Year 10AC9M10A04

Depreciation, Inflation and Currency

Use mathematical modelling to solve applied problems involving growth and decay, including financial contexts.

Growth and decay use the same formula. Only the sign inside the bracket changes.

Builds onYr 10 · Calculate Investment Interest

Method

Growth

A = P(1+i)^n

Decay

A = P(1-i)^n

Depreciation and reducing-balance both use this.

Straight-line depreciation is different

A = P - (\text{amount per year} \times n)

The same amount comes off each year, not the same percentage.

Inflation is compound growth

Prices rise on the new price each year, not the original.

Currency: multiply one way, divide the other

Check the rate direction before calculating.

e.g.

\text{A } \$30\,000 \text{ car depreciates at } 15\% \text{ per year on reducing balance. Find its value after 4 years.}

Identify the type. Reducing balance means compound decay. $A = P(1-i)^n$
List what you have. $P = 30\,000 \quad i = 0{,}15 \quad n = 4$
Substitute. $A = 30\,000(0{,}85)^4$
Work out the bracket first. $(0{,}85)^4 = 0{,}52200625$
Multiply. $A \approx \$15\,660$
Sanity check: about half after four years at 15%. $\text{reasonable} \quad\checkmark$

Identify growth or decay, and reducing balance or straight line, before substituting.

\$20\,000 \text{ depreciating } 20\% \text{ p.a. for } 3 \text{ years.}

Answer

\$10\,240

\text{A } \$500 \text{ item with } 5\% \text{ inflation for } 6 \text{ years.}

Answer

\approx \$670

\text{Straight-line: } \$12\,000 \text{ losing } \$1\,500 \text{ a year. Value after 5?}

Answer

\$4\,500

\text{Convert } \$400 \text{ at } 1 \text{ AUD} = 0{,}65 \text{ USD}.

Answer

\$260 \text{ USD}

\text{Convert } 260 \text{ USD back at the same rate.}

Answer

\$400 \text{ AUD}

\text{Why does reducing balance never reach zero?}

Answer

\text{You always keep a percentage of what is left.}