Hard GCSE Iterative growth processes Questions

Write the iterative model

The same annual multiplier acts every year.

Substitute the two-year data

$P = 5000$, $A = 7200$ and $n = 2$.

Isolate the power of r

The two-year multiplier is $1.44$.

Take the square root

The nth root with $n = 2$ is the square root; $1.2^{2} = 1.44$ exactly.

Interpret the rate

The colony grows by exactly $20\%$ each year.

Round-trip check the rate

The recovered rate reproduces the given two-year figure exactly.

Reject the divide-by-n error

Halving the total $44\%$ rise gives $22\%$, which is too big — growth compounds.

Set up the target inequality

Measure $n$ from the ORIGINAL start, not from the two-year point.

Divide by the starting value

The colony must become more than four times its original size.

Estimate n with logarithms

Taking logs brings the exponent down.

Round up and prepare to test

A non-integer estimate must be rounded up for a "more than" target, then checked on both sides.

Test $n = 7$

After $7$ years there are about $17\,916$ bacteria — still below $20\,000$.

Test $n = 8$

After $8$ years there are about $21\,499$ bacteria, which exceeds $20\,000$.

Confirm the crossing

Year $7$ is below the target and year $8$ is above it, so $8$ complete years are needed.

State the answer

Counting from the start, the colony first exceeds $20\,000$ after $8$ complete years.

Write the hourly multiplier

Losing $30\%$ each hour leaves $70\%$ of the mass.

Write the iterative model

Here $t$ is measured in hours and need not be a whole number.

Write the half-life condition

The half-life is the time taken for the mass to fall to exactly half.

Cancel the starting mass

The half-life does not depend on how much you start with.

Take logarithms of both sides

The log of a power brings the exponent down as a coefficient.

Rearrange for t

Divide by $\log 0.7$. Both logs are negative, so $t$ comes out positive.

Evaluate

Keep the full display before rounding.

Round to 2 decimal places

The third decimal is $3$, so round down.

Check the answer is sensible

After $1$ hour $70\%$ remains and after $2$ hours only $49\%$, so the half-life must lie between $1$ and $2$ hours — and just under $2$.

Round-trip check

Raising $0.7$ to the unrounded half-life returns exactly $0.5$.

Check the rounded value

The rounded $1.94$ hours leaves $50.06\%$ — very close to half, as expected from a $2$ d.p. answer.

Reject the divide-into-50 error

That assumes a constant loss of $30\%$ of the ORIGINAL mass per hour; in fact each hour removes $30\%$ of a smaller amount.

Test the wrong answer

After $1.67$ hours over $55\%$ still remains, so that time is too short.

Write the general rule

This converts any decay multiplier into its half-life.

State the answer

The half-life of the sample is about $1.94$ hours.

Write the exact model

One exact evaluation of the power gives $19\,487.171$, which rounds to $19\,487$.

Look at the first rounding

Year $1$ happens to be exact, so no error yet.

Follow the exact chain

The true running values are not whole numbers from year $5$ onwards.

See where rounding first bites

Rounding to a whole person at year $5$ throws away $0.1$ of a person.

Follow the rounded chain forwards

The rounded year-5 figure is now the INPUT to year $6$, and $17\,715.5$ rounds up to $17\,716$ — the error has changed sign and grown.

Compare the two answers

The year-by-year rounded chain finishes at $19\,488$, one person above the exact answer of $19\,487$.

Identify the cause

The error is not just displayed — it becomes the input to the next step and is multiplied again.

State the correct method

Evaluate the whole power at full precision and round only the final answer.

Check the correct method

A single exact calculation avoids all intermediate rounding.

Note when it matters most

The more steps there are, the more roundings compound — and near a boundary the drift can flip a "first year above" answer.

Rule out a different explanation

The formula itself is right; only the arithmetic method is at fault.

Rule out the multiplier being wrong

A $10\%$ rise really is a multiplier of $1.1$.

Rule out the exponent being wrong

Seven years means seven applications of the multiplier.

Summarise the principle

Every displayed intermediate value should be a rounding of the true running value, never a re-multiplication of a rounded one.

State the explanation

The year-by-year rounding introduces an error that is then multiplied again in every later year.

Write the iterative model

A $7\%$ hourly rise gives a multiplier of $1.07$.

Set up the inequality

The colony must exceed $2000$ bacteria.

Divide by the starting value

The colony must become more than four times its original size.

Estimate n with logarithms

Taking logs brings $n$ down from the exponent.

Round the estimate up

The count of complete hours must be a whole number, and the estimate must be checked on both sides.

Evaluate the power for $n = 20$

The exact twenty-hour multiplier is just under $4$.

Test $n = 20$

After $20$ hours there are $1934$ bacteria — still below $2000$.

Evaluate the power for $n = 21$

The exact twenty-one-hour multiplier is above $4$.

Test $n = 21$

After $21$ hours there are $2070$ bacteria, which exceeds $2000$.

Confirm the crossing

Hour $20$ is below the target and hour $21$ is above it, so hour $21$ is the first to exceed $2000$.

Check the doubling structure

The colony doubles about every $10.25$ hours, so it needs about $20.5$ hours to quadruple — consistent with the answer.

Reject the divide-by-rate error

Adding $7\%$ of the ORIGINAL size each hour would take about $43$ hours to reach $2000$; compounding gets there in half that.

Reject the off-by-one error

Rounding the estimate $20.49$ down to $20$ would give a colony of $1934$, which fails the condition.

Note the exact value at the answer

At the answer hour the colony is comfortably past the target.

State the answer

The colony first exceeds $2000$ bacteria after $21$ complete hours.

Write the quarterly multiplier

A $4\%$ rise each quarter multiplies the concentration by $1.04$.

Count the quarters in a year

There are four $3$-month quarters in a year.

Combine the four quarters

The annual multiplier is the quarterly multiplier raised to the power $4$.

Evaluate the power exactly

This is exact — evaluate the power in one go rather than multiplying rounded values four times.

Subtract 1

The excess over $1$ is the proportional annual increase.

Convert to a percentage

Multiply by $100$.

Round to 2 decimal places

The third decimal is $5$ followed by more digits, so round up from $16.98$ to $16.99$.

Reject the multiply-by-4 error

Multiplying the quarterly rate by $4$ ignores the growth on growth within the year.

Measure the compounding effect

Compounding within the year adds almost a whole percentage point.

Check with a concrete value

A concentration of $1000$ units rises to $1169.86$ units in a year.

Check against the naive figure

The naive $16\%$ figure understates the true concentration by nearly $10$ units.

Show the danger of rounding early

Rounding at each quarter accumulates error; only the exact power gives the right final value.

Recover the quarterly rate from the annual one

The fourth root of the annual multiplier returns the quarterly multiplier exactly, confirming the two rates are equivalent.

Note the general rule

This converts any rate per period into the equivalent rate for a longer period.

State the answer

The equivalent annual increase is $16.99\%$.