Interactive practice questions
$\$7400$$7400 is invested for three years at a rate of $10%$10% p.a., compounding annually.
Complete the table row for the Third Year.
| Balance + interest | Total balance | Interest earned | |
|---|---|---|---|
| First year | $-$− | $\$7400$$7400 | $\$740$$740 |
| Second year | $\$7400+\$740$$7400+$740 | $\$8140$$8140 | $\$814$$814 |
| Third year | $\$8140+\$$$8140+$$\editable{}$ | $\$$$$\editable{}$ | $\$$$$\editable{}$ |
| Fourth year | $\$$$$\editable{}$ | $\$$$$\editable{}$ | $-$− |
Complete the table row for the Fourth Year to determine the final value of the investment.
| Balance + interest | Total balance | Interest earned | |
|---|---|---|---|
| First year | $-$− | $\$7400$$7400 | $\$740$$740 |
| Second year | $\$7400+\$740$$7400+$740 | $\$8140$$8140 | $\$814$$814 |
| Third year | $\$8140+\$$$8140+$$814$814 | $\$$$$8954$8954 | $\$$$$895.40$895.40 |
| Fourth year | $\$8954$$8954$+$+$\$$$$\editable{}$ | $\$$$$\editable{}$ | $-$− |
Using the values in the table, how can we calculate the total interest earned over three years?
Select the two correct answers.
Sum up the values in the "Interest earned" column.
Calculate $10%$10% of the "Total balance" of the fourth year.
Use the greatest value in the "Interest earned" column.
Subtract the "Total balance" of the first year from the fourth year.
Calculate the total interest earned over the three years.
$\$3500$$3500 is invested for three years at a rate of $10%$10% p.a., compounding annually.
$\$3200$$3200 is invested for three years at a rate of $6%$6% p.a., compounding annually.
Tara borrows $\$5000$$5000 at a rate of $4.5%$4.5% p.a, compounding annually.