- PowerScore Staff
- Posts: 8950
- Joined: Feb 02, 2011
- Wed Aug 09, 2017 3:25 pm
#38109
Setup and Rule Diagram Explanation
This is a Grouping: Defined-Moving, Balanced, Numerical Distribution game.
The game scenario establishes that seven candidates—F G H I K L M—are being assigned to four positions (four HR officers)—R S T U. Each candidate is assigned exactly once, and each officer is assigned at least one candidate. That gives us R S T U as the base, and a numerical distribution of 7 into 4. There are three unfixed distribution possibilities at this stage:
4-1-1-1
3-2-1-1
2-2-2-1
When the rules are considered in situations like this it is often the case that one or more of those distributions will be impossible, or at least fixed/limited in some way, so let’s run through the rules and see if that occurs. Regardless, these numerical assignment options tend to be the key to games like this, so they should be at the forefront of your focus going forward.
The first rule assigns G to U, so that’s easy enough.
The second rule creates an FL block, although we don’t know where FL is assigned at this point. While this doesn’t limit the distributions—and doesn’t rule out the 4-1-1-1!; be careful, just because FL go together doesn’t mean they aren’t part of a 4-person group—it does apply to them: there’s at least one single person each time, so FL must avoid that spot.
Rule 3 creates two not-blocks: H and I do not go together, and M and I do not go together. So I is the most powerful of these three variables, as it moves two other people out of its group. Again, we don’t know where these people go just yet (any could go with FL for instance, or with G on U), so we have to simply note the rule and move on.
The fourth rule makes K a single. So we can’t eliminate any distributions yet (each has at least one single), but we can make a Not Law under U for K, since it cannot go with the G that is there.
Finally, the last rule is numerical in nature and tells us that S evaluates more people than T. So S > T, which can be represented within your diagram itself if you choose (just include a “>” sign between the S and T in the base). This means that S must get either 2, 3, or 4 people, and T can only get 1, 2, or 3 people, with the number on one potentially determining the number on the other. It also gives another Not Law: since K is a single then K cannot be assigned to S.
So with all of that in-hand, what can we determine? Well, less than you might suspect unfortunately. That is, there are seven possible fixed distributions still allowed (I’ll show them below with notes on each, but only to satisfy the curious; I would not bother attempting to list these all out on the test seeing as there are so many!), our FL block can still be placed in any of the four base positions (even on U with G), our single K can go to either R or T and doing one or the other does little to perfectly restrict the distributions/placements themselves...so if I’m a savvy test taker at this point I’m moving to the questions, cognizant of the fact that there may be more to learn or add, but I’m better off doing that in response to questions/answers than continuing to hunt broadly for inferences at the outset!
I'll explain what I mean by that last statement above in a second, but first, here are those seven distributions, each fixed respectively for R, S, T, and U:
Now, to my point about moving intelligently to the questions!
If you’re feeling a bit uneasy with the skeletal nature of the diagram, find a Local question, ideal a Local Must Be True (since you get absolute inferences/answers), and use it to give you a concrete sense of how these rules/restrictions are going to play out. For instance, a question like #8 or #9 might provide both a clarification on the mechanics of the game itself since you’re working from a defined starting point and will produce absolute outcomes, and also allow you to confirm/validate your setup and inferences to this point: if you can confidently remove four answers and prove a single choice correct that bodes very well for the setup you’ve created, whereas if you remove all five answers, or if more than one answer seems to work then something is wrong...either in how you read the question or, more probably, in the setup you’ve built. So you’d know to cycle back to your initial diagram and inferences and double-check them
I’ll show you what I mean. Let’s do question 8 quickly here.
8. We’re told that T gets exactly two people. Let’s think about what that means. We know that S gets more than T, so if T has 2 then the only distribution that works is the 3-2-1-1, but fixed for R S T U as: 1 – 3 – 2 – 1. How do you know that it’s fixed? Between S and T we have five people, with S > T so S = 3 and T = 2. That only leaves two people to split between R and U. Who are those people? K at R, and G at U. So now we have the FL block, and H, I, and M where I has to avoid both H and M. How can we keep HI and MI apart? Put I with FL as a group of 3 at S, and M and H as a group of 2 at T! And there you have it, the whole layout for R S T U: K FLI HM G.
Now just run through the five answers removing four that Must Be True:
I hope this helps!
This is a Grouping: Defined-Moving, Balanced, Numerical Distribution game.
The game scenario establishes that seven candidates—F G H I K L M—are being assigned to four positions (four HR officers)—R S T U. Each candidate is assigned exactly once, and each officer is assigned at least one candidate. That gives us R S T U as the base, and a numerical distribution of 7 into 4. There are three unfixed distribution possibilities at this stage:
4-1-1-1 3-2-1-1 2-2-2-1When the rules are considered in situations like this it is often the case that one or more of those distributions will be impossible, or at least fixed/limited in some way, so let’s run through the rules and see if that occurs. Regardless, these numerical assignment options tend to be the key to games like this, so they should be at the forefront of your focus going forward.
The first rule assigns G to U, so that’s easy enough.
The second rule creates an FL block, although we don’t know where FL is assigned at this point. While this doesn’t limit the distributions—and doesn’t rule out the 4-1-1-1!; be careful, just because FL go together doesn’t mean they aren’t part of a 4-person group—it does apply to them: there’s at least one single person each time, so FL must avoid that spot.
Rule 3 creates two not-blocks: H and I do not go together, and M and I do not go together. So I is the most powerful of these three variables, as it moves two other people out of its group. Again, we don’t know where these people go just yet (any could go with FL for instance, or with G on U), so we have to simply note the rule and move on.
The fourth rule makes K a single. So we can’t eliminate any distributions yet (each has at least one single), but we can make a Not Law under U for K, since it cannot go with the G that is there.
Finally, the last rule is numerical in nature and tells us that S evaluates more people than T. So S > T, which can be represented within your diagram itself if you choose (just include a “>” sign between the S and T in the base). This means that S must get either 2, 3, or 4 people, and T can only get 1, 2, or 3 people, with the number on one potentially determining the number on the other. It also gives another Not Law: since K is a single then K cannot be assigned to S.
So with all of that in-hand, what can we determine? Well, less than you might suspect unfortunately. That is, there are seven possible fixed distributions still allowed (I’ll show them below with notes on each, but only to satisfy the curious; I would not bother attempting to list these all out on the test seeing as there are so many!), our FL block can still be placed in any of the four base positions (even on U with G), our single K can go to either R or T and doing one or the other does little to perfectly restrict the distributions/placements themselves...so if I’m a savvy test taker at this point I’m moving to the questions, cognizant of the fact that there may be more to learn or add, but I’m better off doing that in response to questions/answers than continuing to hunt broadly for inferences at the outset!
I'll explain what I mean by that last statement above in a second, but first, here are those seven distributions, each fixed respectively for R, S, T, and U:
1 – 4 – 1 – 1 (In this case G and K are both singles, with K at either R or T. S gets the FL group, as well as M and H so that neither is grouped with I. That makes I the other single, at whichever of R or S is open [not taken by K])
1 – 3 – 2 – 1 (Here K and G are singles at R and U, respectively. To keep the FL block together, and to keep H, M, and I apart, we need to put I with F and L at S, and group H and M together at T. So this distribution places everyone definitively)
1 – 3 – 1 – 2 (With this setup the FL block must be a part of the 3-person group at S, but we don’t know who with. Similarly, K must go to R or T, but we can’t tell which one. I, H, and M can be placed anywhere here)
2 – 3 – 1 – 1 (This is very similar to the 1-3-2-1 layout above: K is a single at T, HM are a block at R, and FLI a 3-person group at S)
1 – 2 – 1 – 3 (In this case there are two possible arrangements: we put the FL block at S and add H and M to U with G; we put the FL block on U with G and an HM block at S. Either way our singles are K and I)
3 – 2 – 1 – 1 (Once again, we have a 3-2-1-1 distribution, and then one is just like the 1-3-2-1 and the 2-3-1-1 options above: K and G are singles, FLI are our triple, and H and M are our double)
2 – 2 – 1 – 2 (This is the most “even” distribution, which are often the most uncertain but in this case we’re fairly restricted: FL goes to either R or S, with the other of R or S getting an HM block; K is single at T; I joins G at U)
Now, to my point about moving intelligently to the questions!
If you’re feeling a bit uneasy with the skeletal nature of the diagram, find a Local question, ideal a Local Must Be True (since you get absolute inferences/answers), and use it to give you a concrete sense of how these rules/restrictions are going to play out. For instance, a question like #8 or #9 might provide both a clarification on the mechanics of the game itself since you’re working from a defined starting point and will produce absolute outcomes, and also allow you to confirm/validate your setup and inferences to this point: if you can confidently remove four answers and prove a single choice correct that bodes very well for the setup you’ve created, whereas if you remove all five answers, or if more than one answer seems to work then something is wrong...either in how you read the question or, more probably, in the setup you’ve built. So you’d know to cycle back to your initial diagram and inferences and double-check them
I’ll show you what I mean. Let’s do question 8 quickly here.
8. We’re told that T gets exactly two people. Let’s think about what that means. We know that S gets more than T, so if T has 2 then the only distribution that works is the 3-2-1-1, but fixed for R S T U as: 1 – 3 – 2 – 1. How do you know that it’s fixed? Between S and T we have five people, with S > T so S = 3 and T = 2. That only leaves two people to split between R and U. Who are those people? K at R, and G at U. So now we have the FL block, and H, I, and M where I has to avoid both H and M. How can we keep HI and MI apart? Put I with FL as a group of 3 at S, and M and H as a group of 2 at T! And there you have it, the whole layout for R S T U: K FLI HM G.
Now just run through the five answers removing four that Must Be True:
A. R has 1 person (K) so this is true.Attempting a question like this early on provides a lot of insight into how the game operates, and success on it demonstrates that our setup and inferences are likely correct! So be smart not just in your initial diagramming, but also in how you choose to tackle the questions
B. U has 1 person (G) so this is true.
C. F goes with L and I to S, not T, so this is NOT TRUE and is our correct answer.
D. FLI go to S so this is true.
E. FLI go to S so this is true.
I hope this helps!
