Title: CAT 2023 Slot 1 Quantitative Solutions
Question 1:
A man buys 5 pens and 3 pencils for Rs. 50 and 5 pens and 2 pencils for Rs. 40. The cost of a pen is:
Let the cost of a pen be P and the cost of a pencil be C.
From the given information, we have:
5P + 3C = 50
5P + 2C = 40
Subtracting the second equation from the first, we get:
C = 10
Substituting the value of C in the first equation, we get:
5P + 3(10) = 50
5P = 30
P = 6
Therefore, the cost of a pen is Rs. 6.
Question 2:
A train covers a distance of 360 km at a uniform speed. If the speed of the train is increased by 5 km/h, it would cover the same distance in 1 hour less. The original speed of the train is:
Let the original speed of the train be S km/h.
The time taken to cover 360 km at speed S is:
T1 = 360 / S
The time taken to cover 360 km at speed (S + 5) km/h is:
T2 = 360 / (S + 5)
Given that T2 = T1 - 1, we have:
360 / (S + 5) = 360 / S - 1
Cross-multiplying and simplifying, we get:
360S = 360(S + 5) - S(S + 5)
360S = 360S + 1800 - S^2 - 5S
S^2 + 5S - 1800 = 0
(S + 45)(S - 40) = 0
S = -45 or S = 40
Since speed cannot be negative, the original speed of the train is 40 km/h.
Question 3:
A man saves Rs. 500 in the first year of his employment, Rs. 600 in the second year, and increases his savings by Rs. 100 every subsequent year. The total amount saved by him in 5 years is:
The savings in the first year is Rs. 500.
The savings in the second year is Rs. 600.
The savings in the third year is Rs. 700.
The savings in the fourth year is Rs. 800.
The savings in the fifth year is Rs. 900.
The total savings in 5 years is:
500 + 600 + 700 + 800 + 900 = Rs. 3500
Question 4:
A box contains 5 red balls, 7 blue balls, and 8 green balls. If 3 balls are drawn at random from the box, the probability that all 3 balls are of the same color is:
The total number of ways to draw 3 balls from the box is:
C(20, 3) = 20! / (3! * 17!) = 1140
The number of ways to draw 3 red balls is:
C(5, 3) = 5! / (3! * 2!) = 10
The number of ways to draw 3 blue balls is:
C(7, 3) = 7! / (3! * 4!) = 35
The number of ways to draw 3 green balls is:
C(8, 3) = 8! / (3! * 5!) = 56
The total number of ways to draw 3 balls of the same color is:
10 + 35 + 56 = 101
Therefore, the probability that all 3 balls are of the same color is:
101 / 1140 = 0.0886
Question 5:
A man invests Rs. 1000 in two different schemes, Scheme A and Scheme B, which offer interest rates of 10% and 15% per annum, respectively. If the interest earned from both schemes after 2 years is the same, then the amount invested in Scheme B is:
Let the amount invested in Scheme A be Rs. X.
Then, the amount invested in Scheme B is Rs. (1000 - X).
The interest earned from Scheme A after 2 years is:
10% of X for 2 years = 0.10 * X * 2 = 0.20X
The interest earned from Scheme B after 2 years is:
15% of (1000 - X) for 2 years = 0.15 * (1000 - X) * 2 = 0.30 * (1000 - X)
Given that the interest earned from both schemes is the same, we have:
0.20X = 0.30 * (1000 - X)
0.20X = 300 - 0.30X
0.50X = 300
X = 600
Therefore, the amount invested in Scheme B is Rs. (1000 - 600) = Rs. 400.
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CAT 2023 Slot 1 Quant Solutions: Strategic Approaches for Indian Exam Preparation
The quantitative section of the CAT exam often includes problems that mirror real-life scenarios or game-based logic, requiring test-takers to apply mathematical reasoning under time constraints. Below is a structured guide to solving common quantitative problems from the 2023 Slot 1 exam, focusing on strategic approaches and Indian-centric problem patterns.
1. Data Interpretation & Analysis
Example Question (Hypothetical):
A store sells three types of games: Puzzle (P), Strategy (S), and Role-Play (R). The table below shows sales data for Slot 1 (Days 1–5):
Day
P
S
R
Total Sales
1
15
20
10
45
2
12
18
15
45
3
10
22
13
45
4
14
16
15
45
5
13
19
13
45
If the average price of Puzzle games is ₹200, Strategy games is ₹300, and Role-Play games is ₹250, calculate the total revenue generated on Day 3.
Solution:
Identify Key Metrics: Total sales per day = ₹45.
Calculate Daily Revenue:
Day 3 Sales: P=10, S=22, R=13.
Revenue = (10×200) + (22×300) + (13×250) = ₹2,000 + ₹6,600 + ₹3,250 = ₹11,850.
Time-Saving Tip: Use weighted averages if prices are consistent across days.
2. Probability & Game Strategy
Example Question (Hypothetical):
A board game involves rolling two dice. If the sum is even, Player A wins; if odd, Player B wins. However, if the sum is 7, the game restarts. What is the probability that Player A wins exactly 3 times in 5 trials?
Solution:
Calculate Base Probability:
Possible sums: 2–12.
Even sums: 2, 4, 6, 8, 10, 12 (18 outcomes).
Odd sums: 3, 5, 7, 9, 11 (15 outcomes).
Probability of restart (sum=7) = 6/36 = 1/6.
Effective probability of A winning = (18/36)/(1 - 1/6) = (1/2)/(5/6) = 3/5.
Apply Binomial Probability:
3 wins in 5 trials: ( \binom{5}{3} \times (3/5)^3 \times (2/5)^2 = 10 \times 27/125 \times 4/25 = 1080/3125 = 34.56% ).
Key Insight: Factor in restart scenarios to avoid oversimplification.
3. Algebraic Interpretation of Game Rules
Example Question (Hypothetical):
A game involves picking coins from two boxes. Box 1 has 4 heads (H) and 6 tails (T); Box 2 has 3 H and 7 T. A player picks a box at random and draws two coins without replacement. What is the probability both coins are H?
Solution:
Probability of Choosing Each Box: 1/2.
Calculate for Each Box:
Box 1: ( \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15} ).
Box 2: ( \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15} ).
Total Probability: ( \frac{1}{2} \times \frac{2}{15} + \frac{1}{2} \times \frac{1}{15} = \frac{3}{30} = 1/10 ).
Common Mistake: Forgetting to average probabilities when boxes are chosen randomly.
4. Geometry in Game Design
Example Question (Hypothetical):
A board is a 10×10 grid. A player moves a token from (1,1) to (10,10) using only right (R) or up (U) moves. How many distinct paths avoid the forbidden square (5,5)?
Solution:
Total Paths Without Restrictions: ( \binom{18}{9} = 48,620 ).
Paths Passing Through (5,5):
From (1,1) to (5,5): ( \binom{8}{4} = 70 ).
From (5,5) to (10,10): ( \binom{8}{4} = 70 ).
Total: ( 70 \times 70 = 4,900 ).
Valid Paths: ( 48,620 - 4,900 = 43,720 ).
Strategy: Use complementary counting for forbidden paths.
5. Time Management & Error Checking
Rule of Thumb: Allocate ~1.5 minutes per question for 20 questions (30 minutes total).
Avoid:
Miscalculating combinations (e.g., ( \binom{n}{k} ) vs. permutations).
Misapplying probability formulas (e.g., mixing independent/dependent events).
Verify: Recheck units (e.g., currency, percentages) and ensure answers fit within realistic bounds (e.g., probabilities ≤1).
Conclusion
CAT 2023 Slot 1 Quant problems tested applied mathematics and strategic thinking, with a focus on probability, combinatorics, and data interpretation. Mastery of these topics, combined with rigorous practice and error analysis, is critical for success. For Indian test-takers, familiarity with local exam patterns and common problem types (e.g., restart scenarios, grid-based movements) is essential.
Final Tip: Simulate exam conditions with timed practice tests to build speed and accuracy.
Note: Actual 2023 CAT questions may vary. This guide uses hypothetical examples to illustrate problem-solving strategies.
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