(1.4) Applications of Arithmetic and Geometric Patterns
Percentage Increase
$100\%$ of $400$ is
.
$50\%$ of $400$ is
.
To increase $400$ by $50\%$ we find
$\%$ of $400$.
$150\%$ of $400$ is
.
$25\%$ of $400$ is
.
To increase $400$ by $25\%$ we find
$\%$ of $400$.
$125\%$ of $400$ is
.
$1\%$ of $400$ is
.
To increase $400$ by $1\%$ we find
$\%$ of $400$.
$101\%$ of $400$ is
.
To increase $400$ by $100\%$ we find
$\%$ of $400$.
$200\%$ of $400$ is
.
Compound Interest
When you put money in the bank, it earns interest.
When you borrow money or take out a loan, you have to pay it back with interest.
Consider a bank that pays $50\%$ interest (i.e. value increases by $50\%$) per annum (year).
Fill in the blanks if we invest $\$200$:
$$200\overset{\times 1.5}{\overset{\Huge\frown}{\phantom{A},\phantom{B}}}\underset{\text{after 1 year}}{\text{_______}}\overset{\times 1.5}{\overset{\Huge\frown}{\phantom{A},\phantom{B}}}\underset{\text{after 2 years}}{\text{_______}}\overset{\times 1.5}{\overset{\Huge\frown}{\phantom{A},\phantom{B}}}\underset{\text{after 3 years}}{\text{_______}}\overset{\times 1.5}{\overset{\Huge\frown}{\phantom{A},\phantom{B}}}\underset{\text{after 4 years}}{\text{_______}}\phantom{A},\quad\ldots$$
After 0 years
After 1 year
After 2 years
After 3 years
After 4 years
$\cdots$
$\$200$
$\$$
$\$$
$\$$
$\$$
$\cdots$
$u_1$
$u_2$
$u_3$
$u_4$
$u_5$
$\cdots$
Notice how "after $n$ years" corresponds to $u_{n+1}$.
Find the value of your investment after $10$ years to 2 decimal places. $\$$
Compound Interest over Different Time Periods
Some banks will pay interest monthly or quarterly or multiple times in a year.
Fill in the blanks:
Biannually or semi-annualy means
times per year.
Quarterly means
times per year.
Monthly means
times per year.
Weekly generally means
times per year.
Daily generally means
times per year.
Consider a bank that pays $50\%$ interest per annum, compounded quarterly.
Every quarter, the amount in the bank would increase by
$\%$.
Fill in the blanks if we invest $\$200$:
$$200\overset{\times 1.125}{\overset{\Huge\frown}{\phantom{A},\phantom{B}}}\underset{\text{after 1 quarter}}{\text{_______}}\overset{\times 1.125}{\overset{\Huge\frown}{\phantom{A},\phantom{B}}}\underset{\text{after 2 quarters}}{\text{_______}}\overset{\times 1.125}{\overset{\Huge\frown}{\phantom{A},\phantom{B}}}\underset{\text{after 3 quarters}}{\text{_______}}\overset{\times 1.125}{\overset{\Huge\frown}{\phantom{A},\phantom{B}}}\underset{\text{after 4 quarters}}{\text{_______}}\overset{\times 1.125}{\overset{\Huge\frown}{\phantom{A},\phantom{B}}}\underset{\text{after 5 quarters}}{\text{_______}}\phantom{A},\quad\ldots$$
After 0 quarters
After 1 quarter
After 2 quarters
After 3 quarters
After 4 quarters (1 year)
After 5 quarters
$\cdots$
$\$200$
$\$$
$\$$
to 2 decimal places
$\$$
to 2 decimal places
$\$$
to 2 decimal places
$\$$
to 2 decimal places
$\cdots$
$u_1$
$u_2$
$u_3$
$u_4$
$u_5$
$u_6$
$\cdots$
Find the value of your investment after $3$ years to 2 decimal places. $\$$
General Formula for Compound Interest
If the bank pays:
$24\%$ interest per annum compounded monthly, the amount in the bank would be multiplied by
every month.
$26\%$ interest per annum compounded weekly, the amount in the bank would be multiplied by
every week.
$73\%$ interest per annum compounded daily, the amount in the bank would be multiplied by
every day.
$r\%$ interest per annum compounded $k$ times per year, the amount in the bank would be multiplied by
$1+$
$r$/
, $k$ times per year.
If $P$ is invested at $r\%$ interest per annum compounded $k$ times per year,
Value after $n$ years$=P\Bigl($
$\Bigl)$
Example
On January 1, 2020, Joanne invests $€3000$ in a bank account that pays $2\%$ interest per annum compounded monthly. If she does not deposit or withdraw any money, how much will be in the account on July 1, 2023?
$€$
Example
Travis wants to invest money now so that he will have $£10000$ in $10$ years time. The investment pays $2\%$ per annum compounded quarterly. How much must Travis invest?