Compound interest application in engineering economics: If ₹10,000 is invested today at an annual compound interest rate of 10%, what will be the accumulated amount at the end of 5 years (rounded to the nearest rupee)?

Difficulty: Easy

Correct Answer: ₹16,105

Explanation:


Introduction / Context:
Many time-value-of-money problems in engineering economics use compound interest to project a present investment to a future amount. This question checks your ability to apply the basic future-worth formula correctly and round sensibly to whole rupees.


Given Data / Assumptions:

  • Present amount (P) = ₹10,000.
  • Annual interest rate (i) = 10% per year.
  • Compounding period = annually.
  • Number of years (n) = 5.
  • Rupee rounding to the nearest whole rupee.


Concept / Approach:
The future value F of a single present sum under annual compounding is given by F = P * (1 + i)^n. This multiplies the present sum by the compound factor over n years.


Step-by-Step Solution:
Write the formula: F = P * (1 + i)^nSubstitute numbers: F = 10,000 * (1 + 0.10)^5Compute the factor: (1.10)^5 = 1.61051Multiply: F = 10,000 * 1.61051 = 16,105.1Round to the nearest rupee: ₹16,105


Verification / Alternative check:
Use year-by-year accumulation: after 1st year ₹11,000; 2nd: ₹12,100; 3rd: ₹13,310; 4th: ₹14,641; 5th: ₹16,105. This matches the formula result.


Why Other Options Are Wrong:
₹15,000 assumes simple interest (10% of 10,000 * 5). ₹18,105 uses an incorrect factor. ₹12,500 corresponds to 5 years at 5% or other incorrect basis.


Common Pitfalls:
Mixing simple and compound interest, miscounting periods, or rounding intermediate steps instead of the final result.


Final Answer:
₹16,105

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