Games Test – Solutions
Click the links immediately below to view the other LSAT diagnostic tests.
Test Questions
Questions 1 and 2
Seven disks–G, H, L, O, P, S, U–are being inserted in a CD player. The order in which the disks are played is subject to the following restrictions:
L must be played before both O and U.
Exactly two disks must be played between G and P one of which must be L.
H cannot be played first.
1. If G is played third, which one of the following must be played second?
(A) G
(B) H
(C) L
(D) O
(E) N
Correct Answer: (B)
Solution: Since G is played third, the condition “Exactly two disks must be played between G and P” forces P into space 6:
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| G | P |
The condition L between G & P forces L into space 4 or 5. But L cannot be in space 5 since that would leave no room for the condition “L must be played before both O and U.” Thus, L must be in space 4:
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| G | L | O/U | P |
Finally, since “H cannot be played first,” H must be played second. The answer is (B).
2. If L and O are played consecutively, which one of the following cannot be true?
(A) S is played second
(B) G is played second
(C) L is played third
(D) O is played forth
(E) H is played sixth
Correct Answer: (A)
Solution: Suppose S is played second.
1 2 3 4 5 6 7
S
Since “L and O are played consecutively” and L is between G & P, the disks G, L, O, and P must be played consecutively. Now, G, L, O, and P cannot be placed in spaces 4, 5, 6, and 7 since that would violate the condition “L must be played before both O and U.” So G, L, O, and P must be placed in spaces 3, 4, 5, and 6:
1 2 3 4 5 6 7
S G L O P U
(Note, P and G can be interchanged in the diagram.) However, this diagram leaves no room for H (H cannot be played first). Hence, S cannot be played second and the answer is (A).
If you like this material, you’ll love the course!
This interactive, comprehensive self-study course presents the equivalent of over 600 pages of printed material, including hundreds of LSAT examples and problems, and feedback from LSAT experts to your questions. In addition, the powerful learning engine StudyDesk increases your learning efficiency by monitoring your progress and directing you to areas where you need further study. All for only $50! Click here to view the course.
Course Features:
- Ask Questions! Our instructors monitor StudyDesk to answer your questions. StudyDesk also records the step where you make a mistake or ask a question. This is just one of many powerful educational tools in StudyDesk.
- Highly Interactive: You can take notes, view solutions, and view reports, etc.
- Versatile: You can access the course from any computer at any time.
- Statistics: Your performance on the exercises and tests is saved and you may review your performance and check solutions at any time. You can also check your ranking based on all students taking the course. How cool is that!
- Guarantee: If, at the end of the course, you do not feel sufficiently prepared for the test, you may repeat the course for free — with full access to our instructors.
Questions 3 and 4
Three committees are formed from eight people–F, G, H, I, J, K, L, M. Two of the committees have three members, and one of the committees has only two members.
G serves with M.
L serves with only one other person.
F does not serve with M.
3. If K, J, and I serve on different committees, which one of the following must be true?
(A) K serves with G.
(B) I serves on a committee of two.
(C) J serves on a committee of two.
(D) H serves with F.
(E) J serves with F.
Correct Answer: (D)
Solution: We shall construct counter-examples for four of the answer-choices; the one for which we cannot construct a counter-example will be the answer. Start with (A). Suppose K serves on Committee I and G serves on Committee II. Then from the condition “G serves with M,” we know that M must also serve on Committee II. And the remaining people can be placed without violating any initial condition as follows:
| Com I | Com II | Com III |
| KHF | JGM | LI |
This diagram is a counter-example not only to (A) but to (C) and (E) as well. This eliminates (A), (C), and (E). Next, test choice (B). Suppose that I serves on Committee I, with G and M. Then the remaining people can be grouped as follows:
| Com I | Com II | Com III |
| IGM | FHJ | LK |
This diagram does not violate any initial condition, so it is a counter-example to (B). Hence, by process of elimination, the answer is (D).
4. Which one of the following conditions is inconsistent with the given conditions? (A) K serves on a committee of three.
(B) M serves with H.
(C) M, H, and I serve together.
(D) F does not serve with G.
(E) H serves with L.
Correct Answer: (C)
Solution: The first counter-example in Question 3 shows that K can serve on a three-person committee. This eliminates (A). Next, turning to choice (B), suppose M serves with H on Committee I. This forces G to also serve on Committee I. Now place F on Committee III and the remaining people as follows:
| Com I | Com II | Com III |
| MHG | JKI | LF |
This diagram does not violate any initial condition, so “M serves with H” is consistent with the initial conditions. This eliminates (B). Next, turning to choice (C), suppose M, H, and I serve together on Committee I. But since M must serve with G, there would then be four people on Committee I. The same result occurs when M, H, and G are on Committee II. Hence (C) is inconsistent with the initial conditions, and the answer is (C).
Questions 5 and 6
In a secret code that uses only the letters A, B, C, and D, a word is formed by applying the following rules:
Rule 1: A B C D is the base word.
Rule 2: If C immediately follows B, then C can be moved to the front of the word.
Rule 3: One letter of the same type can be added immediately after an A, a B, or a C.
5. Which of the following letters can start a word?
I. A
II. B
III. C
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
Correct Answer: (E)
Solution: A can start a word since it starts the base word. This eliminates choices (B) and (C) since they don’t contain I. C can start a word since C A B D is formed from the base word by using Rule 2. This eliminates (A) and (D) since they don’t contain III. Hence the answer is (E), and there is no need to check Statement II.
6. The word C A A B C C D can be formed from the base word by applying the rules in which one of the following orders?
(A) 22333
(B) 23232
(C) 32233
(D) 3223
(E) 3233
Correct Answer: (E)
Solution: This question is hard, because we don’t know to which letter(s) in the base word to apply the rules. Furthermore, there is more than one way to generate the word–but, of course, only one of those ways is listed as an answer-choice. We can, however, narrow the number of answer-choices by analyzing the word C A A B C C D. Notice that C occurs three times and A two times. So Rule 3 must have been applied three times, twice to C and once to A. This eliminates choices (B) and (D) since neither has three 3’s. Next, turning to choice (A), we apply Rule 2 to the base word giving C A B D. Now Rule 2 cannot be applied to this word again, since C is not immediately after B. This eliminates (A). Next, choice (C) seems at first glance to be plausible. It begins the same way as does choice (E). But notice in choice (C) that Rule 2 is applied twice in a row. A little fiddling shows that if this is done, then two C’s in a row would come at the beginning of the word. So eliminate (C). Hence, by process of elimination, the answer is (E). As a matter of test taking strategy this would be sufficient analysis of the question. However, it is instructive to verify that the answer is (E). To that end, apply Rule 3 to C in the base word A B C D which gives A B C C D. Next, apply Rule 2 which gives C A B C D. Finally, apply Rule 3 to A and then to C which gives C A A B C C D.
