The following table shows the sales (in Rs. lakh) of three products (A, B, C) across four regions (North, South, East, West) in 2011. 
- Step 1: Find total sales in East. From the given table/data:
Product A = 20 Rs. lakh, Product B = 40 Rs. lakh, Product C = 50 Rs. lakh.
Total East sales = 20 + 40 + 50 = 110 Rs. lakh.
- Step 2: Find Product C sales in East.
Product C sales in East = 50 Rs. lakh.
- Step 3: Write the percentage formula.
Percentage share of Product C = (Product C sales ÷ Total East sales) × 100.
- Step 4: Substitute values. = (50 ÷ 110) × 100.
- Step 5: Compute step-by-step.
50 ÷ 110 = 5 ÷ 11 ≈ 0.4545.
Multiply by 100 ⇒ 0.4545 × 100 = 45.45%.
- Step 6: Verify calculation.
Another way: (50 × 100) ÷ 110 = 5000 ÷ 110 = 45.4545…%, which rounds to 45.45%.
- Step 7: Match with options.
Given options: (1) 40%, (2) 45.45%, (3) 50%, (4) 55.55%.
Our value (45.45%) exactly matches Option (2).
- Step 8: Cross-check logic.
The fraction 5/11 is a well-known decimal ≈ 0.4545, confirming the percentage.
- Step 9: Conclusion.
The percentage share of Product C in East = 45.45%, so the correct answer is Option (2).





For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: