If you haven’t taken the time to try the LSAT Practice Sequencing Game 2 on your own before reading these explanations, please click back there and give it a try first. These are the explanations for the second sequencing game in a series of lesson articles about LSAT sequencing games. The series includes:
- LSAT Sequencing Games: an Introduction
- Sequencing Game Questions
- LSAT Practice Sequencing Game 1
- LSAT Practice Sequencing Game 2
- LSAT Practice Sequencing Game 3
Sequencing Game 2 Explanations
How do you know this is a sequencing game (besides the fact that it is featured in an article called “sequencing game explanations”)? At first you might think it is a grouping game because you are assigning animals to pens.
You know this is sequencing and not grouping because of the rule in the setup “Each pen can only house a single giraffe.” Since the pens are numbered 1 through 6 in a row, you’re just ordering the giraffes from 1 to 6.
So the first part of your sketch should be six lines with the numbers 1 through 6 under them. You also want to add the giraffes names, just the first initials, so you have a list to work with:
That’s as far as you can get in your sketch before dealing with the rules.
Rules and Deductions
When dealing with the rules in Analytical Reasoning, it’s best to paraphrase them directly in your sketch if possible. You then try to make deductions and think through how a typical game would deal with those rules.
Rule one is “Annabelle’s pen is to the immediate left of Frodo’s pen.” You could paraphrase this with:
A —> F
Rule two states, “Enoch is in pen 1, 2 or 3.” Add this directly to your sketch (see below).
Add rule three, “Beatrice’s pen is number 4,” and four, “Darlene is not in pen 1,”directly to your sketch as well.
Your sketch should now look like this:
There aren’t a lot of easy deduction to make here, so spend a few seconds thinking about any constraints the sketch poses. Where you can fit the Annabelle-Frodo pair will be constrained by Beatrice in pen 4 and where Enoch ends up in pens 1-3. For example, if Enoch (or anyone besides Annabelle and Frodo) end up in pen 2, you have to fit the A-F pair in pens 5 and 6.
Question 1 is a “could be true,” style of question. The right answer only has to be possible. When you get one of these in Analytical Reasoning, you’re often just dealing with rules. Believe it or not, you don’t have to plug a bunch of scenarios into your sketch. Just grab one of the rules and see if you can rule out an answer choice:
1. Which one of the following could be the order of the giraffes’ pens from 1 through 6 respectively?
(A) Enoch, Beatrice, Annabelle, Frodo, Casper, Darlene
(B) Enoch, Annabelle, Frodo, Beatrice, Casper, Darlene
(C) Enoch, Frodo, Annabelle, Beatrice, Darlene, Casper
(D) Annabelle, Frodo, Casper, Beatrice, Enoch, Darlene
(E) Darlene, Casper, Enoch, Beatrice, Annabelle, Frodo
Rule one keeps Annabelle to the immediate right of Frodo and that doesn’t happen in (C), so cross it off.
Rule two has Enoch in pen 1, 2 or 3 and (D) violates the rule. Cross it off.
Rule three has Beatrice in pen 4 so cross off (A).
Rule four says Darlene cannot be in pen 1, so cross off (E).
You’re left with (B).
Question 2 presents you with hypothetical information:
2. If Enoch is in pen 2, then which giraffe must be in pen 1?
Plug Enoch into pen 2 (sounds familiar, so we know we’ll be dealing with the A-F pairing). If E is in pen 2, the only place to put A-F are pens 5 and 6. B is already in 4 and you’re left with D and C. D cannot go into pen 1, so the answer has to be (C), Casper.
You get more hypothetical information in the third question:
3. If Frodo is in pen 3, then which one of the following CANNOT be true?
(A) Darlene is in a pen next to Beatrice.
(B) Enoch is in pen 1.
(C) Annabelle is in a pen next to Enoch.
(D) Casper is in a pen next to Frodo.
(E) Casper is in pen 6.
Be careful, don’t just plug in and go. This is a CANNOT bet true question, so the right answer will look like a wrong answer. The right answer is impossible, the wrong answers are possible but do not have to be must-be-trues. Answer (D) cannot be true since at least Beatrice will separate Casper and Frodo.
If we plug Frodo in 3, a lot happens. You know you have to put Annabelle in 2 which only leaves pen 1 for Enoch. D and C can float between 5 and 6.
Next you get a “complete and accurate” question:
4. Which one of the following is a complete and accurate list of the pens that could house Annabelle?
(A) 1 and 5
(B) 2 and 5
(C) 3 and 6
(D) 1, 3, and 5
(E) 1, 2, and 5
Where could Annabelle go? You know you’ll deal with the A-F pair which limits you to putting Annabelle in 1, 2 or 5. It’s just that simple. Answer (E) is correct.
The last two questions are the most difficult in the game. This is not because of the inferences involved. After dealing with the four previous questions, your understanding of the A-F constraint makes the deductions for these questions easy. The problem is the questions lack a hypothetical, which means there are ten situation to test. If you’re lucky and land on the right answer quickly, you can stop testing, mark the answer and move on. But these will take more time than usual, so you should take a quick look at your watch to see if you have the time to spend or decide to move on.
5. Which one of the following must be true?
(A) If Annabelle is next to Beatrice, then Enoch is next to Darlene
(B) If Frodo is in pen 3, then Beatrice and Casper are in adjacent pens
(C) If Darlene is in pen 2, then Beatrice and Enoch are in adjacent pens
(D) If Casper is in pen 6, then Annabelle is in pen 1
(E) If Casper is in pen 1, then Frodo is in pen 6
This is a “must-be,” so the right answer has to be true and the wrongs answers could be true or have to be false. To eliminate wrong answers, just try to make them false if possible.
(A) has A next to B, which means A has to be in pen 5 and F in pen 6. But we could still put E in 1 and D in 3, so it is not a “must-be” true.
(B) has F in 3, so A has to be in 2 and E has to be in 1. E cannot be next to D no matter what, so cross this one out.
(C) has D in 2, so A and F are in 5 and 6. But E could still go in 1 so this doesn’t have to be true. Cross (C) off.
(D) has C in 6 but the A-F pair could still be in 2 and 3, so cross it off.
At this point, you should mark (E) and move on. There has to be a right answer, so it the first for are wrong, (E) has to be right. You can always make a mark in your test book to come back and double-check it – if you have time.
(E) Must be true because if C and E are in two of the first three pens, the A-F pair has to go in pens 5 and 6.
6. Which one of the following CANNOT be true?
(A) If Enoch is in pen 3, then Annabelle is in pen 2.
(B) If Annabelle is in pen 5, then Casper is in pen 1.
(C) If Darlene is in pen 2, then Enoch is in pen 1.
(D) If Annabelle is in pen 1, then Enoch is in pen 3.
(E) If Frodo is in pen 2, then Casper is in pen 6.
The right answer in this case is impossible and the wrong answers might or must be true. You may want to consult some of your previous sketches and the answer to question 1, since these all represent could be trues.
We lucked out with answer (A). If E in 3, A cannot be in 2 because of the A-F pairing. Mark (A) and move on – you don’t need to read the other answers so save the time for the rest of the section.
(B) could be true because C doesn’t have any rules to worry about.
(C) could be true as we’ve seen in a couple of sketches.
(D) could be true since only pen 2 is limited by the A-F rule.
(E) could be true since we can still fit A in pen 1, E in 3 and C has no real rules to worry about.
Now get some more practice with LSAT Practice Sequencing Game 3.