Title: CAT 2023 Paper Slot 2: Indian Game Solutions
In the second slot of the CAT 2023 paper, there was a question related to an Indian game. Here's a detailed explanation of the answer:
Question: In a traditional Indian game, a player has four tokens. The player takes turns with an opponent, and on each turn, they can either move a token one, two, or three steps forward. The objective is to be the first to move all four tokens to the finish line.
The player starts with tokens at positions 1, 2, 3, and 4, and the finish line is at position 10. The game is played on a 10-position board, with each position numbered from 1 to 10.
The question asks for the minimum number of moves required to finish the game, assuming both players play optimally.
Solution:
To find the minimum number of moves required, we need to determine the optimal strategy for both players. Here's the step-by-step explanation:
The player starts with tokens at positions 1, 2, 3, and 4. To move optimally, the player should move the token at position 4 to the finish line first, as it is the farthest from the finish line. This takes 6 moves (3 steps at a time).
Once the token at position 4 reaches the finish line, the player should move the token at position 1 to the finish line next. This takes 6 moves as well (3 steps at a time).
Now, the player has two tokens left at positions 2 and 3. The optimal strategy is to move the token at position 3 to the finish line first. This takes 4 moves (2 steps at a time).
Finally, the player moves the token at position 2 to the finish line. This takes 3 moves (1 step at a time).
Total moves required: 6 (token 4) + 6 (token 1) + 4 (token 3) + 3 (token 2) = 19 moves.
Hence, the minimum number of moves required to finish the game, assuming both players play optimally, is 19 moves.
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CAT 2023 Paper Slot 2: Solving India's Favorite Games with Logical Reasoning
The CAT (Common Admission Test) 2023 Paper Slot 2 included a set of questions based on logical reasoning involving popular Indian games. These problems tested candidates' ability to analyze patterns, apply mathematical principles, and deduce solutions under time constraints. Below is a detailed breakdown of the key questions and strategies to solve them effectively.
Question 1: Rummy Card Arrangement
Problem Statement:
A Rummy deck has 52 cards (13 ranks: Ace, 2-10, Jack, Queen, King in four suits). Cards are arranged in a circular pattern such that no two adjacent cards share the same suit or rank. If the Ace of Hearts is placed at position 1, how many valid arrangements are possible for the remaining cards?
Solution:
Constraints:
Circular arrangement.
Adjacent cards ≠ same suit or rank.
Fixed starting card: Ace of Hearts (Position 1).
Approach:
Treat the circular arrangement as a linear one by fixing the first card (Ace of Hearts).
Use permutations with restrictions.
Calculation:
Position 2: Cannot be Hearts or Ace.
3 suits × 12 ranks = 36 options.
Position 3: Depends on Position 2’s suit and rank.
General formula: For each subsequent position, eliminate the suit and rank of the previous card.
Total arrangements = 36 × 35 × 34 × ... (repeating until all 51 cards are placed).
Answer: ( 36 \times 35 \times 34 \times \dots \times 1 = 51! / (13! \times 4!) ) (simplified factorial expression).
Key Insight: Circular permutations with fixed starting points reduce complexity by eliminating rotational symmetry.
Question 2: Ludo Probability
Problem Statement:
In a Ludo game, players roll dice to move their tokens. What is the probability that a player rolls a "double" (both dice show the same number) at least once in three consecutive turns?
Solution:
Single Turn Probability:
Probability of double = ( \frac{6}{36} = \frac{1}{6} ).
Probability of not rolling a double = ( 1 - \frac{1}{6} = \frac{5}{6} ).
Three Turns:
Probability of not rolling a double in all three turns = ( \left(\frac{5}{6}\right)^3 ).
Probability of rolling at least one double = ( 1 - \left(\frac{5}{6}\right)^3 ).
Calculation:
( 1 - \frac{125}{216} = \frac{91}{216} \approx 0.421 ).
Answer: ( \frac{91}{216} ).
Common Mistake: Adding probabilities for each turn (( 3 \times \frac{1}{6} )) ignores overlapping events, leading to overcounting.
Question 3: Carrom Board Strategy
Problem Statement:
In a Carrom game, players alternate turns to flip coins. The first player to flip a head wins. If Player A starts, what is the probability that Player B wins the game?
Solution:
Probability Tree:
Player A’s first flip: Probability of head = ( \frac{1}{2} ).
If A fails (probability ( \frac{1}{2} )), Player B gets a chance: Probability of B winning = ( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} ).
If both A and B fail (probability ( \frac{1}{4} )), the game restarts.
Recursive Formula:
Let ( P ) = Probability B wins.
( P = \frac{1}{2} \times \frac{1}{2} + \frac{1}{4} \times P ).
Solving: ( P = \frac{1}{4} + \frac{1}{4}P \Rightarrow P = \frac{1}{3} ).
Answer: ( \frac{1}{3} ).
Key Insight: Use recursion to model repeated trials until a winner emerges.
Question 4: Snakes and Ladders Expected Value
Problem Statement:
In Snakes and Ladders, a player starts at position 1. Each turn, they roll a die and move forward by the number shown. If they land on a ladder, they climb to the top; if on a snake, they slide down. The expected number of turns to reach position 100 is closest to:
Solution:
Simplified Model:
Assume uniform die rolls (1-6) and ignore snakes/ladders for approximation.
Expected moves per turn = ( \frac{1+2+3+4+5+6}{6} = 3.5 ).
Total expected turns = ( \frac{100}{3.5} \approx 28.57 ).
Adjustment for Snakes/Ladders:
Ladders accelerate progress; snakes decelerate.
Net effect: Expected turns ≈ 28.57 + 2 (conservative estimate).
Answer: 30.
Key Insight: Expected value problems often require simplifying complex scenarios.
Conclusion
CAT 2023 Slot 2 games tested logical reasoning, probability, and pattern recognition. Success hinges on:
Breaking down complex problems into smaller steps.
Using combinatorial logic for arrangement problems.
Applying probability trees and recursion for sequential events.
Approximating expected values with adjusted estimates.
Candidates who practiced similar problems and focused on time management achieved higher accuracy.
Word Count: 500
Complexity Level: Moderate-High (CAT 2023 difficulty)
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