Apply the simple interest formula I = Prn and the amount A = P + I, rearrange it to solve for the principal, the rate or the time, and calculate interest over part-years measured in months or days

What this dot point is asking

NESA, the board that sets the HSC, wants you to work with simple interest, also
called flat interest. This is interest that is charged or earned on the
original amount only. It is the same dollar figure every period, never on a
growing balance. You need to use the formula $I = Prn$ to find the interest. You then
add it to the principal with $A = P + I$ to get the final amount. Just as often, you
must rearrange the formula to find a missing principal, rate or time. One part of
the dot point that trips students is interest over part of a year. Here the time
has to be written as a fraction of a year before it goes into the formula.

This is strictly Year 11 Money Matters (MS-F1). Compound interest, where interest
earns interest, is a Year 12 topic and is out of scope here; everything on this page
keeps the interest fixed on the original principal.

The answer

Simple interest rests on two short formulas, and every question is either a direct
substitution into them or a rearrangement of them.

• The interest is the principal times the rate times the number of periods:

• The final amount (the value of an investment, or the total owed on a loan) is
  the principal plus the interest:

Here $P$ is the principal (the amount invested or borrowed), $r$ is the
rate per period written as a decimal (so $6\%$ is $0.06$), and $n$ is the
number of periods that match the rate. Almost always the rate is "per annum"
(per year), so $n$ must be a number of years.

To find a missing quantity, rearrange $I = Prn$ by dividing by whatever multiplies
it:

For interest over part of a year, write the time as a fraction of a year before
substituting: $n = \dfrac{\text{months}}{12}$ for a number of months, or
$n = \dfrac{\text{days}}{365}$ for a number of days.

A term deposit of $8000 at $4.3\%$ per annum for $3$ years earns
$I = 8000 \times 0.043 \times 3 = 1032$, i.e. $1032.00, and is worth
$A = 8000 + 1032 = 9032$, i.e. $9032.00, at the end.

Finding the interest and the final amount

The everyday calculation is the direct one: you know $P$, $r$ and $n$, and you want
the interest and the final amount. Convert the rate to a decimal, multiply the three
together for $I$, then add $I$ to $P$ for $A$. A $12,000 investment at $5\%$ per
annum for $2$ years earns $I = 12\,000 \times 0.05 \times 2 = 1200$, i.e.
$1200.00, and is worth $A = 12\,000 + 1200 = 13\,200$, i.e. $13,200.00. On a
loan the same arithmetic gives the total owed: borrow $12,000 on the same
terms and you repay the $12,000 plus $1200 interest, a total of
$13,200.00. Investment "final value" and loan "total owed" are the same
calculation, $A = P + I$.

Why the amount grows as a straight line

Because the interest each year is the same fixed amount, the value of the account
goes up in equal steps. Plot the amount $A$ against time $n$ and you get a
straight line. It starts at the principal $P$ when $n = 0$, since no time has
passed, so there is no interest yet. From there it climbs by the same amount,
$Pr$ dollars, every year. The starting principal is the vertical intercept, the
point where the line meets the up-and-down axis. The yearly interest $Pr$ is the
constant slope, the steepness of the line. The line below is $5000 invested at
$6\%$ per annum, which adds $5000 \times 0.06 = 300$, i.e. $300.00, each year.

After $n$ years the amount is $A = 5000 + 300n$, so the dots sit at $5300.00,
$5600.00, $5900.00, $6200.00 and $6500.00 for $1$ to $5$ years, each
exactly $300 above the last. That constant step is the signature of simple
interest, and it is why the graph is a straight line rather than a curve. The next
dot point, simple interest graphs, uses this line to compare different rates by their
gradients; for now the picture explains the formula: $A = P + Prn$ is a straight line
with intercept $P$ and slope $Pr$.

Rearranging to find the principal, rate or time

Many questions give you the interest and ask for one of the inputs. The move is
always the same: write $I = Prn$, substitute the two values you know, then divide to
isolate the unknown.

• Solve for the principal: $P = \dfrac{I}{rn}$. Divide the interest by the
  product of the rate and the time.
• Solve for the rate: $r = \dfrac{I}{Pn}$, then multiply by $100$ to state it as
  a percentage.
• Solve for the time: $n = \dfrac{I}{Pr}$, in the same period as the rate
  (usually years), then convert to months or days if asked.

The single most common error is dividing by only one of the two known quantities.
If $9000-worth of interest came from a rate and a time, you must divide by
both the rate and the time multiplied together, because in $I = Prn$ the unknown
is multiplied by that whole product.

Interest over part of a year

The rate $r$ and the time $n$ must use the same period. Australian interest
rates are quoted "per annum", so $n$ has to be in years. When a question gives a
time in months or days, convert it to a fraction of a year first:

For example, $6500 borrowed at $9\%$ per annum for $8$ months uses
$n = \tfrac{8}{12}$, giving $I = 6500 \times 0.09 \times \tfrac{8}{12} = 390$, i.e.
$390.00. Keep the fraction $\tfrac{8}{12}$ in the working rather than rounding it
to a decimal early, so the final cent is exact. The exam uses a $365$-day year for
days unless it says otherwise, and counts the days between two dates without
double-counting the start and end day.

How exam questions ask about this

Each wording points at one rearrangement of $I = Prn$. Learn the translation:

• "Calculate the simple interest / the flat interest earned (or charged)." Use
  $I = Prn$ directly.
• "Find the final value / the amount owed / the total to repay." Find $I$ first,
  then $A = P + I$.
• "How much was invested / borrowed?" The principal is unknown: $P = \dfrac{I}{rn}$.
• "What (interest) rate ...?" or "the minimum rate to earn at least ..." The
  rate is unknown: $r = \dfrac{I}{Pn}$, as a percentage; for "minimum to earn at
  least", set the interest equal to the target and round the rate up if needed.
• "For how long / what is the term of the loan?" The time is unknown:
  $n = \dfrac{I}{Pr}$; convert a decimal of a year to months ($\times 12$) if asked.
• "... for $9$ months / for $90$ days / for $2$ years and $6$ months." Write the
  time as a fraction of a year ($\tfrac{9}{12}$, $\tfrac{90}{365}$, $2.5$) before
  substituting.
• "Calculate the monthly (or fortnightly) repayment on a flat-rate loan." Find
  the total owed $A = P + I$, then divide by the number of repayments.

Edge case: matching the rate and time periods

The formula only works when $r$ and $n$ use the same time period, and Standard 2
sometimes tests this directly. If a rate is given per month, then $n$ must be a
number of months, not years. If it is per annum, that is per year, then $n$ is in
years. A rate of $4.8\%$ per annum is $\tfrac{4.8}{12} = 0.4\%$ per month or
$\tfrac{4.8}{4} = 1.2\%$ per quarter. You may convert the rate to match the period
given rather than converting the time, as long as you are consistent. The safest
habit for the HSC is to keep the per annum rate and put the time in as a fraction
of a year, because that is how almost every question is framed. Note too that simple
interest assumes nothing is added to or taken from the account during the term. The
interest is always on the original principal.