Interest rate is i per annum. Begin with $100 in a bank account. The money accumulates for two years and then $50 is withdrawn. How many more years are needed for the amount in the account to accumulate to $200? (The answer should be a function of i.)

• [[math]] \frac{\log(2)}{\log(1+i)} -1.5[[/math]]
• [[math]]\frac{\ln (200)-\ln \left(100(1+i)^2-50\right)}{\ln (1+i)} [[/math]]
• [[math]]\frac{\ln (150)-\ln \left(100(1+i)^2\right)}{\ln (1+i)}[[/math]]
• [[math]](\frac{200}{50 + 200i}-1)/i[[/math]]
• [[math]]\frac{\ln(2)}{1+i}-2[[/math]]

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.

Solution: B

We want [math]\left(100(1+i)^2-50\right)(1+i)^n=200[/math]. Thus [math](1+i)^n=\frac{200}{\left(100(1+i)^2-50\right)}[/math] so

[[math]] n=\frac{\ln (200)-\ln \left(100(1+i)^2-50\right)}{\ln (1+i)}. [[/math]]

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.