Growth and depreciation

A town's population climbing by 5% a year and a machine's value falling by 10% a year look like opposite stories, but underneath they are the same story told with opposite signs. Both are compound interest in disguise — a quantity changing by a fixed percentage of its own current value, year after year, so that growth accelerates while decline decelerates. This article extends the ICSE Class 9 compound interest formula to non-monetary growth and depreciation problems: how one flipped sign turns $A = P(1+r/100)^n$ into a decay formula, why depreciation eats a shrinking rupee amount every year even at a constant rate, and how to read a word problem to know which sign to use.

The core insight

Growth and depreciation both use the compound interest formula $A = P(1 \pm r/100)^n$ applied to any repeatedly-changing quantity, with $+$ for a quantity that increases each year and $-$ for one that decreases.

The core idea

No new formula — just a sign choice. The compound interest formula was built for money, but nothing in its derivation actually requires money: it only requires a quantity that changes by a fixed percentage of its latest value, every fixed time period. A population, a machine's resale value, the number of bacteria in a culture, or an antique's worth all qualify, as long as the percentage rate stays the same from year to year.

Growth uses the familiar form:

A = P\left(1+\dfrac{r}{100}\right)^n

where: $P$ is the initial value, $r\%$ is the annual growth rate, $n$ is the number of years, and $A$ is the value after $n$ years.

Depreciation flips the sign inside the bracket, since the value is shrinking rather than growing:

A = P\left(1-\dfrac{r}{100}\right)^n

Here $r\%$ is the annual rate of depreciation. Each year the value is multiplied by a factor $\left(1-\dfrac{r}{100}\right)$ that is slightly less than 1, instead of the growth factor that is slightly more than 1 — mathematically, this is compound interest run with a negative rate.

Total change is a separate final step. Whichever direction the quantity moves, the total growth or total depreciation is not $A$ itself but the difference between $A$ and $P$ :

\text{Total growth} = A - P \qquad \text{or} \qquad \text{Total depreciation} = P - A

Reading the problem correctly is the actual skill being tested here. The formula never tells you which sign to use — the wording does. Words like "increases," "grows," or "rises" signal $+$ ; words like "depreciates," "decreases in value," "wears out," or "falls" signal $-$ . If a problem gives different rates for different years, the direct formula cannot be used at all, and the year-by-year (step-by-step) method must be applied instead, exactly as with variable-rate compound interest.

Figure

Bar chart comparing a growing population of forty thousand rising to forty four thousand one hundred over two years against a depreciating machine value of fifty thousand rupees falling to thirty six thousand four hundred fifty rupees over three years

Derivation AWhy depreciation loses a shrinking amount each year at a constant rate

The goal is to show, using nothing beyond the compound interest formula and simple arithmetic, that a machine losing a fixed 10% of its value every year loses a smaller number of rupees in each successive year, even though the percentage rate never changes — the property that most confuses students who expect a constant percentage to mean a constant rupee loss.

Step 1. Set up the value at the start of each year. Take $P = 50000$ and $r = 10\%$ , so each year the value is multiplied by $\left(1-\dfrac{10}{100}\right) = 0.9$ . The value at the start of Year 1 is $50000$ .

Step 2. Compute the fall during Year 1. The value at the end of Year 1 is $50000 \times 0.9 = 45000$ , so the fall is $50000 - 45000 = 5000$ .

Step 3. Compute the fall during Year 2 using the reduced base. The value at the end of Year 2 is $45000 \times 0.9 = 40500$ , so the fall during Year 2 is $45000 - 40500 = 4500$ — smaller than Year 1's fall, because 10% is now being taken of a smaller number.

Step 4. Compute the fall during Year 3 the same way. The value at the end of Year 3 is $40500 \times 0.9 = 36450$ , so the fall during Year 3 is $40500 - 36450 = 4050$ — smaller again than Year 2's fall.

Step 5. Read the pattern and check against the direct formula. The yearly falls are $5000, 4500, 4050$ — each one is exactly $90\%$ of the previous fall, because each year's 10% is computed on a progressively shrinking base, the exact mirror image of how compound interest grows faster each year. The final value $36450$ matches $A = P(1-r/100)^n = 50000(0.9)^3 = 36450$ found by the direct formula, confirming the year-by-year method and the formula agree, as they must for a constant rate.

Worked example

A machine is bought for ₹50,000. If its value depreciates at the rate of 10% per annum, find its value after 3 years, and the total depreciation.

Step 1. Identify this as a depreciation problem, so the formula uses a minus sign: $A = P\left(1-\dfrac{r}{100}\right)^n = 50000\left(1-\dfrac{10}{100}\right)^3 = 50000(0.9)^3$ .

Step 2. Evaluate the power: $(0.9)^3 = 0.9 \times 0.9 \times 0.9 = 0.729$ .

Step 3. Multiply to find $A$ : $A = 50000 \times 0.729 = ₹36450$ .

Step 4. Compute the total depreciation as the separate final step: $\text{Total depreciation} = P - A = 50000 - 36450 = ₹13550$ . Answer: the machine's value after 3 years is ₹36,450, and the total depreciation over the 3 years is ₹13,550.

Competitive stretch

The population of a town is 40,000. If it increases at the rate of 5% per annum, find the population after 2 years, and the increase in population.

Step 1. This is growth, so use the plus sign: $A = 40000\left(1+\dfrac{5}{100}\right)^2 = 40000(1.05)^2$ .

Step 2. Evaluate $(1.05)^2 = 1.1025$ , so $A = 40000 \times 1.1025 = 44100$ . Answer: the population after 2 years is 44,100, an increase of $44100 - 40000 = 2100$ people.

Common traps

Using $+$ instead of $-$ for depreciation. This is the single most common error; always check the wording — "increases/grows" versus "depreciates/decreases in value/wears out" — before writing the formula.
Applying a flat percentage of the original value every year, like simple interest, instead of recalculating the percentage on the reduced value each year — depreciation compounds on the latest value, not the original one.
Reporting only the final value $A$ when the question asks for total growth or total depreciation, which needs the extra step $A - P$ or $P - A$ .
Applying one direct formula when the rate changes across years. A problem with different rates for different years must be solved year by year, not with a single $A = P(1 \pm r/100)^n$ .

Ask the tutor: Want to check whether a word problem is describing growth or depreciation, or try a few more population and machine-value questions with the AI tutor? Just ask, and it will walk through the reasoning with you.

How do I know from the wording whether a problem wants growth or depreciation?

Look for the verb describing the change. Words like "increases," "grows," "appreciates," or "rises" mean the quantity is going up, so use the plus sign, $A = P(1+r/100)^n$ . Words like "depreciates," "decreases in value," "wears out," or "falls" mean it is going down, so use the minus sign, $A = P(1-r/100)^n$ . The formula itself never announces which case you are in — only the sentence does, so read it carefully before writing anything.

Why doesn't a machine lose the same rupee amount every year if the depreciation rate is constant?

Because the rate is a percentage of the current value, not the original value, and the current value keeps shrinking. In the worked example, 10% of ₹50,000 is ₹5,000 in Year 1, but by Year 2 the base has already fallen to ₹45,000, so 10% of that is only ₹4,500, and so on. A constant percentage on a shrinking base always produces a shrinking rupee amount — the derivation above shows this step by step and confirms it matches the direct formula.

What if a problem gives a different growth or depreciation rate for each year?

Then the single formula $A = P(1 \pm r/100)^n$ cannot be used, because it assumes the same rate every year. Instead, apply the percentage change one year at a time, using the previous year's answer as the new base for the next year — exactly the year-by-year method used to double-check the direct formula in the derivation above. This is slower but always correct, and it is the only method available when the rate itself is not constant.

Growth and depreciation

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads