December 1991 LSAT Section 1 Question 21

# If Bob and Anna fly on the same plane, which one of the following must be true?

Replies

Eve on July 11, 2017

I am confused about this entire question and would like to see the set up and explanationHeedo on February 27, 2019

please explain this questionRavi on March 1, 2019

@hyperboleandahalf and @YHD,Happy to give you guys an explanation for this game.

At first glance, this game looks really tricky, but as long as you

stay calm and collected and make sure you're properly diagramming the

rules, the questions can be answered without much difficulty. Let's go

through the steps:

We're told that planes 1, 2, 3, and 4, and no others, are available to

fly in an air show. We're also told that pilots A, B, and C are all

aboard planes that are flying in the air show and they are the only

qualified pilots in the show.

Based on these first two sentences, we know that four planes are

available to fly, but we also know that all four planes do not

necessarily have to be flying. We also know that A, B, and C are all

aboard planes. This means that A, B, and C cannot be in two planes

since they're all aboard one, so they are only in one plane each.

We're then told that copilots D, E, and F are also all aboard planes,

which means that D, E, and F cannot be in two planes since they're all

aboard one, so they are only in one plane each.

Game pieces:

A B C

D E F

We're then told that no plane flies in the show without a qualified

pilot aboard. Since we already know that all of the pilots are aboard

a plane and a pilot can't be in two planes at the same time, we know

that there is a maximum of three planes that can be flying in the

show. We also now know that each plane that is flying in the show has

to have a pilot in it.

The game then states that no one but qualified pilots and copilots

flies in the show. this means that only A, B, C, D, E, and F can fly

in a plane.

We're then told that A will only fly in either 1 or 4

A - >1 or 4

Lastly, we're told that D will only fly in 2 or 3

D - ->2 or 3

This game seems very open-ended, but let's see if we can figure out

any more inferences before going into the questions. Let's look at the

rule involving D. We know it's in 2 or 3. We also know that every

plane that's flying in the show has to have a pilot in it. We know

that A can't go with D since A has to be in 1 or 4 and D in 2 or 3, so

that means that either B or C must always be with D.

B/C always with D in 2 or 3

None of the rules address E or F, so we know those copilots are

floating game pieces that can go wherever, as long as they're in the

presence of a pilot.

Let's take a look at the questions now.

Question one says if A flies in plane 4 and D flies in plane 2, which

one of the following must be true?

This is an inference question, so let's see what we can figure out

before looking at the answer choices. Based on the inference we made

with D, we know that B or C must always be with D. Let's look for an

answer that captures this inference.

Answer B says that if C flies in 3, B must fly in 2. This must be true

because we know that D must go with B or C. We're told that D is in 2,

and if C is in 3, then that means that B must also go into 2 with D.

This is the correct answer.

Question two says if B and A fly on the same plane, which one of the

following must be true?

If B is with A, then that means C must go with D in 2 or 3. Answer C

says D flies with C, and we know this must be true in this scenario,

so it's our correct answer.

Question three says if C and F are the only people in one of the

planes, which one of the following must be true?

We know C and F are by themselves, so that means that B must go with D

because D has to go with either B or C and C is already taken with F

in another plane. Answer D says D flies with B, and this is what must

be true, so it's our correct answer.

Question four says if plane 1 is used, its crew could consist of...

Answer A is out because we know that B or C has to fly with D, so we

can get rid of it.

Answer B looks like it might work (it has not immediately identifiable

problems), so let's eliminate the others before picking it.

Answer C is incorrect because we know that D must go with B or C, and

both B and C are in this answer choice, but D is not. Additionally, we

also know that D can't go in 1. This answer can be eliminated.

Answer D is incorrect because we know that D can't go in 1. We can get

rid of this.

Answer E is incorrect because we know that D can't go in 1. We can get

rid of this.

Going back to Answer B, if we put A, B, E, and F in 1, we could put D

and C in 2 and be fine. This works, so answer B is the correct answer

choice.

Question 5 says if as many of the pilots and copilots as possible fly

in plane 4, that group will consist of...

We already know that D can't go into 4 (it has to go in 2 or 3), and D

must be accompanied by B or C. Let's put B and D together in 2. That

gives us four remaining pieces that we could put in 4. A must go in 1

or 4, so that's no problem. Now, if we put C, E, and F in 4, that's

fine too, as this doesn't contradict any of our rules. A, C, E, and F

in 4 makes for a total of four people, so we know that if we pack as

many people into plane 4, there will be four people total.

Answer C says four people, so it's our answer choice.

As you can see, this was kind of a weird game and was very open, but

in diagramming our rules well and making the key inference that D must

go with B or C in 2 or 3, we were able to answer the questions

confidently.

Does this make sense? Let us know if you have any more questions on

this game or anything else!

on October 8, 2020

Thank you so much for breaking this down Ravi! This makes total sense now! I appreciate it a lot!