LRDI Practice Set 27

The Action:

We’re presented with seven judges, each of which voted either for or against a petition. One way to look at this is that each entity is either in or out, which strongly suggests grouping; specifically, a grouping game of selection. The entities are two conservatives, two moderates, and three liberals. The Key Issues appear to be:

1) How did each judge vote—for or against the petition?

2) How many judges voted for, and how many voted against the petition?

It’s nothing more complicated than that, and the rules, of course, will help us to narrow down the possible voting scenarios.

The Initial Setup:

First, get the basics down on the page. A simple but effective way to proceed is to list the seven entities, under which we’ll endeavor to place a ✔ for “for” and an ✗ for “against.”

The Rules:

4) A quick scan of the rules tells us that Rule 4 is the most concrete of the bunch, and we can put this information right into the sketch: At least one conservative gets an ✗.

3) While we’re at it, we may as well take care of Rule 3, and save the two conditional rules for last.

pretty much sums it up.

1) Now the if-thens: If both conservatives and at least one liberal voted the same way, then both moderates voted that way as well. Thanks to Rule 4, we can make this a little more specific, since we know from that rule that at least one conservative said “nay.” If both conservatives and at least one liberal voted the same way, then that way must be “against.” The rule then tells us that under these circumstances, the two moderates will vote against as well:

Now you may have taken this rule even further, or you may have waited until Step 4 of the Kaplan Method. If you handled the rules in order from 1 to 4, that’s probably what you did, but since we’ve already incorporated Rules 3 and 4, let’s continue. Since we still need two votes in favor of the petition (Rule 3), in this situation the two remaining liberals must have voted for it. So Rule 1, if enacted (remember, it’s conditional based on the if-then) really boils down to this:

If both C’s ✗ and 1 L ✗, then:

If you thought this through in advance, a little sketch off to the side to reflect the implications of this rule is in order. If not, that’s okay—you no doubt had the opportunity to work through all this at some point in the course of answering the questions.

2) If the three liberals voted the same, then neither conservative voted that way, which is to say that both conservatives voted the other way. You can shorthand this any way that makes sense to you. Here’s one possibility:

If L’s the same, then C’s voted the other way.

Let’s take a further look at the implications of this rule in “Key Deductions.”

Key Deductions:

We’ve already worked through a few implications of the rules by combining the more concrete Rules 3 and 4 with the first two if-then rules. But a little extra thought will reveal one more major thing—did you see it? It’s derived from a combination of Rules 2 and 4. If the three liberals voted the same, then both conservatives voted the opposite, and we know from Rule 4 that at least one conservative voted against. So the situation works if all three liberals voted for the petition, in which case both conservatives must have voted against. But the other option is impossible: If all the liberals voted against the petition, then Rule 2 would require that both conservatives vote for it.

But that would violate Rule 4, so it’s impossible for all three liberals to vote against the petition. Which means, driving this thought to completion, that at least one liberal must have voted for the petition. This insight bags Qs. 14 and 15 directly, and is just useful knowledge throughout. Write it into your final sketch (and we can make our shorthand for Rule 2 more specific thanks to this deduction; see shorthand below). Now we’re ready to attack the questions.

The Final Visualization: Here’s what a good final sketch might look like:

The Questions:

1. (B)

If the moderates split their votes, then one voted for and the other voted against. Quickly jot this down in a new sketch:

The contrapositive of Rule 1, which you may have noted earlier in the setup stage, is that if the moderates split their votes, then it cannot be true that both conservatives and at least one liberal voted the same way. But following this train of thought gets a little messy, because there are a few different ways to ensure this doesn’t happen. The better method here, since we’re dealing with a “could be true” question, is to test the choices against the sketch just created, until we find one that works.

(A) The voting arrangement that would result from (A) would violate Rule 1: We would have both conservatives and one liberal voting against the petition, which, according to Rule 1, would force both moderates to vote against as well. But the stem has it otherwise.

(B) works. If you have trouble seeing it in your mind, by all means get the info down on your page:

No problem with this scenario, so (B) could be true and is correct.

(C) violates Rule 2—if all three liberals voted for the petition, then both conservatives would have to vote against, but (C) has one of the conservatives voting in favor of the petition.

(D) and (E) both violate Rule 4, which requires at least one “nay” among the conservatives.

2. (C)

Here’s the reward for the Key Deduction described above—we deduced that at least one liberal must have voted for the petition. That’s because if all the liberals voted no, Rule 2 would have both conservatives voting yes. But Rule 4 disallows that, which means that it cannot be true that all the liberals voted against, which in turn means that at least one liberal must have voted in favor of the petition, choice (C).

(A), (B) No, all the liberals could vote in favor of the petition, in which case both conservatives would vote against.

(D), (E) The moderates can swing all possible ways: As we saw in Q. 13, they can split their votes. On the other hand, both can vote for, or both can vote against, so (D) and (E) both could be true, but need not be.

3. (E)

Now, whether you made the Key Deduction regarding the liberals early on, or just now in the previous question, you can and should use this information to bag Q. 15. If all three liberals voted the same way . . . yup, that’s right, they must all vote for, because it’s impossible for all of them to vote against, for reasons elaborated just above and in the Key Deductions section. And there’s the point in choice (E).

With the liberals all voting for the petition, Rule 2 mandates that both conservatives vote against. So here’s what we know of the voting so far:

(A), (B) and (D) As Rule 1 is not in effect here, the moderates are totally free to vote any old way: both for, both against, or one each way. So these three choices all could be true, but need not be.

(C) is impossible—as shown above, both conservatives have to vote “nay” in this situation.

4. (A)

Our job in this one is to figure out how we can fit exactly two ✗’s (which means, of course, five ✔’s) into the sketch. We already know one definite ✗, a conservative thanks to Rule 4. And we’ve seen, most recently in Q. 15, the case in which both conservatives vote against the petition. Filling out the rest in line with the stem’s mandate results in this arrangement:

No problems there; but is this the only way to get a two/five breakdown of “nays” to “yeas”? Giving the second “against” to one of the moderates doesn’t work—then all three liberals vote for, which according to Rule 2 means that both conservatives would have to vote against. But if the second (and therefore last) ✗ goes to a moderate, the other conservative would have to vote in favor, which cannot be. So it’s impossible for a moderate here to get the second ✗, which means both moderates must vote for the petition, choice (A).

You may have waited until you worked out the other possible scenario before moving to the choices, in which case you would have worked out this arrangement:

This is the only other way to satisfy the stem, which again confirms that (A) must be true. As for the wrong choices, they all could be true, but need not be: (C) and (E) are true in the first possible scenario above, but not in the second. Conversely, (B) and (D) are true in the second, but not in the first.

5. (E)

We’re looking for the choice that contains the list of judges that cannot be a complete and accurate list of judges in favor of the petition. Before jumping into complicated scenarios, let’s see if the answer can be derived by checking the choices against the concrete rules, Rules 3 and 4. Rule 4 is easy to check—there must always be at least one conservative voting against the petition, so if any choice contains “two conservatives,” that must be the answer (remember, the entities in the choices represent judges voting for the petition). No luck. But Rule 3 does the trick. Are there any choices that would result in a lack of two judges on either side? Well, yeah: (E) consumes every judge but one conservative, which would result in six “yea’s” and only one “nay.” Just add ‘em up—six judges voting in favor of the proposal is a no-no, a direct violation of Rule 3.

6. (B)

Take it one step at a time: The conservatives voted the same way, which means, thanks to Rule 4, that they both voted no. Next we’re told that the liberals didn’t all vote the same way. We deduced (and saw later in Q. 14) that at least one liberal must vote for the petition, so now we know that at least one must vote against. So far then, this is how the votes went:

This situation triggers Rule 1—both conservatives and at least one liberal voted the same way, against—which means that both moderates must vote against the petition as well. But wait—we may as well complete the voting: With only one “for” vote out of six so far, we must satisfy Rule 3 by placing a ✔ under the remaining liberal, resulting in:

Every choice conforms to this arrangement except for (B), which is impossible (and therefore correct): Under these circumstances, both moderates must have voted against, not for the petition.