GCSE Iterative growth processes Practice Questions

Start from 100 percent

An increase of $8\%$ leaves the original $100\%$ plus another $8\%$.

Convert the percentage to a decimal

Divide by $100$ to turn a percentage into a multiplier.

State the multiplier

Each step multiplies the quantity by $1.08$, so after $n$ steps the model is $A = P \times 1.08^{n}$.

Write the multiplier

Losing $30\%$ leaves $70\%$, so multiply by $0.7$ each week.

Apply it for two weeks

Two weeks means the multiplier is applied twice: $0.7^{2} = 0.49$.

State the area

The bloom covers $980\text{ m}^2$ after $2$ weeks.

Model the two years properly

Each year multiplies by $0.9$, so two years multiply by $0.9^{2} = 0.81$.

Find the true overall change

Only $81\%$ of the original remains, so $19\%$ has been lost, not $20\%$.

See where the error comes from

The second year loses $10\%$ of the reduced amount $0.9P$, which is only $9\%$ of the original.

Add the two losses

Total loss $= 19\%$ of the original, confirming the multiplier calculation.

Test with a number

A concrete check: $\pounds 200$ falls to $\pounds 162$, a $19\%$ drop.

State the verdict

Percentage changes never simply add — they multiply.

Write the total multiplier for P

Twenty years of $5\%$ growth combine into one multiplier.

Write the total multiplier for Q

Ten years of $10\%$ growth combine into one multiplier.

Evaluate the multiplier for P

Process P multiplies the starting value by about $2.653$.

Evaluate the multiplier for Q

Process Q multiplies the starting value by about $2.594$.

Compare the two multipliers

Process P ends up larger, even though its rate is half as big and its time is only twice as long.

Quantify the gap

Process P finishes about $2.3\%$ ahead of process Q.

Test the tempting shortcut

Adding the yearly rates makes the two processes look identical, but that ignores compounding entirely.

Explain the difference

P compounds twenty times rather than ten, and those extra compounding events more than make up for the smaller rate.

Sanity-check with a starting value

From a start of $100$, P reaches about $265.33$ and Q about $259.37$.

State the conclusion

Process P produces the larger final value.

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.