Here’s a structured solution to a hypothetical Quantitative (Quant) problem related to an Indian board game, inspired by common themes in competitive exams like the GRE or GMAT. This example assumes a game involving probability and combinatorial logic:
Title: 2021 Slot 3 Quant – Analyzing Probability in a Traditional Indian Game
Problem Statement:
In the game "Chaturanga" (a medieval Indian strategy game), players roll two six-sided dice to determine movement. If the sum of the dice is a prime number, the player advances; if it’s a perfect square, they stay; otherwise, they move backward. What is the probability that a player advances or moves backward in a single turn?
Solution:
Identify Possible Outcomes
Total outcomes = 6 × 6 = 36.
Prime Sum Outcomes
Prime sums between 2 and 12: 2, 3, 5, 7, 11.
Count:
2: (1,1) → 1
3: (1,2), (2,1) → 2
5: (1,4), (2,3), (3,2), (4,1) → 4
7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6
11: (5,6), (6,5) → 2
Total prime outcomes = 1 + 2 + 4 + 6 + 2 = 15
Probability of advancing = ( \frac{15}{36} = \frac{5}{12} ).
Perfect Square Sum Outcomes
Perfect squares between 2 and 12: 4, 9.
Count:
4: (1,3), (2,2), (3,1) → 3
9: (3,6), (4,5), (5,4), (6,3) → 4
Total perfect square outcomes = 3 + 4 = 7
Probability of staying = ( \frac{7}{36} ).
Non-Prime/Non-Square Outcomes (Move Backward)
Total outcomes - (Prime + Perfect Square) = 36 - (15 + 7) = 14.
Probability of moving backward = ( \frac{14}{36} = \frac{7}{18} ).
Probability of Advancing or Moving Backward
Since these are mutually exclusive events:
[
P(\text{advance or move back}) = \frac{15}{36} + \frac{14}{36} = \frac{29}{36} \approx 80.56%
]
Answer: ( \boxed{\dfrac{29}{36}} )
Key Takeaways:
Mastering combinatorial counting for dice sums.
Distinguishing prime numbers (2,3,5,7,11) from perfect squares (4,9).
Calculating mutually exclusive probabilities.
Let me know if you’d like to explore a different game or problem type!
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