Lesson Problems on continuously compounded accounts
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<H2>Problems on continuously compounded accounts</H2> <H3>Problem 1</H3>Parents of a newborn baby are given a gift of Php 50,000 and will choose between two options to invest for their child’s college fund. Option 1 is to invest the gift in a fund that pays an average annual interest rate of 8% compounded semiannually; option 2 is to invest the gift in a fund that pays an average annual interest rate of 7.75% compounded continuously. Which is the better option, assuming each investment has a term of 18 years? <B>Solution</B> <pre> Let's calculate one year coefficients of growth: it is enough to make a selection/conclusion. Option 1 has one year growing coefficient {{{(1+0.08/2)^2}}} = {{{1.04^2}}} = 1.0816. Option 2 has one year growing coefficient {{{e^0.0775}}} = {{{2.71828^0.0775}}} = 1.08058 (rounded). Comparing, it is clear that option 1 is better (without making long calculations for 18 years). </pre> <H3>Problem 2</H3>Mia invests $2,000 in a money market account that earns 5% annual interest compounded continuously. Approximately how many years will it take her money to grow to the $4,500 she needs for her small business start-up? <B>Solution</B> <pre> Using the formula for continuously compounded account, write 4500 = {{{2000*e^(0.05*t)}}}. Divide both sides by 2000 {{{4500/2000}}} = {{{e^(0.05*t)}}}. It is the same as 2.25 = {{{e^(0.05*t)}}}. Take natural logarithm of both sides ln(2.25) = 0.05*t, t = {{{ln(2.25)/0.05}}} = 16.22. 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