Explore a diverse range of topics and get expert answers on IDNLearn.com. Our platform provides trustworthy answers to help you make informed decisions quickly and easily.

Because Bernard has some health issues, he must pay [tex]$15\%$[/tex] more for life insurance. About how much more annually will a [tex]$\$[/tex]115,000[tex]$ 10-year term insurance at age 35 cost Bernard than someone of the same age without health issues?

\begin{tabular}{|l|l|l|}
\hline
\multirow{2}{*}{Age} & \multicolumn{2}{c|}{Annual Insurance Premiums (per \$[/tex]5,000 of face value)} \\ \cline{2-3}
& Male & Female \\
\hline
35 & 1.40 & 1.59 \\
\hline
40 & 1.64 & 2.01 \\
\hline
45 & 2.07 & \\
\hline
\end{tabular}

a. [tex]$\$[/tex]161[tex]$
b. $[/tex]\[tex]$185$[/tex]
c. [tex]$\$[/tex]1,073[tex]$
d. $[/tex]\[tex]$24$[/tex]

Please select the best answer from the choices provided:
A
B


Sagot :

To find how much more Bernard will pay annually due to his health issues, follow these steps:

1. Determine the number of [tex]$5,000 face value units for the $[/tex]115,000 term insurance:
[tex]\[ \text{Number of units} = \frac{115,000}{5,000} = 23 \][/tex]

2. Calculate the annual premium for a healthy 35-year-old male:
[tex]\[ \text{Annual premium (no health issues)} = \text{Number of units} \times \text{Rate per unit} \][/tex]
Given the rate per unit for a 35-year-old male is [tex]$1.40: \[ \text{Annual premium (no health issues)} = 23 \times 1.40 = 32.20 \] 3. Calculate the additional cost due to health issues (15% more): \[ \text{Additional cost percentage} = 0.15 \] \[ \text{Additional annual cost} = \text{Annual premium (no health issues)} \times \text{Additional cost percentage} = 32.20 \times 0.15 = 4.83 \] So, the additional cost Bernard has to pay annually due to his health issues is approximately $[/tex]\[tex]$4.83$[/tex].

From the choices provided, the closest amount is [tex]\( \$185 \)[/tex], though it’s evident there might be an error in rounding or reporting the choices. Should you refer to the exact computation, it would indeed be [tex]$\$[/tex]4.83$.