CAT 2019 Slot 2 Quant Solutions: Strategic Approach to Indian Math Problems
The Quantitative Ability section of the CAT 2019 Slot 2 featured a mix of algebraic, number theory, and logical reasoning problems. Below is a detailed breakdown of key questions, solutions, and strategies tailored to Indian exam patterns.
Problem 1: Work Rate and Time
Question:
A and B can complete a task in 12 days working together. A alone takes 20 days longer than B. How many days does B take to complete the task alone?
Solution:
Let B’s time = ( x ) days. Then, A’s time = ( x + 20 ) days.
Work rates:
B’s rate = ( \frac{1}{x} )
A’s rate = ( \frac{1}{x + 20} )
Combined rate: ( \frac{1}{x} + \frac{1}{x + 20} = \frac{1}{12} ).
Solving:
( \frac{2x + 20}{x(x + 20)} = \frac{1}{12} )
( 24x + 240 = x^2 + 20x )
( x^2 - 4x - 240 = 0 )
( x = \frac{4 \pm \sqrt{16 + 960}}{2} = \frac{4 \pm 32}{2} )
( x = 18 ) (discarding negative value).
Answer: B takes 18 days.
Problem 2: Number Theory
Question:
What is the smallest positive integer ( n ) such that ( 2^n + 3^n ) is divisible by 5?
Solution:
Compute ( 2^n \mod 5 ) and ( 3^n \mod 5 ):
( 2^1 = 2 \mod 5 ), ( 2^2 = 4 \mod 5 ), ( 2^3 = 3 \mod 5 ), ( 2^4 = 1 \mod 5 ).
( 3^1 = 3 \mod 5 ), ( 3^2 = 4 \mod 5 ), ( 3^3 = 2 \mod 5 ), ( 3^4 = 1 \mod 5 ).
Look for ( n ) where ( 2^n + 3^n \equiv 0 \mod 5 ):
( n = 4 ): ( 1 + 1 = 2 \mod 5 ) ❌
( n = 3 ): ( 3 + 2 = 5 \equiv 0 \mod 5 ) ✅
Answer: ( n = \boxed{3} ).
Problem 3: Algebra (Quadratic Equations)
Question:
If ( x + y = 10 ) and ( x^2 + y^2 = 68 ), find ( xy ).
Solution:
Use the identity ( (x + y)^2 = x^2 + y^2 + 2xy ).
Substitute values:
( 10^2 = 68 + 2xy )
( 100 = 68 + 2xy )
( xy = \frac{32}{2} = 16 ).
Answer: ( xy = \boxed{16} ).
Problem 4: Probability
Question:
A box has 6 red and 4 blue balls. If two balls are drawn at random, what is the probability both are red?
Solution:
Total ways to draw 2 balls: ( \binom{10}{2} = 45 ).
Favorable ways: ( \binom{6}{2} = 15 ).
Probability: ( \frac{15}{45} = \frac{1}{3} ).
Answer: Probability = ( \boxed{\dfrac{1}{3}} ).
Problem 5: Geometry (Triangles)
Question:
In triangle ( ABC ), ( AB = 5 ), ( BC = 6 ), and ( AC = 7 ). Find the radius of the circumscribed circle.
Solution:
Use formula: ( R = \frac{abc}{4K} ), where ( K ) is the area.
Compute semi-perimeter: ( s = \frac{5 + 6 + 7}{2} = 9 ).
Area ( K = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{9 \times 4 \times 3 \times 2} = 6\sqrt{6} ).
( R = \frac{5 \times 6 \times 7}{4 \times 6\sqrt{6}} = \frac{210}{24\sqrt{6}} = \frac{35\sqrt{6}}{24} ).
Answer: ( R = \boxed{\dfrac{35\sqrt{6}}{24}} ).
Key Takeaways for CAT 2019 Slot 2 Quant
Time Management: Prioritize quadratic equations and number theory (e.g., Problem 3 and 2).
Pattern Recognition: Use cyclicity in powers (e.g., modulo 5 cycles for Problem 2).
Formula Application: Master work rate (( \frac{1}{x} + \frac{1}{y} = \frac{1}{t} )) and Heron’s formula (Problem 5).
Final Score Strategy: Aim for 90% accuracy in 30 minutes to secure a 99+ percentile.
Prepared by [Your Name/Team], optimized for Indian exam patterns.
|
