Quiz 3         MAT 118, Spring 2003

PRINT your Name: Solution

  1. You will need $2400 in cash two years from now. Your parents tell you that if you give them some amount of money now, they will pay you 10% annual simple interest on it, with no compounding. How much money do you need to give them in order to have the $2400 in two years?

    Solution: First, we recall that for simple interest, $\displaystyle {F = P ( 1 + r\cdot{}t) }$. In our case, we know the future value $F$ is $2400, that the annual rate is 10%, and the time is 2 years. We want to know the principle $P$. Since both the rate and the time are given in years, all our units match and there is no need for conversion. Thus, we need to solve

    \begin{displaymath}2400 = P ( 1 + (.10)(2) ) = 1.2 P\end{displaymath}

    for $P$, giving

    \begin{displaymath}P = \frac{2400}{1.2} = 2000 \end{displaymath}

    So we need to give them $2000 now to have $2400 in two years.

  2. If you invest $1000 in a bank account that pays 8% annual interest, compounded monthly, how much will there be in the account after 3 years?

\displaystyle {\$ 1000 \left( 1 + \fr...
...aystyle {\$ 1000 \left( \frac{8}{12} \right)^{3} }

    Solution: Our principle is $1000. Since the account is compounded monthly, our periodic interest rate is $\frac{.08}{12}$ (there are 12 months in a year). We also need to express our time in months, and 3 years is 36 months. Thus, the amount is expressed as

    \begin{displaymath}1000 \left( 1 + \frac{.08}{12} \right)^{36} \end{displaymath}

  3. If you invest $1000 at 8% annually, compounded monthly, how many months will it be until you double your money?

\displaystyle { \log{(1000)} \left( 1 + \f...
...}{12} \log{\left( 1 + \frac{.08}{12} \right) }} \\

    Solution: Since we want to double our money, the future value should be $2000. As above, the periodic rate is $\frac{.08}{12}$, the principle is $1000, and the time is in months. Thus, we need to solve

    \begin{displaymath}2000 = 1000\left( 1 + \frac{.08}{12} \right)^t \end{displaymath}

    for $t$. First, divide both sides by 1000 to get

    \begin{displaymath}2 = \left( 1 + \frac{.08}{12} \right)^t \end{displaymath}

    and then take the logarithm of both sides. Using the fact that $\log\left(b^x\right) = x\log b$, we get

    \begin{displaymath}\log 2 = t \log \left( 1 + \frac{.08}{12} \right) \end{displaymath}

    Now divide to get

    \begin{displaymath}t= \frac{\log{2}}{\log \left( 1 + \frac{.08}{12} \right)}\end{displaymath}

    This is 104.31 months, that is, just over 8 years and 8 months.

    (Note that on the original quiz, the correct answer had a typo, so everyone got full credit on this problem, no matter which choice they picked. Duh.)

Scott Sutherland 2003-02-14