Depreciation and Appreciation 💹📉

students, think about a phone, a car, or even a house. Some things become worth more over time, while others lose value. A new laptop may cost a lot today, but after a few years it may sell for much less. A piece of land may do the opposite and rise in value. This lesson explains depreciation and appreciation using the language of Number and Algebra. You will see how these ideas connect to percentages, sequences, exponentials, and financial modelling in IB Mathematics: Applications and Interpretation SL.

What are depreciation and appreciation? 💡

Depreciation means a decrease in value over time. Appreciation means an increase in value over time. These changes are usually expressed as a percentage of the current value, not the original value.

For example, if a car is worth $20000$ and depreciates by $10\%$ in one year, then after one year its value becomes

$$20000(1-0.10)=18000$$

If a house is worth $300000$ and appreciates by $4\%$ in one year, then after one year its value becomes

$$300000(1+0.04)=312000$$

The key idea is that each year’s change is based on the value at the start of that year. This is called compound change because the new value is used again and again.

Important vocabulary

Initial value: the starting value, often written as $V_0$.
Rate of depreciation: the percentage decrease per time period.
Rate of appreciation: the percentage increase per time period.
Multiplier: the factor used each period, such as $0.9$ for $10\%$ depreciation or $1.04$ for $4\%$ appreciation.
Residual value: the value after a certain time.

These ideas are important in real life because very few assets keep the same value forever. A company, government, or family may need to predict future value for planning, budgeting, or investment decisions.

Using algebraic models for change 📈

Depreciation and appreciation are often modeled with exponential expressions. If the initial value is $V_0$ and the rate per period is $r$, then the value after $n$ periods is:

$$V_n = V_0(1+r)^n$$

for appreciation, where $r$ is positive.

For depreciation, the model is:

$$V_n = V_0(1-r)^n$$

where $r$ is the depreciation rate as a decimal.

These formulas are examples of geometric sequences. Each new term is found by multiplying the previous term by the same ratio. For appreciation, the common ratio is $1+r$. For depreciation, the common ratio is $1-r$.

Example 1: car depreciation

Suppose a car costs $25000$ and loses $15\%$ of its value each year. Then

$$V_n = 25000(0.85)^n$$

After $3$ years:

$$V_3 = 25000(0.85)^3$$

$$V_3 = 25000(0.614125)$$

$$V_3 = 15353.125$$

So the car is worth about $15353.13$ after $3$ years.

Notice that the value does not decrease by exactly $3750$ each year. Instead, the decrease gets smaller because the percentage is applied to a smaller amount each time.

Example 2: house appreciation

Suppose a house is worth $180000$ and appreciates by $6\%$ per year. Then

$$V_n = 180000(1.06)^n$$

After $5$ years:

$$V_5 = 180000(1.06)^5$$

$$V_5 \approx 180000(1.3382255776)$$

$$V_5 \approx 240880.61$$

So the house is worth about $240881$ after $5$ years.

This type of model is often realistic when the rate is stable. However, real values can also be affected by inflation, demand, wear and tear, repairs, and market conditions.

Sequences, compound change, and technology 🧮

In IB Mathematics: Applications and Interpretation SL, it is useful to recognize depreciation and appreciation as sequences. If the first term is $V_0$, the next terms are created by repeatedly multiplying by a constant factor.

For depreciation:

$$V_0,\; V_0(1-r),\; V_0(1-r)^2,\; V_0(1-r)^3,\dots$$

For appreciation:

$$V_0,\; V_0(1+r),\; V_0(1+r)^2,\; V_0(1+r)^3,\dots$$

This pattern makes it easy to use a calculator or spreadsheet. Technology can quickly compute many values, create tables, and show graphs.

A graph of $V_n$ against $n$ for depreciation usually slopes downward and gets flatter over time. A graph for appreciation slopes upward and becomes steeper over time. These graphs help students interpret long-term trends.

Example 3: comparing two models

A scooter costs $600$. It depreciates by $20\%$ in the first year, then by $10\%$ in each later year. This is not a constant-rate model, so each year must be calculated separately.

Year 1:

$$600(0.80)=480$$

Year 2:

$$480(0.90)=432$$

Year 3:

$$432(0.90)=388.8$$

If the depreciation rate changes, the model is no longer a simple geometric sequence. This is a reminder that algebraic modelling depends on the situation.

Why technology matters

With a spreadsheet, students can enter the formula $V_n = V_0(1+r)^n$ or $V_n = V_0(1-r)^n$ and fill down many rows. Technology also helps compare different rates, estimate when a value falls below a certain amount, and identify patterns in financial data.

For example, to find when a $10000$ item depreciating by $12\%$ per year falls below $5000$, you can solve

$$10000(0.88)^n < 5000$$

Taking logarithms gives

$$n > \frac{\log(0.5)}{\log(0.88)}$$

This is about

$$n > 5.42$$

So it falls below $5000$ after $6$ years.

Interpreting and solving real-world problems 🔍

Word problems about depreciation and appreciation often test whether you can choose the correct multiplier and understand what the time period means. Always check whether the rate is per year, per month, or per day. A $12\%$ annual rate is very different from a $12\%$ monthly rate.

Strategy for solving problems

Identify the initial value $V_0$.
Decide whether the value is increasing or decreasing.
Convert the percentage rate to a decimal.
Choose the correct formula.
Substitute values carefully.
Round only at the end if needed.

Example 4: monthly appreciation

A piece of digital art is valued at $800$ and appreciates by $3\%$ each month. What is its value after $8$ months?

Use

$$V_n = 800(1.03)^n$$

Then

$$V_8 = 800(1.03)^8$$

$$V_8 \approx 800(1.266770081)$$

$$V_8 \approx 1013.42$$

So its value is about $1013.42$ after $8$ months.

Example 5: solving for the rate

A machine bought for $12000$ is worth $9000$ after $2$ years. If the depreciation rate is constant, find the yearly rate.

Start with

$$9000 = 12000(1-r)^2$$

Divide by $12000$:

$$0.75 = (1-r)^2$$

Take the square root:

$$1-r = \sqrt{0.75}$$

$$1-r \approx 0.8660$$

$$r \approx 0.1340$$

The machine depreciates by about $13.4\%$ per year.

This shows how algebra can be used to work backwards from a final value to the rate.

Connection to Number and Algebra 🧠

Depreciation and appreciation sit at the center of Number and Algebra because they combine several major ideas:

Percentages and ratios: rates are written as percentages but calculated as multipliers.
Sequences: repeated change creates geometric sequences.
Algebraic manipulation: formulas are rearranged to solve for unknown values.
Exponentials: the power $n$ shows repeated multiplication.
Financial modelling: the math is used for assets, investments, inflation, and business planning.

This topic also supports mathematical communication. students must be able to explain why $1.08$ means $8\%$ appreciation and why $0.92$ means $8\%$ depreciation. A strong answer should include units, clear interpretation, and correct rounding.

Depreciation and appreciation are also closely linked to other syllabus ideas such as compound interest, growth models, and data interpretation. The main difference is the context: some problems describe money saved or invested, while others describe the changing value of objects or property.

Conclusion ✅

Depreciation and appreciation describe how values change over time. Depreciation is a decrease in value, while appreciation is an increase. Both are usually modeled with exponential formulas and geometric sequences. In IB Mathematics: Applications and Interpretation SL, students should be able to calculate future value, interpret the meaning of a rate, solve for unknowns, and use technology to explore patterns. This topic is a powerful example of how Number and Algebra helps describe the real world with precision.

Study Notes

Depreciation means a value decreases over time, and appreciation means a value increases over time.
Use the multiplier $1-r$ for depreciation and $1+r$ for appreciation, where $r$ is the rate as a decimal.
The model for depreciation is $V_n = V_0(1-r)^n$.
The model for appreciation is $V_n = V_0(1+r)^n$.
These models form geometric sequences because the value changes by the same ratio each period.
The initial value is $V_0$, and the value after $n$ periods is $V_n$.
Always check the time period, such as years or months, before calculating.
Technology such as calculators and spreadsheets can help generate tables, graphs, and solutions.
To find a rate from two values, rearrange the exponential equation and use logarithms if needed.
Depreciation and appreciation connect Number and Algebra to real-life financial modelling, asset value, and economic interpretation.