36 exam-style questions on Financial applications: compound interest, depreciation — topic 1.4 of Number and Algebra in the IB Mathematics: Analysis and Approaches syllabus — with answers and explanations. Samples below; the full set is free with an account.
Calculate the value of an investment of 1000 USD after 2 years at 5% interest per annum, compounded annually.
Answer: 1102.50 USD
FV = 1000 × 1.05² = 1102.50 USD. The distractor 1100 comes from using simple interest instead of compound interest.
2000 EUR is invested at a nominal annual rate of 6%, compounded monthly, for 5 years. Determine the correct expression for the future value.
Answer: 2000(1 + 0.06/12)⁶⁰
With monthly compounding the rate per period is 0.06/12 and the number of periods is 12 × 5 = 60, so FV = 2000(1 + 0.06/12)⁶⁰.
A car bought for 20 000 USD loses 15% of its current value each year. Calculate its value after 2 years.
Answer: 14 450 USD
Value = 20 000 × 0.85² = 20 000 × 0.7225 = 14 450 USD. The distractor 14 000 assumes the same dollar loss each year (linear depreciation).
An amount is invested at 10% per annum compound interest. Determine the least number of complete years needed for the investment to at least double in value.
Answer: 8 years
We need 1.1ⁿ ≥ 2: since 1.1⁷ ≈ 1.949 < 2 and 1.1⁸ ≈ 2.144 ≥ 2, the answer is 8 complete years.
A deposit grows from 800 GBP to 968 GBP in 2 years with interest compounded annually. Determine the annual interest rate.
Answer: 10%
(1 + r/100)² = 968/800 = 1.21, so 1 + r/100 = 1.1 and r = 10%. Dividing the total 21% growth by 2 gives the distractor 10.5%.
An asset depreciates at 12% per year. The yearly values form a geometric sequence. State its common ratio.
Answer: 0.88
Losing 12% each year means keeping 88%, so each value is multiplied by 1 − 0.12 = 0.88.
That was 6 of 36 Financial applications: compound interest, depreciation questions.
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