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?

2 Answers
Mar 25, 2016

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.

Mar 25, 2016

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