Question

Emma takes a job with a starting salary of $42,000. Her salary increases by 4% at the beginning of each year. What will be Emma's salary, to the nearest thousand dollars, at the beginning of year 10?

Answer 1

Alternative solution:

Explanation:

We can use geometric sequences to calculate this.

In a geometric series, the formula for `t_n` is `t_n = a xx r^(n - 1)`. "r" is the rate of change, n is the number of terms and a is the first term. Reading the question, we find the following.

`r = 1.04` (since 100% + 4%)

`a = $42 000`

`n = 10`

Therefore, we are solving for `t_n`

`t_10 = 42000 xx 1.04^9`

`t_10= 59779.10`

Emma's salary would be `$59779.10` after 10 years.

Practice exercises:

John gets a job where the base salary is of `$54 322`. His salary increases by 5.7% each year, until a maximum of 10 years. Find his salary after 16 years.

Answer 2

Here is the third solution.
`60,000` Rounded to nearest thousand dollar

Explanation:

Let Emma's salary at the beginning of `10th` year be `=$x`.
Rate of increase per year `4%=(1+4/100)`

General expression for salary at the beginning of `nth` year is given as `x=`
`"Initial Salary"xx(1+"% increase per year"/100)^"no of completed years"`

Number of completed years at the beginning of `10th` year `=9`
Inserting given values we obtain `x=42000xx(1+4/100)^9`
or `x=42000xx(1.04)^9`
or `x=59779.10` rounded to nearest penny.
`x=60000` Rounded to nearest thousand dollar