Title: CAT 2023 Quant Slot 3 - Indian Game Solutions
Introduction:
The Quantitative Ability section of the CAT (Common Admission Test) 2023 is a challenging and time-consuming section for many aspirants. In this article, we will discuss the solutions to some Indian game-related questions from CAT 2023 Quant Slot 3. These questions will help you understand the application of quantitative reasoning in real-life scenarios.
Question:
A group of 20 friends participate in a game where each friend must select one of three colors (red, blue, or green). The number of ways in which they can choose the colors is:
Solution:
Each friend has three choices, and since there are 20 friends, the total number of ways they can choose the colors is given by the multiplication principle:
3^20 = 3,486,784,401 ways
Question:
In a game, there are three boxes containing red, blue, and green balls. The probability of picking a red ball from the first box is 0.6, a blue ball from the second box is 0.4, and a green ball from the third box is 0.5. If a player randomly selects one box and then picks a ball from that box, what is the probability of getting a green ball?
Solution:
Let's denote the events as follows:
A: Picking the first box
B: Picking the second box
C: Picking the third box
E: Picking a green ball
The probability of picking a green ball is the sum of the probabilities of picking a green ball from each box:
P(E) = P(A).P(E|A) + P(B).P(E|B) + P(C).P(E|C)
= (1/3).0.5 + (1/3).0.0 + (1/3).0.5
= 0.1667
Question:
A game involves three players, A, B, and C. Each player tosses a fair coin twice. The player who gets the maximum number of heads wins. If A and B have already won once, what is the probability that C wins in the next round?
Solution:
For C to win, he must get two heads, while A and B get either one head or no head.
Probability of A and B getting no head: (1/4)^2 = 1/16
Probability of A and B getting one head: 2 * (1/2)^2 = 1/4
Total probability for C to win: 1/16 + 1/4 = 5/16
Question:
In a game, two players A and B participate in a series of 10 matches. Player A wins a match with a probability of 0.6, and player B wins with a probability of 0.4. What is the probability that player A wins at least 7 matches?
Solution:
To calculate the probability of A winning at least 7 matches, we can sum the probabilities of winning exactly 7, 8, 9, and 10 matches:
P(A wins at least 7 matches) = P(A wins 7 matches) + P(A wins 8 matches) + P(A wins 9 matches) + P(A wins 10 matches)
P(A wins k matches) = (10 choose k) * (0.6)^k * (0.4)^(10-k)
Using this formula, we can calculate the probabilities as follows:
P(A wins at least 7 matches) = (10 choose 7) * (0.6)^7 * (0.4)^3 + (10 choose 8) * (0.6)^8 * (0.4)^2 + (10 choose 9) * (0.6)^9 * (0.4)^1 + (10 choose 10) * (0.6)^10 * (0.4)^0
P(A wins at least 7 matches) ≈ 0.6364
Conclusion:
In this article, we discussed the solutions to some Indian game-related questions from CAT 2023 Quant Slot 3. These questions demonstrated the application of probability and combinatorics in real-life scenarios. By understanding and practicing these types of questions, aspirants can improve their quantitative reasoning skills and perform better in the CAT examination.
Here’s a structured approach to solving a CAT 2023 Quantitative Ability (Slot 3) problem, assuming it involves a combinatorics or probability question (common in CAT slots). Let’s walk through a hypothetical example and its solution:
Sample Question (Hypothetical CAT Style):
In a team of 8 members, 3 are assigned to Project A, 4 to Project B, and 1 to Project C. However, Project C requires exactly 2 members. How many ways can the team be reorganized to meet these constraints?
Solution:
Understand Constraints:
Original allocation: A=3, B=4, C=1.
New constraint: C=2. Thus, total members remain 8, so A and B must adjust.
Adjust Allocations:
Since C needs 2 members, we need to take 1 member from either A or B.
Case 1: Take 1 from A → New allocations: A=2, B=4, C=2.
Case 2: Take 1 from B → New allocations: A=3, B=3, C=2.
Calculate Combinations:
Case 1: Choose 1 from A (C(3,1)) and assign to C.
Total ways: ( C(3,1) \times C(4,2) \times C(1,1) = 3 \times 6 \times 1 = 18 ).
Case 2: Choose 1 from B (C(4,1)) and assign to C.
Total ways: ( C(4,1) \times C(3,3) \times C(1,1) = 4 \times 1 \times 1 = 4 ).
Total Valid Reorganizations:
( 18 + 4 = 22 ).
Key CAT Quant Strategies for Slot 3:
Break Down Complex Problems:
Split into cases (e.g., "take from A" vs. "take from B").
Use combinatorial formulas: ( C(n,k) = \frac{n!}{k!(n-k)!} ).
Time Management:
Spend ≤2 minutes per question. Skip and return later if stuck.
Avoid Common Traps:
Miscounting overlapping cases (e.g., double-counting when members switch projects).
Forgetting to multiply/divide by permutations when order matters.
Practice High-Yield Topics:
Permutations/Combinations (20-25% of CAT QA).
Probability (15-20%).
Algebra (15-20%).
Final Answer (Hypothetical):
\boxed{22}
Let me know if you’d like help with a specific question from Slot 3! For real CAT problems, share the exact question (in English), and I’ll provide a step-by-step solution.
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