CAT 2025 Slot 1 concluded recently, and aspirants have been discussing the most notable questions from the exam. Based on student reactions and expert analysis, these five questions stood out for their uniqueness, difficulty, and conceptual focus. Solving these questions effectively required a mix of concept clarity, logical reasoning, and time management.
Table of Contents
CAT 2025 Slot 1 Memory-Based Questions with Answers
1. Quadratic Equation Integer Roots
Question: The number of non-negative integer values of kkk for which the quadratic equation x2−5x+k=0x^2 - 5x + k = 0x2−5x+k=0 has only integer roots is?
Answer: 3
Explanation: For integer roots, the quadratic x2−5x+k=0x^2 - 5x + k = 0x2−5x+k=0 should factor as (x−p)(x−q)=x2−(p+q)x+pq(x - p)(x - q) = x^2 - (p+q)x + pq(x−p)(x−q)=x2−(p+q)x+pq.
- Compare: p+q=5p + q = 5p+q=5, pq=kpq = kpq=k
- Non-negative integer pairs (p,q)(p,q)(p,q) summing to 5: (0,5), (1,4), (2,3), (3,2), (4,1), (5,0)
- Corresponding k=pq=0,4,6,6,4,0k = pq = 0,4,6,6,4,0k=pq=0,4,6,6,4,0 → Unique non-negative integer values: 0,4,6
- Number of values = 3
2. Acid-Water Problem
Question: A 200 L solution contains 30% acid and 70% water. If 20% of the solution is replaced with water and 10% is replaced with acid, what is the approximate percentage of acid in the final solution?
Answer: Approximately 34%
Explanation:
- Initial acid: 200 × 30% = 60 L
- 20% replaced with water ,40 L solution removed (acid fraction = 0.3) , acid removed = 40 × 0.3 = 12 L , remaining acid = 60 – 12 = 48 L
- 10% replaced with acid, 20 L solution added as acid, acid = 48 + 20 = 68 L
- Total solution = 200 L (since volume removed + volume added = 200)
- Acid % = 68 / 200 × 100 ≈ 34%
3. Class Students Problem
Question: In a class, there were 10 more boys than a certain number of girls. After 40% of girls and 60% of boys left, the remaining number of girls was 8 more than the remaining boys. Find the possible total number of students initially.
Answer: Let girls = GGG, boys = B=G+10B = G + 10B=G+10
- Remaining girls: 0.6G
- Remaining boys: 0.4B
- 0.6G = 0.4B + 8 → 0.6G = 0.4(G + 10) + 8 → 0.6G = 0.4G + 4 + 8 → 0.2G = 12 → G = 60
- B = 60 + 10 = 70
- Total = 130
4. Stock Portfolio Problem
Question: Stocks A, B, and C are priced at 120, 90, and 80 INR respectively. A trader holds 60 shares of A, and 20 shares of B and C together. If the total value is 3300 INR, find the number of shares of B and C.
Answer:
- Value of 60 A shares: 60 × 120 = 7200
5. Shopkeeper Discount Problem
Question: A shopkeeper allows 22% discount on market price of each chair. He sells 13 chairs at the price of 12 to earn 26% profit. If cost price (CP) of each chair = 100 INR, find the marked price of each chair.
Answer:
- CP of 1 chair = 100 INR → Total CP for 13 chairs = 1300 INR
- Profit = 26% → Total SP = 1300 × 1.26 = 1638 INR
- SP per chair = 1638 / 13 ≈ 126 INR
- Given discount = 22% → SP = MP × (1 – 0.22) → MP = 126 / 0.78 ≈ 161.54 INR
6. Algebra Inequality
Question: x ≥ y ≥ 3, and x + y ≤ 14. Find the number of integer solutions.
Answer:
- y can be 3 to 7 (integer), for each y, x = y to 14 – y
- Count solutions:
- y = 3 x = 3 to 11, 9 values
- y = 4 x = 4 to 10 ,7 values
- y = 5 x = 5 to 9 , 5 values
- y = 6 x = 6 to 8 , 3 values
- y = 7 x = 7 , 1 value
Total = 9 + 7 + 5 + 3 + 1 = 25
For Detailed Answer Key Refer CAT 2025 Answer Key
