Most of my LSAT students identify the “substitute constraint” or “substitute condition” question type as the most difficult type on LSAT logic games.
At first glance, students attempt to test each scenario presented in the answer choices, wasting precious time, and often spinning their wheels. And, even if they “get lucky” discovering the correct answer, they remain unsure of themselves and worry they have missed some detail that actually prevents the substitute constraint from fully replacing the original.
But there is good news. There is a reason that this question type is invariably reserved for the last question in a particular game. If you follow my method for LSAT logic games, you will successfully create the new rule yourself simply by working methodically through the earlier questions, and the correct answer choice will be obvious.
As always, it’s best to use an example, the following from Section 4 of LSAT #75, administered in June 2015:
This is a straightforward linear game, though certainly not an easy one. The first step is to set up your diagram (N.B.: I like to begin with a column referencing the applicable question number). Then, next door, translate your Rules. Here, the rules are all in the same direction (“earlier”) and easy to translate.
Librarians: F, G, H, K, L, M, Z
Q# | M | Tu | W | Th | F | Sa1 | Sa2 |
Now is also a good time to look at how the given Rules interact with each other:
- Rules 2 and 3 acting together tell us that that H, F, and M must all precede G (H, F, M < G). This means G can be scheduled at the earliest on the fourth day, Thursday, and cannot be scheduled Monday through Wednesday.
- Rules 3 and 4 acting together tell us that F must precede all of M, G, K, and Z (F < M, G, K, Z). This means that F can be scheduled at the latest on the third day, Wednesday, and cannot be scheduled Thursday through Saturday.
Add these hybrids to your Rules, and add the two newly-discovered constraints to your table:
(no G) | (no G) | (no G) | (no F) | (no F) | (no F) | (no F) | |
Q# | M | Tu | W | Th | F | Sa1 | Sa2 |
Now to the questions:
As always, the first question is a simple test of the Rules. Follow your Rules methodically (you cannot get the first question wrong!), and enter the correct answer in your table.
Choice (E) is eliminated by Rule 1; (C) by Rule 3; (D) by Rule 4; and (B) by Rule 5. The correct choice is (A). Check it one more time against all your Rules (it works out!) and enter it into your table of correct possibilities as below. [Note also that the answer choices showed a predilection for scheduling F and H early on, and G and Z later on, which is consistent with your Rules and good to keep in mind as you go forward].
(no G) | (no G) | (no G) | (no F) | (no F) | (no F) | (no F) | |
Q# | M | Tu | W | Th | F | Sa1 | Sa2 |
12 | F | H | M | K | Z | G | L |
Your table (example #12) already tells you Hill can be on Tuesday, so (B) is wrong. Now let’s look at Rule 4: Only F and K must precede Z, and there is no other constraint on Z. If two Librarians must precede Z, there is no way Z can be scheduled before the third day, so Zahn CANNOT be scheduled on Tuesday. The correct choice is (E). Add this new piece of information to your table, and while you’re at it, I always encourage my students to produce a confirming example that follows the Rules and is sufficiently different from the first example, so you grow the databank of examples you can use later on:
(no G)
(no Z) |
(no G)
(no Z) |
(no G) | (no F) | (no F) | (no F) | (no F) | |
Q# | M | Tu | W | Th | F | Sa1 | Sa2 |
12 | F | H | M | K | Z | G | L |
13 | H | F | K | M | G | Z | L |
The question demands that K < M, but where should we begin in testing answer choices? Well, we created an example #13 in our table where K < M: does that eliminate any answer choices? Yes! (A), (C), and (D) are all true in example #13, so can be eliminated. We need to choose between (B) and (E). Looking at (B), I am wary because I learned earlier that G needs to come pretty late in the schedule. If we want G < K, and the question tells us K < M, we would combine to G < K < M, which in turn violates Rule 3 (M < G). The correct choice is (B), which CANNOT be true. If you want to be sure, add an example where choice (E) is viable, which only increases your databank of possibilities:
(no G)
(no Z) |
(no G)
(no Z) |
(no G) | (no F) | (no F) | (no F) | (no F) | |
Q# | M | Tu | W | Th | F | Sa1 | Sa2 |
12 | F | H | M | K | Z | G | L |
13 | H | F | K | M | G | Z | L |
14 | F | H | K | Z | M | G | L |
We just created an example #14 where Z is on desk duty on Thursday [N.B., this was no accident: I intentionally looked forward to the coming questions and saw that this would be useful]. From that example, we can eliminate (B). From the other choices, I would try (D) first because it adds a second constraint on Z, and specifically I will try to contradict the “must be true” requirement by putting Z < H. I also know from Rules 1 and 2 that H < L and H < G, slotting L and G for Saturday. Rule 7 puts F < M and F < K, so F must be Monday and K and M are interchangeable between Tuesday and Wednesday. The example works, so it is not accurate that (D) must be true (we’ve shown it false). It is also now obvious from the same example that (C) need not be true (M < H in the example), and (E) need not be true (interchangeability of K and M). The correct choice is (A), and at least we have added to our databank example #15(D):
(no G)
(no Z) |
(no G)
(no Z) |
(no G) | (no F) | (no F) | (no F) | (no F) | |
Q# | M | Tu | W | Th | F | Sa1 | Sa2 |
12 | F | H | M | K | Z | G | L |
13 | H | F | K | M | G | Z | L |
14 | F | H | K | Z | M | G | L |
15(D) | F | M/K | K/M | Z | H | G | L |
If you’re interested, the reason (A) must be true is because F and K must both precede Z, and specifically, F must precede K as well (both from Rule 4). If we want to contradict choice (A), we need L < F, so L would need to be on Monday and F on Tuesday. But that violates Rule 1 (there is no room for H before L), so L must be scheduled on Saturday (Rule 5) and therefore F must precede L.
Good news, again: we have now done all the hard work, and questions 16 through 18 become easy, relying on our databank and the knowledge we have acquired.
We already have an example #15(D) with Moore on desk duty on Tuesday. In that example, neither H nor K is scheduled on Thursday, and Z is on Thursday, not Friday or Saturday. Eliminate (A), (B), (D), and (E). The correct choice is (C), and I’m not going to waste precious time proving it.
F < H in three of our databank examples. We can eliminate (A), (B), and (C) from example #15(D) [example #12 also disproves (C)]; and we can eliminate (E) from example #14. The correct choice is (D), and again I’m not going to waste time proving it.
Finally, we reach the “substitute constraint” question:
We no longer have the rule that F < K and F < M (original constraint 3). We used this constraint, in conjunction with original constraints 2 and 4, to create our Rule 7, namely that F < M, G, K, Z. In other words, we used it to determine (and noted in our table) that F could be scheduled at the latest on Wednesday. If we want to maintain that constraint, answer choice (C) does it perfectly in conjunction with our Rules 1 and 5, by forcing Monday, Tuesday, and Wednesday to be H, L, and F respectively. The correct choice is (C).
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Brian Lizotte offers test preparation and admissions counseling services for students of all ages. He also tutors in mathematics, writing and literature. He has scored perfectly on the PSAT, SAT and ACT, as well as the ISEE and SSAT, and in the 99th percentile on the GRE, GMAT and LSAT. Brian graduated summa cum laude from Yale (3.95 GPA), earning simultaneous B.A./M.S. degrees in Psychology, and at graduation Yale’s faculty awarded him its top academic prize. He also holds an Ed.M. from Boston University (4.00 GPA) and a J.D. from Yale Law School. He has taught mathematics at the Groton School, served as Associate Provost for the Humanities at Yale, and was Assistant Head of the Brearley School in Manhattan. Brian’s method teaches students how to think, not memorize, and his broad knowledge of learning styles allows him to adapt to your own learning needs. Brian offers both one-on-one tutoring at competitive hourly rates and small-group classes that follow a syllabus.