The
game section of the LSAT consists of four games, each of
which has about six questions--sometimes 5 and sometimes
7. Thus, there are usually twenty-four questions. The section
is 35 minutes long.
LSAT
TEST GAME FORMATS
Game 1 (about 6 questions)
Game 2 (about 6 questions)
Game 3 (about 6 questions)
Game 4 (about 6 questions)
This
means that you have a little less than nine minutes for
each game. Or if you skip the most difficult game, as most
people should, then you have a little less than 12 minutes
for each game. If this sounds fast-paced, you're right.
The LSAT is a highly "timed" test, and the game section
is the most highly "timed" part.
SKIPPING
A GAME
Because
games are difficult and time consuming, you should consider
skipping the hardest one.
ORDER
OF DIFFICULTY
Unlike
most standardized tests, the questions on the LSAT are not
listed from the easiest to the hardest. If they were, then
deciding which game to skip would be easy--skip the last
game. However, this much can be said: the first game will
not be the hardest and the last game will not be the easiest;
do not, therefore, skip the first game. This is also true
of the question-set to a game.
THE
THREE MAJOR TYPES OF LSAT TEST GAMES
LSAT
Ordering Games
These
games require you to order elements, either in a line or
around a circle. The criteria used to determine order can
include size, time, rank, etc. Ordering games are the easiest
games on the LSAT. Luckily, they are also the most common.
LSAT
TEST Grouping Games
Grouping
games, as the term implies, require you to separate elements--typically
people--into groups. Some conditions of the game can apply
to entire groups only, some to elements within a group only,
and some to both. This added complexity makes grouping games,
in general, harder than ordering games.
Assignment
Games
These
games involve assigning characteristics to the elements,
typically people. The most common task in these games is
to assign a schedule. You probably have had some experience
with schedules; you may have written the weekly work-schedule
for a business. If so, you know how difficult the task can
become, even when only a few conditions are placed on the
employees: Bob will work Monday, Tuesday, or Friday only.
Susan will work evenings only. Steve will not work with
Bob. Add to this that the company must have a full staff
weekdays, but only three people can work weekends. Scheduling
games on the LSAT are similar to this.
LSAT
TEST LOGICAL CONNECTIVES
While
no training in formal logic is required for the LSAT, essentially
it is a logic test. So some knowledge of formal logic will
give you a definite advantage.
To
begin, consider the seemingly innocuous connective "if...,
then...." Its meaning has perplexed both the philosopher
and the layman through the ages. The statement "if A,
then B" means by definition "if A is true, then B must be
true as well," and nothing more. For example, we know
from experience that if it is raining, then it is cloudy.
So if we see rain falling past the window, we can validly
conclude that it is cloudy outside.
There
are three statements that can be derived from the implication
"if A, then B"; two are invalid, and one is valid.
From
"if A, then B" you cannot conclude "if B, then A." For example,
if it is cloudy, you cannot conclude that it is raining.
From experience, this example is obviously true; it seems
silly that anyone could commit such an error. However, when
the implication is unfamiliar to us, this fallacy can be
tempting.
Another,
and not as obvious, fallacy derived from "if A, then B"
is to conclude "if not A, then not B." Again, consider the
weather example. If it is not raining, you cannot conclude
that it is not cloudy--it may still be overcast. This fallacy
is popular with students.
Finally,
there is one statement that is logically equivalent to "if
A, then B." Namely, "if not B, then not A." This
is called the contrapositive, and it is very important.
If
there is a key to performing well on the LSAT, it is the
contrapositive.
To
show the contrapositive's validity, we once again appeal
to our weather example. If it is not cloudy, then from experience
we know that it cannot possibly be raining.
We
now know two things about the implication "if A, then B":
1)
If A is true, then B must be true.
2) If B is false, then A must be false.
If
you assume no more that these two facts about an implication,
then you will not fall for the fallacies that trap many
students.
We
often need to rephrase a statement when it's worded in a
way that obscures the information it contains.
On
the LSAT, as in everyday speech, two negatives make a positive--they
cancel each other out.
not(not
A) = A
Example:
"It
is not the case that John did not pass the
LSAT" means the same thing as "John did pass the
LSAT."
The
statement "if A, then B; and if B, then A" is logically
equivalent to "A if and only if B." Think of "if and only
if" as an equal sign: if one side is true, then the other
side must be true, and if one side is false, then the other
side must be false.
(If
A, then B; and if B, then A) = (A if and only if B)
| A |
B |
| True |
True |
| False |
False |
Example:
"If
it is sunny, then Biff is at the beach; and if Biff is at
the beach, then it is sunny" is logically equivalent to
"It is sunny if and only if Biff is at the beach."
"A
only if B" means that when A occurs, B must also occur.
That is, "if A, then B."
A
only if B = if A, then B
Example:
"John
will do well on the LSAT only if he studies hard" is logically
equivalent to "If John did well on the LSAT, then he studied
hard."
(Note:
Students often wrongly interpret this statement to mean
"if John studies hard, then he will do well on the LSAT."
There is no such guarantee. The only guarantee is that if
he does not study hard, then he will not do well.)
The
statement "A unless B" means that A is true in all cases,
except when B is true. In other words if B is false, then
A must be true. That is, if not B, then A.
A
unless B = if not B, then A
Example:
"John
did well on the LSAT unless he partied the night before"
is
logically equivalent to
"If
John did not party the night before, then he did well on
the LSAT."
The
two statements "if A, then B" and "if B, then C" can be
combined to give "if A, then C." This is called the transitive
property.
("If
A, then B" and "if B, then C") = ("if A, then C")
Example:
From
the two statements "if John did well on the LSAT, then he
studied hard" and "if John studied hard, then he did not
party the night before the test" you can conclude that "if
John did well on the LSAT, then he did not party the night
before the test."
DIAGRAMMING
Virtually
every game can be solved more easily and efficiently by
using a diagram. Unless you have a remarkable memory and
can process reams of information in your head, you must
draw a diagram. Because of the effectiveness of diagrams,
games are the best candidates for improvement. A well-constructed
diagram can change a convoluted, unwieldy mass of information
into an easily read list. In fact, from a well-constructed
diagram, you can often read-off the answers without any
additional thought.
Symbols
The
ability to symbolize sentences is one of the most important
skills you need to develop for the LSAT.
Five
basic symbols are used throughout this discussion. They
are
| Symbol |
Meaning |
| & |
and |
| or |
or |
| ~ |
not |
| --> |
If...,
then... |
| (
) |
parentheses |
LSAT
TEST ORDERING GAMES
Ordering
games are the easiest games, and fortunately they also appear
the most often.
Line-up
Game
There
are five people--Bugsy, Nelson, Dutch, Clyde, and Gotti--in
a police line-up standing in spaces numbered 1 through 6,
from left to right. The following conditions apply:
There
is always one empty space.
Clyde is not standing in space 1, 3, or 5.
Gotti is the third person from the left.
Bugsy is standing to the immediate left of Nelson.
"Clyde
is not standing in space 1, 3, or 5" is symbolized as C
= not(1, 3, 5). "Gotti is the third person from the
left" is naturally symbolized as G = 3rd. Note: the
fact that Gotti is third does not force him into space 3--he
could stand in spaces 3 or 4. "Bugsy is standing to the
immediate left of Nelson" is symbolized as BN. The
diagram will consist of 6 dashes:
_1_
_2_ _3_ _4_ _5_
_6_
When
placing the elements on the diagram, first look for a condition
that fixes the position of an element. There is none. Next,
we look for a condition that limits the position of an element.
The second condition, "Gotti is the third person from the
left," limits Gotti to spaces 3 and 4. This condition, as
often happens with ordering games, generates two diagrams:
one with the empty space to Gotti's left and one with the
empty space to his right:
Diagram
I ___ ___ ___ _G_
___ ___
Diagram
II ___ ___ _G_ ___
___ ___
Next,
we look for a condition that connects two or more people.
The last condition, BN, connects B with N. However,
at this stage we cannot place it on the diagram. Finally,
we look for a condition that states where a person cannot
be standing. The first condition states that Clyde cannot
be standing in space 1, 3, or 5. Noting this on the diagram
yields
Diagram
I _~C_ ___ _~C_ _G_
_~C_ ___
Diagram
II _~C_ ___ _G_ ___
_~C_ ___
(Note:
D is "wild" because the conditions do not refer to him.
Thus D can stand in more positions than any other person.)
This
diagram is self-contained. There is no need to refer to
the original problem. If possible, avoid rereading the problem.
1.
Nelson CANNOT stand in which one of the following spaces?
(A)
2 (B) 3 (C)
4 (D) 5 (E)
6
The
method of solution to this problem is rather mechanical:
We merely place Nelson in one of the spaces offered. Then
check whether it is possible to place the other people in
the line-up without violating any initial condition. If
so, then we eliminate that answer-choice. Then place Nelson
in another space offered, and repeat the process.
To
that end, place Nelson in space 2 in Diagram II:
_~C_
_N_ _G_ ___ _~C_
___
From
the condition BN, we know that B must be in space
1:
_B_
_N_ _G_ ___ _~C_
___
Now
D could stand in space 4, and C could stand in space 6--both
without violating any initial condition:
_B_
_N_ _G_ _D_ _X_
_C_ (Where X means "empty.")
This
diagram is consistent with the initial conditions. So N
could stand in space 2. This eliminates choice (A).
Next,
place Nelson in space 4. Then Diagram I is violated since
G is already in space 4, and Diagram II is also violated
since there is no room for the condition BN:
___
___ _G_ _N_ ___
___
The
answer is (C).
2.
Which one of the following spaces CANNOT be empty?
(A)
1 (B) 2 (C) 3 (D) 4
(E) 5
Assume
that space 1 is empty. Then in Diagram I, the condition
BN can be placed in spaces 2 and 3, D can be placed
in space 5, and C can be placed in space 6--all without
violating any initial condition:
_X_
_B_ _N_ _G_ _D_
_C_
Thus
space 1 could be empty. This eliminates (A).
Next,
assume that space 2 is empty. In Diagram I, this forces
BN into spaces 5 and 6:
___
_X_ ___ _G_ _B_
_N_
However,
this diagram does not leave room for C [recall C = not(1,
3, 5)]. Diagram I is thus impossible when space 2 is
empty. Turning to Diagram II, we see immediately that space
2 cannot be empty, for this would make G second, violating
the condition G = 3rd. Hence Diagram II is also impossible
when space 2 is empty. Thus space 2 cannot be empty, and
the answer is (B).
PATHS
AND FLOW CHARTS
Because
this type of game typically has many conditions, the chart
can easily get out of control. Charting is an art. However,
there are some guidelines that help:
1.
Look for a condition that starts the "flow" or that contains
a lot of information.
2. Look for an element that occurs in many conditions.
3. Keep the chart flexible; it will probably have to
evolve with the changing conditions.
As
you work through these games be alert to any opportunity
to apply the contrapositive rule of logic. Often negative
conditions can be expressed more clearly by rewording them
in the contrapositive. For example, the statement
"if
it is not sunny, then Biff is not going to the beach"
can
be reworded more directly as
"if
Biff is going to the beach, then it is sunny."
We
need to review two common fallacies associated with the
contrapositive. From the statement "if A, then B" we can
conclude, using the contrapositive, "if not B, then not
A." It would be fallacious, however, to conclude either
"if not A, then not B" or "if B, then A." Also note that
some means "at least one and perhaps all."
Flow
Chart
Six
debutantes--Alison, Bridgette, Courtney, Dominique, Emily,
Francine--meet at a party. During the time they have been
at the party some girls have come to like certain other
girls.
Amiable
Alison likes every girl at the party.
Aloof, yet popular Bridgette likes no one at the party,
but everyone likes her.
Courtney likes only two girls, one of whom is Dominique.
Dominique likes three girls, none of whom are Courtney or
Francine.
Emily and Francine each like only one girl.
Alison likes every girl, so we start the flow chart with
her:
Next,
every girl likes Bridgette, but she does not like any of
them. So we end the "flow" with Bridgette. (Note how the
diagram evolves.)
Next,
since Dominique likes three girls, two of whom are neither
Courtney nor Francine, she must like both Alison and Emily,
in addition to Bridgette. Adding this result plus the third
condition, "Courtney likes Dominique," to the diagram gives
Finally,
since Emily and Francine each like only one girl and everyone
likes Bridgette, Emily and Francine each must like Bridgette
only. So there is nothing else to add to the diagram.
Note
A,C,D forms a "loop", because from A the arrows can be followed
all the way around the "loop" back to A. But A,D,E does
not form a loop, because from A you cannot get back to A,
whether you go first to D, or first to E.
1.
A "click" is a group of two or more girls who like one another.
How many clicks are formed amongst the six girls?
(A)
0 (B) 1 (C)
2 (D) 3 (E)
4
There
is only one. In the chart, a two-way arrow connects A and
D, so they form a click. The loop A,C,D does not form a
click because it's not two-way: A likes C, but that feeling
is not reciprocal. The answer is (B).
2.
How many girls at the party like at least one girl whose
feelings are not reciprocal?
(A)
2 (B) 3 (C)
4 (D) 5 (E)
6
In
the chart, there are 5 arrows pointing to B, so 5 girls
like B. There are no arrows emanating from B, so none of
those feelings are reciprocal. The answer is (D).
GROUPING
GAMES
Because
grouping games partition elements into sets, the number
of elements is often an issue. Counting may have been one
of man's first thought processes; nevertheless, counting
possibilities is deceptively hard. This tends to make grouping
games more difficult than ordering games.
Pay
close attention to the maximum or minimum number of elements
in a group; this is often the heart of the game.
Selection
Game
The
starting line-up for the Olympic basketball "Dream Team"
is chosen from the following two groups:
Group
A: Johnson, Drexler, Bird, Ewing
Group B: Laettner, Robinson, Jordan, Malone, Pippen
The
following requirements must be meet:
Two
players are chosen from Group A, and three from Group B.
Jordan starts only if Bird starts.
Drexler and Bird do not both start.
If Jordan starts, then Malone does not.
Exactly 3 of the four fast-break specialists--Johnson, Bird,
Jordan, Pippen--must be chosen.
It
is best to solve this problem without a diagram; however,
we will still symbolize the conditions for clarity and easy
reference. The condition "Jordan starts only if Bird starts"
implies only that if Jordan is starting then Bird must be
starting as well. So we symbolize it as Jordan-->Bird.
The condition "Drexler and Bird do not both start" means
that if one starts then the other does not. So we symbolize
it as Drexler-->~Bird. Students often misinterpret
this condition to mean that neither of them starts. To state
that neither starts, put both at the beginning of the sentence:
Both Drexler and Bird do not start.
The
condition "If Jordan starts, then Malone does not" is naturally
symbolized as Jordan-->~Malone. It tells us that
if J starts then M does not, but tells us nothing when M
does not start. Such a condition, where the two parts of
an if-then statement do not similarly affect each other,
is called a nonreciprocal condition. On the other hand,
a condition such as Jordan<-->~Malone affects J and M
equally. In this case, we are told that if J starts then
M does not as before, but we are told additionally that
if M does not start then J does. It is important to keep
the distinction between reciprocal and nonreciprocal relations
clear; a common mistake is to interpret a nonreciprocal
relation as reciprocal. The remaining conditions cannot
be easily written in symbol form, but we will paraphrase
them in the schematic:
Jordan-->Bird
Drexler-->~Bird
Jordan-->~Malone
2 from Group A
3 from Group B
fast-break specialists: Johnson, Bird, Jordan, Pippen
3 fast-break specialists
Ewing, Laettner, Robinson are "wild"
Note:
Ewing, Laettner, and Robinson are independent because there
are no conditions that refer directly to them. We now turn
to the questions.
1.
If Jordan starts, which of the following must also start?
(A)
Malone or Johnson
(B) Drexler or Laettner
(C) Drexler or Johnson
(D) Johnson or Pippen
(E) Malone or Robinson
From
the condition Jordan-->Bird, we know that if Jordan
starts, then Bird must start as well. Now both Jordan and
Bird are fast-break specialists, and three of the four fast-break
specialists must start. So at least one of the remaining
fast-break specialists--Johnson or Pippen--must also start.
The answer is (D).
2.
All of the following pairs of players can start together
EXCEPT:
(A)
Ewing and Drexler
(B) Jordan and Johnson
(C) Robinson and Johnson
(D) Johnson and Bird
(E) Pippen and Malone
Begin
with choice (A). Both Ewing and Drexler are from Group A,
so the remaining 3 starters must be chosen from Group B.
Additionally, they must all be fast-break specialists since
neither E nor D is--there are exactly 3 fast-break specialists.
But Jordan and Pippen are the only fast-break specialists
in Group B. So the third fast-break specialist cannot be
chosen. The answer therefore is (A). This type of question
can be time consuming because you may have to check all
the answer-choices--save these questions for last.
3.
If Malone starts, which one of the following is a complete
and accurate list of the players from Group A any one of
whom could also start?
(A)
J (B) J, D (C)
J, B (D) J, D, B (E)
J, E, B
Jordan
cannot start with Malone according to the condition Jordan-->~Malone.
To play three fast-break specialists, therefore, Johnson,
Bird, and Pippen are all required to start. Since both Johnson
and Bird are from Group A and exactly two players from that
group start, these two players comprise the complete list
of starters from Group A when Malone also starts. The answer
is (C).
4.
Which one of the following players must start?
(A)
Pippen (B) Johnson (C)
Jordan (D) Malone (E)
Bird
Suppose
Bird does not start. Then the 3 fast-break specialists must
be Johnson, Jordan, and Pippen. But if Jordan starts, then
from the initial conditions Bird must also start. Hence
Bird must always start. The answer is (E).
ASSIGNMENT
GAMES
Assignment
games match a characteristic with an element of the game.
For example, you may be asked to assign a schedule: Bob
works only Monday, Tuesday, or Friday. Or you may be told
that a person is either a Democrat or a Republican.
Because
the characteristics are typically assigned to groups of
elements, assignment games can look very similar to grouping
games. Additionally, in grouping games the groups are often
identified by their characteristics. However, in assignment
games you pair each element with one or more characteristics,
whereas in grouping games you partition the elements into
two or more groups.
Many
assignment games can be solved very efficiently by using
an elimination grid. An example will illustrate this method
of diagramming.
Elimination
Grid
Dean
Peterson, Head of the Math Department at Peabody Polytech,
is making the fall teaching schedule. Besides himself there
are four other professors--Warren, Novak, Dornan, and Emerson.
Their availability is subject to the following constraints.
Warren
cannot teach on Monday or Thursday.
Dornan cannot teach on Wednesday.
Emerson cannot teach on Monday or Friday.
Associate Professor Novak can teach at any time.
Dean Peterson cannot teach evening classes.
Warren can teach only evening classes.
Dean Peterson cannot teach on Wednesday if Novak teaches
on Thursday, and Novak teaches on Thursday if Dean Peterson
cannot teach on Wednesday.
At any given time there are always three classes being taught.
We
indicate that a teacher does not work at a particular time
by placing an X on the elimination grid. Placing the conditions
on a grid yields
To
answer the following questions, we will refer only to the
table, not the original problem.
1.
At which one of the following times can Warren, Dornan,
and Emerson all be teaching?
(A)
Monday morning
(B) Friday evening
(C) Tuesday evening
(D) Friday morning
(E) Wednesday morning
The
table clearly shows that all three can work on Tuesday night.
The answer is (C).
2.
For which day will the dean have to hire a part-time teacher?
(A)
Monday
(B) Tuesday
(C) Wednesday
(D) Thursday
(E) Friday
Dornan
and Novak are the only people who can work Monday evenings,
and three classes are always in session, so extra help will
be needed for Monday evenings. The answer is (A).
3.
Which one of the following must be false?
(A)
Dornan does not work on Tuesday.
(B) Emerson does not work on Tuesday morning.
(C) Peterson works every day of the week except Wednesday.
(D) Novak works every day of the week except Wednesday.
(E) Dornan works every day of the week except Wednesday.
The
condition "Dean Peterson cannot teach on Wednesday if Novak
teaches on Thursday, and Novak teaches on Thursday if Dean
Peterson cannot teach on Wednesday" can be symbolized as
(P not W)<-->(N = TH). Now, if Novak works every
day of the week, except Wednesday, then in particular he
works Thursday. So from the condition (P not W)<-->(N
= TH), we know that Dean Peterson cannot work on Wednesday.
But from the table this leaves only Novak and Emerson to
teach the three Wednesday morning classes. Hence the answer
is (D).
4.
If Novak does not work on Thursday, then which one of the
following must be true?
(A)
Peterson works Tuesday morning.
(B) Dornan works Tuesday morning.
(C) Emerson works on Tuesday.
(D) Peterson works on Wednesday.
(E) Warren works on Tuesday morning.
If
you remember to think of an if-and-only-if statement as
an equality, then this will be an easy problem. Negating
both sides of the condition (P not W)<-->(N = TH)
gives (P = W)<-->(N not TH). This tells us that
Dean Peterson must work on Wednesday if Novak does not work
on Thursday. The answer, therefore, is (D).
Caution:
Not all scheduling games lend themselves to an elimination
grid. It's sweet when this method can be applied because
the answers typically can be read directly from the table
with little thought. Only one-third of the assignment games,
however, can be solved this way. Most often the game will
require a more functional diagram, and you will need to
spend more time tinkering with it.
When
you first read an assignment game, you need to quickly
decide whether or not to use an elimination grid. You
may decide to use a table. Then spend three minutes trying
to set it up, only to realize you have taken the wrong
path and have wasted three minutes. Unfortunately, exact
criteria cannot be given for when to use an elimination
grid. But this much can be said: if only two options (characteristics)
are available to the elements--yes/no, on/off, etc.--then
an elimination grid is probably indicated.
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