6  Percent and rates

Expressing change as a fraction of something

Your phone battery drains 3% per hour on standby. A savings account pays 4.5% annual interest. A streaming service puts its prices up by 8%. Your favourite team’s win rate this season is 60%.

All four are the same operation: a fraction expressed out of a hundred, applied to a quantity. The difference is what the quantity is, and whether the percent is applied once or repeatedly.

6.1 What this chapter helps you do

Symbols to keep handy

These are the bits of notation you'll see a lot. If a line of symbols feels like a fence, read it out loud once, then keep going.

  • (1 + r)^n: a quantity grown at rate r for n periods

  • r: rate of change per period

Here is the main move this chapter is making, in plain terms. You do not need to be fast. You just need to keep the thread.

  • Coming in: You can express any two quantities as a ratio. You know what it means for a number to be a fraction of a whole.

  • Leaving with: Percent is ratio with a fixed denominator of 100. A rate is a ratio of two different units. Both express change or comparison in a form that makes different-sized things comparable. Compound growth and decay are percent applied repeatedly.

6.2 What the notation is saying

Start with a familiar example. If 25 students out of 100 passed a test, we say 25%. That means 25 out of every 100.

Percent means per hundred. p\% is the same as \frac{p}{100}.

This means percent is just a special kind of fraction.

To find p\% of a quantity Q: multiply Q by \frac{p}{100}.

\text{amount} = Q \times \frac{p}{100}

Here, Q is the starting quantity and p is the percent. This means the formula finds the part of the whole named by the percent.

Percent change compares a change to the original:

\text{percent change} = \frac{\text{new} - \text{original}}{\text{original}} \times 100

This means we compare the change to where we started, not just the size of the change by itself.

A rate is a ratio with different units: speed is km/h, data use is MB/day, interest is $/year. The structure is the same as percent — one quantity compared to another.

This means a rate tells us “how much for each unit of time, distance, mass, or something else.”

Compound growth applies a rate repeatedly. After n periods at rate r (expressed as a decimal), a quantity Q_0 becomes:

Q_n = Q_0 \times (1 + r)^n

Here, Q_0 is the starting amount, r is the rate written as a decimal, and n is the number of periods.

This means we multiply by the same growth factor again and again.

Compound decay uses 1 - r instead:

Q_n = Q_0 \times (1 - r)^n

The exponent n is why compound growth and decay are not linear — they accelerate in a way that simple percent change does not.

This means the amount changes on top of the new amount each time, not just the original amount.

6.3 The method

Simple percent of a quantity

p\% of Q = Q \times \frac{p}{100}

Convert percent to decimal first: 7\% = 0.07; 120\% = 1.20.

This means “of” in percent problems usually tells you to multiply.

Percent increase / decrease

To increase Q by p\%: multiply by (1 + \frac{p}{100}) — one step does the same job as finding the percent and then adding it.

To decrease Q by p\%: multiply by (1 - \frac{p}{100}).

Here, (1 + \frac{p}{100}) and (1 - \frac{p}{100}) are called multipliers. This means one multiplier can do the whole change in a single step.

Finding the original after a percent change

If after a p\% increase the value is V, work backwards by dividing by the multiplier rather than multiplying:

Q_0 = \frac{V}{1 + \frac{p}{100}}

This means reversing a percent increase uses division, not subtraction.

Compound growth

Q_n = Q_0 \times (1 + r)^n where r = \frac{p}{100}.

This means we are applying the same percent again and again over time.

Why this works

Each compounding period, the new starting point is the previous result. Multiplying by (1+r) once applies the rate for one period. Applying it again gives (1+r)^2 — the rate is now applied to the already-grown quantity. The exponent is just repeated multiplication. This is why compound growth eventually overtakes any flat trend: the base grows, so the next period’s growth is larger than the last.


Interactive: Percent change visualiser. Adjust the original value and the percent change. The bar shows the result and the single-step multiplier used.


6.4 Worked examples

Example 1 — Savings: compound interest. You put $2,000 into a savings account at 3.5% annual interest, compounded every year. What is the balance after 5 years?

Set up the compound formula with Q_0 = 2000, r = 0.035, n = 5:

Q_5 = 2000 \times (1.035)^5

Work out (1.035)^5 — that is 1.035 multiplied by itself five times. Using a calculator: (1.035)^5 = 1.1877 (to 4 d.p.).

Q_5 = 2000 \times 1.1877 = 2375.40

So after 5 years, the value is $2,375.40.

Compare to simple interest: 5 \times 3.5\% \times 2000 = 350, giving $2,350. The compounded version is $25.40 more — the gap grows larger the longer you leave it.

This means compound interest pays interest on earlier interest as well.


Example 2 — Phone plan. A mobile plan costs $25 per month. The provider raises prices by 12%. What is the new monthly cost? How much extra do you pay over a year?

Use the increase multiplier:

25 \times (1 + 0.12) = 25 \times 1.12 = \$28

Extra per month:

28 - 25 = \$3

Extra over a year:

3 \times 12 = \$36

This means a small monthly increase can become a bigger yearly cost.


Example 3 — Geography: population. A city has a population of 850,000. The annual growth rate is 1.4%. What is the projected population in 10 years?

P_{10} = 850{,}000 \times (1.014)^{10}

(1.014)^{10} = 1.1495 (to 4 d.p.)

P_{10} = 850{,}000 \times 1.1495 \approx 977{,}000

Here, P_{10} means the population after 10 years. This means the city grows by the same percentage each year, but the number added each year gets larger.


Example 4 — Science: radioactive decay. A sample contains 5,000 atoms of a radioactive isotope that decays at 12% per hour. How many atoms remain after 4 hours?

This is compound decay — each hour, 12% is lost, so 88% remains.

N_4 = 5000 \times (1 - 0.12)^4 = 5000 \times (0.88)^4

(0.88)^4 = 0.5997

N_4 = 5000 \times 0.5997 \approx 2,999 \text{ atoms}

In 4 hours, roughly 40% of the atoms have decayed.

This means decay is the same structure as growth, but the multiplier is less than 1.


Interactive: Compound growth explorer. Adjust the annual rate and number of years. The bars show the compounded value year by year. The dashed line shows what simple interest would give over the same period.


6.5 Where this goes

This chapter builds directly on ratio and proportion. First we used comparisons. Now we use a comparison out of 100.

When percent is applied once, it is still a proportion problem. When it is applied again and again, it becomes exponential growth or decay. Exponents and logarithms (Volume 4, Chapter 1) studies that repeated growth more carefully.

Statistics and probability (Chapter 6) uses rates constantly: probability is a rate between 0 and 1, and relative frequency is a percent.

In finance, virtually every calculation — present value, future value, yield, annuity — is a variant of the compound formula. In engineering, efficiency, signal attenuation, and structural safety factors are all percent-based ratios. The formula here is simple. The consequences are not.

Where this shows up

  • A savings account paying 4% a year is using the compound formula — the interest each year is calculated on the new, larger balance.
  • A streaming service raising prices by 8% is applying a percent increase to a base price.
  • A sports commentator quoting a 65% win rate is expressing a fraction as a percent.
  • A phone showing “battery 23%” is telling you what fraction of capacity remains.

The arithmetic is identical. The stakes vary.

6.6 Exercises

  1. A laptop costs $1,149. It is on sale at 15% off. What is the sale price?

  2. A city had 320,000 residents in 2015 and 387,000 in 2025. What was the total percent change over that decade?

  3. An investment grows from $4,500 to $5,200 over 3 years. What was the approximate annual growth rate? (Estimate; you do not yet need logarithms — try 3–4 percent and refine.)

  4. A water tank contains 12,000 litres. It leaks at a rate of 5% of its current volume per day. How much water remains after 1 week?

  5. A car’s drivetrain transfers engine power to the wheels with 87% efficiency. If the engine produces 110 kW, what power reaches the wheels? What percentage is lost?

  6. A country’s annual inflation rate is 2.8%. A loaf of bread costs $1.40 today. What will it cost in 8 years at this rate?

  7. A business owner offers a 10% early-payment discount. A customer receives an invoice for $3,400. They pay early. How much do they pay? What percent of the original price was the discount in pounds?