Placement Papers - MCQ Practice Questions

Let numbers be 3x and 4x. Sum = 7x = 140, so x = 20. Larger number = 4×20 = 80

Using y - y₁ = m(x - x₁): y - 3 = 4(x - 2) → y = 4x - 5

Savings = 100% - 40% - 30% = 30%. 30% of 50,000 = 15,000

SI = (P × R × T)/100 = (5000 × 8 × 3)/100 = 1200. Total = 5000 + 1200 = 6200

Number = Divisor × Quotient + Remainder = 7 × 8 + 3 = 56 + 3 = 59

Total work = 3 × 8 = 24 man-days. Men needed = $\frac{24}{6}$ = 4

Favorable outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6. Total outcomes = 36. Probability = $\frac{6}{36}$ = $\frac{1}{6}$

TCS 2024-25 placement drive prioritizes Java and Python as primary programming languages, reflecting current industry trends

Let smaller number = x. Larger = x+5. Sum: x + (x+5) = 25 → 2x = 20 → x = 10. Larger = 15

2x + 3 = 11 → 2x = 8 → x = 4

TCS extensively uses Agile methodologies combined with DevOps practices for faster delivery and continuous integration.

Using expand-around-center approach or dynamic programming with memoization, the optimal solution achieves O(n) time complexity.

The nested loop runs: 1+2+3+4+5 = 15 times. This is the sum of first 5 natural numbers.

Support = 100% - 45% - 30% = 25% of 120 = 30 employees.

Heaps provide O(log n) insertion and deletion, making them optimal for priority queues compared to other structures.

10110₂ = 1×16 + 0×8 + 1×4 + 1×2 + 0×1 = 16 + 4 + 2 = 22

Merge sort requires O(n) extra space for creating temporary arrays during the merge process.

TCS's digital transformation strategy emphasizes AI/ML, cloud computing, and cybersecurity as core competencies.

O(2^n) represents exponential time complexity, which becomes impractical for large input sizes (n > 40).

In circular queue: elements = (rear - front + size) % size = (2 - 7 + 10) % 10 = 5. But accounting properly: 7,8,9,0,1,2 = 6 elements.