Puzzle - 1 - Free MCQ Practice Test with solutions, CAT Logical Reasoning

Answer: Option D - F.

Number positions from 1 (front) to 6 (back). The clues are: A before C; B immediately after D; E not at positions 1 or 6; F ahead of D and not at position 1; and |pos(A) - pos(F)| = 2 (exactly one person between A and F).

Eliminate impossible D positions quickly: D = 1 is impossible because B must be immediately after D; D = 2 is impossible because then B = 3 and F would have to be at position 1 to be ahead of D, but F cannot be first; D = 3 fails because B = 4 and the only allowed positions for F ahead of D (not 1) lead to a conflict with the A-F separation.

Try D = 4. Then B = 5, and F must be at 2 or 3 (ahead of D but not 1). F = 3 allows A = 1, leaving E = 2 (not an end) and C = 6. This gives a valid line with F at position 3.

Try D = 5. Then B = 6, and F can be 2, 3 or 4. The only choices that satisfy A before C, E not at ends, and the A-F separation lead to valid arrangements where F is again at position 3 (for example, A, E, F, C, D, B or A, C, F, E, D, B).

No other D positions produce valid arrangements. In every valid arrangement that satisfies all clues, F occupies position 3. Therefore the third person is F, i.e., Option D.

Label days as: Mon(1), Tue(2), Wed(3), Thu(4), Fri(5).
From “L’s appointment is exactly two days after H’s”, possible pairs (H,L) are:
(Mon, Wed), (Tue, Thu), (Wed, Fri).
“J’s appointment is not on Wednesday” excludes J = 3.
Also, G is later than J but earlier than H: J < G < H.
Try each possible H:
If H = Mon (1), L = Wed (3). Then J < G < 1, impossible as no days earlier than Monday.
If H = Tue (2), L = Thu (4). Then J < G < 2 ⇒ J and G must be on Mon only (impossible for two different people).
If H = Wed (3), L = Fri (5). Now J < G < 3 and J ≠ 3. Possible (J,G) = (Mon, Tue). K cannot be Mon or Fri, so K must be Thu. This uses all five days exactly once and satisfies all constraints.
Thus the only consistent choice is H = Wednesday.

List implications:
A cannot be W or X, so A ∈ {Y, Z}.
C ∈ {X, Z}.
B’s ward letter must be alphabetically before D’s.
Consider possibilities:
If D were in Z, then “no doctor is assigned to ward Y” would make Y unused, contradicting “one doctor per ward”. So D cannot be Z.
Thus D ∈ {W, X, Y} but not Z.
Now, since C ∈ {X, Z} and D ≠ Z, C cannot be Z at the same time as D being Z, so C can be X or Z.
Testing all consistent assignments with one doctor per ward and the alphabetical rule shows there are only two valid complete assignments; in both of them B is always in ward W.
Therefore, B must be assigned to ward W.

Check each option against the rules.
Option A: P – T – R – S – Q
P left of Q: yes (1st vs 5th).
R immediately next to S: yes (3rd and 4th).
T not at an end: yes (2nd).
Q not in the middle seat (3rd): Q is 5th, so fine.
All conditions satisfied ⇒ possible.
Option B: T – P – R – S – Q
T not at an end: violated (T is 1st).
Option C: P – Q – T – R – S
Q not in the middle seat: violated (Q is 2nd? Wait, middle is 3rd. But also R and S are not adjacent? R is 4th, S 5th – they are adjacent, so that part is okay. However P must be left of Q – true. The key issue is T not at an end: T is 3rd so fine. But Q is not forbidden at 2nd; the explicit middle is 3rd. The problem is actually that R must be immediately next to S – they are (4th, 5th), so also fine. So this arrangement appears to satisfy all conditions, which would make the question non-unique.)
Re-evaluating all options systematically:
Option A: satisfies all rules.
Option B: T at an end – rule broken.
Option C:
P left of Q: true.
R next to S: true.
T not at an end: true.
Q not 3rd (middle): true.
So Option C is also valid, which would make there more than one correct answer.
Therefore this specific question would not be acceptable in a UCAT-style test because it has more than one valid seating order.

Let the order of shifts be Morning < Afternoon < Evening < Night.
From “Exactly one of Ben or Dan works the Evening shift”, consider cases:
Suppose Ben works Evening.
Then Dan cannot be Evening, and Dan cannot be Morning. So Dan is either Afternoon or Night.
Cara must be later than Ben (since Ben earlier than Cara), but there is no shift later than Evening except Night. However Night would then be taken by Cara, forcing Dan into Afternoon (permissible).
Suppose Dan works Evening.
Then Ben must be earlier than Cara. With Dan at Evening and Cara not in Afternoon, to be later than Ben, Cara must be Night.
In all valid assignments, Cara is always the one on Night. Anna never works Night by rule; Ben and Dan are tied up with Evening/other earlier shifts.

Case analysis:
Since K is exactly two seats to the right of J, J cannot be at the right end. So J must be at the left end (position 1), and then K must be at position 3.
The other conditions (I to the left of H, L immediately left of M) can be satisfied in more than one way, but none of them affect K being fixed at position 3.
Thus the third seat from the left must always be occupied by K.

Check each option against the rules.

Rule recap:

A ∉ {P, Q}.

B ∈ {P, R}.

C’s hospital letter must be alphabetically after B’s.

D ≠ S.

Option A: A → Q (invalid: A cannot be Q).

Option B: B → R (allowed), A → S (allowed), D → P (allowed). But C → Q must be alphabetically after B’s hospital R; ‘Q’ comes before ‘R’ alphabetically, so this breaks the third rule.

Option C: A → R, B → P, C → S, D → Q.

A is R (not P or Q) – allowed.

B is P (P or R) – allowed.

C is S, which is after P alphabetically – allowed.

D is Q (not S) – allowed. All rules satisfied and each hospital used exactly once.

Option D: B → Q (invalid: B must be P or R).

Only Option C satisfies all constraints, so it is the only possible assignment.

From “exactly one of P or R takes module D”:

If R takes D, then P cannot take D and must take a different module that is not A. But Q must be B or C, and all four modules A–D must be used once. It turns out that when all constraints are satisfied in full assignments, R never ends up with D.

Checking all consistent assignments systematically gives only two valid patterns:

P → D, Q → B, R → A, S → C

P → D, Q → C, R → A, S → B

In both valid cases, R always takes module A. Therefore, R must choose module A.

Let times be numbered 1–5 for 9:00–13:00 respectively.

From “there is exactly one interview time between W and Y”, the time numbers of W and Y differ by 2 (e.g. 1 & 3, 2 & 4, 3 & 5).

Y cannot be 1 or 5, so Y ∈ {2, 3, 4}.

Then W must be two steps away from Y.

Z must be earlier than both X and T, and X must be later than W. T is not at 13:00.

Checking all arrangements that satisfy all five conditions, every valid schedule places X at time 5 (13:00). No other time for X is compatible with all constraints simultaneously.

Therefore, X is interviewed at 13:00.

Since A insists on travelling with C, A and C must be in the same car.

B will not travel with C, so B cannot be in C’s car.

C will not travel with D, so D also cannot be in C’s car.

But A is fixed with C, so the car containing C has exactly the pair A and C. The remaining car must then contain B and D.

Thus A and C must travel together in every valid arrangement.