This is an explanation of the fourth logic game from Section II of LSAT Preptest 63, the June 2011 LSAT.
A street entertainer has six boxes, stacked on top of each other. They are numbered 1 through 6, from the bottom to the top. Each box has a single colored ball. There are three possible colors – green, red and white (G, R, W).
Game Setup
This is a linear game, but it says the balls are above/below each other. So here’s how most people draw this game:
But here’s how you’ve drawn dozens of games that are very much like this one:
Why would you totally switch the way you make a diagram, just because the setup says above/below? None of the questions use “above/below”. They just use the normal numbers you’ve seen on dozens of other linear games.
Draw this one left-to-right, like the rest.
The rules themselves are nothing special. I couldn’t combine them. Here’s how I drew the three rules:
Rules two and three are both “at least one”, so the ‘1’ in both rules is a reminder. I just memorized the fact that it was “at least one” and not “only one”.
(The “at least one” is implied by “there is a”. Unless a rule specifies “only one”, the rule means “at least one”.)
The first rule is the one worth thinking about. There are more red balls than white balls. The possibilities are rather limited. If there are two white balls, there are three red balls, and one green ball.
If there is one white ball, then things are more flexible. There could be as many as four red balls, and at least one green ball. Or there could also be as many as three green balls, and only two red balls.
The other two rules you simply have to memorize. This game is a lot faster if you always remember that there’s a green ball before any red balls, and that there’s at least one white directly before a green.
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Major says
I am struggling with the last condition where “a white ball is immediately BELOW a box that contains a green ball” .
G – above W
W – below G
Please help me understand why this is not correct?
TutorLucas (LSAT Hacks) says
This rule means that we have at least one instance where a white ball is immediately below a box that contains a green ball. That doesn’t mean that all white balls are located immediately underneath green balls.
MemberDena Rubanowitz says
Hi,
What is wrong with setting the game up vertically?
FounderGraeme says
Nothing, if your mind works that way. I’ve found, for me and for most people, that a horizontal setup here works better. Try both and see what fits.
AD says
Hi,
Why can’t mix rules two and three?
FounderGraeme Blake says
Because they’re not referring to the same thing. The second rule refers to G being before *any* white ball. The third rule refers to WG happening at least once.
I admit the diagrams are slightly ambiguous, but that’s the sort of distinction I can hold in my head while doing the game. It worked for me, but you can always modify the diagrams if something like that is confusing.
Aray says
Graeme, I believe you meant to say that the second rule refers to G being before *any* red ball.
Anyway, thanks so much for all the detailed explanations — LSAT Hacks is terrific!
l.boy says
thanks for this I’m stuck on #19 because the explanation says “Rules two and three are both “at least one”, so the ‘1’ in both rules is a reminder. I just memorized the fact that it was “at least one” and not “only one”. ” that to me means its possible to have GR once meaning that G could come after R as long as there’s 1 box that has G before R. Reading this comment made it clear that G can never go after R.
Also thanks you so much for all the explanations — LSAT Hacks is awsome!