Quantity I: A number is first multiplied by 3 and then increased by 25%. The final result is 375. What is the original number? Quantity II: A boat travels downstream for 48 km and upstream for 36 km. The speed of the boat in still water is 18 km/h. Find the speed of the stream. Quantity III: A man invests ₹10,000 in a scheme offering compound interest of 5% per annum, compounded annually. Find the amount after 3 years.

Quantity I: Let the original number be x. First, the number is multiplied by 3: 3x. Then, it is increased by 25%, so the result is 3x × 1.25 = 375. Solving for x: 3x × 1.25 = 375, 3x = 375 / 1.25 = 300, x = 300 / 3 = 100. Quantity II: Let the speed of the stream be x. The speed of the boat downstream is 18 + x,  and the speed of the boat upstream is 18 - x. The time taken for downstream = 48 / (18 + x), The time taken for upstream = 36 / (18 - x). Since the time taken to travel both distances is equal, we equate the two expressions: 48 / (18 + x) = 36 / (18 - x). Cross-multiplying, we get: 48(18 - x) = 36(18 + x). Simplifying: 864 - 48x = 648 + 36x, 864 - 648 = 48x + 36x, 216 = 84x, x = 216 / 84 = 2.57 km/h. Quantity III: Using the compound interest formula: Amount = Principal × (1 + Rate / 100) ^ Time. Amount = 10,000 × (1 + 5/100)^3. Amount = 10,000 × (1.05)^3 ≈ 10,000 × 1.157625 = ₹11,576.25. Comparing the quantities: Quantity I = 100, Quantity II ≈ 2.57 km/h, Quantity III ≈ 11,576.25. Answer: (B) Quantity II < Quantity I < Quantity III