Title: CAT 2023 Slot 2 Paper - Solutions to Indian Game-related Problems
Introduction
The CAT 2023 exam included questions referencing traditional Indian games. Below are solutions to hypothetical examples involving strategic reasoning and mathematical applications related to these games. Note: Specific question details may vary, so this is a generalized guide.
Example Question 1: Kabaddi Team Formation
Problem Statement
In a Kabaddi tournament, 4 teams (A, B, C, D) each have 6 players. For the first round, each team selects 1 captain and 1 player (other than the captain) to form a "Kabbadi unit." How many unique Kabbadi units can be formed if captains cannot team up with players from their own队伍?
Solution
Select Captains:
4 teams → 4 captains (1 per team).
Select Players:
Captains cannot choose players from their own team.
Each captain has 6 players in total, but excludes their own 5 players.
Thus, each captain has 6 - 1 (captain) - 5 (own team) = 0 players? Wait, this logic is flawed.
Revised Approach:
Each captain must pick 1 player from the remaining 3 teams.
Total players per captain = 6 (total) - 1 (captain) = 5 players.
Players available = 5 (own team’s non-captains) + 3 teams × 6 players = 5 + 18 = 23? No. Wait, clarification:
Each captain cannot pick from their own team. Total players excluding their team:
3 other teams × 6 players = 18 players.
However, each captain must choose 1 player from these 18.
So, for each captain: 18 choices.
Total Units:
4 captains × 18 players = 72 unique Kabbadi units.
Example Question 2: Probability in Kho Kho Tournaments
Problem Statement
In a Kho Kho match, 2 teams of 12 players each compete. Each player has a 70% chance of scoring a point. What is the probability that exactly 8 players from one team score points, given that both teams have 8 scorers?
Solution
Conditions:
Team A and Team B each need exactly 8 scorers.
Each player’s success is independent (70% probability).
Binomial Probability:
For one team: Probability of exactly 8 scorers = C(12,8) × (0.7)^8 × (0.3)^4.
Since both teams must satisfy this condition:
Total Probability = [C(12,8) × (0.7)^8 × (0.3)^4]^2.
Calculation:
C(12,8) = 495.
Probability ≈ (495 × 0.05764801 × 0.0081)^2 ≈ (0.2315)^2 ≈ 0.0536 or 5.36%.
Example Question 3: Board Game Strategy (Rummy)
Problem Statement
In Rummy, players aim to form sets of 3 cards with the same number. If you have 10 cards: 3×1, 2×5, 4×9, and 1×7, how many unique valid sets can you form?
Solution
Forming Sets:
Set 1: Use 3×1 → 1 valid set.
Set 2: Use 2×5 + 1×9 → Not valid (must be same number).
Set 3: Use 4×9 → Can form 1 set of 3, leaving 1×9.
Set 4: Remaining cards (2×5, 1×7, 1×9) → No valid sets.
Total Unique Sets:
1 (from 1s) + 1 (from 9s) = 2 valid sets.
Conclusion
For precise solutions, please provide the exact CAT 2023 questions. This guide illustrates strategies for common Indian game-related problems in CAT, focusing on combinatorics, probability, and logical reasoning.
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