Ratings, Rankings and Ordered Comparisons Practice Questions for CAT: 241+ Solved Questions with Step-by-Step Solutions

Solve 241+ Ratings, Rankings and Ordered Comparisons practice questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Ratings, Rankings and Ordered Comparisons

Ratings, Rankings and Ordered Comparisons

Master the art of structured evaluation and relative ordering.

Topic 01 - High Weightage

Ratings and Evaluation Matrices

Decode multi-parameter grids, distinct integer constraints, and cumulative scoring systems.

Topic 02 - Moderate Weightage

Ordered Size Comparisons

Master transitive logic and physical constraints when comparing sizes and capacities.

Topic 03 - Moderate Weightage

Sports Rankings and Performance Phases

Track dynamic rank changes, phase-based eliminations, and performance thresholds.

Ratings and Evaluation Matrices

Turning subjective scores into airtight logical grids.

What you will master

Structuring multi-evaluator and multi-parameter matrices
Applying distinct integer and sum constraints
Optimizing scores for bonuses and thresholds
Tracking cumulative ratings across time phases

Ratings, Rankings and Ordered Comparisons: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [1-5]
Five athletes — A, B, C, D, and E — compete in a triathlon consisting of three events: Swimming, Cycling, and Running.
In each event, athletes are ranked 1st to 5th with no ties. Points are awarded as follows: 1st = 5, 2nd = 4, 3rd = 3, 4th = 2, 5th = 1.
The overall winner is the athlete with the highest total points. In case of a tie, the athlete with the better (lower) rank in Running wins.
The following facts are known:
1. Athlete A finished 1st in Swimming.
2. Athlete B finished 2nd in Cycling.
3. Athlete C finished 3rd in Running.
4. The sum of points for Athlete D was exactly 10.
5. Athlete E did not finish last in any event.
6. No two athletes had the same total score.
7. Athlete A won the triathlon.
What was the minimum possible total score of the runner-up (2nd place)?

A.
9
B.
10
C.
11
D.
12

Correct Answer:

Step-by-Step Solution

Key idea: This is a multi-constraint optimization problem combining rating matrices and ordered comparisons. We must maximize the winner's score while minimizing the runner-up's score, subject to strict distinctness and rank constraints.

Step 1: Analyze Total Points.

Total points per event = $5 + 4 + 3 + 2 + 1 = 15$ .

Total points in triathlon = $15 \times 3 = 45$ .

Step 2: Determine A's Score.

A won the triathlon. To allow the runner-up to be as low as possible, we should check if A can have a "low" winning score. However, A must have the highest score.

Let the scores be $S_{A} > S_{B} > S_{C} > S_{D} > S_{E}$ (assuming order for now, but we know D=10).

We know $S_{D} = 10$ .

Since scores are distinct integers, and D=10, the other scores must be distinct from 10 and each other.

Step 3: Analyze Constraints on D.

D has 10 points. Possible combinations for 3 events (ranks 1-5):

5+3+2 (1st, 3rd, 4th)
5+4+1 (1st, 2nd, 5th) - Impossible because A was 1st in Swimming, so D cannot be 1st in Swimming if A is. Wait, D could be 1st in another event.
4+3+3 (Impossible, distinct ranks per event? No, an athlete can have same rank in different events? No, "In each event... ranked 1st to 5th". An athlete gets one rank per event. Can D get 3rd in two events? Yes. But points are 3+3+4=10.
4+4+2 (2nd, 2nd, 4th)
3+3+4 (3rd, 3rd, 2nd)

Step 4: Minimize the Runner-Up.

Let the runner-up be $S_{2 n d}$ . We want to minimize $S_{2 n d}$ .

We know $S_{A} > S_{2 n d}$ .

Also $S_{2 n d} > S_{3 r d} > S_{4 t h} > S_{5 t h}$ .

One of these is D=10.

Case 1: D is the runner-up ( $S_{2 n d} = 10$ ).

Then $S_{A} > 10$ . Min $S_{A} = 11$ .

Remaining scores for 3rd, 4th, 5th must be $< 10$ and distinct.

Sum of all scores = 45.

$S_{A} + 10 + S_{3} + S_{4} + S_{5} = 45$ .

If $S_{A} = 11$ , then $11 + 10 + S_{3} + S_{4} + S_{5} = 45 ⟹ S_{3} + S_{4} + S_{5} = 24$ .

Max possible sum for three distinct integers $< 10$ is $9 + 8 + 7 = 24$ .

So this distribution is possible: 11, 10, 9, 8, 7.

Here Runner-Up = 10.

Is this valid?

We need to check if A can get 11 and D can get 10 simultaneously with other constraints.

A=11: e.g., 5 (Swim) + 4 + 2. Or 5+3+3.

D=10: e.g., 4+3+3.

Constraint: A finished 1st in Swim (5 pts).

Constraint: E did not finish last (1 pt) in any event. So E's min score is $2 + 2 + 2 = 6$ .

In the set {11, 10, 9, 8, 7}, the lowest is 7. This is $> 6$ , so E can be 7.

Constraint: C finished 3rd in Running (3 pts).

Constraint: B finished 2nd in Cycling (4 pts).

Let's try to construct this scenario:

Scores: A=11, D=10, X=9, Y=8, Z=7.

Who is who?

A=11. D=10.

Remaining athletes: B, C, E. Scores 9, 8, 7.

E cannot have 1s. Min score 6. 7 is ok.

C has 3 in Run.

B has 4 in Cycle.

Let's assign ranks to achieve these scores.

A: 5 (Swim), 4 (Cycle), 2 (Run). Total 11.

D: 4 (Swim), 3 (Cycle), 3 (Run). Total 10.

Check conflicts:

Swim: A=1(5), D=2(4). Remaining ranks 3,4,5 for B,C,E.

Cycle: D=3(3), A=2(4). Wait, A=4 in Cycle means A was 2nd. B was 2nd in Cycling? Conflict!

"Athlete B finished 2nd in Cycling." So B gets 4 pts in Cycle.

My assignment for A had 4 pts in Cycle. Conflict.

Let's retry A's score composition.

A=11. A=5 (Swim). Needs 6 more from Cycle/Run.

Options for 6: 4+2, 3+3, 2+4, 1+5(impossible, A already 1st Swim).

If A=5, 3, 3: A is 1st Swim, 3rd Cycle, 3rd Run.

B is 2nd Cycle (4 pts).

D=10.

Let's check if Runner-Up can be 9.

If $S_{2 n d} = 9$ , then $S_{A} \geq 10$ .

Since D=10, if $S_{A} = 10$ , tie? No, distinct scores. So $S_{A} \geq 11$ .

If $S_{A} = 11$ and $S_{2 n d} = 9$ , then D(10) must be $S_{A}$ or $S_{2 n d}$ ? No, D=10.

So scores must include 10.

If $S_{2 n d} = 9$ , then 10 must be $S_{A}$ ? But $S_{A} > S_{2 n d}$ . So $S_{A}$ could be 10? No, D=10. If A=10, D=10, tie. Not allowed.

So if D=10, and scores are distinct, someone else has 10? No, only D has 10.

So 10 is one of the scores.

If Runner-Up is 9, then the scores above 9 are just the Winner.

So Winner > 9. And 10 is in the set.

If Winner > 10, say 11. Then 10 is 2nd? No, we assumed Runner-Up is 9.

Contradiction. If 10 is in the set, and Winner > 10, then 10 is at least 2nd.

So Runner-Up cannot be less than 10 if D=10 is in the set and D is not the winner.

Could D be the winner?

If D=10 is winner, then all others < 10.

Max sum of 4 distinct integers < 10 is $9 + 8 + 7 + 6 = 30$ .

Total sum = $10 + 30 = 40 < 45$ . Impossible.

So D cannot be the winner.

Therefore, Winner > 10.

Therefore, 10 is at most the 2nd highest score.

So Runner-Up $\geq 10$ .

Can Runner-Up be 10?

We found a potential conflict earlier. Let's resolve it.

Scores: 11, 10, 9, 8, 7.

A=11. D=10.

B, C, E take 9, 8, 7.

E $\geq 6$ . 7 is ok.

Assignments:

A: 5 (Swim), 4 (Cycle), 2 (Run). -> 1st Swim, 2nd Cycle, 4th Run.

Conflict: B is 2nd Cycle. A cannot be 2nd Cycle.

Try A: 5 (Swim), 3 (Cycle), 3 (Run). -> 1st Swim, 3rd Cycle, 3rd Run.

B: 4 (Cycle).

D: 10. Needs 10.

If D takes 2nd Swim (4), 1st Cycle (5)? No, A is 1st Swim.

D=1st Cycle (5), D=2nd Run (4), D=1st Swim? No, A is 1st.

D=1st Cycle (5), D=2nd Run (4), D=1st Swim (Impossible).

D=1st Cycle (5), D=4th Run (2), D=1st Swim (Imp).

D=1st Cycle (5), D=5th Run (1), D=4th Swim (4)? 5+1+4=10.

So D: 4th Swim, 1st Cycle, 5th Run.

Current Grid:

Swim: A=1, D=4. Rem: 2,3,5 for B,C,E.

Cycle: D=1, B=2, A=3. Rem: 4,5 for C,E.

Run: A=3, D=5, C=3? Conflict! A is 3rd Run, C is 3rd Run.

So A cannot be 3rd Run if C is 3rd Run.

Retry A: 5 (Swim), 2 (Cycle), 4 (Run)?

A: 1st Swim, 4th Cycle, 2nd Run.

Grid:

Swim: A=1.

Cycle: B=2, A=4. Rem: 1,3,5 for C,D,E.

Run: C=3, A=2. Rem: 1,4,5 for B,D,E.

D=10.

D's ranks from available:

Swim: 2,3,5.

Cycle: 1,3,5.

Run: 1,4,5.

Combinations for 10:

5+4+1? (1st Swim? No, A=1. 1st Cycle? Yes. 1st Run? Yes.)

If D=1st Cycle (5), D=1st Run (5)? No, distinct events, but can D be 1st in both? Yes.

But if D=1st Cycle, who is 1st Swim? A.

D=1st Cycle (5), D=2nd Run? No, A=2nd Run.

D=1st Cycle (5), D=4th Run (2)? Rem Swim needs 3.

D=3rd Swim (3). 3+5+2=10.

So D: 3rd Swim, 1st Cycle, 4th Run.

Update Grid:

Swim: A=1, D=3. Rem: 2,4,5 for B,C,E.

Cycle: D=1, B=2, A=4. Rem: 3,5 for C,E.

Run: A=2, C=3, D=4. Rem: 1,5 for B,E.

Remaining Athletes: B, C, E.

Remaining Scores needed: 9, 8, 7.

B's current points: Cycle=4(2nd). Needs 5 or 4 or 3 more?

B's ranks avail: Swim(2,4,5), Run(1,5).

If B=1st Run (5), B=2nd Swim (4)? 4+5+4=13? No, B Cycle=4.

B total = Swim + 4 + Run.

If B=2nd Swim (4) and 1st Run (5): Total 13. Too high.

If B=5th Swim (1) and 5th Run (1)? No, E cannot be last.

Let's check C.

C's current points: Run=3(3rd). Needs Swim + Cycle + 3.

C's ranks avail: Swim(2,4,5), Cycle(3,5).

Let's check E.

E's ranks avail: Swim(2,4,5), Cycle(3,5), Run(1,5).

E cannot be 5th in any.

Let's assign remaining ranks.

Cycle rem: 3,5 for C,E.

Run rem: 1,5 for B,E.

Swim rem: 2,4,5 for B,C,E.

E cannot be 5th.

In Run, avail 1,5. E cannot be 5. So E=1st Run (5).

Then B=5th Run (1).

In Cycle, avail 3,5.

In Swim, avail 2,4,5.

E has Run=5(1st).

E cannot be 5th in Swim or Cycle.

B has Run=1(5th).

B needs total 9, 8, or 7.

B = Swim + 4(Cycle) + 1(Run) = Swim + 5.

If B=9, Swim=4(2nd).

If B=8, Swim=3(Impossible, A=1,D=3 taken? No, D=3 Swim. Avail 2,4,5).

If B=7, Swim=2(3rd? No, 2nd place is 4 pts. 3rd place is 3 pts).

Wait, Rank 2 = 4 pts. Rank 3 = 3 pts.

Avail Swim ranks: 2,4,5. Points: 4,2,1.

If B=2nd Swim (4 pts), B Total = 4+4+1=9.

If B=4th Swim (2 pts), B Total = 2+4+1=7.

If B=5th Swim (1 pt), B Total = 1+4+1=6.

So B can be 9 or 7.

Now E.

E=1st Run (5 pts).

E needs Swim and Cycle.

Avail Cycle: 3,5. Points: 3,1.

Avail Swim: 2,4,5 minus B's choice.

E cannot be 5th.

Case 1: B=9 (B=2nd Swim).

Rem Swim for C,E: 4,5. Points: 2,1.

E cannot be 5th (1 pt). So E=4th Swim (2 pts).

Then C=5th Swim (1 pt).

Rem Cycle for C,E: 3,5. Points: 3,1.

E has Swim=2, Run=5. Total so far 7.

E needs Cycle.

If E=3rd Cycle (3 pts), E Total = 2+3+5=10. Conflict with D=10.

If E=5th Cycle (1 pt), E Total = 2+1+5=8.

So E=8.

Then C gets remaining Cycle: 3rd Cycle? No, E took 5th?

Avail Cycle: 3,5.

If E=5th Cycle (1 pt), C=3rd Cycle (3 pts).

C Total: Swim=1(5th) + Cycle=3(3rd) + Run=3(3rd) = 7.

Scores: A=11, D=10, B=9, E=8, C=7.

Distinct? Yes.

E not last? E ranks: 4th Swim, 5th Cycle, 1st Run. E was 5th in Cycle. Violation!

"E did not finish last in any event."

So E cannot be 5th in Cycle.

So Case 1 fails.

Case 2: B=7 (B=4th Swim).

B=4th Swim (2 pts). B Total = 2+4+1=7.

Rem Swim for C,E: 2,5. Points: 4,1.

E cannot be 5th (1 pt). So E=2nd Swim (4 pts).

Then C=5th Swim (1 pt).

Rem Cycle for C,E: 3,5. Points: 3,1.

E has Swim=4, Run=5. Total 9.

E needs Cycle.

E cannot be 5th.

If E=3rd Cycle (3 pts), E Total = 4+3+5=12.

If E=5th Cycle (1 pt), Violation.

So E=12?

But A=11. E=12 > A. A must win.

So this distribution makes E the winner. Invalid.

Thus, Runner-Up cannot be 10.

Next possible Runner-Up: 11?

If Runner-Up=11, Winner $\geq 12$ .

D=10.

Scores: 12, 11, 10, 9, 8? Sum=50 > 45.

Scores: 13, 11, 10, 6, 5? Sum=45.

Let's try Winner=12, Runner-Up=11.

Sum remaining = $45 - 12 - 11 = 22$ .

D=10. Rem sum 12 for 2 people.

Distinct, <10. 7+5=12? Or 8+4?

Set: 12, 11, 10, 7, 5.

E min 6. 5 is invalid for E.

Set: 12, 11, 10, 8, 4. E min 6. 4 invalid.

Set: 12, 11, 10, 9, 3. Invalid.

Try Winner=13, Runner-Up=11.

Sum rem = $45 - 13 - 11 = 21$ .

D=10. Rem sum 11.

Set: 13, 11, 10, 6, 5. E min 6. E=6 ok. Other=5.

Is 5 valid for non-E? Yes.

Scores: 13, 11, 10, 6, 5.

Runner-Up is 11.

Can we construct this?

A=13. D=10. B,C,E take 11, 6, 5.

E=6 (2,2,2).

This seems possible.

So minimum Runner-Up is 11.

Answer: C

Question 2 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [7-9]
Three analysts (A1, A2, A3) rated four stocks (S1, S2, S3, S4) on a scale of 1 to 4. Each analyst assigned distinct ratings to the four stocks, so each column is a permutation of {1, 2, 3, 4}.
The following facts are known:
1. S1 received a total score of 9.
2. S2 received a total score of 6.
3. A1 gave S3 a rating of 4.
4. A2 gave S4 a rating of 1.
5. No stock received the same rating from more than one analyst.
Based on these facts, which of the following scenarios is IMPOSSIBLE?

A.
A3 gave S1 a rating of 4.
B.
A1 gave S2 a rating of 1.
C.
A3 gave S3 a rating of 2.
D.
A2 gave S1 a rating of 3.

Correct Answer:

Step-by-Step Solution

Key idea: This is an Impossibility / Contradiction problem. Fixed row sums plus column permutations create a rigid structure. Identify the unique rating sets for S1 and S2, then assume each option is true and look for a collision.

Step 1: Determine the rating sets for S1 and S2.

Each row uses distinct integers from {1,2,3,4}.

S1 total = 9. The only distinct triple summing to 9 is {2,3,4}.

S2 total = 6. The only distinct triple summing to 6 is {1,2,3}.

Step 2: Note the fixed cells.

A1(S3) = 4 and A2(S4) = 1.

Step 3: Test Option A by assuming A3(S1) = 4.

S1's set is {2,3,4}. If A3 supplies the 4, then A1(S1) and A2(S1) must be {2,3}.

So A2(S1) is 2 or 3, never 4.

Step 4: Track A2's rating of 4.

A2 must place a 4 somewhere among S1, S2, S3, S4.

S1 cannot take 4 (it needs 2 or 3 from A2).
S2 cannot take 4 (its set is {1,2,3}).
S4 cannot take 4 (A2(S4) = 1).

Therefore A2(S3) must be 4.

Step 5: Find the collision.

We already know A1(S3) = 4.

So S3 would receive 4 from both A1 and A2.

This violates Fact 5: no stock receives the same rating from more than one analyst.

Hence Option A is impossible.

Step 6: Sanity check the other options.

Options B, C, and D can each be embedded in a valid grid without forcing such a collision, so they remain possible.

Answer: A

Question 3 · Data Interpretation and Logical Reasoning NAT

Common Description: Instructions [9-12]
Six trucks — T1, T2, T3, T4, T5, and T6 — have distinct weights. They are tested on three bridges — B1, B2, and B3 — each with a distinct weight limit. If a truck's weight is less than or equal to the bridge's limit, it crosses safely. If it is strictly greater, it is barred.
The following facts are known: 1. T1 and T2 cross B1 safely, but T3, T4, T5, and T6 are barred. 2. T1, T2, T3, and T4 cross B2 safely, but T5 and T6 are barred. 3. Only T1 crosses B3 safely; all other trucks are barred. 4. T6 is the heaviest truck. If a new bridge B4 has a weight limit that is strictly greater than B1's limit and strictly less than B2's limit, how many of the 6 trucks can DEFINITELY cross B4 safely?

Correct Answer:

Step-by-Step Solution

Key idea: This is a threshold method question that requires bounding. It is recognisable because it asks about a new threshold (B4) defined relative to existing thresholds, requiring you to find the extreme possible values for the trucks.

Step 1: Establish the truck order.

Clue 3: Only T1 crosses B3 $⟹ T 1 \leq B 3 < T 2$ . So T1 is the lightest.
Clue 1: T1, T2 cross B1, T3 barred $⟹ T 2 \leq B 1 < T 3$ .
Clue 2: T1..T4 cross B2, T5 barred $⟹ T 4 \leq B 2 < T 5$ .
Clue 4: T6 is heaviest $⟹ T 5 < T 6$ .

Full truck order: $T 1 < T 2 < T 3 < T 4 < T 5 < T 6$ .

Step 2: Bound the new bridge B4.

We are given: $B 1 < B 4 < B 2$ .

From Step 1, we know: $T 2 \leq B 1$ and $B 2 < T 5$ .

Therefore: $T 2 \leq B 1 < B 4 < B 2 < T 5$ .

This simplifies to: $T 2 < B 4 < T 5$ .

Step 3: Determine which trucks DEFINITELY cross B4.

A truck definitely crosses B4 if its weight is strictly less than the minimum possible value of B4.

The minimum possible value of B4 is just above B1. Since $B 1 \geq T 2$ , the absolute minimum for B4 is just above T2.

Thus, any truck $\leq T 2$ will definitely cross.

T1 and T2 are $\leq T 2 < B 4$ . They definitely cross.
T3 could be greater than B4 (if B4 is close to B1 and B1 is close to T2). So T3 is not definite.
T4 and T5 are $> B 4$ . They definitely fail.

Step 4: Count the definite trucks.

Only T1 and T2 can definitely cross. Total = 2.

Answer: 2

Question 4 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [1-5]
Five athletes — A, B, C, D, E — compete in a decathlon-style event consisting of three rounds: R1, R2, and R3.
In each round, athletes are ranked 1st to 5th with no ties. Points awarded per round: 1st = 5, 2nd = 4, 3rd = 3, 4th = 2, 5th = 1.
The Cumulative Score (CS) after any round is the sum of points earned in all rounds up to that point.
After each round, athletes are re-ranked based on their CS. In case of a tie in CS, the athlete who achieved a better (lower) rank in the *most recent* round gets the higher overall rank.
The following facts are known:
1. Athlete A won Round 1.
2. Athlete B finished 2nd in Round 2.
3. After Round 2, Athlete C was ranked 1st overall.
4. Athlete D’s Cumulative Score after Round 2 was exactly 7.
5. In Round 3, Athlete E improved their position relative to their Round 2 overall rank by exactly 2 spots (e.g., from 3rd to 1st).
6. No two athletes had the same Final Cumulative Score after Round 3.
7. Athlete A finished 3rd in Round 3.
Who finished 1st in Round 2?

A.
A
B.
B
C.
C
D.
D

Correct Answer:

Step-by-Step Solution

Key idea: This is a dynamic ranking problem requiring state tracking across rounds. We must reconstruct the Round 1 and Round 2 results to determine who finished 1st in R2, using the cumulative score constraints and tie-breaker rules.

Step 1: Analyze Round 1 (R1).

A won R1, so $R 1_{A} = 1$ . Points $P 1_{A} = 5$ .

Since ranks are 1-5, other athletes got ranks 2,3,4,5 and points 4,3,2,1.

Step 2: Analyze Round 2 (R2) and Cumulative Scores (CS2).

We know $R 2_{B} = 2$ , so $P 2_{B} = 4$ .

We know C was Rank 1 overall after R2.

We know $C S 2_{D} = 7$ .

Let's determine possible scores for D.

$C S 2_{D} = P 1_{D} + P 2_{D} = 7$ .

Possible pairs $(P 1_{D}, P 2_{D})$ from ${1, 2, 3, 4, 5}$ :

$(2, 5) ⟹ R 1_{D} = 4, R 2_{D} = 1$
$(3, 4) ⟹ R 1_{D} = 3, R 2_{D} = 2$ (Impossible, B was 2nd in R2)
$(4, 3) ⟹ R 1_{D} = 2, R 2_{D} = 3$
$(5, 2) ⟹ R 1_{D} = 1$ (Impossible, A was 1st in R1)

So D is either Case X: $(R 1 = 4, R 2 = 1)$ or Case Y: $(R 1 = 2, R 2 = 3)$ .

Step 3: Determine who was 1st in R2.

Let $W_{2}$ be the winner of R2 ( $P 2_{W 2} = 5$ ).

Consider Case X: $R 2_{D} = 1$ . Then $D$ won R2.

If D won R2, $P 2_{D} = 5$ . Then $P 1_{D} = 2 ⟹ R 1_{D} = 4$ .

Remaining R1 ranks for B,C,E: 2,3,5 (since A=1, D=4). Points: 4,3,1.

Remaining R2 ranks for A,C,E: 3,4,5 (since B=2, D=1). Points: 3,2,1.

Calculate CS2 for C. C is Rank 1 overall.

$C S 2_{C} = P 1_{C} + P 2_{C}$ .

To be Rank 1, C must have the highest CS2.

Max possible CS2 is $5 + 5 = 10$ . Min is $1 + 1 = 2$ .

Let's check if D can be Rank 1 in Case X.

$C S 2_{D} = 7$ .

If C is Rank 1, $C S 2_{C} > 7$ OR ( $C S 2_{C} = 7$ and C beat D in R2 tie-breaker).

In Case X, $R 2_{D} = 1$ . If $C S 2_{C} = 7$ , C must have $R 2_{C} < 1$ ? Impossible. So if tie, D wins. Thus C must have $C S 2_{C} > 7$ .

Possible CS2 for C: 8, 9, 10.

Let's look at remaining points in R1 for B,C,E: ${4, 3, 1}$ .

Remaining points in R2 for A,C,E: ${3, 2, 1}$ .

$P 1_{A} = 5$ . $P 2_{A} \in {3, 2, 1}$ .

$P 1_{B} \in {4, 3, 1}$ . $P 2_{B} = 4$ .

$P 1_{E} \in {4, 3, 1} ∖ {P 1_{B}, P 1_{C}}$ .

$P 2_{E} \in {3, 2, 1} ∖ {P 2_{A}, P 2_{C}}$ .

If C is Rank 1, C likely has high scores.

Max $P 1_{C} = 4$ (if B,E take 3,1 or 1,3). Max $P 2_{C} = 3$ (if A,E take 2,1 or 1,2).

Max $C S 2_{C} = 4 + 3 = 7$ .

But we established C needs $> 7$ to beat D (who has 7 and R2=1).

Contradiction. So Case X is impossible. D did not win R2.

Consider Case Y: $R 2_{D} = 3$ . Then $P 2_{D} = 3$ . And $P 1_{D} = 4 ⟹ R 1_{D} = 2$ .

So D was 2nd in R1, 3rd in R2.

Who won R2? Not B (2nd), Not D (3rd).

Winner $W_{2} \in {A, C, E}$ .

R1 Ranks used: A=1, D=2. Remaining: 3,4,5 for B,C,E. Points: 3,2,1.

R2 Ranks used: B=2, D=3. Remaining: 1,4,5 for A,C,E. Points: 5,2,1.

$C S 2_{D} = 7$ .

C is Rank 1 overall.

Let's test candidates for $W_{2}$ (Rank 1 in R2, 5 pts).

Hypothesis 1: A won R2.

$P 2_{A} = 5$ . $R 2_{A} = 1$ .

Remaining R2 points for C,E: ${2, 1}$ .

Remaining R1 points for B,C,E: ${3, 2, 1}$ .

$C S 2_{A} = 5 + 5 = 10$ .

If A has 10, can C be Rank 1?

Only if C has 10 and beats A on tie-breaker.

Max $P 1_{C} = 3$ (from rem 3,2,1). Max $P 2_{C} = 2$ (from rem 2,1).

Max $C S 2_{C} = 3 + 2 = 5$ .

C cannot beat A. So A would be Rank 1. But C is Rank 1. Contradiction.

So A did not win R2.

Hypothesis 2: E won R2.

$P 2_{E} = 5$ . $R 2_{E} = 1$ .

Remaining R2 points for A,C: ${2, 1}$ .

Remaining R1 points for B,C,E: ${3, 2, 1}$ .

$C S 2_{E} = P 1_{E} + 5$ .

$P 1_{E} \in {3, 2, 1}$ . So $C S 2_{E} \in {8, 7, 6}$ .

C is Rank 1.

Remaining R2 for A,C: one gets 2, one gets 1.

Remaining R1 for B,C,E: distinct from each other.

If C is Rank 1, C must have highest CS2.

Max $P 1_{C} = 3$ .

If $P 2_{C} = 2$ , $C S 2_{C} = 5$ .

If $P 2_{C} = 1$ , $C S 2_{C} = 4$ .

Can C be Rank 1 with max 5?

Check others.

$C S 2_{D} = 7$ . D has 7. C has max 5. C cannot be Rank 1. Contradiction.

So E did not win R2.

Hypothesis 3: C won R2.

$P 2_{C} = 5$ . $R 2_{C} = 1$ .

Remaining R2 points for A,E: ${2, 1}$ .

Remaining R1 points for B,C,E: ${3, 2, 1}$ .

$C S 2_{C} = P 1_{C} + 5$ .

$P 1_{C} \in {3, 2, 1}$ . So $C S 2_{C} \in {8, 7, 6}$ .

We need C to be Rank 1.

Check D: $C S 2_{D} = 7$ .

If $C S 2_{C} > 7$ , i.e., $C S 2_{C} = 8$ , then C is Rank 1 (unless someone else has >8).

If $C S 2_{C} = 8$ , then $P 1_{C} = 3$ .

Then remaining R1 for B,E: ${2, 1}$ .

Let's verify if this works.

$P 1_{C} = 3, P 2_{C} = 5 ⟹ C S 2_{C} = 8$ .

$C S 2_{D} = 7$ .

Who else?

A: $P 1_{A} = 5$ . $P 2_{A} \in {2, 1}$ .

If $P 2_{A} = 2, C S 2_{A} = 7$ . Tie with D.

Tie-breaker: Better rank in R2.

$P 2_{A} = 2 ⟹ R 2_{A} = 4$ (since pts 5,4,3,2,1 correspond to ranks 1,2,3,4,5. Wait. Pts: 1st=5, 2nd=4, 3rd=3, 4th=2, 5th=1).

So $P 2 = 2 ⟹ R 2 = 4$ .

D had $R 2_{D} = 3$ .

D (Rank 3) beats A (Rank 4) in tie-breaker.

So if $C S 2_{A} = 7$ , Rank order: C(8), D(7), A(7)... D is 2nd, A is 3rd.

If $P 2_{A} = 1, C S 2_{A} = 6$ .

B: $P 1_{B} \in {2, 1}$ (since C=3). $P 2_{B} = 4$ .

If $P 1_{B} = 2, C S 2_{B} = 6$ .

If $P 1_{B} = 1, C S 2_{B} = 5$ .

E: $P 1_{E} \in {2, 1} ∖ {P 1_{B}}$ . $P 2_{E} \in {2, 1} ∖ {P 2_{A}}$ .

We need C to be Rank 1. With $C S 2_{C} = 8$ , and max others 7, C is Rank 1.

This scenario is valid.

Therefore, C won Round 2.

Answer: C

Question 5 · Data Interpretation and Logical Reasoning MCQ

When reverse-engineering a tournament's final standings, if you know the Gold medalist's final score is 90.0m and the rules state the gap between Gold and Silver is exactly 1.5m, what is the most direct deduction for the Silver medalist's score?

A.
88.5m
B.
91.5m
C.
90.0m
D.
Cannot be determined

Correct Answer:

Step-by-Step Solution

Key idea: This is a "reverse engineering" question, recognisable because it asks to deduce a missing score from a known final rank and a specific gap constraint.

Step 1: Understand the ranking. Gold is Rank 1, Silver is Rank 2.

Step 2: Understand the gap. In a distance-based event, Gold has a higher score than Silver. The gap is Gold - Silver = 1.5m.

Step 3: Calculate Silver's score. Silver = Gold - 1.5m = 90.0m - 1.5m = 88.5m.

Answer: A

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