Iteration, growth and decay – Year 11 GCSE maths revision notes to download as PDF

Worked example 1: compound interest

£500 is invested for 3 years at 4% a year compound interest. Find the final value and the interest.

Multiplier \(=1.04\). \(A=500\times1.04^3=500\times1.124864=562.432\). The value is £562.43 and the interest is \(562.43-500=\)£62.43.

Worked example 2: depreciation

A car worth £12 000 loses 20% of its value each year. Find its value after 3 years.

Multiplier \(=0.8\). \(12000\times0.8^3=12000\times0.512=\)£6144.

Worked example 3: different rates each year

£800 earns 3% in year 1 and 2% in year 2. Find the final value.

Multiply the separate multipliers: \(800\times1.03\times1.02=\)£840.48.

Worked example 4: working backwards

A sum grows by 10% a year for two years to £1210. Find the original. £5000 grows to £5618 in two years: find the rate. How many years for £1000 at 5% to pass £1200?

Original \(=\dfrac{1210}{1.1^2}=\dfrac{1210}{1.21}=\)£1000. For the rate, \(r^2=\dfrac{5618}{5000}=1.1236\), so \(r=1.06\), a rate of 6%. For the years: \(1.05^3=1.1576\) gives £1157.63, not enough; \(1.05^4=1.2155\) gives £1215.51, so it takes 4 years.

Worked example 5: "is she correct?"

Someone says 5% a year for 2 years is the same as 10%. Is that right?

No: \(1.05^2=1.1025\), a 10.25% increase. A decrease of 10% followed by an increase of 10% has multiplier \(0.9\times1.1=0.99\), so the quantity ends 1% lower.

Worked example 6 (Higher): iterating

\(x_{n+1}=\sqrt{x_n+2}\) with \(x_0=1\). Find \(x_1\), \(x_2\), \(x_3\).

\(x_1=\sqrt3=1.7321\). \(x_2=\sqrt{3.7321}=1.9319\). \(x_3=\sqrt{3.9319}=1.9829\). The values are heading towards 2, which solves \(x^2=x+2\) (that is \(x^2-x-2=0\)).

Worked example 7 (Higher): showing a root exists

Show that \(x^3+x-3=0\) has a solution between 1 and 2, then find it to 2 decimal places.

Let \(f(x)=x^3+x-3\). \(f(1)=-1\) (negative) and \(f(2)=7\) (positive). The sign change shows a root between 1 and 2, because the graph is continuous. Rearranging gives \(x^3=3-x\), so \(x_{n+1}=\sqrt[3]{3-x_n}\). With \(x_0=1.2\): \(x_1=1.2164\), \(x_2=1.2127\), \(x_3=1.2136\). To 2 decimal places the root is 1.21.

Worked example 8: Increasing by a percentage repeatedly

A town of 25 000 people grows by 3% each year. Find the population after 4 years, to the nearest 100.

The multiplier is 1.03. After 4 years the population is \(25\,000\times1.03^4=25\,000\times1.1255=28\,138\). To the nearest 100 that is 28 100. Note that this is more than \(25\,000+4\times3\%\) of \(25\,000=28\,000\), because the percentage is applied to a bigger number each year.

Worked example 9: Finding the original value after a decrease

A television is on sale at £420 after a discount of 30%. What was the original price?

The sale price is 70% of the original, so the multiplier is 0.7. Original price \(=420\div0.7=600\). A common mistake is to find 30% of 420 and add it on; this gives the wrong answer because the 30% was taken from the original, not from the sale price.

Worked example 10: Equating a decay to a target

A radioactive sample falls by 10% each day. Starting with 200 g, after how many days is less than 100 g left?

Use \(200\times0.9^n\). After 6 days: \(200\times0.531=106.3\) g. After 7 days: \(200\times0.478=95.7\) g. So it takes 7 days. This is a trial-and-improvement method: the mass first drops below half after 7 days.

Worked example 11: Using the multiplier for two different changes

A house price rises by 8% in the first year and then falls by 5% in the second year. By what overall percentage does the price change?

The multipliers are 1.08 and 0.95. The overall multiplier is \(1.08\times0.95=1.026\). This is an increase of 2.6%. Do not simply subtract 5 from 8, because the 5% is taken from the new, higher price.

Worked example 12: An iterative formula for a solution

Show that \(x^3-5x+1=0\) can be rearranged to \(x=\dfrac{x^3+1}5\) and use \(x_0=0\) to find a root to 3 decimal places.

Rearrange: \(5x=x^3+1\), so \(x=\dfrac{x^3+1}5\). Then \(x_1=\dfrac15=0.2\), \(x_2=\dfrac{0.008+1}5=0.2016\), \(x_3=0.2016\) (to 4 d.p.: \(0.20164\)). The root is \(x=0.202\) to 3 decimal places. Check: \(0.202^3-5\times0.202+1=0.008-1.01+1=-0.002\approx0\).

Worked example 13: A compound growth formula for exam questions

Write a formula for the value \(V\) of £\(P\) after \(n\) years at \(r\%\) compound growth, and use it for £800 at 3.5% for 6 years.

\(V=P\left(1+\dfrac r{100}\right)^n\). Substituting \(P=800\), \(r=3.5\) and \(n=6\) gives \(V=800\times1.035^6=800\times1.2293=983.40\). For decay replace the plus with a minus: \(V=P\left(1-\dfrac r{100}\right)^n\).

Worked example 14: Percentage change from start to end

An investment grows from £2 500 to £3 000. Find the percentage increase and the yearly rate if this happened over 3 years.

Percentage increase \(=\dfrac{500}{2500}\times100=20\%\). The overall multiplier is \(1.2\). The yearly rate satisfies \(m^3=1.2\), so \(m=\sqrt[3]{1.2}=1.0627\): about 6.3% per year. This is a little less than \(20\div3=6.7\%\) because of compounding.