Q19

[mmcvaney]

Could someone please explain #19 to me? I cannot seem to figure this out and the number (1 at top, or 6 at top) really seems to be throwing me off. Help!

[ohthatpatrick]

When the test hands you a vertical ordering game, you should decide for yourself whether it's easier to create a vertical ordering diagram or whether you'd prefer sticking with the traditional horizontal ordering diagram and just translating the vertical ideas into horizontal ones.

I've tried it both ways, and overall I think it's easier to just continue to diagram these horizontally. We just have to label our 1st spot and our last spot according to which is the bottom and which is the top.

The setup paragraph tells you that 1 is the bottom and 6 is the top.

So I would have a diagram that looks something like this:

_ _ _ _ _ _
1 2 3 4 5 6
(btm) . . (top)

I label the bottom and top so that I'm reminded that something HIGHER goes to the right, something LOWER goes to the left.

What's different about this game, as opposed to most ordering games, is that we only have three elements to work with: G, R, W

Whenever you don't have a 1:1 ratio of elements / spaces, you need to ask yourself questions about how much / little these letters can be used.

For example, can I just have 6 greens? Do I have to use any of the letters? What's the min/max I can use each of them?

Rule 1: R > W
(this tells me that I may want to think in terms of quantity possibilities, once I've seen all the rules)

Rule 2: (at least 1) G -- (any) R

So R can't be first. The first spot will always be W/G.

Rule 3: WG chunk (remember BELOW means to the left of, so this says that there is a W immediately to the left of a G)

Is there anywhere this chunk CAN'T go? Hard to say. Seems like it could go anywhere.

From the 3 rules, I know I have to use R, W, and G at least once a piece. What's the minimum and maximum I could use each one? Let's work with W, which has to be a smaller number than R.

If I have the minimum W, which is 1, then the minimum R I can have is 2. That leaves 3 G's to fill out our 6 spots. If I keep increasing the number of R's until I hit the maximum, I see I could have 2, 3, or 4 R's (with a different number of G's for each scenario).

W R G
1 2 3
1 3 2
1 4 1

The maximum W's I can have is 2, because that forces R to be at least 3, leaving only 1 G.

W R G
2 3 1

So my whole list of quantity possibilities looks like this:

W R G
1 2 3
1 3 2
1 4 1
2 3 1

When I do Q19, and I see there are G's in 5 and 6, I ask myself whether I know which quantity scenario I'm in. I definitely can't be doing the last two possibilities, in which there is only 1 G.

So I have to consider either:
1W, 2R, and 3G
or
1W, 3R, and 2G

If I do that second option, then I only get 2 G's. Could the two G's in spots 5 and 6 be my only G's?

_ _ _ _ G G
1 2 3 4 5 6

No. I need at least one more G, because I have this rule:

(at least 1) G --- (any) R

So I need to put another G into this scenario, somewhere early on, so that it can come before my R's.

I now know definitely that I will be working with this quantity scenario:
1W, 2R, 3G

My other rule says that I need a WG chunk.

So now I have to think about where that WG chunk could occur.

One way would be:
_ _ _ W G G

That means that my final G and my 2 R's have to go in the first three spots:
G R R W G G

Any other ways to make the WG chunk work? I could try starting off with that:
W G _ _ G G

That means that my 2 R's would go in the middle:
W G R R G G

Any other ways? What if I put the WG chunk 2nd and 3rd?
_ W G _ G G

That means that my 2 R's would go in the remaining spots, but I can see that that would be breaking a rule, since R can never be first.

So my two options for this question stem's condition are
W G R R G G
and
G R R W G G

The only possible answer from that list is (C).

If you were having trouble with this game/question, there are a few takeaway suggestions I have:

1. If you're doing an ordering game, but your ranking things according to terms like "higher/lower", "better/worse", or something else that's non-numerical, make sure you label the extremes of your diagram so that it's easy to see what being more to the left or more to the right means in terms of that game.

2. If you're doing an ordering game and there is NOT a 1:1 ratio of elements/spaces, take the time to think through the min/max that you can use each element.

3. When you're spinning your wheels on any question, write out the scenarios in question. Don't try to visualize them all in your head. By the time I was done with this game, I had about 12 scenarios written on my page. It only takes a few seconds to write out 6 letters, and having them written in front of me allows my brain to analyze whether or not any of them break any rules.

Hope this helps.

[cdjmarmon]

I'm very curious about Rule #2. How it translates as well as an issue I have between what Im reading on here vs. what Kaplan is saying.

I find this rule equally confusing to the 1st rule in game 2 of PT 38

Kaplan says, "There must be atleast one green ball lower than the lowest red ball" And it continues to say, "Make sure you dont read more into this rule than what is here. There could be a green ball above a red ball in the sequence but there must be atleast one green one lower than the lowest red."

However, from what Im reading in this post and in the diagram is yall are saying rule 2 means any greens must be below any/all reds that are placed.

Which further confuses me on why a Red could not go in the first spot.

why couldnt I have:

R
R
G
G
W
R

[timmydoeslsat]

We know that we must have a G below the lowest R.

In your hypothetical, your lowest R (#1) will not allow for a G to go below it. That hypothetical cannot work here.

What the explanation is really saying is that we can have an R be below a G, however, even with that, you better have a G beneath the lowest R you place.

[cdjmarmon]

I suppose I am confused because I do not see how the rule says a G must to go below the "lowest" R.

Why can't a G just go below any of the R's?

[timmydoeslsat]

G can go below any of the R's.

We just cannot have the lowest R be below a G.

Quick hypotheticals of similiar situation:

G
R
G
R

That one is no good! The lowest R is not above a G.

R
G
R
G

That is good. Notice the lowest R, it has a G below it.

R
R
G
G

That is good. We have the lowest R followed by a G.

[CharlesS800]

I began this question by creating a diagram and placing G/G in boxes five and six.

I realized that if there were only these two Gs in this diagram, the game would not work, as you could not include both a W/G chunk AND still have a green ball that is located beneath all red balls. So, there must be three green balls, two red balls, and one white ball in this scenario.

The one white ball and the remaining green ball must be a W/G chunk to satisfy the third rule of the game and this chunk must end up below both remaining red balls to satisfy the second rule of the game.

This left me with this for a diagram:

G
G
R
R
G
W

Using this diagram, I could eliminate every answer expect for answer choice C.