Objection’s LSAT Tips – “More/Less” Ordering Games

Question Type: More/Less Ordering Games
Section: Logic Games

Introduction

Simple ordering games are just that – simple. There is a 1:1 ratio of variables to the number of slots. But you will often encounter a game where there are more variables than slots (e.g., six workers for four days) orfewer variables than slots (e.g., five babies for seven cribs). These can be tricky and complicated, but often have the potential to be broken wide open.

Method

As I said in the simple ordering games article, the two keys to logic games are making clear diagrams and making inferences. The “more/less” ordering games have one more key step: distributions. You must work out the possible numerical distributions for your setups.

For example, if there are six workers for four days of work, and a rule states that each person works only one day, and each day has at least one person working, you would have the following potential distributions:

3-1-1-1
2-2-1-1

The first distribution means you could have three workers on one day and one worker on the other three. The second says you could have two workers on two days and one worker on each of the other two.

Rules often restrict your potential distributions. Let’s look at a second example.

If there are five babies for seven cribs and the rules say nothing limiting the number of babies in each crib, you could have the following distribution of babies to cribs:

5-0-0-0-0-0-0
4-1-0-0-0-0-0
3-2-0-0-0-0-0
3-1-1-0-0-0-0
2-2-1-0-0-0-0
2-1-1-1-0-0-0
1-1-1-1-1-0-0

That is seven possible distributions. However, if the rules say no more than two babies per crib, you restrict the potential distributions to the bottom three distributions. Make sure you read all the rules prior to creating your distributions.

Games are more easily taught by example. Let’s do an example of both a more game and a less game.

Examples

Four people – Q, X, Y, Z – have formed a party club to party together six days a week from Monday through Saturday. Each day, a party is hosted by exactly one of the people. The hosting schedule for any week must meet the following conditions:

Each person hosts a party on at least one day.
No person hosts a party on two consecutive days.
Z hosts a party on Wednesday or Saturday or both, and he may also host on other days.
If X hosts a party on Monday, then Z does not host a party on Saturday.
Q does not host a party on Monday.

The first two rules should jump out you as restricting your distributions. In this case, our distributions will be number of times someone can host per week.

Since each person must host at least once, you have the following two distributions: 3-1-1-1 (one person hosts three times, the other three host once) 2-2-1-1 (two people host twice, the other two host once)

The rest of the rules are pretty simple to diagram.

Here is the diagram I used:

 

 

 

 

 

 

Diagram rules (DR#) explanations:

DR1. The distributions.
DR2. No consecutive hosting.
DR3. If X hosts Monday, Z cannot host Saturday and this rule’s contrapositive.
DR4. Z must be on Wednesday, Saturday, or both (but can still be on others).
DR5. Q cannot be on Monday.

Now, on to the questions!

1. Which one of the following could be a schedule of hosts for one week, for the days Monday through Saturday, respectively?

A) X, Q, Z, Y, X, X
B) X, Q, Z, Y, Q, Z
C) Y, Q, X, Z, Y, Q
D) Y, X, Z, Q, Y, Q
E) Y, X, Z, Y, Z, X

How to Solve: Process of Elimination

(Note: The rule numbers (DR#) correspond to the labels on my diagram.)

A – Incorrect. Violation of DR2.
B – Incorrect. Violation of DR3 (X host Monday means Z can’t host Saturday).
C – Incorrect . Violation of DR4 (Z must be host at least one of Wednesday or Saturday).
D – Correct. No rules violations.
E – Incorrect. Violation of DR1. Everyone must host at least once (missing Q).

2. If during one week, Z hosts on Wednesday and Saturday only, which one of the following must be true of the week?

A) Q hosts on Tuesday.
B) X hosts on Friday.
C) Y hosts on Monday.
D) Q hosts on exactly two days.
E) Y hosts on exactly two days.

How to Solve: Mini-diagram plus Inference

 

 

 

 

 

Draw a mini-diagram and put Z on Wednesday and Saturday. What does this mean? For starters, you’re looking at a 2-2-1-1 distribution, but none of the answer choices deal with this.

DR3 shows us that if Z is hosting Saturday, X cannot host on Monday. Since Z is only allowed to host on Wednesday and Saturday in this mini-diagram, Z cannot host on Monday either. DR5 says that Q cannot host Monday. This means that Y must host on Monday. This is answer choice C.

3. If during one week X hosts on Monday and Saturday only, which one of the following must be true of that week?

A) One other person besides X hosts on exactly two days.
B) The person who hosts Wednesday does not host Friday.
C) Y hosts on a day immediately before a day on which Q hosts.
D) Either Q or Y hosts on Friday.
E) Either Y or Z hosts on Tuesday. How to Solve: Inference

If X hosts on exactly two days, this means that you are looking at a 2-2-1-1 distribution (DR1). This means that besides X, one person will host on two days. And… you can stop there. Answer choice A says this.

4. Which one of the following could be true of one week’s schedule of hosts?

A) Q hosts on both Wednesday and Saturday.
B) X hosts on both Monday and Wednesday.
C) Z hosts on both Tuesday and Friday.
D) X hosts on Monday and Z hosts on Thursday.
E) Z hosts on Wednesday and X hosts on Saturday.

How to Solve: Process of Elimination

(Note: “Could be true” means the wrong answers must be false.)

A – Incorrect. Must be false because of DR4.
B – Incorrect. Must be false because of DR3+4. X hosting Monday means Z can’t host Saturday. But X hosting on Wednesday means that Z must host on Saturday (DR4). Z must and must not host on Saturday. This is clearly impossible.
C – Incorrect. According to DR4, Z must host on Wednesday and/or Saturday. If he hosts both Tuesday and Friday, placing him on either Wednesday or Saturday would violate DR2 against consecutive hosting.
D – Incorrect. According to DR3+4, if X hosts Monday, Z must host Wednesday (he cannot host Saturday). We again have a violation of DR2, as Z is required to host on Wednesday and Thursday.
E – Correct. No rules violations.

5. Which one of the following CANNOT be true of one week’s schedule of hosts?

A) Q hosts on Tuesday and X hosts on Friday.
B) X hosts on Monday and Z hosts on Tuesday.
C) X hosts on Monday and Z hosts on Friday.
D) Y hosts on Monday and Z hosts on Tuesday.
E) Y hosts on Tuesday and Z hosts on Friday.

How to Solve: Inference

This one is tough. Process of elimination would be lengthy because of the difficulty of eliminating “could be true” answers.

What you want to do is look for and start with answer choices that deal with two variables that tend to restrict each other or are used in a rule with each other. Answer choices B and C deal with variables X and Z, which are used together in DR3.

In answer choice B, X is hosting on Monday (which, by DR3, means Z cannot host Saturday, which by DR4 means Z must host Wednesday). Answer choice B also specifies that Z is hosting Tuesday. We now have Z hosting on Tuesday and Wednesday, which is a violation of DR2. B is therefore the correct response.

6. If during one week Q hosts exactly twice but he hosts on neither Tuesday nor Wednesday, which one of the following could be true of that week?

A) One person hosts exactly three times during the week.
B) Three people host exactly one time each during the week.
C) Z hosts on no day that is immediately before a day on which Q hosts.
D) X hosts on Wednesday.
E) Z hosts on Friday.

How to Solve: Process of Elimination plus Mini-Diagram plus Inference

If you haven’t noticed by now, it is usually easiest to solve “could be true” problems by process of elimination, while it is usually easiest to solve “must be true” problems by finding the answer directly using inferences.

This problem is also aided by using a mini-diagram:

 

 

 

 

If Q does not host on Tuesday or Wednesday, and by DR5 he cannot host on Monday, Q has three consecutive days (Thursday, Friday, and Saturday) to host twice. But since DR2 says he cannot host on consecutive days, Q must host on Thursday and Saturday. Further, Z must host on Wednesday, since by DR4 he hosts on at least one of Saturday and Wednesday.

Remember, in “could be true” questions, the wrong answers all must be false.

A – Incorrect. This must be false because if Q hosts exactly twice, the distribution is 2-2-1-1. There is no room for someone to host three times.
B – Incorrect. Again, this must be false because if Q hosts exactly twice, the distribution is 2-2-1-1. Someone else must host twice.
C – Incorrect. Remember, Z must host Wednesday and Q must host Thursday. A quick glance at our mini-diagram confirms this.
D – Incorrect. Only one person can host per day and we’ve already established that Z must host on Wednesday.
E – Correct. Z could host on Friday. DR4 says he must host on Wednesday and/or Saturday, but he can also host on other days. There are no other rules violations.

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Now let’s do a “less” game:

During a single week, from Monday through Friday, tours will be given at three different houses – X, Y, and Z. Exactly five tours will be given that week, one each day. The schedule of tours for the week must conform to the following restrictions:

Each home is toured at least once.
X is not toured on Monday.
Z is toured on two consecutive days, and on no other days.
Y is not toured on Wednesday.
If X is toured on Thursday, then Y is toured on Friday.

This is a “less” game because you have three variables to fit into five days. Z is the limiting variable here. Before you read on, you should stop and try and figure out why this is key…

The reason Z is key is twofold. First, Z can only be used twice. This means your distribution is 2-2-1. S and one other home can be toured twice and the third home only once.

The second reason Z is key – and this really breaks the game wide open – is that Z must be toured consecutively; it must be in a block. This limits the placement of Z to four different two-day blocks: MT, TW, WTh, ThF. It also limits the entire game to four general templates.

Here is the diagram:

 

 

 

 

 

Diagram rules explanations:

DR1. This is the only possible distribution.
DR2. If X is Tuesday, Y is Friday (and its contrapositive).
DR3. The ZZ consecutive block.
DR4. X cannot be Monday.
DR5. Y cannot be Wednesday.

Here are the four possible templates:

 

 

 

Template explanation:
1. If the ZZ block is on Monday and Tuesday, then X must be Wednesday because (1) only X or Z can be on Wednesday, and (2) Z can only be used twice.
2. If the ZZ block is on Tuesday and Wednesday, Y must be on Monday since (1) Monday is either Y or Z, and (2) Z can only be used twice.
3. If the ZZ block is on Wednesday and Thursday, Y must be on Monday for the same reasons as in template two. Therefore, Tuesday and Friday are both either X or Y.
4. If the ZZ block is on Thursday and Friday, Y must be on Monday for the same reasons as in template two. X must be Wednesday for the same reasons as in template one. Tuesday is either X or Y since Z can only be used twice.

Off we go…

1. If in the week’s tour schedule the home that is toured on Tuesday is also toured on Friday, then for which one of the following days must a tour of home Y be scheduled?

A) Monday
B) Tuesday
C) Wednesday
D) Thursday
E) Friday

How to Solve: Refer to Templates

The home that toured Tuesday can also be on Friday in template three only. In template three, Y must be Monday. The answer is A.

2. Which of the following CANNOT be true of the week’s tour schedule?

A) The home toured Monday is also toured Tuesday.
B) The home toured Monday is also toured Friday.
C) The home toured Tuesday is also toured Thursday.
D) The home toured Wednesday is also toured Friday.
E) The home toured Thursday is also toured Friday.

How to Solve: Look for Template Violations

You can answer this question pretty quickly by going through each answer choice and looking for clear violations of our rules and/or templates. Answer choice C can never happen in any of our templates. In templates one and two, Z is on Tuesday but cannot be on Thursday. In templates three and four, Z is on Thursday but cannot be on Tuesday.

3. If in addition to home Z, one other home is toured on two consecutive days, then it could be true of the week’s tour schedule both that

A) Y is toured on Monday and X is toured on Thursday.
B) Y is toured on Tuesday and Z is toured on Wednesday.
C) X is toured on Tuesday and Y is toured on Friday.
D) Z is toured on Monday and X is toured on Friday.
E) Z is toured on Wednesday and Y is toured on Friday.

How to Solve: Process of Elimination

Remember, the wrong answers must be false.

A – Incorrect. Y is toured on Monday in templates two, three, and four. In the third and fourth, Z is on Thursday. In the second, if X were to be Thursday, Y would be on Friday (as per rule two) meaning nothing else would be consecutive.
B – Correct.
C – Incorrect. X can be on Tuesday and Y can be on Friday only in template three. That would make for the following (in which nothing other than ZZ is consecutive): YXZZY.
D – Incorrect. Z is toured Monday only in template one. This means X is on Wednesday and, per this answer choice, Friday. Since everything must be toured once, Y would be on Thursday, giving us ZZXYX. Again, nothing consecutive besides Z.
E – Incorrect. Z is on Wednesday only in templates two and three. If Y is on Friday, then we have either YZZXY or YXZZY. Neither meets the criteria for this problem.

4. If in the week’s tour schedule the home that is toured on Tuesday is also toured on Wednesday, then which one of the following must be true of the week’s tour schedule?

A) Y is toured on Monday.
B) X is toured on Tuesday.
C) Z is toured on Wednesday.
D) Z is toured on Thursday.
E) Y is toured on Friday.

How to Solve: Templates

At this point you’ve probably recognized how much easier the templates have made this game.

The templates where the home toured Tuesday can be the home toured Wednesday are two and four. In both of those setups, Y is toured Monday. Answer choice A is correct.

5. If in the week’s tour schedule the home that is toured on Monday is not the home that is toured on Tuesday, then which one of the following could be true of the week’s schedule?

A) A tour of Z is scheduled for some day earlier in the week than is any tour of Y.
B) A tour of X is scheduled for some day earlier in the week than is any tour of Y.
C) Z is toured Monday.
D) Y is toured Tuesday.
E) X is toured Wednesday.

How to Solve: Process of Elimination

Remember, wrong answers must be false.

A – Incorrect. This must be false because Y is on Monday in each of the three possible templates (two, three, and four). (In template one, the ZZ block violates the rule introduced in the question.) Therefore, Z cannot come before Y.
B – Incorrect. See previous explanation.
C – Incorrect. In the three templates where the home toured Monday isn’t necessarily the home toured Tuesday, Y is on Monday, not Z. In addition, Z on Monday means Z on Tuesday as per rule three.
D – Incorrect. The only templates that don’t violate the rule introduced in the question have Y on Monday. Y cannot also then be on Tuesday.
E – Correct.

Notice that we didn’t really use our distribution rule in the second game as much as in the first, but it’s still nice to have.

Final Notes

“More” and “less” games are a little tougher than simple ordering games with their 1:1 ratio of variables to possible slots, but they are still manageable. Focus, find inferences, limit the game when you can. Look for those game-breaking rules. Tackle each problem carefully. You will be fine.

As usual, if you have any questions, please visit the TLS LSAT Prep forum at http://www.top-law-schools.com/forums/viewforum.php?f=6 .