Skip to content

Fundamentals of Business Mathematics and Statistics · Time Value of Money and Annuity - Simple and Compound Interest

Present Value, Future Value and Depreciation Problems

Updated 10 October 2026 · Fact-checked

Future value is what a sum today grows to: FV = P(1 + r)^n. Present value is what a future sum is worth today: PV = A ÷ (1 + r)^n. Depreciation on the reducing balance uses V = C(1 − d)^n. Convert the rate to a decimal, match rate to period, then compute.

Understand Present Value, Future Value and Depreciation

Money today is worth more than the same money later. You can invest it and earn interest. This is the time value of money.

Compounding moves money forward in time. You start with a present sum and find what it becomes after n periods when interest is added to the principal each period. The result is the future value (FV).

Discounting moves money backward in time. You know a sum due in the future and want its worth today. The result is the present value (PV). It is the same compound formula turned upside down, so you divide instead of multiply.

Depreciation and growth use the same idea. When an asset loses a fixed percentage of its current value every year, the value shrinks by the factor (1 − d) each year. This is the reducing balance method. When a population or sales figure grows by a fixed percentage each year, it multiplies by (1 + g) each year.

In every problem, ask three things: is the money moving forward or backward, what is the rate per period, and how many periods are there?

Key formulas to remember

Future value (compound interest)
A = P × (1 + r)^n
P is the present sum, r is the rate per period as a decimal, n is the number of periods.
Present value
P = A ÷ (1 + r)^n
Discounting. The present value is always less than the future sum when r > 0.
Compounding more than once a year
A = P × (1 + r/m)^(m × n)
r is the annual rate, m is the compounding periods per year (half-yearly 2, quarterly 4). Use rate r/m and m × n periods.
Depreciation (reducing balance)
V = C × (1 − d)^n
C is the cost, d is the yearly rate as a decimal, V is the value after n years. Total depreciation = C − V.
Growth at a fixed rate
Pₙ = P₀ × (1 + g)^n
Used for population, sales, or output growing at g per period.
Straight-line depreciation (for contrast)
Yearly depreciation = (Cost − Scrap value) ÷ n
This is a fixed amount each year. It is not the compound formula.

How to solve Present Value, Future Value and Depreciation questions

Use this method for any question on present value, future value, growth or depreciation.

  1. 1Read what is asked: future value, present value, value after depreciation, or total depreciation.
  2. 2Decide the direction. Forward in time means multiply by (1 + r)^n. Backward means divide by (1 + r)^n. Falling value means use (1 − d)^n.
  3. 3Convert the rate to a decimal (10% = 0.10) and make it a rate per period.
  4. 4If interest is compounded half-yearly or quarterly, divide the annual rate by 2 or 4 and multiply the years by 2 or 4.
  5. 5Calculate the factor, such as 1.1³ = 1.331, step by step.
  6. 6Multiply or divide the sum by the factor.
  7. 7Check the answer for sense: FV should exceed PV, and a depreciated value should be below cost.
  8. 8If the question asks for depreciation, subtract: cost − value after n years.

Quickest way: Memorise common factors and eliminate by direction

When to use it: Use it in MCQs where rates are simple, such as 5%, 10%, 20% and periods of 2 or 3 years.

  1. Learn the common factors: 1.1² = 1.21, 1.1³ = 1.331, 1.2² = 1.44, 1.05² = 1.1025, 0.9² = 0.81, 0.9³ = 0.729, 0.8² = 0.64, 0.8³ = 0.512.
  2. Look at the options first. For a future value or growth question, drop any option below the starting sum. For PV or depreciation, drop any option above the given sum.
  3. Check whether the sum is a neat multiple of the factor. For example, 1,21,000 ÷ 1.21 divides cleanly to 1,00,000.
  4. Multiply or divide once, and pick the matching option. There is no negative marking, so always mark an answer.

Common mistakes in Present Value, Future Value and Depreciation

  • Putting r = 10 instead of 0.10 in the formula.

    The rate is written as a percentage and students copy it directly.

    Fix: Always write 10% as 0.10 before bringing it into (1 + r).

  • Multiplying instead of dividing when finding present value.

    Students remember only A = P(1 + r)^n and apply it in both directions.

    Fix: Ask which way the money moves. A future sum brought to today is divided by (1 + r)^n.

  • Not adjusting for half-yearly or quarterly compounding.

    The annual rate and the number of years are used as given.

    Fix: Divide the rate by m and multiply the years by m. 12% for 2 years compounded half-yearly is 6% for 4 periods.

  • Calculating depreciation as C × d × n.

    Students mix reducing balance with straight-line or simple interest.

    Fix: Under reducing balance, each year's loss is on the previous year's value. Use C(1 − d)^n.

  • Giving the value after depreciation when the question asks for the depreciation amount, or the reverse.

    Students stop at the formula result without reading the last line of the question.

    Fix: Underline what is asked. Depreciation = Cost − V. The book value is V.

  • Mistakes in multiplying out powers such as 1.1³.

    Students rush and square or cube the digits incorrectly.

    Fix: Multiply one step at a time: 1.1 × 1.1 = 1.21, then 1.21 × 1.1 = 1.331.

Worked examples

Example 1

What is the present value of ₹1,21,000 due after 2 years, if money earns 10% per year compounded annually?

Show the solution
  1. The money moves backward in time, so divide: P = A ÷ (1 + r)^n.
  2. A = ₹1,21,000, r = 0.10, n = 2.
  3. (1 + r)^n = 1.1² = 1.21.
  4. P = 1,21,000 ÷ 1.21 = 1,00,000.
  5. Check: 1,00,000 × 1.21 = 1,21,000, and the present value is less than the future sum.

Answer: ₹1,00,000

Example 2

A machine bought for ₹5,00,000 depreciates at 20% per year on the reducing balance. Find its value after 3 years and the total depreciation.

Show the solution
  1. Use V = C × (1 − d)^n with C = 5,00,000, d = 0.20, n = 3.
  2. 1 − d = 0.8, and 0.8³ = 0.8 × 0.8 × 0.8 = 0.512.
  3. V = 5,00,000 × 0.512 = 2,56,000.
  4. Total depreciation = 5,00,000 − 2,56,000 = 2,44,000.
  5. Check: the value is below the cost. Straight 20% a year on the original cost would wrongly give 3,00,000 of depreciation.

Answer: Value after 3 years = ₹2,56,000; total depreciation = ₹2,44,000

Exam tips

  • Look at the wording. Words like 'worth today', 'discounted' or 'due after' signal present value. Words like 'amounts to' signal future value.
  • Questions use friendly rates and periods, so learn the common factors and save calculation time.
  • Check the compounding frequency in the question before you start. Half-yearly and quarterly cases are a common trap.
  • For 'depreciates' or 'decreases' questions, the answer is always below the starting figure. Use this to eliminate options.
  • If the question gives the future value and the present value, you may be asked for r or n. Compute the ratio A ÷ P and match it to a known factor.

Practice questions from Time Value of Money and Annuity - Simple and Compound Interest

Present Value, Future Value and Depreciation in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Present Value, Future Value and Depreciation: frequently asked questions

What is the formula for present value and future value?

Future value is A = P(1 + r)^n, where r is the rate per period as a decimal and n is the number of periods. Present value is P = A ÷ (1 + r)^n. The second is the first rearranged.

How do I calculate present value of a sum?

Convert the rate to a decimal, find (1 + r)^n for the given periods, and divide the future sum by it. If interest is compounded half-yearly or quarterly, adjust the rate and the number of periods first.

How is depreciation linked to compound interest?

Under the reducing balance method, the asset loses a fixed percentage of its current value each year. This is the compound formula with a negative rate, so V = C(1 − d)^n.

Is the compound formula the same for growth problems?

Yes. For a quantity growing at a fixed percentage each period, use Pₙ = P₀(1 + g)^n. Only the meaning changes, such as population or sales instead of money.