Q14

[zhangstagangsta]

I found the logic chain set-up to be helpful for this game, but I am wondering how to answer a question such as number 14. Would one look at previously worked questions, or is there a deduction to be made?

[aileenann]

Hi there,

This is a must be true question, so I'd say you have highlighted one method of attack. That is, to figure out what must be true, you can look at all the possible combinations you have already generated and see what they all have in common - if one of those matches up with an answer choice *and* you cannot disprove it, that is your answer to a must-be-true question.

However, that method is probably not the faster. If something must be true, it will often show up as a direct inference of the rules - sometimes it is difficult to see though, so it may not be a first order inference you made from your diagram. If it is not an inference in your diagram, looking at an answer choice and deciding whether there is a logical reason why it must always be true - or alternately finding you cannot disprove it - may be a little faster because it is working a little more abstractly/theoretically.

All that goes in general. More specifically with respect to the logic chain, I generally find on a standard logic chain setup that just about anything that must be true will also be laid out in the logic chain in one form or another, though it may mean following all the arrows to see the inference.

I hope this helps. As always, feel free to followup with questions and comments.

[a_ezekiel]

I am wondering how to answer a question such as number 14. Would one look at previously worked questions, or is there a deduction to be made?

I'm not an Atlas employee, nor do I play one on TV but I can tell you how I solved problem #14.

Choices A, B, and C were clearly wrong. Why? First, note that the logic chain diagram for this problem is very sparse. There are no arrows/dependencies at all for six elements including two stones of each type: F and G (rubies), J and K (sapphires), X and Y (topaz). To put it another way:

• Any solution using G could equally well use F.


• Any solution using J could equally well use K.


• Any solution using X could equally well use Y.

Thus one of these numerous "floater" elements cannot be required in every possible solution unless at least TWO of the six elements are fixed in every possible solution... and that would make the game way too easy for the LSAT. (Besides which, our logic chain is pretty obviously not that restrictive -- in fact, it's very sparse!)

That leaves us with only two possible answers: D or E. But which is right?

To be sure of answer choice D we must find a solution with a 1-stone group, but not including any 3-stone groups.

To be sure of answer choice E we must find a solution with a 3-stone group, but not including any 1-stone groups.

But note that all our groups have either 3 or 4 stones, and we must choose six total stones! To select answer choice D the only possible breakdown between the three groups is 1-1-4. That's the only way we can do it... and 1-1-4 isn't possible. The only group with four stones is topaz, and the logic chain won't allow us to choose all four topaz stones. W forces out Z, and Z forces out W, which means that 3 is the largest number of topaz stones we can ever choose.

That eliminates answer choice D, leaving answer choice E as the only remaining solution.

To confirm that E is correct: we need a solution with a 3-stone group, but no 1-stone group. The only way to do that is 3-3-0. Is there such a solution? Sure! In fact there are two such solutions. Topaz XYZ plus ruby FGH, or Topaz WXY plus sapphire JKM. Thus the 3-3-0 split is possible, and answer choice E is correct.

Hope this helps!

[noah]

Great thinking on the first 3 eliminations. On the last two, you actually would disprove an answer by showing that you do NOT need the set-up it suggests. For example, for (D), you would look to see if you can make a selection that does not include a stone type of which exactly one stone is selected. This is possible. You could do Z X Y & F G H. By the way, it's possible to do what (D) suggests, just not necessary: W X Y J K F

For (E), you're efforts should be focused on constructing a selection that does not include a threesome of some stone type. This proves impossible. You would have to do a 2 - 2 - 2 set-up, which triggers the 2S --> R rule, or a 4 - 1 - 1, and the only type that has 4 options is Topaz, but you W and Z are mutually exclusive, so only 3 are possible.

The important point here is that on must be true questions, we generally want to disprove answers, not prove them correct. If you try to prove (D) correct, you can mistake the fact that it is possible for the idea that it is necessary.

I hope that helps.

BTW, here's my diagram:

[a_ezekiel]

The important point here is that on must be true questions, we generally want to disprove answers, not prove them correct. If you try to prove (D) correct, you can mistake the fact that it is possible for the idea that it is necessary.

Noah is correct, of course. My solution assumed that one of the remaining choices had to be correct, which was true in this specific case; but in general, it's better to think the other way, as he says.


In reviewing this problem, however, I actually see a direct logical solution for Q14, i.e. one that doesn't require process of elimination to eliminate the four wrong answers. Here it is.

RULE: The only four-stone group is topaz; but we can never choose all 4 because of the W/Z exclusion rule...
****INFERENCE: Thus the largest allowable selection from any group is 3.

RULE: We must choose 6 stones from 3 groups.
RULE: 2 Sapphire --> 1 Ruby...
****INFERENCE: Thus an even split (2-2-2 between the groups) is not allowed.

SOLUTION: If we can't split 2-2-2, but still need a total of 6, then at least one group must be larger than 2 and at least one group must be smaller than 2. But "larger than 2" means 3, because that is the biggest allowable selection from any group. Thus all possible groupings must contain at least one 3, which gives us answer choice (E) by direct count-based logic.

[noah]

Alan, Good thinking. On another note about this question, I usually tell students that "number answers", like (D) and (E) in #14, are more likely to be the answer than "straight answers". Where I came up with this terminology, I have no idea. It's worth your time to re-order how you evaluate answers, starting from the "number answers" and moving up the line from those.

- N

[interestedintacos]

In my opinion all the methods previously discussed are not ideal--there is an ideal way to solve that question more quickly.

CLEARLY, from the set-up and rules we can see this game will have a number distribution element (i.e. that the game will have a limited number of potential distributions of Rs, Ss and Ts). This is very typical for a closed binary grouping game--certainly if we are given rules like the first two (which make it obvious that there will be a limited set of distributions).

Aside from the logic chain, which is part (but just part) of solving this problem, we need to create a list of the potential number distributions of T, R and S.

We know there are at least two Ts, and we cannot have all 4 Ts because W and Z cannot both be selected (another given rule). So we can fill in the possibilities of there being either 2 Ts or 3Ts.

With 2 Ts there can be either 3 Rs and 1 S or 3 Ss and 1 R. We cannot have 2 of each R and S because that violates another one of our given rules (a previous poster pointed this out). So we have exhausted the possibilities with 2 Ts. Now onto 3 Ts...

For this question we don't need to explore the options with 3 Ts because we already can see that answer choice E is correct. If we were to explore the options with 3 Ts we could come up with a quick set of possibilities that will help on later questions as well (Q 16 and 18).

For instance, on Q 18, which may have been difficult otherwise, knowing that only two sapphires are selected means only one potential distribution of T S and R is possible (3-1-2), and this lets you take care of the question quickly.

[noah]

Good point! I agree that writing out the possible arrangements of T, S, and R is helpful. Actually, I should go back and amend the diagram, as I've been doing that for some time when I teach that.

I usually write this out:

3 2 1
3 3 0

And that represents the general number assignments that are possible. It leaves open which group gets which - and obviously there are some arrangements that won't work. It could be useful to write out all possibilities (i.e. T T T S S R, T T T S R R, etc.), but it's more work than I usually like to do.

[brian.atchley]

Why does Noah say, when describing Q14, that if we are to consider that G must be in, then we have to consider that F must be in as well?

He uses the term "logic twins" to describe G and F, and as such, they must have the same conditions. But why?

Couldn't we have G, K,M W,X,Y??

[ManhattanPrepLSAT1]

Good question! Is there any rule that applies to G that does not also apply to F? Since there is no rule that applies to one but not the other of them, they operate as equivalents. Anything that must /could/cannot be true of one, must/could/cannot be true of the other.

Couldn't we have G, K,M W,X,Y??

Couldn't the solution also be F, K, M, W, X, and Y? F and G are simply good substitutes for each other (twins).