September 2019 LSAT Section 1 Question 8

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

3 Replies

Skylar on May 18 at 01:14AM

@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 at 02:13AM

@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 at 02:27AM

@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.