Chapter 4 of 7

All 7 chapters
  1. 1Percentage increase and decrease, readunderstood
  2. 2Finding the original amount, readunderstood
  3. 3Simple and compound interest, readunderstood
  4. 4Growth and decay, readunderstood
  5. 5Harder ratio problems, readunderstood
  6. 6Proportion as a formula, readunderstood
  7. 7Graphs of proportion, readunderstood

M4.2.4 Chapter 4

Growth and decay

The fish in a lake fall by 15% every year. Will the lake ever have no fish at all?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Use the multiplier 0.85 for a 15% fall each year
  • Model a goat herd that grows by 10% a year

Ana studies a lake. It has 8 000 fish. Each year the number falls by 15%. How many fish are left after 3 years?

A fall of 15% leaves 85% of the fish. As a decimal that is 0.85. Each year the lake keeps 0.85 of what it had. So each year you multiply by 0.85 again.

Draw it first

Watch the fish fall

  1. A graph with the x axis time (years) and the y axis fish (thousands). Step 1 shows one dot at year 0, at 8. Step 2 adds dots at years 1, 2 and 3, each lower than the one before. Step 3 draws a curve named fish through them, falling fast at first and then more slowly.123456123456780time (years)fish (thousands)

    1 At the start the lake has 8 000 fish, which is 8 on this scale of thousands.

  2. A graph with the x axis time (years) and the y axis fish (thousands). Step 1 shows one dot at year 0, at 8. Step 2 adds dots at years 1, 2 and 3, each lower than the one before. Step 3 draws a curve named fish through them, falling fast at first and then more slowly.123456123456780time (years)fish (thousands)

    2 After 1, 2 and 3 years the dots are lower each time. The drops get smaller, because 15% of a smaller number is smaller.

  3. A graph with the x axis time (years) and the y axis fish (thousands). Step 1 shows one dot at year 0, at 8. Step 2 adds dots at years 1, 2 and 3, each lower than the one before. Step 3 draws a curve named fish through them, falling fast at first and then more slowly.fish123456123456780time (years)fish (thousands)

    3 The curve joins the dots. It falls fast at first, then more slowly, and it stays above zero.

A quantity that falls by the same percent every year is in decay. A quantity that rises by the same percent every year is in growth. Both are described by a model: a rule that says how the amount changes.

Worked example

A lake has 8 000 fish. The number falls by 15% each year. How many fish are there after 3 years?

  1. A fall of 15% leaves 85%, so the yearly multiplier is 0.85.
  2. After year 1: 8 000 × 0.85 = 6 800.
  3. After year 2: 6 800 × 0.85 = 5 780.
  4. After year 3: 5 780 × 0.85 = 4 913.

Answer: 4 913 fish

A table with the head Year, Fish. Rows: 0, 8 000; 1, 6 800; 2, 5 780; 3, 4 913; 4, 4 176; 5, 3 550; 6, 3 017.YearFish08 00016 80025 78034 91344 17653 55063 017
Each row is 8 000 × 0.85 once for every year, rounded to the nearest whole fish. The fall is 1 200 in year 1 but only 533 in year 6.

Growth works the same way, with a multiplier above 1. A herd of 100 goats grows by 10% each year. The yearly multiplier is 1 + 0.1 = 1.1.

A graph with the x axis time (years) and the y axis goats. A curve named herd starts at 100 and bends upward. A dot at year 5, named (5, 161), has dashed lines down to the x axis and across to the y axis.(5, 161)herd24681040801201602002400time (years)goats
Growth bends upward. Each year the herd gains more than the year before.

Worked example

A herd of 100 goats grows by 10% each year. How many goats are there after 5 years, to the nearest goat?

  1. A rise of 10% gives a yearly multiplier of 1.1.
  2. Five years means multiplying by 1.1 five times: 1.1⁵ = 1.61051.
  3. The herd is 100 × 1.61051 = 161.051.
  4. Goats are whole animals, so round to the nearest goat: 161.

Answer: 161 goats

The model gives parts of a goat. We compare the model's numbers with 150 and only round at the end.

Worked example

The same herd starts at 100 goats. In which year does it first pass 150 goats?

  1. Year 3 is below 150: 100 × 1.1³ = 133.1.
  2. Year 4 is below 150: 133.1 × 1.1 = 146.41.
  3. Year 5 is above 150: 146.41 × 1.1 = 161.051.
  4. So the herd first passes 150 goats in year 5. In year 5 the model gives 161.051, which is about 161 goats.

Answer: Year 5

See it three ways

With things

A herd of 100 goats. After one year add a tenth: 110. After year two add a tenth of that: 121.

As a picture

A box 100 joins to a box 110 by an arrow labelled × 1.1. That box joins to a box 121 by an arrow labelled × 1.1.100× 1.1110× 1.1121

In symbols

100 × 1.1 × 1.1 = 121.

Try it

Tear a strip of paper in half

You need: a strip of paper.

  1. Let the whole strip stand for 100%. Tear it in half and keep one piece.
  2. Tear the piece in half again and keep one piece. Do this three times in all.

What you will see: The pieces are 50%, 25% and 12.5% of the strip. Each tear multiplies by 0.5, a fall of 50%. In the model you can tear for ever and never reach nothing. A real strip becomes too small to tear.

If you cannot try it, this is what happens.
Four bars. The first is labelled Whole strip, with the label 100%. The second is labelled After 1 tear, with the label 50%. The third is labelled After 2 tears, with the label 25%. The fourth is labelled After 3 tears, with the label 12.5%. Each is half as long as the one before.Whole strip100%After 1 tear50%After 2 tears25%After 3 tears12.5%

A wrong idea many people have

A fall of 15% a year takes about 7 years to wipe out the fish, because 7 × 15% is about 100%.

Why it feels right: Seven falls of 15 take off about 100. That is right for 15 fish a year, but not for 15 percent.

The test that breaks it: A fall of 15% takes 15% of what is left, which gets smaller every year. After 7 years the lake still has about 2 565 of its 8 000 fish, which is 32%.

What is true: Multiply by 0.85 once for every year: 0.85 × 0.85 and so on. The number keeps falling but never reaches zero.

A wrong idea many people have

A rise of 10% a year for 2 years is a rise of 20%.

Why it feels right: Two lots of 10% add up to 20%, and the sum looks like the rule for simple interest.

The test that breaks it: Work it out for 100 goats. After 2 years the herd is 121, not 120. The second rise is 10% of the bigger herd.

What is true: Multiply the multipliers: 1.1 × 1.1 = 1.21. That is a rise of 21%.

Key words

decay
a fall by the same percent every period. The multiplier is below 1.
growth
a rise by the same percent every period. The multiplier is above 1.
model
a rule that says how an amount changes, such as multiplying by 0.85 every year.

Back to your guess

The answer is B: No. Each year it keeps 85% of the fish, so some always remain.

Each year the lake keeps 85% of the fish it had. Multiplying a number above zero by 0.85 never gives zero. In the model the fish get fewer and fewer, but never none. A real lake has whole fish, so the model is a guide while there are many.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. Which multiplier gives a fall of 8% each year?

    Tap your answer.

    1. Not this one. This is 8% of the amount, which is the size of the fall. The amount that is left is 100% − 8% = 92%. Go to M4.2.1. Try another.

    2. Not this one. This is the multiplier for a rise of 8%. A fall uses a multiplier below 1. Go to M4.2.1. Try another.

    3. Not this one. This is 80%, a fall of 20%. Read the decimal carefully: 8% leaves 92%, which is 0.92. Go to M4.2.1. Try another.

    4. Yes, that is right. A fall of 8% leaves 100% − 8% = 92%, which is 0.92 as a decimal.

    See the answer

    0.92

    A fall of 8% leaves 100% − 8% = 92%, which is 0.92 as a decimal.

    • If you chose A: This is 8% of the amount, which is the size of the fall. The amount that is left is 100% − 8% = 92%. Go to M4.2.1.
    • If you chose B: This is the multiplier for a rise of 8%. A fall uses a multiplier below 1. Go to M4.2.1.
    • If you chose C: This is 80%, a fall of 20%. Read the decimal carefully: 8% leaves 92%, which is 0.92. Go to M4.2.1.

  2. A herd of 200 goats grows by 10% each year. How many goats are there after 2 years?

    Tap your answer.

    1. Yes, that is right. 200 × 1.1 = 220 after one year, and 220 × 1.1 = 242 after two.

    2. Not this one. You added 10% of 200 twice. The second rise is 10% of the new herd, 220, not of 200. Go to M4.2.4. Try another.

    3. Not this one. This is the herd after one year. The question asks for two years, so multiply by 1.1 again. Go to M4.2.4. Try another.

    See the answer

    242

    200 × 1.1 = 220 after one year, and 220 × 1.1 = 242 after two.

    • If you chose B: You added 10% of 200 twice. The second rise is 10% of the new herd, 220, not of 200. Go to M4.2.4.
    • If you chose C: This is the herd after one year. The question asks for two years, so multiply by 1.1 again. Go to M4.2.4.

  3. An amount of 500 falls by 20% each year. What is it after 2 years?

    Tap your answer.

    1. Not this one. You took 20% of 500 off twice. The second fall is 20% of what is left after year 1, which is only 400. Go to M4.2.4. Try another.

    2. Not this one. This is the amount after one year. The question asks for two years, so multiply by 0.8 again. Go to M4.2.4. Try another.

    3. Yes, that is right. 500 × 0.8 = 400 after one year, and 400 × 0.8 = 320 after two.

    See the answer

    320

    500 × 0.8 = 400 after one year, and 400 × 0.8 = 320 after two.

    • If you chose A: You took 20% of 500 off twice. The second fall is 20% of what is left after year 1, which is only 400. Go to M4.2.4.
    • If you chose B: This is the amount after one year. The question asks for two years, so multiply by 0.8 again. Go to M4.2.4.

  4. The graph shows frogs in a pond. The number falls by 10% each year from 600. About how many frogs are there at point A, in year 4?

    A graph with the x axis time (years) and the y axis frogs. A falling curve named frogs starts at 600 and bends flat. A dot named A at year 4 has dashed lines down to the x axis and across to the y axis.Afrogs24681002003004005006000time (years)frogs
    Tap your answer.

    1. Not this one. You took 10% of 600 off four times, which is 60 each time. Each fall is 10% of what is left, so the falls get smaller. Go to M4.2.4. Try another.

    2. Yes, that is right. The multiplier is 0.9. Four years: 600 × 0.9⁴ = 393.66, which is about 394 frogs.

    3. Not this one. This is the number after one year. Point A is in year 4, after four falls. Go to M4.2.4. Try another.

    4. Not this one. This is the number after three years. Point A is in year 4, so multiply by 0.9 once more. Go to M4.2.4. Try another.

    See the answer

    394

    The multiplier is 0.9. Four years: 600 × 0.9⁴ = 393.66, which is about 394 frogs.

    • If you chose A: You took 10% of 600 off four times, which is 60 each time. Each fall is 10% of what is left, so the falls get smaller. Go to M4.2.4.
    • If you chose C: This is the number after one year. Point A is in year 4, after four falls. Go to M4.2.4.
    • If you chose D: This is the number after three years. Point A is in year 4, so multiply by 0.9 once more. Go to M4.2.4.

  5. Fish in a lake fall by 15% each year in the model. Will the number of fish ever reach exactly zero?

    Tap your answer.

    1. Yes, that is right. Each year the number is multiplied by 0.85. A number above zero multiplied by 0.85 is still above zero. So the model never reaches zero.

    2. Not this one. Seven falls of 15% do not add up to 100%. Each fall is 15% of what is left, which is smaller each year. After 7 years about a third of the fish remain. Go to M4.2.4. Try another.

    3. Not this one. There is no year when the multiplier 0.85 makes the number zero. A number above zero times 0.85 is still above zero. Go to M4.2.4. Try another.

    See the answer

    No. It gets smaller each year but never reaches zero.

    Each year the number is multiplied by 0.85. A number above zero multiplied by 0.85 is still above zero. So the model never reaches zero.

    • If you chose B: Seven falls of 15% do not add up to 100%. Each fall is 15% of what is left, which is smaller each year. After 7 years about a third of the fish remain. Go to M4.2.4.
    • If you chose C: There is no year when the multiplier 0.85 makes the number zero. A number above zero times 0.85 is still above zero. Go to M4.2.4.

  6. From an earlier chapter

    Work out the simple interest on 2 500 coins at 4% a year for 3 years.

    See the answer

    300

    One year: 4% of 2 500 = 100. For 3 years: 3 × 100 = 300.

  7. A herd of 300 goats grows by 10% each year. How many goats are there after 2 years?

    See the answer

    363

    300 × 1.1 = 330 after one year, and 330 × 1.1 = 363 after two.

  8. The number of fish falls by 20% each year from 1 000. How many fish are there after 3 years?

    See the answer

    512

    The multiplier is 0.8. 1 000 × 0.8 = 800, then 800 × 0.8 = 640, then 640 × 0.8 = 512.

Keep it

Copy this into your exercise book.

A fall of p% a year multiplies by 1 − p/100 every year. A rise multiplies by 1 + p/100. Decay stays above zero. Growth bends upward.

Draw: A falling curve that flattens and a rising curve that steepens, each with its multiplier written beside it.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

No internet tomorrow? Save the whole book as one file (743 KB). It opens on any phone.