Logic Challenge #22 - Poetry

[monica.ann.mark]

Here's my crack at it (though I fear number six may have slammed me):
1. D
2. C
3. E
4. A
5. E
6. C

[pablomoretto]

1. D
2. C
3. E
4. A
5. D
6. A

I'm not sure I understood 6 correctly...

[pablomoretto]

I've been trying to win this contest for months, so let's see if I can do it this time:

A group of students in a poetry class is being assigned readings. Each of three students _ Francis, Grace, and Hu _ will read at least one of four poems _ Walkabout, Xtravagance, Yonder Cloudbreak, and Zoolandia.

"¢ No student reads both Yonder Cloudbreak and Zoolandia.
"¢ Hu reads more poems than Grace and Francis.
"¢ Any student that reads Xtravagance also reads Walkabout.
"¢ Any poem that is read by Francis is also read by Grace.

Representation of the rules:

s | F G H
p | W X Y Z

1. ~(YpZp) _ I actually drew a circle around YZ and then put a slash through it.
2. Hs > Gs, Fs
3. Xp --> Wp
4. Fs --> Gs

Note that I tracked the groups that each of the variables was in when writing out the rules (e.g. writing "YpZp" instead of "YZ"). I find that this helps me internalize the groups faster (i.e. I learn which variables correspond to which groups). It’s a method worth trying; if it doesn’t work for you, don’t use it.

1. Which of the following could be a complete and accurate list of the poems read by the students?
A. Francis: Xtravagance, Walkabout; Grace: Xtravagance, Walkabout, Yonder Cloudbreak;
Hu: Walkabout, Yonder Cloudbreak
B. Francis: Walkabout, Xtravagance; Grace: Walkabout, Yonder Cloudbreak; Hu:
Walkabout, Xtravagance, Yonder Cloudbreak
C. Francis: Walkabout; Grace: Walkabout, Xtravagance; Hu: Yonder Cloudbreak, Zoolandia
D. Francis: Yonder Cloudbreak; Grace: Yonder Cloudbreak; Hu: Walkabout, Xtravagance
E. Francis: Walkabout; Grace: Walkabout; Hu: Xtravagance, Yonder Cloudbreak

This is a typical "complete and accurate" question; it is solved (as usual) by taking each rule and looking for violations in the answer choices ("ACs").

Rule 1 eliminates C.
Rule 2 eliminates A and C.
Rule 3 eliminates E.
Rule 4 eliminates B.

The answer is thus D. After answering the question I wrote out and boxed the hypo that the question had given me:

F | Y
G | Y
H | W X

I drew a box around the hypo (not shown here) as shorthand to let my future self know that this is a valid hypo. I find that this helps me get a feel for what a correct hypo will look like. It also can be used to eliminate ACs for future questions.

2. Which of the following must be true?
A. Grace reads more poems than Francis.
B. If Grace reads Xtravagance, Francis reads Xtravagance.
C. If Francis reads Xtravagance, Grace reads Walkabout.
D. If Hu reads Yonder Cloudbreak, Francis and Grace do not.
E. If Hu reads Zoolandia, Francis and Grace do not.

In the past I used to hesitate with these questions, wondering if there was some trick way to answer it without having to eliminate each AC. Now I know better: when you see a question like this, dive right in and start creating hypos (hypothetical variable assignments). If there is a trick to the question that allows you to solve it quickly, you are more likely to spot it quickly in the course of creating a hypo.

Some more notes:
- In this case you are going to try to create hypos to disprove each AC.
- When at all possible, re-use old hypos to create new ones; see if you can make a small modification that will eliminate another AC and not break any rules.

The hypo from #1 disproves A:
F | Y
G | Y
H | W X

The modification below of #1’s hypo disproves B. I switched G’s Y for an X, added a W (Rule 3), and then added a Y/Z to H (Rule 2).
F | Y
G | X W
H | W X Y/Z

Soon after starting to work on C I knew it was correct. If Fs reads Xp, then Gs has to read Xp (Rule 4). And if Gs reads Xp, then Gs has to read Wp (Rule 3). Thus, if F reads X, then G reads W; that's exactly what C says. At this point there is no point trying to eliminate D or E; circle C and move on to the next question.

3. If Hu does not read Yonder Cloudbreak or Zoolandia, then all of the following could be true
EXCEPT:
A. Francis reads Zoolandia.
B. Grace reads Yonder Cloudbreak.
C. All three students read Walkabout.
D. Exactly two students read Zoolandia.
E. Exactly two students read Xtravagance.

I started by creating a graph:

F |.................~X
G |.................~X
H | W X [~Y, ~Z]

I realized that because H didn’t have Y or Z, it would need to have both W and X in order to have more poems than F and G (Rule 2). I also knew that F and G would only have one poem each (Rule 2 again). I then realized that because of Rule 3 (X --> W), neither F nor G could read poem X; they could only read one poem each, and it would have to be the same poem. I hoped that this would give me enough information to answer the question and skimmed the ACs, looking for a likely choice. E turns out to be just what I was looking for.

I was able to answer this question the way I did because I’d done so many logic games that I could quickly internalize and combine the rules of a new game. Don’t expect to be able to make these kinds of inferences without practice; it’s like riding a bicycle.

4. Which of the following must be true?
A. Francis cannot read Xtravagance unless Hu does.
B. Francis and Grace always read at least one poem that Hu reads.
C. Hu can only read Zoolandia if Grace does not.
D. Hu can read three poems only if Francis reads two poems.
E. Zoolandia can be read by at most two students.

Dive right in with answer choice A:

F | X W
G | X W
H | W X Y/Z

If F reads X then F also reads W (Rule 3). G reads what F reads (Rule 4). So the only way for H to have more poems than F and G (Rule 2) is for H to have more than two poems. But because Y and Z can’t be together (Rule 1), H MUST have W, X, and then either Y or Z. So A must be true: if F reads X, then H reads X. Circle and move on.

5. If every poem is read by at least one student, each of the following could be true EXCEPT:
A. At least one poem is read by all three students.
B. Francis reads exactly the same poems as Grace.
C. Grace reads more poems than Francis.
D. Francis and Hu read two of the same poems.
E. Xtravagance is read by either Francis or Grace.

Dive in and start creating hypos:

This hypo shows A, B, and C are possible:

F | W (Y) _ The parentheses mean that Y doesn’t have to be there.
G | W Y
H | W X Z

I can’t see how either D or E are possible. Here’s the problem as I see it:

If you try to create a hypo that shows D is possible, you end up with a problem: if F and H read two of the same poems, then you end up not being able to have both Y and Z read.

F | W Y
G | W Y
H | X W Y (but what about Z?)

If you try to create a hypo that shows E is possible, you run into the same problem. E is really a sub-set of D:

G reads X:
F | W ?
G | X W
H | X W Y/Z (but what about Z/Y?)

[Note: F can't read Z/Y because then G will have to as well (Rule 4). But G can't read 3 poems (Rules 2 and 1).]

or

F reads X:
F | X W
G | X W
H | X W Y/Z (but what about Z/Y?)

And this next one is more slog than you might want...

6. How many different ways can the poems be assigned such that each poem is read by at least
one student?
A. 4
B. 5
C. 6
D. 7
E. 8

I thought to myself, "H seems to be the big determinant in this game; if H has three vars, then F and G can have either one or two vars. If H has two vars, then F and G can only have one var. So I’m going to create two different groups of graphs: one with H having two vars, and one with H having three vars. Then I’ll add up the possibilities."

But it turns out that H can’t have only two vars under the stipulation of the question:

F | Z
G | Z
H | W Y (but what about X?)

So I scrapped that and looked only at situations in which H had three vars. I realized that H’s three vars would have to be X, W, and either Y or Z:
F | ?............~X
G | ?............~X
H | X W Y/Z?

Either F or G or both would need to read the Y or Z that H did not read. This means that they could not read X, because that would force them to read three poems, violating Rule 2. F and G could have either one or two vars, but the second var could only be W. So I drew a graph:

F.......G.......H
_____________
..............X
..............W

Possibility #1
Z.......Z.......Y

Possibility #2
Y.......Y.......Z

Possibility #3
Z.......Z.......Y
W......W

Possibility #4
Y.......Y.......Z
W.......W

[tnoziglia]

1D
2C
3E
4B
5E
6B

[tad.kim]

1. d
2. c
3. e
4. a
5. e
6. c

[klpayan03]

Answers
1.D
2.C
3.E
4.A
5.C
6.A

Explainations
1. Rule 2 eliminates (A) and (C). Rule 3 eliminates (E) and Rule 4 eliminates (B) This leaves (C) as the correct Answer.

2. (A) The answer to question 1 proves this choice wrong
(B) Only poems F reads must be read by G. There is no rule that stating the reverse.
(C) According to rule 3 if F reads X he must also read W and since rule 4 states that any poem F reads is also read by G this statement is true.
(D) (E) No rule prohibitting either F or G from reading anything that H reads.

3. If H does not read Y or Z and he has to read a minimum of 2 Books (rule 2) that means he must read both X and W (also forced in rule 3). Because rule 3 states that both X and W are read together and rule 2 states H reads more than F or G, H is the only person who can read either X or W making (E) the only false statement.

4. Due to rule 3 X and W must be read together and rule 2 making H read more than F. The only way F can read two poems is if H reads 3. This makes (A) the correct answer.

5. Drawings help with this one.

F=XW
G=XW
H=XWY
This drawing eliminates (A) (B) and (D)
F=Z
G=Z
H=YXW
This drawing elimiates (E)
This leaves (C) as the correct answer.

6. If we take the drawings above that gives us two senarios already. Then we can add the next two
F=XW
G=XW
H=XWZ

F=Y
G=Y
H=XWZ

any other senario would leave out a poem and the added rule to 6 was that each poem must be read by at least one person.

[rippinradio]

The modification below of #1’s hypo disproves B. I switched G’s Y for an X, added a W (Rule 3), and then added a Y/Z to H (Rule 2).
F | Y
G | X W
H | W X Y/Z

This actually breaks the fourth rule.

There's been a lot of disagreement about # 6. Here are the eight possibilities:

F: W Z
G: W Z
H: W X Y

F: W Y
G: W Y
H: W X Z

F: W
G: W Z
H: W X Y

F: Z
G: W Z
H: W X Y

F: W
G: W Y
H: W X Z

F: Y
G: W Y
H: W X Z

F: Y
G: Y
H: W X Z

F: Z
G: Z
H: W X Y

[pablomoretto]

The modification below of #1’s hypo disproves B. I switched G’s Y for an X, added a W (Rule 3), and then added a Y/Z to H (Rule 2).
F | Y
G | X W
H | W X Y/Z

This actually breaks the fourth rule.

There's been a lot of disagreement about # 6. Here are the six possibilities:
[...]

Excellent work. I'm not surprised I made mistakes; this happens to me almost every time! I know I'll be ready for the LSAT when I can correctly answer all of the questions in these games!

[noah]

I just wanted to say, for anyone who struggled with #6, I HATED solving that question. It's far from a good example of an LSAT question, because it's a real case of slogging through the scenarios; that's why we put it in a separate category.

Looks like a good conversation so far -- nice work. I'll let Brian B., the game's author, come on and chime in a week or so once you all have hashed this out.

[rippinradio]

I have amended my post above. I had neglected the possibilities resulting from H being assigned 3 poems and F and G each being assigned 1 poem. Mapping out the possible slot structures takes some of the uncertainty out of the game.

F: __
G: __
H: __ __ __

F: __
G: __ __
H: __ __ __

F: __ __
G: __ __
H: __ __ __

[rippinradio]

Two other inferences are worthy of note:

• Whenever H is assigned three poems, they must be W, X, and one of Y and Z, due to the first rule.


• If either F or G (or both) is assigned one poem, it cannot be X, due to the third rule.

[tayjones]

As far as mapping out the possible numbers, I agree. That seems to be an important anchor in these grouping games. Good call.

[cydiego]

I got ...

1.D
2.C
3.E
4.A
5.E
6.D

When do we find out who won?

[noah]

We're going to post the winners on Monday...Stay tuned.

[bbirdwell]

Hey guys! Thanks so much for checking out the game and participating in the forums. You have all brought to attention something that I, in my logic-puzzle-writing frenzy, overlooked!

Question #5 does indeed have two valid solutions: D and E. I totally goofed

pablomoretto, I enjoyed your detailed explanation (and your perseverance). I liked your question-by-question narration of the game and its solutions. There is one thing, I think, that would improve your explanation, and perhaps even your game performance, and that is to always, always, always make inferences after diagramming the game and symbolizing the constraints - BEFORE proceeding to the questions!

Making obvious and especially "2nd level" inferences before running to the questions will save you time and greatly improve your efficiency and accuracy, so be sure to make that part of your gameplan, and your future explanations!

Thanks again for checking out the game and hanging out on the forums with us. I'll be publishing another one shortly, and you can be sure it will be rid of dbl-solutions! (and i'll leave off the "how many possible arrangements" type of question that requires quite a bit of slogging here.

Rippinradio, thanks for your explanation of the 8 possibilties.

Cheers,
^b

P.S. Since we goofed and had two correct answers to question 5, so we went back and changed it so that there is only one correct answer. So, the correct answers are now: D C E A E E

[Jjia]

Hi everyone,

I'm curious as to how there are 8 possibilities? Aren't there 6 possibilities?

I agree with 3 of Rippinradio's diagrams, but not the first one. If H is to have 2 poems and F/G each have one, then that means H can either have XW, or X +Y/Z. If the former is the case, then F or G must have Y and Z, but since they can only have one slot each and whatever is in F must repeat in G, this won't work.

For the latter, we are missing W as well as either Y or Z. The same logic as last time tells us that we can't have two different poems appearing in the individual slots of F and G.

Am I missing something here, or is the answer actually supposed to be C?

edit: I also forgot, there is another option for H, which is W +Y/Z, but this leaves us with X and either Y/Z to be used, and we most certainly cannot use X in either of F or G's individual slots since it will bring along W.

[rippinradio]

Good eye! The 1, 1, 2 allocation is indeed invalid. If you check the post on the previous page, I mapped out all 8 possibilities.

[Jjia]

Awesome. Thanks for mapping it out as well!