Revision as of 10:31, 18 November 2023 by Admin (Created page with "An investor’s retirement account pays an annual nominal interest rate of 4.2%, convertible monthly. On January 1 of year y, the investor’s account balance was X. The investor then deposited 100 at the end of every quarter. On May 1 of year (y + 10), the account balance was 1.9X. Determine which of the following is an equation of value that can be used to solve for X. <ul class="mw-excansopts"> <li><math display = "block">\frac{1.9X}{\left(1.0105\right)^{\frac{124}{...")
ABy Admin
Nov 18'23
Exercise
An investor’s retirement account pays an annual nominal interest rate of 4.2%, convertible monthly. On January 1 of year y, the investor’s account balance was X. The investor then deposited 100 at the end of every quarter. On May 1 of year (y + 10), the account balance was 1.9X.
Determine which of the following is an equation of value that can be used to solve for X.
- [[math]]\frac{1.9X}{\left(1.0105\right)^{\frac{124}{3}}}+\sum_{k=1}^{42}\frac{100}{\left(1.0105\right)^{k-1}}=X [[/math]]
- [[math]]X+\sum_{k=1}^{42}{\frac{100}{(1.0035)^{3(k-1)}}}={\frac{1.9X}{(1.0035)^{124}}} [[/math]]
- [[math]]X+\sum_{k=1}^{41}\frac{100}{(1.0035)^{3k}}=\frac{1.9X}{(1.0035)^{124}}[[/math]]
- [[math]]X+\sum_{k=1}^{41}{\frac{100}{(1.0105)^{k-1}}}={\frac{1.9X}{(1.01105)^{\frac{124}{3}}}} [[/math]]
- [[math]]X+\sum_{k=1}^{42}{\frac{100}{(1.0105)^{k-1}}}={\frac{1.9X}{(1.0105)^{\frac{124}{3}}}}[[/math]]
ABy Admin
Nov 18'23
Solution: C
The monthly interest rate is 0.042/12 = 0.0035. The quarterly interest rate is 1.00353-1 = 0.0105. The investor makes 41 quarterly deposits and the ending date is 124 months from the start. Using January 1 of year y as the comparison date produces the following equation
[[math]]
X+\sum_{k=1}^{41}{\frac{100}{1.0105^k}}={\frac{\textstyle1.9X}{\textstyle1.0035^{124}}}
[[/math]]