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Study Guide12th GradeLevel 3 · Challenge

Free 12th Grade Annuities, Loans and Depreciation

A senior study guide to the financial models met in senior math: straight-line and declining-balance depreciation, the future and present value of an annuity, calculating loan repayments and total interest, and recurrence relations for a reducing-balance loan. Worked examples use realistic car, savings and home-loan figures, with a six-question self-check and answers.

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Grade
12th Grade
Subject
Math
Topic
Financial Math
Difficulty
Level 3 · Challenge
Estimated time
30 minutes
Curriculum
Texas grade levels (TEKS)
Answers
Not applicable
Format
PDF (A4) + print

Students will practice

  • calculating the value of an asset under straight-line and declining-balance depreciation
  • finding the future value and present value of an annuity
  • calculating the regular repayment and total interest on a reducing-balance loan
  • using a recurrence relation to track a loan or investment balance period by period

Curriculum: Texas grade levels (TEKS). We show specific outcome codes only where they have been verified against the official curriculum document.

What's next?

Completed: Annuities, Loans and Depreciation

  1. 1Compound Interest: Effective Rates and Inflation
  2. 2Reducing-Balance Loans: Repayment Tables and Extra Repayments
Study guide

How to use this study guide

  1. Read it together first, pausing at each worked example to try the step before reading the answer.
  2. Attempt the "Check yourself" questions at the end without looking back.
  3. Then practice with a worksheet from the pathway above and finish with the topic test.

Common questions

Who is this study guide for?

12th Grade students (typically ages 17–18) working on financial math. It is pitched at level 3 · challenge.

Are the answers included?

This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.

How long does it take?

About 30 minutes. Short, regular sessions work best: two or three a week beats one long one.

Do I need to sign up to download?

No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.

What should we do next?

Try Compound Interest: Effective Rates and Inflation.

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Success TutoringWorksheets12th Grade · Mathematics · Financial Mathematics

Annuities, Loans and Depreciation

What you need to know

Depreciation is the loss in value of an asset over time. With straight-line (flat-rate) depreciation the same amount D is lost each period: V = V₀ − Dn. With declining-balance (reducing-balance) depreciation the same percentage r is lost each period: V = V₀(1 − r)n, so the value falls quickly at first and never quite reaches zero.

An annuity is a series of equal payments M made at regular intervals with compound interest at rate r per period. Its future value is what the payments grow to (a savings plan); its present value is the single amount today that is worth the same as all the payments (a loan). A reducing-balance loan is an annuity in reverse: each period, interest is added to the balance, then the repayment is subtracted. The same step written as a recurrence relation is Vₙ₊₁ = Vₙ(1 + r) − M.

ModelFormulaNotes
Straight-line depreciationV = V₀ − DnD = amount lost per period
Declining-balance depreciationV = V₀(1 − r)nr = rate per period as a decimal
Future value of an annuityFV = M × payments at the end of each period
Present value of an annuityPV = M × also the amount that can be borrowed
Loan repaymentM = rearranged from the PV formula
Recurrence (loan)Vₙ₊₁ = Vₙ(1 + r) − M, V₀ = amount borrowedinterest first, then repayment

A $30,000 car depreciates by $4500 per year (straight line). Find its value after 4 years, and when it is written off.

  1. V = 30,000 − 4500 × 4 = 30,000 − 18,000 = $12,000
  2. Written off when V = 0: 30,000 ÷ 4500 = 6.67, so during the 7th year (after 6 years it is still worth $3000).

The same $30,000 car depreciates at 20% per year (declining balance). Find its value after 4 years.

  1. V = 30,000 × (1 − 0.2)4 = 30,000 × 0.84 = 30,000 × 0.4096 = $12,288
  2. Compare: straight-line gave $12,000 after 4 years, but declining balance loses more early on ($6000 in the first year versus $4500) and less later.

Savings plan: $500 is deposited at the end of every month for 10 years at 6% per year compounded monthly. Find the future value.

  1. r = 0.06 ÷ 12 = 0.005; n = 10 × 12 = 120
  2. FV = 500 × = 500 × = 500 × 163.879…
  3. FV = $81,939.67. The deposits total $60,000, so $21,939.67 is interest.

Home loan: $400,000 is borrowed at 6% per year compounded monthly over 30 years. Find the monthly repayment and the total interest.

  1. r = 0.005; n = 360
  2. M = = = = $2398.20 per month
  3. Total repaid = 2398.20 × 360 = $863,352
  4. Total interest = 863,352 − 400,000 = $463,352 — more than the amount borrowed.

Recurrence relation: $20,000 is borrowed at 12% per year compounded monthly, with repayments of $600 per month. Find the balance after 3 months.

  1. r = 0.12 ÷ 12 = 0.01, so Vₙ₊₁ = 1.01Vₙ − 600 with V₀ = 20,000.
  2. V₁ = 1.01 × 20,000 − 600 = 20,200 − 600 = 19,600
  3. V₂ = 1.01 × 19,600 − 600 = 19,796 − 600 = 19,196
  4. V₃ = 1.01 × 19,196 − 600 = 19,387.96 − 600 = $18,787.96
  5. Interest in month 1 was $200 and the principal fell by $400; each month slightly more of the $600 goes to principal.

Present value: how much must be invested now at 5% per year compounded annually to provide $2000 at the end of each year for 5 years?

  1. PV = 2000 × = 2000 × = 2000 × 4.3295…
  2. PV = $8658.95. The payments total $10,000; the rest comes from interest earned on the money still invested.
Common mistake: using the annual rate with monthly periods. A 6% per year loan repaid monthly uses r = 0.005 and n in months. Also keep (1 + r)n unrounded until the end — rounding 1.005360 to 6.02 changes a repayment by cents and the total interest by dollars. Another common mistake: forgetting that total interest = (repayment × number of repayments) − amount borrowed. And with declining-balance depreciation, the value never reaches zero, so a question asking "when is it worth less than $5000?" needs logarithms or a table, not a division.

Check yourself

  1. A $12,000 machine depreciates by $1500 per year. Find its value after 5 years.
  2. A $12,000 machine depreciates at 15% per year (declining balance). Find its value after 3 years.
  3. V₀ = 5000 and Vₙ₊₁ = 1.02Vₙ − 300. Find V₁ and V₂.
  4. Find the future value of $1000 deposited at the end of each year for 3 years at 4% per year
  5. $10,000 is borrowed at 12% per year compounded monthly over 12 months. Find the monthly repayment.
  6. Find the total interest paid on the loan in question 5.

Answers: 1. 12,000 − 7500 = $4500 2. 12,000 × 0.853 = $7369.50 3. V₁ = 5100 − 300 = 4800; V₂ = 4896 − 300 = 4596 4. 1000 × = $3121.60 5. M = = $888.49 6. 888.49 × 12 − 10,000 = $661.88

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Last reviewed
1 October 2026
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