Practice Question

These practice LSAT questions are provided by Get Prepped! Get Prepped offers affordable LSAT prep, exclusively teaches the LSAT and offers tutoring, live classes, and video classes. These questions are from Get Prepped’s Ace the LSAT Logic Games. A detailed solution follows the question.

Tennis Games

On a single day a tennis instructor must schedule lessons with each of six different students—Dunn, Fick, Green, Hines, James, and Xavier. The instructor meets with exactly one student at a time, and each lesson is exactly one hour long. The times available for scheduling the lessons are 9:00 AM, 10:00 AM, 11:00 AM, 12:00 noon, 1:00 PM, and 2:00 PM.

The following conditions apply:

• Xavier’s lesson must be scheduled at some time before Fick’s lesson.
• James’ lesson cannot be scheduled for 12:00 noon or later.
• Hines’ lesson must be scheduled at some time before Xavier’s lesson.
• There must be exactly one lesson scheduled between the lessons of Dunn and Fick.

1.

Which one of the following is an acceptable assignment of students in order from 9:00 AM to 2:00 PM?

(A) Green, Xavier, James, Dunn, Hines, Fick
(B) Hines, Xavier, Fick, Green, Dunn, James
(C) James, Green, Hines, Xavier, Fick, Dunn
(D) Green, Hines, James, Dunn, Xavier, Fick
(E) Fick, James, Dunn, Hines, Xavier, Green

2.
Which one of the following is a complete and accurate list of students, any one of whom could be scheduled for the first lesson of the day?

(A) Dunn, James, Xavier
(B) Dunn, Hines, Green
(C) Green, Hines, James
(D) Fick, Hines, James
(E) Green, James, Xavier

3.

If James’ lesson is scheduled for 11:00 AM and Fick’s lesson is scheduled before Dunn’s lesson, then which one of the following must be true?

(A) Hines’ lesson is scheduled for 9:00 AM.
(B) Green’s lesson is scheduled for 10:00 AM.
(C) Xavier’s lesson is scheduled for 12:00 noon.
(D) Dunn’s lesson is scheduled for 12:00 noon.
(E) Fick’s lesson is scheduled for 1:00 PM.

4.

If Hines’ lesson is scheduled immediately before Green’s lesson, then which one of the following could be true?

(A) Xavier’s lesson is scheduled for 9:00 AM.
(B) Dunn’s lesson is scheduled for 9:00 AM.
(C) James’ lesson is scheduled for 10:00 AM.
(D) Xavier’s lesson is scheduled for 10:00 AM.
(E) Green’s lesson is scheduled for 11:00 AM.

5.

If Xavier’s lesson is scheduled before 12:00 noon, then which one of the following must be true?

(A) Hines’ lesson is scheduled for 9:00 AM.

(B) Xavier’s lesson is scheduled for 10:00 AM.
(C) Dunn’s lesson is scheduled for 12:00 noon.
(D) Green’s lesson is scheduled for 1:00 PM.
(E) Fick’s lesson is scheduled for 2:00 PM.

6.

Which one of the following CANNOT be true?

(A) Green’s lesson is scheduled for 9:00 AM.
(B) Green’s lesson is scheduled for 10:00 AM.
(C) Xavier’s lesson is scheduled for 11:00 AM.
(D) Dunn’s lesson is scheduled for 11:00 AM.
(E) Hines’ lesson is scheduled for 12:00 noon.

Here’s the solution fron Get Prepped:

This easy simple line game has six slots, each to be filled by one member. Further simplifying the game, each member is used exactly once. A line game does not get any easier than this. You should combine Rule 1 and Rule 3 to create the chain of H < X < F. Rule 2 is self-explanatory: J cannot be 12:00, 1:00, or 2:00. (Figure 1) Rule 4 is slightly more complex. Rule 4 could be F __ D or it could be D __ F. Now, make warranted conclusions. Start with the chain of H < X < F. Because F must come after both H and X, F cannot be at 9:00 or 10:00. (Figure 1) Similarly, because H must come before both F and X, H cannot be at 1:00 or 2:00. But, H must also come before D, because only one student separates D and F. So, even if the order were D X F, H would have to be before D. So H cannot be at 12:00 either. What about X? X must, at a minimum, be after H and before F, so X cannot be at 9:00 or 2:00. If you still have difficulty seeing how these blocks affect the diagram, use your fingers to measure out the length of the three-letter block, and place your fingers on the diagram to help you visualize how they take up the spaces. Although you may have made other conclusions, the conclusions that are now on the graph are more than adequate to begin answering the questions.

9	10	11	12	1	2
__	__	__	__ J	__ J	__ J
F	F
			H	H	H
X					X

Fig. 1

1. (D) – For this question, you can use the rules to quickly eliminate four of the answer choices.
   (A) This choice violates Rule 3 because it has X before H.
   (B) This choice violates Rule 2 because it has J at 2:00 PM.
   (C) This choice violates Rule 4 because it has F preceding D by only one hour.
   (D) * This choice does not violate any rules.
   (E) This choice violates Rule 1 because it has F before X.

2. (C) – Before checking the answer choices, consider how this question is written. It asks which of the answer choices contains students, any one of whom could be scheduled for the first lesson. So, all you need to do is eliminate one of the three students in an answer choice and you can eliminate the whole answer choice. Look at figure 1 for guidance. We know that neither F nor X can have the first lesson. Three of the answer choices contain either F or X, so we can eliminate those three answer choices. Only answer choices (B) and (C) remain. H and G are in both of these answer choices, so there is no need to check them. You only need to check D and J.

   (A) X cannot be first. (Figure 1) Nor, as it turns out, can D be first.
   (B) D cannot be first, because if it were, then F would have to be third and J would have to be second. This would not leave space for H and X to precede F. You should add this new information to figure 1 for future reference.
   (C) * Any of these three students are possible for the first lesson.
   (D) F cannot be first. (Figure 1)
   (E) X cannot be first. (Figure 1)

3. (A) – Simply create a new diagram with J at 11:00 AM. Now, consider figure 1 and where F and D may be. H and X must precede F, and F must precede D. So, the only place F will fit is 12:00 noon. D is at 2:00 PM. (Figure 2)
   (A) * H must be at 9:00 AM.
   (B) G must be at 1:00 PM.
   (C) X must be at 10:00 AM.
   (D) D must be at 2:00 PM.
   (E) F must be at 12:00 noon.

   9	10	11	12	1	2
   H	X	J	F J	G J	D J
   F	F
   			H	H	H
   X					X
   D

Fig. 2

4. (E) – H is now immediately before G. Add G to the chain of H < X < F to create a longer chain, H G < X < F, F _ D. What is the latest H and G could be? G could be at 10:00 AM, or G could be at 11:00 AM. G cannot be later than 11:00 AM. Quickly diagram these two permutations. (Figures 3a-b) Because J must be before 12:00 noon, the two permutations are easy to diagram.

   (A) X must be at 1:00 PM.
   (B) D must be at 12:00 noon.
   (C) J must be at 11 :00 AM.
   (D) X must be at 1:00 PM.
   (E) * G could be at 11:00 AM.

   9	10	11	12	1	2
   J	H	G	D	X	F

   Fig. 3b

5. (E) – If X is before 12:00 noon, then X can be either 10:00 AM or 11:00 AM. (Figure 1 reminds us that X cannot be 9:00 AM.) The quickest way to solve this question is to graph the permutations for X at 10:00 AM (figure 4a) and at 11:00 AM (figure 4b). In either case, G must be at 1:00 PM; all the other members are not fixed.

(A) H can be 9:00 AM or 10:00 AM.
(B) X can be 10:00 AM or 11:00 PM.
(C) D can be 12:00 noon or 2:00 PM.
(D) * G must be 1:00 PM.
(E) See (C).

9	10	11	12	1	2
H	X	J	F/D	G	F/D

Fig. 4a

9	10	11	12	1	2
J/H	J/H	X	F/D	G	F/D

Fig. 4b

6. (E) Because this question adds no new information, you should be able to answer it using the initial warranted conclusions, as well as any insights you gleaned while answering earlier questions.

   (A) Figure 1 does not show that G cannot be at 9:00 AM, but a little more confirmation would be nice.
   A quick check of the previous correct answer choices—specifically question 1, answer choice (D)—shows that G can be at 9:00 AM.
   (B) Figure 1 does not show that G cannot be at 10:00 AM, but a little more confirmation would be nice.
   A quick check of the work you did for question 4, answer choice (E) shows that G can be at 10:00 AM.
   (C) In question 5, we saw that X can be at 11:00 AM.
   (D) Figure 1 does not show that D cannot be at 11:00 AM, but none of the previous work we did shows that D can be at 11:00 AM. It is a simple matter to graph a permutation where D is at 11:00 AM. For example: H, J, D, X, F, G.
   (E)
   * As we learned while making the warranted conclusions, H cannot be at 12:00 noon. (Figure 1)

These questions are provided by Get Prepped! Get Prepped offers affordable LSAT prep, exclusively teaches the LSAT and offers tutoring, live classes, and video classes. These questions are from Get Prepped’s Ace the LSAT Logic Games.