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 T.B.Justin
  • Posts: 194
  • Joined: Jun 01, 2018
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#59309
Jon Denning wrote:Hey Sandy,

Thanks for the question! This is a tricky game for a lot of people, but hopefully I can help clear things up a bit.

Numbers first. With 9 students into 6 positions, the game at first feels unbalanced, but the test makers then give you the groups: 3 block with P, 2 block with IL, and 4 singles for the other 4 spaces. We don't know the exact placement of the blocks yet of course, but at least we know the definitive group sizes.

To me the real key to this one is noticing the long chain that forms with K, IL, G, and both O and the P block. That's really powerful, especially once you see that K can't be first (first position must be a Male), since it pushes K into 2 or 3, IL block into 3 or 4, G into 4 or 5, and the P block (which could contain O, so be careful) into 5 or 6.

Consider:

First position is single Male since K is first female but not first student, and since L, P, and O are all pushed further down by our chain then either N or S must be first. That answers question 20 and really helps on 22.

Starting at position 2 the "K > IL > G > Pblock and O" sequence takes over. That answers question 21 and is tested directly in question 23, where K is forced into 2, IL is 3, and G is 4.

The final question puts O between K and G, meaning K is 2, IL and O are 3 and 4 (not sure of order), G is 5, and the P block is 6. Since only the P block has an opening for H, then H must be 6.

And that's it! You could certainly go through and do a bunch of Not Laws, but really just realizing how limited that chain is crushes this game.

I hope that helps!
The last question that suspends G > O, and inserts the rule O > G, which I interpret as O and I > G not K > O and I > G- I ordered the sequence as follows: K > H and IL > G > P, (O>G, and can go before K, H, IL), and N,S are randoms.

How is it that K > O?
 Robert Carroll
PowerScore Staff
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#60970
T,

Read the entire question. It claims that the new rule has K before O. You missed this second half.

Robert Carroll
 T.B.Justin
  • Posts: 194
  • Joined: Jun 01, 2018
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#61276
Robert Carroll wrote:T,

Read the entire question. It claims that the new rule has K before O. You missed this second half.

Robert Carroll
Robert,

Thank you
 T.B.Justin
  • Posts: 194
  • Joined: Jun 01, 2018
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#61277
Is this a Grouping-Linear combination, partially-defined, balanced, numerical distribution game type?
 Brook Miscoski
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#62682
Justin,

I view this game as an advanced linear game--two of the lessons have 2 students, and is a rule that depends on male/female. The game requires numerical distribution (2-2-1-1-1-1, in no particular order) and is overloaded (9 students for 8 slots).
 madisonzill
  • Posts: 9
  • Joined: Mar 16, 2021
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#87233
Why is H/O or both in the 3 person class? How did you make that inference?
 Adam Tyson
PowerScore Staff
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#87242
That inference is based mostly on a process of elimination, madisonzill. We know that P cannot be with K, G, I, or L due to the sequence that places all of those variables somewhere before P. That leaves only H, O, N, and S to choose from, and either N or S must go first so they cannot both be paired with P. That leaves H, O, and just one of N or S as the options from which to select 2 to fill that group of 3. Without H or O, P could only be with one of N or S, meaning the group would have only two variables, or else with both, leaving nothing to go first. So the P group needs at least one of H or O in it!

It's a number of steps to get there, but when you have a group of a fixed size and limited options it's usually a good idea to consider what those options are, since we can usually expect there to be a question that tests that inference. And in fact questions 22, 23, and 24 ALL use that inference to some extent!

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