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IB Mathematics: Applications and Interpretation · Number and Algebra

Financial applications: compound interest, annual depreciation practice questions.

65 exam-style questions on Financial applications: compound interest, annual depreciation — topic 1.4 of Number and Algebra in the IB Mathematics: Applications and Interpretation syllabus — with answers and explanations. Samples below; the full set is free with an account.

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About this topic

Syllabus code
1.4
Subject
IB Mathematics: Applications and Interpretation
Topic
1 · Number and Algebra
Questions
65 with worked answers
Question types
46 fill-in-the-gap questions · 10 multiple-choice questions · 5 true/false questions · 2 flashcards · 1 matching task · 1 structured question
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Sample questions on Financial applications: compound interest, annual depreciation

8000 CHF is invested at a nominal rate of 3.6% per annum, compounded monthly. Calculate the value of the investment after 5 years, to the nearest franc.

  • 9575
  • 9547
  • 9440
  • 8120

Answer: 9575

FV = 8000(1 + 0.036/12)^(12×5) = 8000 × 1.003⁶⁰ ≈ 9575 CHF. Compounding annually instead gives 9547, and simple interest gives 9440.

A machine is bought for 60 000 USD and depreciates at 12% per year. Determine its value after 5 years, to the nearest dollar.

  • 35 982
  • 24 000
  • 31 664
  • 105 740

Answer: 31 664

Value = 60 000 × 0.88⁵ ≈ 31 664 USD. Subtracting 12% of the original value each year (linear depreciation) wrongly gives 24 000, and using 1.12 instead of 0.88 gives 105 740.

A bank offers a nominal interest rate of 6% per annum, compounded monthly. Determine the effective annual rate, to 3 significant figures.

  • 6.00%
  • 6.17%
  • 6.09%
  • 6.30%

Answer: 6.17%

Effective rate = (1 + 0.06/12)¹² − 1 = 1.005¹² − 1 ≈ 0.0617 = 6.17%. More frequent compounding makes the effective rate exceed the nominal rate.

An investment of 3000 GBP grows to 3993 GBP in 3 years with interest compounded annually. Determine the annual interest rate.

  • 9%
  • 12%
  • 11%
  • 10%

Answer: 10%

3000(1 + r)³ = 3993 gives (1 + r)³ = 1.331, so 1 + r = 1.1 and r = 10%. Dividing the total 33.1% growth by 3 (a simple-interest approach) wrongly suggests about 11%.

Laboratory equipment loses 20% of its value each year. Determine after how many complete years its value first falls below half of the original price.

  • 4
  • 3
  • 5
  • 2

Answer: 4

0.8³ = 0.512 is still above 0.5, but 0.8⁴ = 0.4096 < 0.5, so it takes 4 years. Checking the actual powers avoids the tempting answer of 3.

A student deposits 8000 EUR at a nominal annual rate of 5 %, compounded quarterly. Find the value of the deposit after 3 years, in EUR to the nearest cent.

  • 9286.04
  • 9261.00
  • 9200.00
  • 8303.80

Answer: 9286.04

FV = 8000(1 + 0.05/4)^(4x3) = 9286.04. 9261.00 assumes annual compounding, 9200.00 is simple interest, and 8303.80 forgets to multiply the number of periods by 4.

That was 6 of 65 Financial applications: compound interest, annual depreciation questions.

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