Logic Challenge #12-Scrabble

[jakespieg]

The answers to this weeks challenge are:

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

Jake Spiegelhauer

[rojbeto]

Here's my solution to Atlas LSAT's Logic Game #12 _ Scrabble. This was a fun one

Setup:

Games--Player 1--Player 2--Limitations
1 J K/G/H/F
2 G H/F/J No K
3 I K/H/F/G/J
4 K/H/F/G/J K/H/F/G/J If K, no G or H

Rules:
1: 6 players/8 slots
2: 2 players/1 game
3: 1 player =< 1 game
4: I ≠ game 2, I ≠ game 4
5: J = game 1 but J> =1 game
6: K = 1 game, K ≠ G, K ≠ H
7: G = 2 games and G > I once, I = 1 game

The 8 possible slots for each player are listed in the grid. The games are numbered 1-4 and the two players in each game are named player 1 and player 2.

Deductions:
Rule 1 tells us that because there are more slots than players in the games, two of the players must play exactly twice or one of the players must play exactly three times.
Rule 2 tells us there can only be two players per game, not more, not less.
Rule 3 says that players CAN play in more than one game, which we already deducted in Rule 1.
Rule 4 says that I can only play in games 1 or 3 (contrapositive to not being able to play in games 2 or 4).
Rule 5 puts J in game 1 (the player position is irrelevant to this game). I’ve inserted this in player slot 1 in the table (but you can just as easily place it in slot 2, since the positions don’t really matter).
Rule 6 says K can only play in 1 game and that game is not against G or H. This means K has to play against either J, I or F.
Rule 7 tells us G is one player that plays in 2 games. This means there is 1 player that plays twice (we need to fill 2 extra slots and only have 6 players). It also says that one of these games comes before the game I is in. I can only be in games 1 or 3 (rule 4), so this means that in order for G to play in the game BEFORE I, I must play in game 3, thus placing G in game 2, for 1 of the 2 games total G is in (we don’t know anything about the second game G is in yet). This is true because there are no games before game 1, so if I played in game 1, it would be impossible for G to play in a game before I (rule 7).

I’ve mapped out the certain placements in bold and the potentials in brown.

Questions:

1. Which one of the following could be an accurate schedule of games?
(A) first game: Gayle, Jin; second game: Hannah, Issa; third game: Gayle, Kit; fourth game: Farang, Hannah.
(B) first game: Jin, Hannah; second game: Hannah, Farang; third game: Issa, Gayle; fourth game: Farang, Kit.
(C) first game: Jin, Kit; second game: Gayle, Hannah; third game: Issa, Gayle; fourth game: Farang, Kit.
(D) first game: Kit, Gayle; second game: Gayle, Farang; third game: Gayle, Issa; fourth game: Kit, Hannah.
(E) first game: Jin, Farang; second game: Gayle, Hannah; third game: Issa, Gayle; fourth game: Kit, Farang.

Explanation: Let’s see what we can eliminate. A and C violate rule 6 and therefore are out. B places G in one game only so it’s incorrect. D doesn’t have J in game 1, leaving E as the only possible choice.

2. If Hannah plays the third game, then each one of the following could be true EXCEPT
(A) Kit and Jin do not play against each other.
(B) Farang and Kit do not play against each other.
(C) Gayle plays a game after Issa.
(D) Gayle plays a game against Issa.
(E) Hannah and Gayle play a game against each other.

Explanation: Since I plays in game 3, this means H plays against I in game 3. A: there is nothing that prevents K and J from playing against each other. B: K and F don’t have to play against each other, if A is true (K can play against J, for example). Since K can play against J in game 1 and G has to play one more time (and is already playing in game 2), G can certainly play a game after I, however, since H is already plays against I in game 3 and game 3 is the only game I is in, G cannot play a game against I, rendering D the correct choice. E: There is nothing preventing H and G from playing against each other.

3. If Farang’s only game is against Kit, and Jin plays only one game, how many possible arrangements of contests are there?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Explanation: There are only two possibilities if F only plays once (with K) and J plays only one game (be sure you didn't miss that one). That means numbers-wise we have:

GG
HH
I
K
F
J

1: J and G
2: G and H
3: I AND H
4: K and F

And the other one is:

1: J and H
2: G and H
3: I and G
4: K and F

In other words, the open spaces in games 1-3, once we know that games 1-4 must be filled with J, G, I and F+K respectively, are either going to be filled with GHH (in that order, 1-3), or HHG. (G can't play himself.)

4. Which one of the following is a complete and accurate list of the players who could play in two back-to-back games?
(A) Farang, Hannah
(B) Hannah, Gayle
(C) Hannah, Gayle, Issa
(D) Jin, Kit, Farang, Gayle
(E) Jin, Farang, Gayle, Hannah

Explanation: We know from rules 6 and 7 that I and K can only play in one game each. This leaves the rest (J, F, G, H) as potential players for more than one game, which is choice E. No need to even check the wrong choices here, although it is easy to eliminate C and D right away, as they violate rules 6 and 7.

5. If Farang’s only game is against Kit, and Jin plays only one game, how many possible arrangements of contests are there?
(A) One
(B) Two
(C) Three
(D) Four
(E) Five

Explanation: For this question, we can assume with certainty that I, J, K and F only play once. This means G and H have to play twice each (under rule 7 G plays in 2 games). Since G can’t play against herself, H has to fill the second player slot in game 2, leaving the G/H or H/G combo to fill out the opponent’s slots in games 1 and 3, leaving us with two possible arrangements of contests (choice B).

Here is what the setup under this rule may look like:

1 J G/H
2 G H
3 I H/G
4 K F

6. Which of the following, if true, would result in exactly one possible arrangement of contestants to games?
(A) Gayle plays in the first and fourth game.
(B) Hannah plays in the first and fourth game.
(C) Gayle plays in the second and third games.
(D) Hannah plays in the third and fourth games.
(E) Jin plays in the fourth game.

Explanation: We can go back to our initial setup for this question.

1 J K/G/H/F
2 G H/F/J No K
3 I K/H/F/G/J
4 K/H/F/G/J K/H/F/G/J If K, no G or H

Trial and error works best for this type of question.

A: If G plays in games 1 and 4, the grid changes accordingly:

1 J G
2 G H/F/J No K
3 I K
4 H/F/J G If K, no G or H

K has to play in game 3 under this scenario, but there are still quite a few possibilities for the rest of the players.

B: Plugging in H in the player 2 slots for games 1 and 4, forces G into game 4 (G has to play twice and it can’t play against herself; K on the other hand cannot play against H or G, so it is forced to play against I in game 3). We still have to place F and the only spot left is against G in game 2, rendering this the one and only possible arrangements under this scenario. B is the correct answer so we need not check the rest of the choices.

Here is the grid again, under choice B:

1 J H
2 G F No K
3 I K
4 G H If K, no G or H

Final answers:
1E 2D 3D 4E 5B 6B

Bonus Question

7. If Gayle and Hannah are not allowed to play against one another, each of the following contestant arrangements must be false EXCEPT
A) First game: Jin, Issa; second game: Kit, Hannah; third game: Gayle; fourth game: Kit, Farang.
B) Fist game: Jin, Gayle; second game: Gayle, Farang; third game: Issa, Hannah; fourth game: Hannah, Farang.
C) First game: Hanna, Jin; second game: Gayle, Farang; third game: Issa, Hannah; fourth game: Gayle, Jin.
D) First game: Jin, Gayle; second game: Gayle, Farang; third game: Issa, Kit; fourth game: Hannah, Jin.
E) First game: Gayle, Issa; second game: Kit, Hannah; third game: Jin, Farang; fourth game: Farang, Hannah.

[ajs227]

I believe that D works as well for #6.
We know J G I for the first three games.
If H plays in 3 and 4 then we have:

J G I H H

K has to play in game 1 because it can't play against G or H in 2 or 4.

J K G I H H

Leaving F needing to play a game and G needing a second game so F has to play in game 2 because G can't play themselves. Thus G must play in game 4 leaving you with one possible solution:

J K G F I H H G

[stackoutawinner]

I thought at one point in time that I had found a question with two correct answers, but I quickly realized I was wrong. I never posted my solutions because there was no point unless I had the silver bullet to win me BOTH prizes.

Even though I was wrong that first time through, with #6, I never went past choice B.

ajs227 is correct, if you say H goes in game 3 and 4, it forces the set up:

K can't pair with G or H, and therefore is FORCED into game 1.

G has to go twice and since games 1 & 3 are now determined, it must go in game 4. You now have three of the four games determined and F has not played yet. Therefore, F goes into game 2 and the whole set up is solved.

ajs227... you need to write out your answers and solutions, because you should win both prizes as being the only one to spot two right answers.

[ManhattanPrepLSAT2]

Not sure if you saw my previous (now deleted) post, but it was WRONG. Sorry!

Here is the correct explanation for why (D) is NOT a viable answer to #6:

If H is in the third and fourth slots, we've got:

1: J _

2: _ _

3: H, I

4: G, H

The remaining letters are K, F, and G, and they can be arranged in a couple of different ways in the remaining spots.

In retrospect, I think the confusing phrase was , "Gayle plays in exactly two games, exactly one of which comes before the only game that Issa plays." This does not mean Gayle has to play in the immediate game before Issa's, and therefore the second G can be in the first or second game.

[interestedintacos]

It's alright. It should have been obvious by saying "exactly one of which comes before" you mean generally before, not that there is only 1 G instead of 2 G's right before I, since it would make no sense to have the same person competing against himself.

I had no trouble getting that part, but I did mess up initially and assumed that G cannot go with I in game 3. Thus I put a firm G in game 4, which is a mistake. I made the extra, incorrect assumption that if exactly 1 G comes before I, the other G comes after I. Luckily I fixed that when it became clear that the second question revolved around the limitation on G you would get when H fills up the empty spot in game 3.

[vincent.m]

Hi,

So on a previous explanation for #5, if F's only game is against K that doesn't mean that that is K's only game. So, isn't B right because G and H have to play each other in the second game , so games 2 through 4 are set, but H or K can play J in the first game? Is that right? Thank you!

[Athomas87]

Could someone give an explanation for #3, I'm not sure how it's supposed to be 2, I keep getting 4 different combinations. Also, the thorough explanation for #3 above is a different question than what's listed. Any help would be appreciated!

[Athomas87]

Could someone give an explanation for #3, I'm not sure how it's supposed to be 2, I keep getting 4 different combinations. Also, the thorough explanation for #3 above is a different question than what's listed. Any help would be appreciated!

[maryadkins]

There are only two possibilities if F only plays once (with K) and J plays only one game (be sure you didn't miss that one). That means numbers-wise we have:

GG
HH
I
K
F
J

1: J and G
2: G and H
3: I AND H
4: K and F

And the other one is:

1: J and H
2: G and H
3: I and G
4: K and F

In other words, the open spaces in games 1-3, once we know that games 1-4 must be filled with J, G, I and F+K respectively, are either going to be filled with GHH (in that order, 1-3), or HHG. (G can't play himself.)

Thanks for pointing that out about the wrong question being included in the explanation above! Fixing it now.

[alanr739]

regarding maryadkins explanation for #3:

doesn't your first example of listed above violate one of the rules, that G can only come exactly 1 time before I? or am i misreading?