cat 2018 slot 3 quant
Title: CAT 2018 Slot 3: Quantitative Ability Solutions
Content: Detailed Solutions to Key Quantitative Problems from CAT 2018 - Slot 3
Problem 1: Probability and Combinations
Question:
Three friends A, B, and C each pick a number from 1 to 10. If the sum of their numbers is divisible by 3, A wins. If divisible by 5, B wins. Otherwise, C wins. What is the probability that C wins?
Solution:
Total Possible Outcomes:
Each friend has 10 choices, so total outcomes = (10 \times 10 \times 10 = 1000).
A Wins (Sum Divisible by 3):
For sums divisible by 3, the number of valid triples ((a, b, c)) can be calculated using modular arithmetic.
Each number (x) mod 3 can be 0, 1, or 2.
Let (n_0, n_1, n_2) be the count of numbers from 1–10 congruent to 0, 1, 2 mod 3:
(n_0 = 3) (3, 6, 9)
(n_1 = 4) (1, 4, 7, 10)
(n_2 = 3) (2, 5, 8)
Valid combinations for sum ≡ 0 mod 3:
(n_0^3 + n_1^3 + n_2^3 + 3(n_0 n_1 n_2) + 3(n_0^2 n_1 + n_0^2 n_2 + n_1^2 n_0 + n_1^2 n_2 + n_2^2 n_0 + n_2^2 n_1))
Simplified: (3^3 + 4^3 + 3^3 + 3(3 \times 4 \times 3) = 27 + 64 + 27 + 108 = 226).
A’s Winning Cases: 226.
B Wins (Sum Divisible by 5):
Similarly, calculate sums ≡ 0 mod 5.
Numbers 1–10 mod 5:
(n_0 = 2) (5, 10)
(n_1, n_2, n_3, n_4) = 2, 2, 2, 2 respectively.
Valid combinations:
(n_0^3 + \sum_{i=1}^4 n_i^3 + 3(n_0 n_1 n_2 + n_0 n_1 n_3 + ...))
Simplified: (2^3 + 4 \times 2^3 + 3 \times 2 \times 2 \times 2 \times 4 = 8 + 32 + 96 = 136).
B’s Winning Cases: 136.
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Overlap (Sum Divisible by 15):
Use LCM(3,5)=15. Calculate sums ≡ 0 mod 15.
Numbers 1–10 mod 15: Only 5 and 10 are ≡ 5 and 10 mod 15.
Valid combinations:
(2^3 = 8) (all three pick 5 or 10).
Overlap Cases: 8.
C Wins (Neither Divisible by 3 nor 5):
Total C cases = Total - (A + B - Overlap)
(C = 1000 - (226 + 136 - 8) = 1000 - 354 = 646).
Probability:
[
P(C) = \frac{646}{1000} = 0.646 \quad \text{(or 64.6%)}
]
Final Answer: The probability that C wins is (\boxed{\frac{323}{500}}).
Problem 2: Data Interpretation (DI)
Question:
A table shows sales (in ₹ crores) of two companies, X and Y, over four years. Year 1: X=120, Y=80; Year 2: X=150, Y=100; Year 3: X=180, Y=120; Year 4: X=200, Y=140.
Part (a): What is the average annual growth rate of Company X’s sales?
Part (b): By what percentage did Company Y’s sales increase from Year 1 to Year 4?
Solution:
Part (a):
Growth from Year 1 to Year 4:
[
\text{Growth Rate} = \left(\frac{200}{120}\right)^{1/3} - 1 = \left(\frac{5}{3}\right)^{1/3} - 1 \approx 14.47% \quad \text{(Annualized)}
]
Answer: (\boxed{14.47%}).
Part (b):
Total growth for Y:
[
\frac{140 - 80}{80} \times 100 = 75%
]
Answer: (\boxed{75%}).
Key Takeaways:
Probability: Use modular arithmetic and combinatorial counting for overlapping conditions.
DI: Apply CAGR for growth rates and percentage change for straightforward comparisons.
CAT Strategy: Prioritize DI and probability questions for higher marks in Slot 3.
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