7 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.

7.1 What the notation is saying

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

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

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

Percent change compares a change to the original:

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

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.

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

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

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.

7.2 The method

Simple percent of a quantity

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

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

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 \((1 - \frac{p}{100})\).

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:

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

Compound growth

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

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.

Code

{
  const W = 560;
  const H = 160;
  const BAR_Y = 60;
  const BAR_H = 36;
  const LABEL_Y = BAR_Y + BAR_H + 20;
  const MULT_Y = LABEL_Y + 22;
  const PAD_L = 12;
  const PAD_R = 12;

  // Derived values — guard against NaN/zero
  const orig = Number.isFinite(origVal) && origVal > 0 ? origVal : 100;
  const pct  = Number.isFinite(pctChange) ? pctChange : 0;
  const multiplier = 1 + pct / 100;
  const newVal = orig * multiplier;

  // Scale: always show at least max(orig, newVal); min width 1
  const maxVal = Math.max(orig, newVal, 1);
  const usableW = W - PAD_L - PAD_R;
  const scale = usableW / maxVal;

  const origW  = Math.max(orig  * scale, 1);
  const newW   = Math.max(newVal * scale, 1);

  // Colours: increase = teal, decrease = amber, neutral = slate
  const isIncrease = pct > 0;
  const isDecrease = pct < 0;
  const origColour = "#4b5563";  // slate-600
  const newColour  = isIncrease ? "#0d9488"  // teal-600
                   : isDecrease ? "#d97706"  // amber-600
                   : "#4b5563";

  const sign = pct >= 0 ? "+" : "";
  const multSign = multiplier >= 1 ? "×" : "×";

  const multStr = multiplier.toFixed(2);
  const newValStr = Number.isInteger(newVal)
    ? newVal.toString()
    : newVal.toFixed(2);

  const svg = d3.create("svg")
    .attr("viewBox", `0 0 ${W} ${H}`)
    .attr("width", "100%")
    .attr("style", "max-width:" + W + "px; font-family: inherit;");

  // Original bar
  svg.append("rect")
    .attr("x", PAD_L)
    .attr("y", BAR_Y)
    .attr("width", origW)
    .attr("height", BAR_H)
    .attr("fill", origColour)
    .attr("rx", 3);

  // Original label (above bar, left-aligned)
  svg.append("text")
    .attr("x", PAD_L)
    .attr("y", BAR_Y - 6)
    .attr("fill", origColour)
    .attr("font-size", "12px")
    .attr("font-weight", "600")
    .text(`Original: ${orig}`);

  // New value bar (drawn below original as a second row)
  const BAR2_Y = BAR_Y + BAR_H + 10;

  svg.append("rect")
    .attr("x", PAD_L)
    .attr("y", BAR2_Y)
    .attr("width", newW)
    .attr("height", BAR_H)
    .attr("fill", newColour)
    .attr("rx", 3);

  // New value label (above second bar, right of bar if space allows)
  const newLabelX = PAD_L + newW + 6;
  const newLabelAnchor = newLabelX + 120 < W ? "start" : "end";
  const newLabelXAdj   = newLabelAnchor === "end" ? PAD_L + newW - 4 : newLabelX;

  svg.append("text")
    .attr("x", newLabelXAdj)
    .attr("y", BAR2_Y + BAR_H / 2 + 5)
    .attr("fill", newColour)
    .attr("font-size", "13px")
    .attr("font-weight", "700")
    .attr("text-anchor", newLabelAnchor)
    .text(`New value: ${newValStr}`);

  // Multiplier label at bottom
  const changeLabel = isIncrease
    ? `${sign}${pct}% means × ${multStr}  (multiply by ${multStr} in one step)`
    : isDecrease
    ? `${sign}${pct}% means × ${multStr}  (multiply by ${multStr} in one step)`
    : `0% means × 1.00  (no change)`;

  svg.append("text")
    .attr("x", PAD_L)
    .attr("y", BAR2_Y + BAR_H + 20)
    .attr("fill", "#374151")
    .attr("font-size", "12px")
    .text(changeLabel);

  return svg.node();
}

7.3 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.).

\[Q_5 = 2000 \times 1.1877 = £2,375.40\]

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.

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?

New cost: \(25 \times (1 + 0.12) = 25 \times 1.12 = £28\).

Extra per month: \(£28 - £25 = £3\).

Extra over a year: \(£3 \times 12 = £36\).

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.)

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

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.

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.

Code

{
  const W = 560;
  const H = 300;
  const MARGIN = { top: 30, right: 20, bottom: 50, left: 62 };
  const innerW = W - MARGIN.left - MARGIN.right;
  const innerH = H - MARGIN.top - MARGIN.bottom;

  const Q0 = 1000;  // base value — shown as index, labelled clearly
  const r  = (Number.isFinite(annualRate) ? annualRate : 5) / 100;
  const N  = Number.isFinite(numYears) && numYears >= 1 ? Math.round(numYears) : 10;

  // Build data arrays (year 0 to N)
  const years = d3.range(0, N + 1);
  const compound = years.map(n => Q0 * Math.pow(1 + r, n));
  const simple   = years.map(n => Q0 * (1 + r * n));

  const finalCompound = compound[N];
  const finalSimple   = simple[N];
  const extraStr = (finalCompound - finalSimple).toFixed(2);

  const yMax = Math.max(...compound, ...simple) * 1.08;

  const xScale = d3.scaleBand()
    .domain(years)
    .range([0, innerW])
    .padding(0.18);

  const yScale = d3.scaleLinear()
    .domain([0, yMax])
    .range([innerH, 0]);

  const svg = d3.create("svg")
    .attr("viewBox", `0 0 ${W} ${H}`)
    .attr("width", "100%")
    .attr("style", `max-width:${W}px; font-family: inherit;`);

  const g = svg.append("g")
    .attr("transform", `translate(${MARGIN.left},${MARGIN.top})`);

  // Gridlines
  g.append("g")
    .attr("class", "grid")
    .call(
      d3.axisLeft(yScale)
        .ticks(5)
        .tickSize(-innerW)
        .tickFormat("")
    )
    .call(gg => gg.select(".domain").remove())
    .call(gg => gg.selectAll("line").attr("stroke", "#e5e7eb").attr("stroke-dasharray", "3,3"));

  // Compound bars
  g.selectAll(".bar-compound")
    .data(years)
    .enter()
    .append("rect")
      .attr("class", "bar-compound")
      .attr("x", d => xScale(d))
      .attr("y", d => yScale(compound[d]))
      .attr("width", xScale.bandwidth())
      .attr("height", d => innerH - yScale(compound[d]))
      .attr("fill", "#0d9488")
      .attr("rx", 2);

  // Simple interest line
  const lineGen = d3.line()
    .x(d => xScale(d) + xScale.bandwidth() / 2)
    .y(d => yScale(simple[d]));

  g.append("path")
    .datum(years)
    .attr("fill", "none")
    .attr("stroke", "#d97706")
    .attr("stroke-width", 2)
    .attr("stroke-dasharray", "5,4")
    .attr("d", lineGen);

  // Simple interest dots
  g.selectAll(".dot-simple")
    .data(years)
    .enter()
    .append("circle")
      .attr("cx", d => xScale(d) + xScale.bandwidth() / 2)
      .attr("cy", d => yScale(simple[d]))
      .attr("r", 3)
      .attr("fill", "#d97706");

  // X axis
  g.append("g")
    .attr("transform", `translate(0,${innerH})`)
    .call(
      d3.axisBottom(xScale)
        .tickValues(years.filter((_, i) => N <= 10 || i % 2 === 0))
        .tickFormat(d => `Y${d}`)
    )
    .call(gg => gg.select(".domain").attr("stroke", "#9ca3af"))
    .call(gg => gg.selectAll("text").attr("font-size", "11px"));

  // X axis label
  g.append("text")
    .attr("x", innerW / 2)
    .attr("y", innerH + 40)
    .attr("text-anchor", "middle")
    .attr("fill", "#6b7280")
    .attr("font-size", "12px")
    .text("Year");

  // Y axis
  g.append("g")
    .call(d3.axisLeft(yScale).ticks(5).tickFormat(d => `£${d}`))
    .call(gg => gg.select(".domain").attr("stroke", "#9ca3af"))
    .call(gg => gg.selectAll("text").attr("font-size", "11px"));

  // Final value label on last bar
  g.append("text")
    .attr("x", xScale(N) + xScale.bandwidth() / 2)
    .attr("y", yScale(finalCompound) - 6)
    .attr("text-anchor", "middle")
    .attr("fill", "#065f46")
    .attr("font-size", "11px")
    .attr("font-weight", "700")
    .text(`£${finalCompound.toFixed(0)}`);

  // Legend
  const legendY = -18;
  // Compound bar swatch
  g.append("rect")
    .attr("x", 0).attr("y", legendY - 9).attr("width", 12).attr("height", 12)
    .attr("fill", "#0d9488").attr("rx", 2);
  g.append("text")
    .attr("x", 16).attr("y", legendY)
    .attr("fill", "#374151").attr("font-size", "11px")
    .text("Compound growth");

  // Simple line swatch
  g.append("line")
    .attr("x1", 140).attr("y1", legendY - 3)
    .attr("x2", 152).attr("y2", legendY - 3)
    .attr("stroke", "#d97706").attr("stroke-width", 2).attr("stroke-dasharray", "4,3");
  g.append("text")
    .attr("x", 156).attr("y", legendY)
    .attr("fill", "#374151").attr("font-size", "11px")
    .text("Simple interest");

  // Extra label
  g.append("text")
    .attr("x", innerW)
    .attr("y", innerH + 40)
    .attr("text-anchor", "end")
    .attr("fill", "#0d9488")
    .attr("font-size", "12px")
    .attr("font-weight", "600")
    .text(`Compounding adds £${extraStr} extra over ${N} year${N === 1 ? "" : "s"}`);

  return svg.node();
}

7.4 Where this goes

Percent applied once is straightforward proportion. Applied repeatedly, it produces exponential behaviour — and exponents and logarithms (Volume 4, Chapter 1) is where that behaviour is studied properly. Logarithms undo the compounding: if you want to know how long it takes for a quantity to double, you are solving an exponential equation, which requires a logarithm.

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.

7.5 Exercises

A laptop costs £1,149. It is on sale at 15% off. What is the sale price?
A city had 320,000 residents in 2015 and 387,000 in 2025. What was the total percent change over that decade?
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.)
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?
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?
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?
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?