 7th instead of H, let's check in with our answer choices and see if we can already answer the question.
- The question asks what "must be true," and there is only one answer choice that is definite in our diagram - (C). So, (C) is correct and M must be 6th.

QUESTION 4:
We are asked which answer choice could be true. We should start out by looking for past hypotheticals that prove one of the answer choices could work.
- The hypothetical we just drew in Question 3 shows that P could be 1st.
- So, (D) is correct.

QUESTION 5:
Like the previous question, we are asked what could be true and we should first start by checking for past hypotheticals that may correspond with one of the answer choices.
- The hypothetical we drew in Question 3 shows that J could be 1st while S is 3rd.
- Therefore, (E) is correct.

QUESTION 6:
We know that P is 2nd. Let's write this out with our two scenarios for the 7th spot.
<- __ P __ __ __ __ H ->
1 2 3 4 5 6 7
- We know that in order to fulfill Rule #2 and Rule #3, M would have to be 6th.
- However, Rule #4 and Rule #5 say that P, S, and J must all be separated. This is not possible in the current arrangement.
- So, H cannot be 7th. Let's look at the alternate scenario.
<- __ P __ __ __ __ J ->
1 2 3 4 5 6 7
- In order to separate P, S, and J like Rules #4 and #5 call for, S must be placed either 4th or 5th. Remember that the 7th spot is next to the 1st spot.
- Let's try putting S 4th first.
<- __ P __ S __ __ J ->
1 2 3 4 5 6 7
- Rule #3 says that M and H must be next to each other. The only two adjacent open spots for them are 5 and 6.
- This means that either G or N would go in spot 1, and the other would go in spot 3.
- Let's check in with our answer choices at this point.
- M could be 6th, so (D) is correct.

Hope this helps!






Skylar on May 18, 2020

@zia305, for Game #3, please see the message board on the last question of that game. The post history there contains both a walkthrough of each question and a grid to infer all possible combinations.