WACE Maths Applications flashcards

32 cards. Read the term and the definition here. The deck at the top of this page is optional.

Simple interest formula

Interest I = P × r × t, with P the principal, r the annual rate as a decimal, and t the time in years. The accumulated amount is A = P(1 + rt).

Compound interest formula

A = P(1 + r)^n gives the final amount when interest is added to the principal each period. Here r is the rate per period and n the number of periods.

Compounding periods adjustment

When the nominal annual rate is r and interest compounds m times per year, use rate r/m per period and n = m × years. The total number of periods drives the exponent.

Effective annual rate

The effective annual rate converts a nominal rate compounded m times yearly into a single yearly rate: EAR = (1 + r/m)^m − 1. It allows fair comparison between investment options.

Reducing-balance loan recurrence

Each period the balance grows by interest then falls by the repayment: B(n+1) = B(n)(1 + r) − R. Repeated application, or a spreadsheet, gives the outstanding balance after any number of periods.

Straight-line depreciation

Value falls by a fixed amount each period: V = V0 − D × t, where D is the constant depreciation per period. The graph of value against time is a straight line.

Declining-balance depreciation

Value falls by a fixed percentage of its current value each period: V = V0(1 − r)^t. The decay curve is exponential and never reaches zero in the model.

Doubling time rule of 70

For small growth rates, doubling time ≈ 70 ÷ percentage growth rate per period. For example, 5% growth per year doubles a quantity in roughly 14 years.

Arithmetic sequence terms

An arithmetic sequence adds a constant difference d each step: Tn = a + (n − 1)d, where a is the first term. Its graph against n is linear.

Geometric sequence terms

A geometric sequence multiplies by a constant ratio r each step: Tn = a·r^(n−1). Its graph against n is exponential when r exceeds 1 and decays when r is between 0 and 1.

Arithmetic series sum

The sum of the first n terms of an arithmetic sequence is Sn = n/2 × (2a + (n − 1)d), equivalently n/2 × (first term + last term).

Geometric series sum

For ratio r ≠ 1, the sum of the first n terms of a geometric sequence is Sn = a(r^n − 1)/(r − 1). This underpins compound-interest and annuity calculations.

Future value of annuity

Regular equal deposits R made at the end of each period grow to FV = R × ((1 + r)^n − 1)/r. This is the geometric-series sum applied to money.

Present value of annuity

The lump sum equivalent today of n payments of R is PV = R × (1 − (1 + r)^(−n))/r. It is used to compare loan or investment options.

Inflation and purchasing power

Rising prices reduce what money buys: real value = nominal value ÷ (1 + r)^t for inflation rate r over t years. Nominal wages may rise while real wages stagnate.

Solving for the exponent

To find n in A = P(1 + r)^n, divide by P then take logarithms: n = log(A/P) ÷ log(1 + r). The same rearrangement solves doubling-time questions exactly.

Interpreting r as a decimal

A percentage rate must be converted to a decimal before substitution: 6.5% per annum becomes 0.065. Forgetting this conversion is a common source of large errors.

Nominal versus real values

Nominal values are stated in current dollars without adjustment; real values are adjusted for inflation. Comparisons across years should use real values to be meaningful.

Linear growth model

A quantity growing by a constant absolute amount each period follows V = V0 + kt. Successive differences are constant, distinguishing it from exponential growth.

Exponential growth model

A quantity growing by a constant percentage each period follows V = V0(1 + r)^t. Successive ratios are constant, which identifies exponential rather than linear behaviour.

Choosing linear or exponential

Check first differences: constant differences indicate linear change, while constant ratios between successive values indicate exponential change. The pattern determines which model to apply.

Spreadsheet recurrence modelling

Enter the starting value in one cell and a formula referencing the previous cell for the next period, then fill down. This efficiently generates loan balances, depreciation schedules, and growth tables.

Total cost of a loan

Total repayments equal the repayment amount multiplied by the number of payments. Interest paid is total repayments minus the original principal borrowed.

Per-period rate conversion

A nominal annual rate divided by the number of compounding periods gives the rate per period. Weekly compounding at 5.2% annually uses 0.052 ÷ 52 per week.

Time in years from months

Convert months to years by dividing by 12 before substituting into annual formulas. For example, 18 months equals 1.5 years in simple-interest calculations.

Growth rate from data

Estimate the percentage growth rate by dividing the change in value by the previous value, then multiplying by 100. Averaging successive rates approximates the compound rate.

Depreciation total loss

Total depreciation equals initial value minus final value. For straight-line models, dividing this loss by the number of periods gives the annual depreciation amount.

Break-even in savings plans

Set the two model expressions equal and solve for time to find when plans give equal value. The solution identifies the crossover point between competing options.

Rounding in money problems

Round final monetary answers to the nearest cent, but retain full precision in intermediate steps. Early rounding can shift results by several dollars over long periods.

Graph shape recognition

Linear models graph as straight lines; compound-interest and depreciation models graph as curves. The curvature distinguishes exponential from arithmetic change at a glance.

Recurrence for savings

Savings with regular deposits follow B(n+1) = B(n)(1 + r) + D, where D is the deposit. This mirrors loan recurrence but adds rather than subtracts the regular payment.

Comparing investment options

Convert each option to an effective annual rate before comparing. Options with more frequent compounding at the same nominal rate yield higher effective returns.