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cat 2024 slot 1 question paper

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Title: CAT 2024 Slot 1 Question Paper with Indian Game Answers


[Note: The actual CAT question paper is copyrighted material and cannot be provided here. However, I can create a fictional example of what a question from a CAT (Common Admission Test) might look like, along with a hypothetical Indian game-related answer.]



Question 1:
In a game called "Chase the Elephant," players take turns trying to catch an elephant. Each player rolls a six-sided die. If a player rolls a 6, they get to chase the elephant. If they roll a number less than 6, the elephant moves forward one space on a board. The game ends when the elephant reaches the final space. If the player rolls a 6, they can move the elephant backward two spaces. Assuming each player rolls the die twice, calculate the probability that the elephant is caught before it reaches the final space.


Answer:


To solve this question, we need to consider the probabilities of the elephant being caught on each of the player's turns and the sequence of dice rolls.



Probability of rolling a 6: Since each die has 6 faces and each face is equally likely to come up, the probability of rolling a 6 on a single roll is 1/6.



Probability of not rolling a 6: The probability of not rolling a 6 on a single roll is 1 - 1/6 = 5/6.



Probability of the elephant being caught on a turn:


If the player rolls two 6s, the elephant is caught immediately. The probability of this happening is (1/6) * (1/6) = 1/36.
If the player rolls one 6 and one number less than 6, the elephant moves forward one space. The probability of this happening is 2 * (1/6) * (5/6) = 10/36.



Probability of the elephant reaching the final space before being caught:


The elephant must not be caught on the first turn (probability = 35/36), move forward on the second turn (probability = 5/6), and then not be caught again (probability = 34/36).
The probability of this sequence is (35/36) * (5/6) * (34/36).



Overall probability of the elephant being caught before reaching the final space:


The probability is 1 minus the probability of the elephant reaching the final space before being caught.
Overall probability = 1 - [(35/36) * (5/6) * (34/36)]




After calculating the above probabilities, we would find the probability that the elephant is caught before reaching the final space. This is a simplified example, and actual CAT questions would require a more detailed analysis or calculation.



Please note that the above example is purely illustrative and the actual CAT question paper would contain a variety of questions from different topics.


CAT 2024 Slot 1 Question Paper: Sample Data Interpretation & Logical Game Question with Solution


Question 1: Team Assignment Puzzle

Direction:

A company needs to assign 6 employees (A, B, C, D, E, F) to 3 projects (X, Y, Z) with the following constraints:


Each project must have exactly 2 employees.
Employee A cannot work with B.
Employee C must work in the same project as Employee E.
Employee D cannot work with Employee F.


Task:

How many valid team combinations are possible?



Solution:



Total Possible Combinations Without Constraints:


Choose 2 out of 6 for Project X: ( \binom{6}{2} = 15 )
Choose 2 out of remaining 4 for Project Y: ( \binom{4}{2} = 6 )
Remaining 2 form Project Z: ( 1 )
Total: ( 15 \times 6 = 90 )



Applying Constraints:



Constraint 3 (C & E together): Treat C-E as a single unit. Now, we have 5 units: (A, B, D, F, CE).


Assign CE to any project.
Remaining 4 employees (A, B, D, F) need to fill 2 spots in the other two projects.
Valid assignments: ( \binom{4}{2} \times \binom{2}{2} = 6 \times 1 = 6 ).
Since CE can be in X, Y, or Z: ( 6 \times 3 = 18 ).



Constraint 1 (A & B separate): Subtract cases where A and B are together.


If A and B are in the same project:
Fix A-B in one project. CE must be in another.
Remaining employees: D, F. Assign to the third project.
Valid combinations: ( 3 ) (projects for A-B) × ( 2 ) (projects for CE) × ( 1 ) (D-F) = 6.


Subtract invalid cases: ( 18 - 6 = 12 ).





Constraint 4 (D & F separate): Ensure D and F are not in the same project.


If D and F are together:
CE must be in a different project.
Assign CE to one project, D-F to another, and A/B to the third.
Valid combinations: ( 3! = 6 ).


Subtract these: ( 12 - 6 = 6 ).






Final Answer: 6 valid combinations.



Question 2: Data Interpretation (Table Analysis)

Direction:

The table below shows sales (in ₹ crores) of three products (P, Q, R) across four cities (M, N, O, P) in 2023.




Product
M
N
O
P




P
25
30
20
35


Q
40
45
50
55


R
15
20
25
30




Task:

What is the percentage increase in total sales of product Q from 2022 to 2023, given that the total sales of all products in city M in 2022 was ₹12.5 crores?


Assumption: Product sales in 2022 followed the same ratio across cities as in 2023.


Solution:



Total Q Sales in 2023:

( 40 + 45 + 50 + 55 = 190 ) crore.



Total Sales in City M (2023):

( 25 + 40 + 15 = 80 ) crore.



Ratio of Q in City M (2023):

( \frac{40}{80} = 0.5 ) or 50%.



Total Q Sales in City M (2022):

( 12.5 \times 0.5 = 6.25 ) crore.



Total Q Sales in 2022:


Since the ratio is consistent across cities:
Total Q in 2022 = ( 6.25 \times 4 ) (since City M’s 2022 sales = 12.5 crore, and total city sales in 2023 = 80 crore for M, implying a scaling factor of ( \frac{80}{12.5} = 6.4 )).
Alternative: Calculate using ratios for all cities:
2022 sales for each city = ( \frac{2023\text{ sales}}{6.4} ).
Total Q in 2022 = ( \frac{190}{6.4} \approx 29.69 ) crore.





Percentage Increase:

( \frac{190 - 29.69}{29.69} \times 100 \approx 540% ).




Final Answer: 540% increase.



Key Takeaways for CAT 2024:


Logical Games: Prioritize constraint mapping (e.g., using grids or process of elimination).
Data Interpretation: Focus on scaling factors and ratio assumptions.
Time Management: Allocate 20 mins for puzzles and 30 mins for DI.


Let me know if you need further clarification! 📊✨
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