Title: CAT 2024 Slot 3 Quant Solutions & Strategies
Content: Detailed Answer Explanations for Indian Exam Patterns
Question 1: Algebraic Inequalities
Problem:
If ( 2x + 3y = 30 ) and ( x, y > 0 ), find the maximum possible value of ( xy ).
Solution:
Express ( y ) in terms of ( x ):
( 3y = 30 - 2x ) → ( y = 10 - \frac{2}{3}x ).
Formulate ( xy ):
( xy = x\left(10 - \frac{2}{3}x\right) = 10x - \frac{2}{3}x^2 ).
Maximize using calculus or vertex formula (quadratic):
Vertex at ( x = -b/(2a) = -10/(2 \cdot -2/3) = 7.5 ).
Substitute ( x = 7.5 ):
( y = 10 - \frac{2}{3} \times 7.5 = 5 ).
Maximum ( xy = 7.5 \times 5 = 37.5 ).
Key Insight:
Use quadratic optimization or AM-GM (e.g., ( 2x + 3y = 30 ) → split terms for equality).
Common Error:
Forgetting ( x, y > 0 ) constraints.
Question 2: Data Interpretation (DI)
Problem:
Given a bar graph showing monthly sales (in Lakh) of two products (A and B) from Jan–Dec 2023:
Question: Which month had the highest combined sales of A and B?
Solution:
Extract values (example data):
Jan: A=20, B=15 → Total=35
Feb: A=25, B=20 → Total=45
...
Nov: A=40, B=30 → Total=70
Dec: A=35, B=35 → Total=70
Identify maximum:
Nov and Dec both have total 70 (answer depends on exact data).
Key Insight:
Calculate totals for each month and compare.
Common Error:
Miscalculating totals or misreading units (Lakh = 100,000).
Question 3: Geometry
Problem:
A right circular cone has volume 100π. If its height is doubled, what is the new volume if the radius remains the same?
Solution:
Original volume formula:
( V = \frac{1}{3}\pi r^2 h = 100\pi ).
Simplify:
( r^2 h = 300 ).
New height = 2h. New volume:
( V_{new} = \frac{1}{3}\pi r^2 (2h) = 2 \times 100\pi = 200\pi ).
Key Insight:
Volume scales linearly with height if radius is constant.
Common Error:
Confusing volume with surface area formulas.
Question 4: Logical Reasoning
Problem:
Four friends (A, B, C, D) are sitting in a row. B is adjacent to A, and C is not adjacent to D. If A is in the middle, who sits first?
Solution:
Possible positions (1 to 4):
A is in position 2 or 3.
If A is in 2: Positions = [1, A, 3, 4].
B must be in 1 or 3.
C and D cannot be adjacent.
Valid arrangement: B (1), A (2), D (3), C (4).
If A is in 3: Positions = [1, 2, A, 4].
B must be in 2 or 4.
C and D cannot be adjacent.
Valid arrangement: C (1), D (2), A (3), B (4).
Consistent answer: First seat is either B or C (depends on constraints).
Key Insight:
Draw rows and eliminate invalid cases systematically.
Common Error:
Overlooking non-adjacent constraints for C and D.
Question 5: Number Theory
Problem:
Find the smallest positive integer ( n ) such that ( n^2 + 5n ) is divisible by 24.
Solution:
Factor: ( n(n + 5) ).
Divisibility by 24 = 8 × 3:
Divisible by 8: Either ( n ) or ( n+5 ) must be even, and one must be divisible by 4.
Divisible by 3: Either ( n ) or ( n+5 ) must be divisible by 3.
Test values:
n=1: 6 → Not divisible by 8.
n=3: 24 → Divisible by 24.
Answer: ( n = 3 ).
Key Insight:
Use LCM properties and modular arithmetic.
Common Error:
Missing the combination of 8 and 3 divisibility.
Practice Problems
Algebra: If ( x + y = 10 ), find the maximum of ( x^2 + y^2 ).
Answer: 50 (when x=0 or y=0).
DI: Given a pie chart of expenses (Food: 40%, Housing: 30%, etc.), find the month with the lowest total expenses.
Geometry: A cylinder’s radius increases by 10%, height decreases by 10%. What is the % change in volume?
Answer: -1.9% (Volume = πr²h).
Final Tips for CAT 2024:
Prioritize speed using shortcuts (e.g., quadratic vertex formula).
Review unit conversions (Lakh,crore) and data interpretation traps.
Practice time-bound DI sets (15–20 min per 5 questions).
Let me know if you need further clarification!
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