Question

    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. Consider only numeric value.
    A Quantity I = Quantity II < Quantity III Correct Answer Incorrect Answer
    B Quantity II < Quantity I < Quantity III Correct Answer Incorrect Answer
    C Quantity I > Quantity II = Quantity III Correct Answer Incorrect Answer
    D Quantity I < Quantity II = Quantity III Correct Answer Incorrect Answer
    E Quantity I = Quantity II > Quantity III Correct Answer Incorrect Answer

    Solution

    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

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