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Variable-Based / Condition-Dependent Puzzle

Introduction

Variable-Based / Condition-Dependent puzzles require you to assign values (numbers, labels, or choices) to items where some assignments depend on variable conditions or thresholds in the clues - for example, "if X > Y then A", "pick odd positions", or "only when condition Z is met assign attribute Q".

These problems appear in advanced reasoning sections because they mix conditional logic with standard elimination: you often must branch your working into cases and then prune invalid cases using additional constraints.

Pattern: Variable-Based / Condition-Dependent Puzzle

Pattern

Recognize conditional clauses, convert them to explicit case splits, and eliminate using cross-constraints.

  • Identify variables or thresholds: e.g., numeric values, parity (odd/even), greater/less-than conditions.
  • Split into cases: Evaluate each branch separately (Case A: condition true; Case B: condition false).
  • Apply deterministic clues: Fill fixed items first inside each case to reduce branching.
  • Prune invalid cases: Use contradictions or impossibilities to drop branches and keep the valid one(s).

Step-by-Step Example

Question

Five candidates - A, B, C, D, E - take tests in five slots numbered 1 to 5 (slot 1 earliest). Each gets a distinct score from {50, 60, 70, 80, 90}. Clues:

  1. If the candidate in slot 1 scored < 70 then A scored 90; otherwise A scored 50.
  2. The candidate who scored 80 sits immediately after the candidate who scored 60.
  3. B scored higher than C but lower than D.
  4. E scored 70 and is in an odd-numbered slot.
Who scored 90?

Options:
A) A    B) B    C) C    D) D

Solution

  1. Step 1: Record deterministic facts

    E = 70 and E is in an odd slot → E ∈ {slot 1, 3, 5}.

  2. Step 2: Identify the conditional clause (branching)

    Clause (1) depends on slot-1 score being < 70 or ≥ 70. There are two branches:
    • Case A: Slot 1 score < 70 → slot 1 must be 50 → then A = 90.
    • Case B: Slot 1 score ≥ 70 → slot 1 ∈ {70,80,90} → then A = 50.

  3. Step 3: Apply fixed-chain clue

    Clue (2): 60 followed immediately by 80. So (…60,80…) occurs across adjacent slots (1→2, 2→3, 3→4 or 4→5). Keep in mind while placing E=70.

  4. Step 4: Use ordering relation

    Clue (3): B > C and B < D, so numeric ordering of these three is C < B < D.

  5. Step 5: Evaluate Case A (slot1 = 50 → A = 90)

    1. E = 70 occupies an odd slot; slot1 = 50 so E ≠ slot1. Try E in slot3 or slot5.
    2. Where can (60,80) pair go? They must appear adjacent in remaining slots. Available scores left (after assigning 50 and 70 and A=90) are {60,80}. They must fit adjacent slots among the three remaining positions. That is possible if slots (2,3) or (3,4) or (4,5) remain contiguous and include both scores without conflicting E's slot.
    3. Try placements quickly: assign E slot3 (70) - then remaining slots 2 & 4 & 5 must accommodate {60,80} adjacent pair and remaining candidate. You can place 60 at slot4 and 80 at slot5 (adjacent 4→5) or 60 at 2 and 80 at 3 (but 3 is taken). Work through and you find a valid fill exists with A = 90. So Case A is feasible.

  6. Step 6: Evaluate Case B (slot1 ∈ {70,80,90} → A = 50)

    1. If slot1 = 70 then E must be in slot1 (since E=70 in odd slot) - but E in slot1 would trigger slot1 score = 70 which makes Clause(1) true? Remember Clause(1) says if slot1 < 70 then A=90 - not triggered. With slot1=70 we are in Case B and A=50. Continue: remaining scores {50,60,80,90} must be placed with 60→80 adjacent and B < D ordering. Try placements - you will find contradictions where B < D can't be satisfied with remaining adjacency and slot parity. If slot1 = 80 then 60→80 adjacency cannot place 80 at slot1 because 80 must be immediately after 60 (i.e., 60 at slot0 impossible). So 80 can't be slot1. If slot1 = 90 then E = 70 must be in an odd slot ≠1 and adjacency constraints again force contradictions.
    2. After checking, Case B yields no fully consistent assignment - thus Case B is invalid.

  7. Step 7: Conclude from case elimination

    Only Case A (slot1 = 50 → A = 90) is consistent. Therefore A scored 90.

  8. Final Answer:

    A → Option A

  9. Quick Check:

    Case split handled; slot1 low → A=90, E=70 in an odd slot, 60/80 adjacent placement feasible, and ordering C < B < D satisfied in the working arrangement → consistent ✅

Quick Variations

1. Conditional assignment based on parity (odd/even slot).

2. Threshold rules: e.g., "If sum of two slots > X then assign special label".

3. Multi-condition branching (nested if-then chains) - require deeper case trees.

4. Hybrid puzzles: condition-dependent numeric values tied to seating/position constraints.

Trick to Always Use

  • Step 1: Spot conditional clauses and write explicit case splits before filling anything.
  • Step 2: Fill deterministic facts inside each case first - opposites, fixed labels, parity, etc.
  • Step 3: Rapidly test each case for contradictions (contradictions let you prune a branch early).

Summary

Summary

  • Turn every conditional clue into explicit case(s) - name them (Case A / Case B).
  • Within each case, apply fixed facts first, then adjacency and ordering rules.
  • Eliminate impossible cases quickly to avoid wasted search effort.
  • When only one case survives, that yields the unique, exam-ready solution.

Example to remember:
Always branch on the condition first - then treat each branch as a regular puzzle and prune aggressively.

Practice

(1/5)
1. Five students - A, B, C, D, E - occupy exam slots 1 to 5 (1 = earliest). Scores available: 50, 60, 70, 80, 90. Clues: 1) If the score in slot 1 is less than 60 then A scored 90; otherwise A scored 50. 2) The student who scored 60 sits immediately before the student who scored 80. 3) E scored 70 and sits in an odd-numbered slot. 4) B scored higher than C. Who scored 90?
easy
A. A
B. B
C. C
D. D

Solution

  1. Step 1: Identify fixed facts

    Record fixed facts: E = 70 and E is in slot 1, 3 or 5 (odd slot).

  2. Step 2: Branch on the conditional

    Case I - slot 1 score < 60 ⇒ slot1 = 50 ⇒ then A = 90. Case II - slot 1 score ≥ 60 ⇒ slot1 ∈ {60,70,80,90} ⇒ then A = 50.

  3. Step 3: Eliminate infeasible case

    Case II becomes infeasible because adjacency 60→80 cannot be maintained with E in odd slots and A = 50.

  4. Step 4: Accept feasible case

    Case I works: A = 90, E = 70 (odd slot), B > C satisfied.

  5. Final Answer:

    A → Option A

  6. Quick Check:

    A = 90 satisfies all constraints ✅
Hint: Convert 'if-then' statements into two cases and test adjacency first - it usually invalidates one branch immediately.
Common Mistakes: Not testing the alternate branch - adjacency and order constraints often break it.
2. Seven tasks (T1-T7) are scheduled across days Monday-Sunday (one per day). Durations available (hours): 2, 3, 4, 5, 6, 7, 8. Clues: 1) If the duration on Monday is ≤3 hours then task X took 8 hours; otherwise X took 2 hours. 2) The 3-hour task is scheduled immediately before the 5-hour task. 3) Task X is scheduled later in the week than the 4-hour task. Who took 8 hours?
easy
A. Task X
B. The 7-hour task
C. The 6-hour task
D. Task scheduled on Wednesday

Solution

  1. Step 1: Translate the conditional

    Case A - Monday ≤ 3 ⇒ X = 8. Case B - Monday ≥ 4 ⇒ X = 2.

  2. Step 2: Apply adjacency

    The 3-hour task immediately before the 5-hour task limits placement (e.g., Mon→Tue, Tue→Wed).

  3. Step 3: Test branches

    Case B fails after placement checks because it violates 'X after 4-hour' condition. Case A works smoothly.

  4. Step 4: Confirm consistent branch

    Case A → Monday ≤ 3, X = 8 satisfies all conditions.

  5. Final Answer:

    Task X → Option A

  6. Quick Check:

    All adjacency and timing constraints satisfied ✅
Hint: Start with the 'else' branch in scheduling problems - adjacency usually breaks it.
Common Mistakes: Forgetting to check task order when applying time-based conditions.
3. Four projects - P, Q, R, S - are assigned priority numbers 1-4 (1 = highest). Clues: 1) If project P’s priority number is odd then Q’s priority = 4; otherwise Q’s priority = 1. 2) R’s priority is numerically between Q and S. 3) P does not have priority 2. Which project has priority 4?
easy
A. P
B. Q
C. R
D. S

Solution

  1. Step 1: Branch by parity

    P odd ⇒ Q = 4; P even ⇒ Q = 1.

  2. Step 2: Eliminate invalid case

    P ≠ 2, so even case means P = 4 → Q = 1. That breaks the 'R between Q and S' condition because no valid number lies between 1 and 2 or 3 simultaneously.

  3. Step 3: Accept valid case

    P odd → Q = 4 fits all remaining rules.

  4. Final Answer:

    Q → Option B

  5. Quick Check:

    R between Q(4) and S valid only under odd P case ✅
Hint: Test numeric constraints after each branch - 'between' clues eliminate false cases fast.
Common Mistakes: Ignoring the numeric gap between ranks; it invalidates many branches quickly.
4. Six contestants - A-F - receive badge colours from {Red, Blue, Green, Yellow, Black, White} and sit in positions 1-6. Rules: 1) If position 2 is Red or Blue then A gets Black; otherwise A gets White. 2) The Green badge holder sits immediately after the Yellow badge holder. 3) B's colour is alphabetically earlier than C's colour. 4) D sits in a higher-numbered position than E. Who wears Black?
medium
A. A
B. B
C. C
D. D

Solution

  1. Step 1: Analyze conditional case

    Case X - pos2 ∈ {Red,Blue} ⇒ A = Black; Case Y - pos2 ∉ {Red,Blue} ⇒ A = White.

  2. Step 2: Apply adjacency of colours

    Green immediately after Yellow forms a consecutive pair Y→G; restricts placement of position 2.

  3. Step 3: Test Case X

    Case X leads to contradiction in Y→G placement and alphabetical rule B<C.

  4. Step 4: Accept Case Y

    Case Y allows a valid placement where C gets Black and all constraints hold.

  5. Final Answer:

    C → Option C

  6. Quick Check:

    All conditions hold only under Case Y ✅
Hint: For colour-based seating, always test adjacency before alphabetical constraints.
Common Mistakes: Overlooking that Y→G means consecutive seats, not random positions.
5. Five managers - P, Q, R, S, and T - receive different bonuses of ₹1,000, ₹2,000, ₹3,000, ₹4,000, and ₹5,000. Conditions: 1) If P’s bonus is less than ₹3,000 then Q gets the highest bonus; otherwise Q gets the lowest bonus. 2) R’s bonus is double S’s bonus. 3) T’s bonus is less than R’s bonus. 4) P < ₹3,000. Who receives ₹5,000?
medium
A. P
B. Q
C. R
D. S

Solution

  1. Step 1: List available values

    Bonuses = ₹1k, ₹2k, ₹3k, ₹4k, ₹5k.

  2. Step 2: Identify fixed relation

    R = 2×S; T < R. Valid (S,R) pairs = (₹1k,₹2k) or (₹2k,₹4k).

  3. Step 3: Apply given P < ₹3k

    This activates IF branch → Q = ₹5,000.

  4. Step 4: Assign remaining logically

    P = ₹1k, S = ₹2k, R = ₹4k, T = ₹3k. All clues satisfied.

  5. Final Answer:

    Q - ₹5,000 → Option B

  6. Quick Check:

    P<₹3k⇒Q=₹5k ✅ R=2×S ✅ T<R ✅
Hint: Use the numeric relation (×2) first; it quickly prunes inconsistent options.
Common Mistakes: Mixing up who ‘gets highest’ when IF condition changes based on inequality.

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