Logic Challenge #14 - The Basket Game

[rx211]

THE SETUP

Elements
1. The People: S, T, W
2. The Gifts/Weight: H-1, K-1, L-1, M-1, O-2, P-2
3. Conditions: Basket ≦3
T->K
P->S = L+H->T
M+O

W S T
K


1. If the oranges are sent to Wendy, which one of the following must be true?
Senerio 1:
W S T
M+O P K, L+H
3 2 3

Senerio 3:
W S T
M+O L+H K, P
3 2 3

X(A) The macadamia nuts are sent to Tara. (M has to be with O)
X(B) The pears are sent to Stan. (Could be true, not MUST be true)
(C) The havarti cheese and the licorice are sent in the same gift basket. (Both scenarios, L+H have to be together, impossible scenarios are PLK for Tara or PLH for Stan, it would exceed 3 pound limit)
X(D) The kiwis and exactly one other gift are sent to Tara.(Could be true...)
X(E) The kiwis and exactly two other gifts are sent to Tara.(Could be true...)

2. If the pears are sent to Wendy, which one of the following could be true?

W S T
P M+O K L, H are floaters
2 3 1
No M+O, Only position No M+O
Exceed weight for M+O Exceed weight


(A) The pears are the only gift sent to Wendy. (Tara can have K, L, and H, it K+L+H=3)
X(B) The oranges are sent to Tara.( Exceed weight, K+M+O=4)
X(C) The macadamia nuts are sent to Wendy. (Same reason, P+M+O=4)
X(D) The havarti cheese is sent to Stan. (Stan can only receive M+O)
X(E) The licorice is sent to Stan. (Stan can only receive M+O)

3. If the pears are sent to Tara, which one of the following must be true?

W S T
K, P
3
Floaters are M+O, and L, H, but since M+O is already 3 pounds, L & H have to be in a group of 2 pounds. The two pairs can be for either Wendy or Stan

X(A) The oranges are sent to Stan. (It can be for Wendy, could be true, but not MUST)
X(B) The licorice is sent to Wendy. (It can be for Stan, could be true, but not MUST)
X(C) The havarti cheese is sent to Wendy. (It can be for Stan, could be true, but not MUST)
X(D) One basket contains three gifts. ( Impossible, M+O is already 3 pounds, and cannot be seperated to combine a basket with L&H)
(E) Each basket contains exactly two gifts.

4. Which one of the following must be true?

(A) Exactly one gift basket weighs two pounds. (Two 3 pound basket, and one 2 pound basket)
X(B) Exactly one gift basket weighs three pounds.(There are total of 8 pounds, you can't have two 2 pound basket and one 4 pound basket)
X(C) Either the havarti cheese or the licorice is sent with the kiwis. (Proven not true with last question, kiwi can be with pear)
X(D) Either the oranges or the pears are sent to Stan. (Again, proven not true by previous questions)
X(E) Either the oranges or the pears are sent to Wendy.(Again, proven not true by previous questions)

5. Which one of the following, if true, would completely determine the contents of every basket?

X(A) Tara’s basket contains exactly two pounds of items. (L & H are floaters)
X(B) Stan’s basket contains exactly three pounds of items.(L & H are floaters)
X(C) Wendy’s basket contains exactly two pounds of items.(P, L, H are floaters)
X(D) Stan’s basket contains no two-pound items. (Multiple senarios)
X(E) Tara’s basket contains no two-pound items. (Multiple senarios)
FLAWED Question

6. Each of the following could be true EXCEPT:

(A) Wendy’s basket contains exactly three items. (Besides K for Tara, L,H, M+O,P are left, you can't make a three gift combination from these choices for Wendy, they all exceed 3 pounds)
(B) Tara’s basket contains exactly three items.
(C) Tara’s basket contains exactly two items.
(D) Stan’s basket contains exactly two items.
(E) Stan’s basket contains exactly one item.

7. Suppose it is no longer the case that the kiwis weigh one pound, but instead weigh two pounds. If all of the other conditions remain in effect, which one of the following could be true?

Elements
1. The People: S, T, W
2. The Gifts/Weight: H-1, K-2, L-1, M-1, O-2, P-2
3. Conditions: Basket ≦3
T->K
P->S = L+H->T
M+O

W S T
P M+O K Floaters: L, H between Wendy &
2 3 2 Tara
Has to be Only Space
here. If Pear of M+O
goes to Stan,
L&H has to go
to Tara, it would
exceed 3 pounds

X(A) Both the licorice and the havarti cheese are sent to Wendy.(Only one can be sent)
X(B) The oranges are sent to Wendy.(Impossible by diagram)
X(C) The pears are sent to Stan. (Impossible by diagram)
X(D) The licorice is sent to Stan. (Impossible by diagram)
(E) The havarti cheese is sent to Tara.(Possible)

[rippinradio]

Question 5 is not flawed. Take a look at choice D. With no two-pound items in Stan's basket, we know that the MO basket must be Wendy's. The other two-pound item, P, must go with Tara. H and L must complete Stan's basket. The layout is:
H P M
L K O
S T W

[noah]

Alright, let's see another explanation then!

I think there's more that can be inferred before hitting the questions...

[rx211]

Question 5 is not flawed. Take a look at choice D. With no two-pound items in Stan's basket, we know that the MO basket must be Wendy's. The other two-pound item, P, must go with Tara. H and L must complete Stan's basket. The layout is:
H P M
L K O
S T W

muahahahahaha you cant hace Stan have H & L, that would be EXACTLY two pounds! which is the OPPOSITE of what the answer says, with NO two-pound items. i get what you are saying, but the wording is horrible, it the answer had said with no two-pound item, that would suggest no P or O.

[rippinradio]

H and L are not a two-pound item. They are two individual items that are one pound each. D didn't say that Stan's basket doesn't weigh two pounds. There's no ambiguity in the wording. The key distinction is between basket and item.

[rippinradio]

[rippinradio]

8. Suppose, in place of the condition that the weight of each basket cannot exceed three pounds, a new condition specifies that exactly one basket weighs four pounds. If all the other conditions remain in effect, and if Stan's basket contains the oranges, which one of the following could be true?
(A) Wendy's basket contains the licorice
(B) Tara's basket contains the pears
(C) Wendy's basket contains exactly two items
(D) Stan's basket contains the licorice
(E) Tara's basket does not weigh two pounds

[rippinradio]

9. Suppose John decides to keep one of the gifts for himself. If all of the other conditions remain in effect, which one of the following could be true?
(A) Wendy's basket contains two gifts
(B) Stan's basket contains the pears
(C) Tara's basket contains the havarti
(D) Stan's basket contains exactly one gift
(E) Wendy's basket contains the licorice

[orion]

Note for the following (except #7) that since the items for the baskets total 8lbs, this means that two baskets must have 3lbs of items and one basket must have 2lbs of items, since 3+3+2 is the only combination of positive integers less than or equal to 3 which sum to 8, and all baskets must have 3 or fewer pounds of items.

Also note that since the macadamia nuts and oranges must go together, the basket with these items weighs 3lbs already and may not hold any more items, so one person must receive the exact combination MO. Since Tara always receives the kiwis, this person must be either Stan or Wendy.

1. Since Wendy receives the oranges, her basket is completely determined to be W:MO. If Tara receives the pears, her basket is complete at 3lbs since she always gets the kiwis (T:KP) and therefore Stan must receive both the havarti and the licorice (S:HL). Conversely, if Stan receives the pears (S:P), the problem conditions specify that Tara gets both the havarti and the licorice (T:KHL). So either way, the havarti and the licorice go together and the answer is (C).

2. If Wendy receives the pears, she is eliminated as a recipient for the basket consisting only of macadamias and oranges, and therefore Stan must receive it instead (S:MO). As a result choices (B), (C), (D), and (E) are impossible. Tara of course still receives the kiwis, and she may also receive both the havarti and the licorice (since each of these items is 1lb). In this case, Wendy will receive only the pears, so answer (A) is the only possibility.

3. If the pears are sent to Tara her basket is completely determined as consisting only of pears and kiwis (T:KP). One of the other two baskets must be macadamias and oranges (MO), and therefore the remaining basket must be havarti and licorice (HL). So all baskets must contain two items, which is choice (E).

4. As discussed in my introduction, the only combination of positive integers less than or equal to 3 which sums to 8 is 3+3+2, therefore exactly one basket must have two items, which is choice (A).

5. If Stan's basket contains no two-pound items, we immediately know that he cannot have the basket with macadamias and oranges (since the oranges weigh 2lbs), so this must be Wendy's basket (W:MO). Stan also may not have the pears, so since Wendy's basket is now full these go to Tara and fill her basket completely (T:KP). Therefore the remaining items go to Stan, which means that his basket contains the havarti and licorice (S:HL). Therefore option (D) completely determines all basket contents. This is not true for any of the other choices, so the correct answer is (D).

6. Wendy may either have the basket containing macadamias and oranges (MO), or a basket containing items selected from amongst the set {havarti, licorice, pears}. (The remaining item, kiwis, is always in Tara's basket.) She may not have all three of the items in this latter set, as their combined weight is 4lbs. Therefore, she may not have three items in her basket. Since all of the remaining answer choices have actually appeared as possible results in previous problems, the correct answer is therefore (A).

7. If the kiwis weigh 2lbs, the total weight of all the items is now 9lbs, and therefore all baskets must now contain exactly 3lbs of items. Tara must have the kiwis, so she must also receive one of the 1lb items to complete her basket. She may not receive the macadamias (which must still be with the oranges), therefore she receives either the havarti or the licorice (T:KH or KL). Since it is now impossible for Wanda to receive both the havarti and the licorice, Stan may not have the pears, and therefore Wanda receives the pears and an additional 1lb item, which must again be the havarti or the licorice (W:HP or LP). This leaves Stan with the basket containing macadamias and oranges (S:MO). The first four answer choices are therefore impossible, leaving choice (E) as the only correct option.

[superfilms]

(The original game is at http://www.atlaslsat.com/lsat-logic-game-14.cfm.)

This game takes less than 5 minutes total if you begin by making a good diagram. This game can be almost completely diagrammed before tackling the questions.

The Inventory & Clues:


Now we set up the diagram, with S, T & W at the top.

First place K in T. MO can't go in T because that exceeds three pounds, so MO goes in either S (Model 1) or W (Model 2).


Look at Model 1. P can't go in S (it's 3 pounds) so P goes in T or W. Copy Model 1 onto line 3, then place P in T on 1, and P in W on 3.


In Model 1, S & T are both now full, forcing H & L into W.
In Model 3, H & L could go in T, or they could split between W & T. Just note on the right side of your diagram that H & L still need to be placed.


Now look at Model 2. Either P goes in S, or P goes in T. Break out Model 4 on the next line.


In Model 2, since P is in S, H & L both go in T with K. In model 4, T & W are full so H & L are forced into S.


The first 6 questions are now super-easy and very straightforward. Answer them using the diagram.

For question 7, the rules are changed, so we will make a new diagram. On the next two lines, put K in T, and MO in S on the first line and MO in W on the second line.


Since K & P both weigh 2 pounds, they can't be together, thus in 7.1 P is in W, and in 7.2 P is in S.


In 7.1 H & L split between T & W.
In 7.2, P is in S so H & L are in T. That puts 4 pounds in T, which eliminates it as an option. Cross it out.


Now question 7 is straightforward.


The images and writing in this post are the property of Lucas Penick (superfilms@hotmail.com). Please do not redistribute without my permission. Thanks!

[ianjorgeson]

Hey guys. Great explanations. I'm especially happy to see that some of you decided to frame the possible iterations of the game. This is a good example of a game that is constrained enough to make that an effective choice. Superfilms in particular had a good set of frames that made it easier to answer the questions with.

The key to this game is to recognize that this is a standard open assignment game with one twist; instead of the limiting factor per basket being the number of gifts allowed, it is the weight of the objects in each basket. The LSAT is pretty good at taking a standard game and adding a bit of a twist to create something new; in general, the best approach is to recognize the base game type, set your diagram up on that basis, and incorporate the twist in the simplest way possible.

Rippinradio and Orion both caught on to the main deduction, which is that 8 pounds divided into 3 baskets, none of which exceed three pounds, leaves only one arrangement; two 3 pound baskets and a 2 pound basket. With this deduction, it becomes simpler to figure out the different ways the game can work.

Rippinradio, I attempted the bonus questions you posted but I wasn't able to narrow them down to a single answer.

8. Suppose, in place of the condition that the weight of each basket cannot exceed three pounds, a new condition specifies that exactly one basket weighs four pounds. If all the other conditions remain in effect, and if Stan's basket contains the oranges, which one of the following could be true?
(A) Wendy's basket contains the licorice
(B) Tara's basket contains the pears
(C) Wendy's basket contains exactly two items
(D) Stan's basket contains the licorice
(E) Tara's basket does not weigh two pounds

As I read this question, the only rule changed is that one of the baskets weighs four pounds. If we have a single 4 pound basket, then the remaining four pounds of items might be divided up into a 3 pound basket and a 1 pound basket, or two 2 pounders. So, the distribution of weight is either 4-3-1, or 4-2-2. That limits the game a bit, but not by much. Placing the oranges in Sam's basket also places the macadamia nuts in that basket, and the kiwis are always in Tara's. So our starting point is:

M
O K
S T W

Answer choice A works in a number of ways, including the setup below (numbers below are total weight):

H
M P
O K L
S T W
4 3 1

This places the oranges with Stan, and has exactly one basket that weighs 4 pounds. It also shows that A, B, and E each could be true. The licorice and havarti can swap, and that would show that D could be true. The pears can be moved to Wendy's basket, and thus C could be true as well. Am I missing something?

I also was unable to answer #9.

9. Suppose John decides to keep one of the gifts for himself. If all of the other conditions remain in effect, which one of the following could be true?
(A) Wendy's basket contains two gifts
(B) Stan's basket contains the pears
(C) Tara's basket contains the havarti
(D) Stan's basket contains exactly one gift
(E) Wendy's basket contains the licorice

If John keeps a gift for himself, but the rest of the rules are in place, then he keeps the pears, licorice or havarti, since the kiwis are assigned, and he can't keep either the oranges or macadamia nuts without violating the rule that they are together. If he keeps the pears for himself, then there are 6 pounds to distribute; if he keeps l or h for himself, there are 7 pounds left to distribute. If he decides not to keep the pears, he cannot send them to Stan, because that would force him to place the l and h with Tara, and thus keep all items. That eliminates B as a possible answer. But the rest of the answers seem possible.

For example, if we do the following....

O L
M K P
S T W
...then John has kept the havarti for himself, all other rules are followed, and answers A and E could both be true.

Did I misinterpret the question?

[rippinradio]

Oops... I carried over the inference that none of the baskets could weigh under 2 pounds, even though suspending those rules opened up the distribution. Sorry about the mixup.