[tex] \large\mathcal{ANSWER:} [/tex]
[tex] \boxed{\begin{array}{l} 1.)\:\textsf{Php 3,130.91} \\ \\ 2.)\:\textsf{Php 1,471,669.01} \end{array}} [/tex]
[tex] \large\mathcal{SOLUTION:} [/tex]
[tex] \begin{array}{l} \large \textsf{Compound Interest Formula} \\ \\ \quad \quad \displaystyle A = P\left(1 + \dfrac{r}{n}\right)^{nt} \\ \textsf{where:} \\ \begin{aligned} \quad A &= \textsf{final amount or future value} \\ P&= \textsf{principal/original amount} \\ r&=\textsf{interest rate} \\ n&=\textsf{compoundings per period} \\ t&=\textsf{number of periods} \end{aligned} \end{array} [/tex]
[tex] \begin{array}{l} 1.)\: \textsf{After 5 years,} \\ \quad \quad \begin{aligned} \displaystyle A &= 150,000\left(1 + \dfrac{0.03}{12}\right)^{12(5)} \\ A&=\textsf{Php 174,242.52} \end{aligned} \\ \\ \textsf{Josiah accumulated Php 174,242.52 after 5} \\ \textsf{years. Then another 5 years of fixed monthly} \\ \textsf{withdrawals until no money is left.} \end{array} [/tex]
[tex] \begin{array}{l} \textsf{Using the formula for Compound Interest with} \\ \quad \quad \quad \quad \textsf{Fixed Withdrawals,} \\ \\ \footnotesize \displaystyle \quad A = P\negthickspace\left(1 + \dfrac{r}{n}\right)^{nt} - W\negthickspace\left(\dfrac{n}{r}\right)\negthickspace\left[\left(1 + \dfrac{r}{n}\right)^{nt} - 1\right] \\ \textsf{where:} \\ \quad W = \textsf{fixed withdrawal amount} \end{array} [/tex]
[tex] \begin{array}{l} \textsf{Now, solve for}\: W\:\textsf{with}\:\textsf{Php 174,242.52}\:\textsf{as the} \\ \textsf{initial amount and 0 as the final amount.} \\ \\ \quad \footnotesize \displaystyle \ 0 = 174,242.52\left(1 + \dfrac{0.03}{12}\right)^{12(5)} \\ \quad \quad \quad \quad \quad \footnotesize - W\left(\dfrac{12}{0.03}\right)\negthickspace\left[\left(1 + \dfrac{0.03}{12}\right)^{12(5)} - 1\right] \\ \\ \begin{aligned} \quad \quad \: \: \quad 64.6467W &= 202,403.0353 \\ \implies W &= \boxed{\textsf{Php 3,130.91}} \end{aligned} \\ \\ \textsf{Thus, Josiah's monthly withdrawals should be}\\\textsf{Php 3,130.91}. \end{array} [/tex]
[tex] \begin{array}{l} 2.)\:\textsf{Solving for the original amount}\:P\:\textsf{using the}\\ \quad \textsf{same formula for Compound Interest with} \\ \quad \textsf{Fixed Withdrawals,} \\ \\ \footnotesize \displaystyle \quad 0 = P\negthickspace\left(1 +\dfrac{0.12}{12}\right)^{12(25)} \\ \footnotesize \quad \quad \quad \quad - 15,500\left(\dfrac{12}{0.12}\right)\negthickspace\left[\left(1 + \dfrac{0.12}{12}\right)^{12(25)} - 1\right] \\ \\ \begin{aligned} 15,500(1,878.8466) &= 19.78847P \\ \implies P &= \boxed{\textsf{Php 1,471,669.01}} \end{aligned}\\ \\ \textsf{Therefore, the original cost of the house and} \\ \textsf{lot is approximately equal to Php 1,471,669.01.} \end{array} [/tex]
[tex] \texttt \color{cyan} {\#CarryOnLearning} [/tex]
