Chapter 14, Problem 27P

To determine

The difference in salary due to raise in the salary in year $10$, year $20$, year $30$ and year $40$.

Answer to Problem 27P

The difference in salary due to raise in year $10$ is $\$4032$.

The difference in salary due to raise in year $20$ is $\$5411.33$.

The difference in salary due to raise in salary in year $30$ is $\$7290$.

The difference in salary due to raise in year $40$ is $\$9723$.

Explanation of Solution

Given:

The salary in year $1$ is $\$60000$.

The annual raise is $5\%$.

The inflation rate is $3\%$.

Concept used:

Write the expression for future value.

$F=P{\left(1+i\right)}^n$ ...... (I)

Here, the present value is $P$, the inflation rate is $i$ and the time period is $n$.

Write the expression for increased salary due to the raise.

$P^{\prime }=P{\left(1+i^{\prime }\right)}^n$ ...... (II)

Here, the present value of increased salary is $P^{\prime }$, the annual raise is $i^{\prime }$ and the time period is $n$.

Calculations:

Calculate the future worth of $\$60000$ after $10$ years.

Substitute $\$60000$ for $P$ and $3\%$ for $i$ and $10$ for $n$ in Equation (I).

$\begin{array}{c}{F}_{10}=\$60000{\left(1+0.03\right)}^{10}\\ =\$60000\times 1.344\\ =\$80640\end{array}$

Calculate the future worth of $\$60000$ after $20$ years.

Substitute $\$60000$ for $P$ and $3\%$ for $i$ and $20$ for $n$ in Equation (I).

$\begin{array}{c}{F}_{20}=\$60000{\left(1+0.03\right)}^{20}\\ =\$60000\times 1.806\\ =\$108366.67\end{array}$

Calculate the future worth of $\$60000$ after $30$ years.

Substitute $\$60000$ for $P$ and $3\%$ for $i$ and $30$ for $n$ in Equation (I).

$\begin{array}{c}{F}_{30}=\$60000{\left(1+0.03\right)}^{30}\\ =\$60000\times 2.43\\ =\$145800\end{array}$

Calculate the future worth of $\$60000$ after $40$ years.

Substitute $\$60000$ for $P$ and $3\%$ for $i$ and $40$ for $n$ in Equation (I).

$\begin{array}{c}{F}_{40}=\$60000{\left(1+0.03\right)}^{40}\\ =\$60000\times 3.261\\ =\$195720\end{array}$

Calculate the increased salary due to the raise for year $1$.

Substitute $\$60000$ for $P$, $5\%$ for $i^{\prime }$ and $1$ for $n$ in Equation (II).

$\begin{array}{c}{P}^{\prime }=\$60000{\left(1+0.05\right)}^1\\ =\$60000\left(1.05\right)\\ =\$63000\end{array}$

Calculate the future worth of $\$63000$ after $10$ years.

Substitute $\$63000$ for $P$ and $3\%$ for $i$ and $10$ for $n$ in Equation (I).

$\begin{array}{c}{F}_{10}{}^{\prime }=\$63000{\left(1+0.03\right)}^{10}\\ =\$63000\times 1.344\\ =\$84672\end{array}$

Calculate the future worth of $\$63000$ after $20$ years.

Substitute $\$63000$ for $P$ and $3\%$ for $i$ and $20$ for $n$ in Equation (I).

$\begin{array}{c}{F}_{20}{}^{\prime }=\$63000{\left(1+0.03\right)}^{20}\\ =\$63000\times 1.806\\ =\$113778\end{array}$

Calculate the future worth of $\$63000$ after $30$ years.

Substitute $\$63000$ for $P$ and $3\%$ for $i$ and $30$ for $n$ in Equation (I).

$\begin{array}{c}{F}_{30}{}^{\prime }=\$63000{\left(1+0.03\right)}^{30}\\ =\$63000\times 2.43\\ =\$153090\end{array}$

Calculate the future worth of $\$63000$ after $40$ years.

Substitute $\$63000$ for $P$ and $3\%$ for $i$ and $40$ for $n$ in Equation (I).

$\begin{array}{c}{F}_{40}{}^{\prime }=\$63000{\left(1+0.03\right)}^{40}\\ =\$63000\times 3.261\\ =\$205443\end{array}$

The table for computation of difference in salary in year $1$ with the salary in year $10$, year $20$, year $30$ and year $40$ is shown below.

Year	$\$60000$ value for similar purchasing power	Increased salary value for similar purchasing power	Difference
$10$	$\$80640$	$\$84672$	$\$4032$
$20$	$\$108366.67$	$\$113778$	$\$5411.33$
$30$	$\$145800$	$\$153090$	$\$7290$
$40$	$\$195720$	$\$205443$	$\$9723$

Table (1).

Conclusion:

Thus, the difference in salary due to raise in year $10$ is $\$4032$.

The difference in salary due to raise in year $20$ is $\$5411.33$.

The difference in salary due to raise in salary in year $30$ is $\$7290$.

The difference in salary due to raise in year $40$ is $\$9723$.