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.

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: __ __...

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 ...

4) - the rule DEFINES one extra position for us upon knowing one position. a - no, leaves many possibilities b - no, leaves many possibilites at center position c - no, only eliminates a few possibilities of combinations d - no, leaves a few possibilites *e - yes, upon knowing that a woman (not R o...

I just confirmed that the answer to the rule substitution effect question on September's test created the exact same limitations as the original rule did, so unless things change, you should be good on that front.

I actually noticed that it could be completed with templates after the fact. You're probably right that the time saved on the questions wouldn't have justified the time spent upfront, but boy were the questions a breeze once the templates were set up.

I'm guessing the standards to which LSAT questions have to live up to dictate that a credited replacement rule would need to have the exact same effect as the original, no more, no less. I'll have to see the September test to say for sure though.

Rule replacement questions (in which a new rule has the same effect as one of the original rules) first showed up on the June LSAT. That particular rule replacement didn't result in any changes to the possible sequences allowed by the rules. I haven't seen the September test yet, but I've heard that...

6. (no correct answer) We can confirm that there is no correct answer by examining each answer choice and asking if there are any sequences permitted under the old rule which are not permitted under the new rule or vice versa. (A) fails because the new rule does not permit the previously allowed se...

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 exactl...

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...

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.

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

With P out, we know that N is selected (second rule). Due to the third rule, L must also be selected. So far, we have N, L, and one of J and K (first rule) selected. The only variables to look at are M and Q. Due to the fourth rule, if Q is selected, K must be as well. That accounts for one solution...

jdillier Wrote:Does anyone know any other games that are similar to this one?

The third game of PrepTest B in SuperPrep has circular linearity. I heard that there was one on this year's February administration, but obviously, you can't purchase that.

That's a great question! I got B for number 7.

7. If R and T are both in the front row, how many unique pairings of crayons to positions are there?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10