Question

Arrange the simple interest of the following cases in decreasing order-
(A) The simple interest on Rs 6600 at 5% per annum for 2 yrs.
(B) The simple interest on Rs 200 at 6% per annum for 5 yrs.
(C) The simple interest on Rs 840 at 5% per annum for 4 yrs.
(D) The simple interest on Rs 5000 at 12% per annum for 2 yrs.
Choose the correct answer from the options given below:

• A. (A), (B), (C), (D)
• B. (B), (A), (D), (C)
• C. (B), (A), (C), (D)
• D. (D), (A), (C), (B)

Hint:

For quick mental estimation, you can look at the product of R and T (interest for the whole period) or P and R. For (D), \(5000 \times 12% = 600\) per year, so 1200 for 2 years. For (A), \(6600 \times 5% = 330\) per year, so 660 for 2 years. This quick check shows that D is much larger than A, helping you eliminate incorrect options early.

Answer 1

The Correct Option is D

Step 1: Concept Understanding: The problem necessitates the calculation of Simple Interest (SI) for four distinct scenarios and subsequent arrangement of the calculated amounts in descending order.Step 2: Calculation Methodology: The standard formula for Simple Interest is employed:\[ SI = \frac{P \times R \times T}{100} \]Where P represents the Principal amount, R denotes the annual Rate of interest, and T signifies the Time duration in years.Step 3: Detailed Calculations:Case (A): P = 6600, R = 5, T = 2. Calculated SI: \( SI_A = \frac{6600 \times 5 \times 2}{100} = 660 \)Case (B): P = 200, R = 6, T = 5. Calculated SI: \( SI_B = \frac{200 \times 6 \times 5}{100} = 60 \)Case (C): P = 840, R = 5, T = 4. Calculated SI: \( SI_C = \frac{840 \times 5 \times 4}{100} = 168 \)Case (D): P = 5000, R = 12, T = 2. Calculated SI: \( SI_D = \frac{5000 \times 12 \times 2}{100} = 1200 \)Comparing the SI values: \(SI_D = 1200\), \(SI_A = 660\), \(SI_C = 168\), \(SI_B = 60\). The descending order is \(1200>660>168>60\), corresponding to (D)>(A)>(C)>(B).Step 4: Conclusion: The simple interests, arranged in descending order, are represented by the sequence (D), (A), (C), (B).