Howdy, Stranger!
It looks like you're new here. If you want to get involved, click one of these buttons!
Categories
- 32.3K All Categories
- 27.2K LSAT
- 16.6K General
- 5.1K Logical Reasoning
- 1.3K Reading Comprehension
- 1.7K Logic Games
- 73 Podcasts
- 193 Webinars
- 9 Scholarships
- 193 Test Center Reviews
- 2K Study Groups
- 108 Study Guides/Cheat Sheets
- 2.4K Specific LSAT Dates
- 14 June 2024 LSAT
- 3 April 2024 LSAT
- 11 February 2024 LSAT
- 23 January 2024 LSAT
- 37 November 2023 LSAT
- 42 October 2023 LSAT
- 12 September 2023 LSAT
- 38 August 2023 LSAT
- 27 June 2023 LSAT
- 4 August 2024 LSAT
- 30 Sage Advice
- 4.8K Not LSAT
- 4K Law School Admissions
- 11 Law School Explained
- 10 Forum Rules
- 581 Technical Problems
- 276 Off-topic
.
loganaldapegarcia
Alum Member
.
Comments
Chances are there are not two that are correct. I don't say it to be snide, but the LSAC is really good at making sure that there is only one correct answer.
Can you post the PT/Section and question #?
Not in LG. I've never encountered one, at least. I've encountered one incorrect answer choice marked on an answer key, though.
.
Often times, this will come down to a tempting (albeit incorrect) "could be true" answer and a correct "must be true" answer. It isn't always easy to tell what could be true and what must be true apart in the beginning. Telling the difference will come with time. Just keep fool proofing and you will eventually learn to eliminate tempting CBT answers on questions like this.
There will NEVER be two "must be true" answer choices on a MBT question — always just one answer choice that has to be true no matter what in any possible world, and then 4 incorrect CBT/MBF choices.
Another helpful strategy is just to move forward in the game/review previously completed sub-gameboards for other questions in the game. You can usually use this to help you tell which one of your remaining answer choices "must" be true — it will occur in every sub-gameboard you've drawn!
@loganaldapegarcia
Yes I will, but make sure you edit this. We can't have the exact text. Delete it and replace it with the PT# section and question.
Everyone of those can be true except B.
We know that only 4 of the 6 cities will be visited.
Of those 4 cities H and T must be included, but cannot be next to each other.
Answer choice B says;
Based on our rules J must go 3rd. Which places M fourth.
That means H/T have to go first and second, and that puts them next to each other which violates our first rule.
Therefore all are true except B.
Which other one do you think is right?
Thinking about it more, you asked if the question is a "must be true" and that question is not a "must be true" but instead a "must be false" question. Maybe this is what happened?
So in this case we know J has to be 3rd and H on 2nd spot. We can't have T be in 1st spot so T has to be in 4th spot. So what can go in 1st spot? Not M. But one of O or S could go there.
OHJT
In this case J is 3rd and M on 4th position. Since H and T have to be both in they have to occupy 1st and second spot. But this violates our first rule that says H and T cannot be visited consecutively. So this is our right answer as it is Must be False.
O can be 4th, H and T can occupy 1st and 3rd spot and M 2nd spot. So this is fine and could be true.
HMTO
My work in answer choice C proves this is possible.
If H and T are occupying 1st and 4th spot, I can put M in 2 and J in 3.HMJT
So this is also possible.
I hope this helped.
P.S. I believe posting an entire LSAT LG stimulus is not allowed.
Which PT?
@loganaldapegarcia
In this case only C must be true.
We have to put S 4th.
------ / ------- / ------ / ---S---
Since S is in, we have to kick O out. (if O goes in S can't)
Then we know H/T are both in but not next to each other. H/T have to go in slot 1 and 3. H can go 1 or 3 and T can go 1 or 3. We do this so they are not next to each other.
--h or t--- / ------ / --h or t-- / --s--
Since h or t has to go 3rd then J is out. This leaves M as our only letter to separate h and t. So M must go 2nd.
which other answer do you see as correct?
lol just saw this.
If S is 4th we know O is out. We also know H and T must occupy 1st and 3rd position since they cannot be consecutive. So 2nd spot is the only one left open. J cannot go in there. The only one that can go in 2 is M. So answer choice C is correct.
H has to go in 1st or 3rd.
J is out as the 3rd spot is occupied by H or T.
Bingo.
O is out as S is in.
T can be 1st or 3rd. It doesn't have to be in 1st.
We saw letter B as the correct answer because J has to be 3rd.
J has to go 3rd ONLY IF it is included. It can be the case that J is excluded. And since S is 4th we have to exclude J all together. If we put J in, and S is fourth. Then T and H have to go next to each other in 1/2 and that breaks our first rule.
If J is 3rd, and S is 4th and H and T have to be in but they cannot be consecutive, where do you plan to put H and T?
Also, you made a post earlier about what the rule about J means. I believe that's where you are making a mistake - thinking J has to be in and that way the 3rd slot is occupied. J doesn't have to be in and it can be out. The rule only triggers when J is in, when J is not in there rule is useless.
Thank you guys. We got it now (:
@loganaldapegarcia
Oh haha I didn't see that @Sami was answering too. Use her advice, I'd bet she is much more articulate than I am. She'll make way more sense!
lol na aaaa. You are also very articulate and by the way faster than me. I couldn't type fast enough to keep up with you! lol. You beat me to it every time
Haha there is nothing to do at work right now. I am soooooooo bored. Every time I got a little red bubble I'd reply. I didn't even look to see if others were doing it!
Please edit your post! We can't post full LSAT material. Better to just reference the PT # and game #.
That said, in this question, C must be true, and E could be true. The others must be false. So clearly, C is the correct answer.
This is a combo in/out and sequencing game. There are 4 "in" slots and 2 "out" slots.
We know that H & T must always be in but cannot be consecutive. We have this:
__ __ __ S
Since H and T must be in but cannot be consecutive, they must go in slots 1 & 3, but we don't know in which order. So it looks like this:
H/T __ T/H S
Now, we also know that if J is in, then it must be in slot 3. But, oh no! Slot 3 is taken. So, J must be out. We also know if O is in, then S is out. However, we failed that necessary condition because S is in, so we also know that O is out.
(O --> /S and contrapositive S --> /O)
So, this leaves us with 1 empty slot and just the M left over. And now we have this:
H/T M T/H S
So that shows us that C is correct, M must be 2nd. E could be true, because it's possible that T visited first. But those 2 could be switched with H in 1st and T in 3rd. So that makes it a could be true, not must be true.
Hope that helps!
Annnnnd now I just saw that this has already been answered haha. But I'm leaving it up anyway for posterity I guess. No use in deleting my post lol.