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Maria, age 28, wants to pay no more than [tex]$\$300$[/tex] a year in life insurance. What is the face value of the largest 20-year term policy she can buy without spending more than [tex]$\[tex]$300$[/tex][/tex] annually?

\begin{tabular}{|c|c|c|c|c|}
\hline
\multirow[t]{2}{*}{ Age } & \multicolumn{4}{|c|}{ Annual Insurance Premiums (per [tex]$1,000$[/tex] of face value) } \\
\hline & 10-Year Term & 15-Year Term & 20-Year Term & Whole Life \\
\hline
28 & 2.20 & 2.54 & 1.59 & 9.46 \\
\hline
\end{tabular}

a. [tex]$\$234,000$[/tex]

b. [tex]$\[tex]$158,000$[/tex][/tex]

c. [tex]$\$11,000$[/tex]

d. [tex]$\[tex]$567,000$[/tex][/tex]

Please select the best answer from the choices provided:

A

B

C

D

Sagot :

To determine the face value of the largest 20-year term policy that Maria can buy without exceeding her annual budget of \[tex]$300, follow these steps: 1. Identify the relevant annual premium per $[/tex]1,000 of face value: For a 20-year term policy and a 28-year-old female, the annual premium is \[tex]$1.59 per $[/tex]1,000 of face value.

2. Set up the equation: The maximum annual payment Maria wants to make is \[tex]$300. Therefore, the relationship between the face value of the policy and the annual premium can be expressed as: \[ \text{Annual Payment} = \left(\frac{\text{Face Value}}{1000}\right) \times \text{Annual Premium per \$[/tex]1000}
\]
Substitute the given values into the equation:
[tex]\[ 300 = \left(\frac{\text{Face Value}}{1000}\right) \times 1.59 \][/tex]

3. Solve for the face value:
[tex]\[ \frac{\text{Face Value}}{1000} = \frac{300}{1.59} \][/tex]
[tex]\[ \text{Face Value} = \frac{300}{1.59} \times 1000 \][/tex]
[tex]\[ \text{Face Value} \approx 188679.24528301888 \][/tex]

Hence, the face value of the largest policy Maria can buy without spending more than \[tex]$300 annually is approximately $[/tex]\[tex]$188,679. This value matches the calculation closely, which means \( \$[/tex]188,679 \) is the correct answer.

Now let's review the choices provided:
a. \[tex]$234,000 b. \$[/tex]158,000
c. \[tex]$11,000 d. \$[/tex]567,000

The closest value to our calculated figure of approximately \[tex]$188,679 is not exactly listed, but option \( b. \, \$[/tex]158,000 \) seems to be the conservative choice that would definitely not exceed the budget. Nevertheless, our precise and accurate answer, as calculated, is actually very close to \[tex]$188,679. So, despite none of the options fitting perfectly, the best answer from the closest approximation is: a. \$[/tex]188,679 (but given none existing, closest one is not listed).