This question is a logic puzzle involving scheduling seven people (P, Q, R, X, T, A, K) for the gym on different days of the week (Sunday to Saturday). We need to determine the final schedule based on the given clues and then identify who goes to the gym the day before P.

Analyzing the Gym Schedule Constraints

Let's break down the information provided:

• People: P, Q, R, X, T, A, K
• Days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday
• Clue 1: R goes to the gym on the weekend (Saturday or Sunday).
• Clue 2: K goes after X but before Q (Sequence: X ... K ... Q).
• Clue 3: The number of people who go before A is the same as the number of people who go after Q. Let this number be '$n$'. This means A is the $(n+1)^{th}$ person, and Q is the $(7-n)^{th}$ person. Mathematically, if $p_A$ is the position of A (1=Sun, 7=Sat) and $p_Q$ is the position of Q, then $p_A - 1 = n$ and $7 - p_Q = n$. Therefore, $p_A - 1 = 7 - p_Q$, which simplifies to $p_A + p_Q = 8$.
• Clue 4: K and X do not go to the gym on Wednesday.
• Clue 5: Q goes to the gym a few days after A (A is scheduled before Q).
• Clue 6: Only one person goes to the gym between R and X. This means their schedules are separated by exactly one day (R _ X or X _ R).
• Clue 7: P likes to go to the gym on weekdays (Monday to Friday).

Deducing the Schedule Step-by-Step

Step 1: Place R and X based on Clues 1 and 6.

• R is on Saturday or Sunday.
• There's one person between R and X.
  • If R = Sunday, X must be Tuesday (Sun, Mon, Tue).
  • If R = Saturday, X must be Thursday (Thu, Fri, Sat).

Step 2: Integrate Clue 2 (X...K...Q) and Clue 4 (K, X != Wed).

• Consider the case R = Saturday, X = Thursday. The sequence X...K...Q must fit. K must go after Thursday, and Q after K. The available days after Thursday are Friday and Saturday. Since R is on Saturday, K must be Friday. But then there's no day left for Q to go *after* K. This case (R=Sat, X=Thu) is impossible.
• Therefore, R must be Sunday, and X must be Tuesday. (R=Sun, X=Tue).
• Now, fit X...K...Q. X is Tuesday. K must be after Tuesday, Q after K. K cannot be on Wednesday (Clue 4). Available days after Tuesday are Wed, Thu, Fri, Sat. K cannot be Wed.
  • Possibility A: K = Thursday. Then Q can be Friday or Saturday.
  • Possibility B: K = Friday. Then Q must be Saturday.

Step 3: Use Clue 3 ($p_A + p_Q = 8$) and Clue 5 (A before Q).

• Let's test Possibility B: R=Sun, X=Tue, K=Fri, Q=Sat.
  • Positions: R(1), X(3), K(6), Q(7).
  • Check $p_A + p_Q = 8$. Here $p_Q = 7$. So $p_A + 7 = 8$, which means $p_A = 1$. But R is on Sunday (position 1). This is a contradiction. So, Possibility B (K=Fri, Q=Sat) is invalid.
• Let's test Possibility A: R=Sun, X=Tue. K is either Thursday or Friday. Q is after K.
  • Subcase A1: K=Thu, Q=Fri.
    • Positions: R(1), X(3), K(5), Q(6).
    • Check $p_A + p_Q = 8$. Here $p_Q = 6$. So $p_A + 6 = 8$, which means $p_A = 2$. But X is on Tuesday (position 2). This is a contradiction. Hmm, let me recheck the positions based on the R=Sun, X=Tue deduction.
  • Let's re-list days and tentative positions: Sun(1), Mon(2), Tue(3), Wed(4), Thu(5), Fri(6), Sat(7).
  • We established R=Sun (1), X=Tue (3).
  • Constraint X...K...Q. K cannot be Wed (4).
  • Try K=Thu (5), Q=Fri (6). Sequence: R(1), X(3), K(5), Q(6).
  • Check $p_A + p_Q = 8$. With $p_Q = 6$, we need $p_A = 2$. Day 2 is Monday. So A=Mon(2).
  • Check A before Q: A(Mon) is before Q(Fri). Yes.
  • Check people before A = people after Q: Before A(Mon) is R(Sun) - 1 person. After Q(Fri) is ?? - 1 person. This fits.
  • Current assignments: R(Sun), A(Mon), X(Tue), K(Thu), Q(Fri).
  • Remaining days: Wed(4), Sat(7). Remaining people: P, T.
  • Check Clue 4: K, X != Wed. K(Thu), X(Tue). Satisfied.
  • Check Clue 7: P likes weekdays. The remaining weekdays are Wed(4). So P must go on Wednesday.
  • This leaves T for Saturday(7).
  • Final proposed schedule: Sun(R), Mon(A), Tue(X), Wed(P), Thu(K), Fri(Q), Sat(T).
• Let's re-verify all constraints with this schedule:
  • R weekend: Yes (Sun).
  • X...K...Q: X(Tue), K(Thu), Q(Fri). Yes.
  • Before A = After Q: Before A (Mon) is R (1 person). After Q (Fri) is T (1 person). $n=1$. Yes.
  • K, X not Wed: X(Tue), K(Thu). Yes.
  • Q after A: Q(Fri) after A(Mon). Yes.
  • One between R, X: R(Sun), X(Tue). Mon is between. Yes.
  • P weekdays: P(Wed). Yes.
  All constraints are met.

Final Gym Schedule

The deduced schedule is as follows:

Answering the Question

The question asks: Who goes to the gym a day before P?

• From the schedule, P goes to the gym on Wednesday.
• The day immediately before Wednesday is Tuesday.
• Looking at the schedule, X goes to the gym on Tuesday.

Therefore, X goes to the gym a day before P.