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(Created page with "John made a deposit of 1000 into a fund at the beginning of each year for 20 years. At the end of 20 years, he began making semiannual withdrawals of 3000 at the beginning of each six months, with a smaller final withdrawal to exhaust the fund. The fund earned an annual effective interest rate of 8.16%. Calculate the amount of the final withdrawal. <ul class="mw-excansopts"><li>561</li><li>1226</li><li>1430</li><li>1488</li><li>2240</li></ul> {{soacopyright | 2023 }}")
 
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John made a deposit of 1000 into a fund at the beginning of each year for 20 years.
'''Solution: D'''
At the end of 20 years, he began making semiannual withdrawals of 3000 at the beginning of
each six months, with a smaller final withdrawal to exhaust the fund. The fund earned an annual
effective interest rate of 8.16%.


Calculate the amount of the final withdrawal.
The outstanding balance at time 25 is


<ul class="mw-excansopts"><li>561</li><li>1226</li><li>1430</li><li>1488</li><li>2240</li></ul>
<math display = "block">
100(D a)_{\overline{{{25}}|}}=100\frac{25-a_{\overline{{{25}}}|}}{i}.
 
</math>
 
The principle repaid in the 26th payment is
 
<math display = "block">
X=2500-i(100){\frac{25-a_{\overline{{{25}}}|}}{i}}=2500-2500+100a_{\overline{{{25}}|}}=100a_{\overline{{{35}}|}}.
 
</math>
 
The amount borrowed is the present value of all 50 payments,
 
<math display = "block">
2500a_{\overline{{{25}}}|}+{{\nu}^{25}}100(D a)_{\overline{{{25}}}|}
</math>
 
Interest paid in the first payment is then
 
<math display = "block">
\begin{array}{l}{{i\Big[2500a_{\overline{{{25}}}|}+v^{25}100(D a)_{\overline{{{25}}}|}]}}\\ {{=2500(1-v^{25})+100v^{25}(25-a_{\overline{{{25}}}|})}}\\ {{=2500-2500v^{25}+2500v^{25}-v^{25}100a_{\overline{{{25}}}|}}}\\ {{=2500-Xv^{25}.}}\end{array}
</math>


{{soacopyright | 2023 }}
{{soacopyright | 2023 }}

Latest revision as of 00:06, 19 November 2023

Solution: D

The outstanding balance at time 25 is

[[math]] 100(D a)_{\overline{{{25}}|}}=100\frac{25-a_{\overline{{{25}}}|}}{i}. [[/math]]

The principle repaid in the 26th payment is

[[math]] X=2500-i(100){\frac{25-a_{\overline{{{25}}}|}}{i}}=2500-2500+100a_{\overline{{{25}}|}}=100a_{\overline{{{35}}|}}. [[/math]]

The amount borrowed is the present value of all 50 payments,

[[math]] 2500a_{\overline{{{25}}}|}+{{\nu}^{25}}100(D a)_{\overline{{{25}}}|} [[/math]]

Interest paid in the first payment is then

[[math]] \begin{array}{l}{{i\Big[2500a_{\overline{{{25}}}|}+v^{25}100(D a)_{\overline{{{25}}}|}]}}\\ {{=2500(1-v^{25})+100v^{25}(25-a_{\overline{{{25}}}|})}}\\ {{=2500-2500v^{25}+2500v^{25}-v^{25}100a_{\overline{{{25}}}|}}}\\ {{=2500-Xv^{25}.}}\end{array} [[/math]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.