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The Mathematics of Finance

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These are the solutions to Cambridge Ordinary Level (O-Level) past questions on the Mathematics of Finance.
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Formula Sheet
Financial Mathematics Literacy

(1.) Monthly interest payment = Monthly interest rate * Average balance

(2.) Net monthly cash flow = Monthly income − Monthly expenses


Simple Interest

$ (1.)\:\: SI = Prt \\[3ex] (2.)\:\: SI = A - P \\[3ex] (3.)\:\: P = \dfrac{SI}{rt} \\[5ex] (4.)\:\: t = \dfrac{SI}{Pr} \\[5ex] (5.)\:\: r = \dfrac{SI}{Pt} \\[5ex] (6.)\:\: A = P + SI \\[3ex] (7.)\:\: A = P(1 + rt) \\[3ex] (8.)\:\: P = \dfrac{A}{1 + rt} \\[5ex] (9.)\:\: t = \dfrac{A - P}{Pr} \\[5ex] (10.)\:\: r = \dfrac{A - P}{Pt} \\[5ex] (11.)\:\: SI = \dfrac{Art}{1 + rt} $


Compound Interest

$ (1.)\:\: A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[7ex] (2.)\:\: P = \dfrac{A}{\left(1 + \dfrac{r}{m}\right)^{mt}} \\[10ex] (3.)\:\: r = m\left[\left(\dfrac{A}{P}\right)^{\dfrac{1}{mt}} - 1\right] \\[10ex] (4.)\:\: r = m\left(10^{\dfrac{\log\left(\dfrac{A}{P}\right)}{mt}} - 1\right) \\[10ex] (5.)\:\: t = \dfrac{\log\left(\dfrac{A}{P}\right)}{m\log\left(1 + \dfrac{r}{m}\right)} \\[7ex] (6.)\:\: A = P + CI \\[3ex] (7.)\:\: CI = A - P \\[3ex] (8.)\:\: A = P(1 + i)^n \\[4ex] (9.)\:\: P = \dfrac{A}{(1 + i)^n} \\[7ex] (10.)\:\: i = \dfrac{r}{m} \\[5ex] (11.)\:\: n = mt \\[3ex] (12.)\;\; Total\;\;Return = \dfrac{A - P}{P} * 100\% \\[7ex] (13.)\;\; Annual\;\;Return = \left[\left(\dfrac{A}{P}\right)^{\dfrac{1}{t}} - 1\right] * 100\% \\[7ex] $ Future Value (Amount) of Cash Flows (Principal) for Several Years

$ (13.)\:\:At\:\:the\:\:end\:\:of\:\:each\:\:year:\:\: FV = PV\left(1 + \dfrac{r}{m}\right)^{m(last\:\:year - that\:\:year)} \\[7ex] (14.)\:\: Total\:FV = \Sigma FV $

Values of $m$

If Compounded: m =
Annually $1$ ($1$ time per year)
Also means every twelve months
Semiannually 2 (2 times per year)
Also means every six months
Quarterly 4 (4 times per year)
Also means every three months
Monthly 12 (12 times per year)
Also means every month
Weekly 52 (52 times per year)
Daily (Ordinary/Banker's Rule) 360 (360 times per year)
Daily (Exact) 365 (365 times per year)
(1.) (a.) The exchange rate between dollars ($) and Malaysian Ringgits (MYR) is $1 = 4.19 MYR.
The exchange rate between dollars ($) and Pakistani Rupees (PKR) is $1 = 179.12 PKR.
Find the exchange rate between Malaysian Ringgits and Pakistani Rupees.

(b.) Farhad invests $1500 in an account paying compound interest at a rate of 4% per year.
Gulsan invests $1500 in an account paying simple interest at a rate of x% per year.
Farhad and Gulsan have the same amount of money in their accounts at the end of 2 years.
Calculate the value of x


$ (a.) \\[3ex] \$ 1 = 4.19\;MYR \\[3ex] \$ 1 = 179.12\;PKR \\[3ex] \$1 = \$1 \implies \\[3ex] 4.19\;MYR = 179.12\;PKR \\[3ex] 1\;MYR = what\;PJR \\[3ex] \underline{\text{Unity Fraction Method}} \\[3ex] 1\;MYR * \dfrac{...PKR}{...MYR} \\[5ex] 1\;MYR * \dfrac{179.12\;PKR}{4.19\;MYR} \\[5ex] 1\;MYR = 42.74940334\;PKR \\[3ex] 1\;MYR \approx 42.75\;PKR \\[5ex] (b.) \\[3ex] principal = P \\[3ex] rate = r \\[3ex] \text{number of compounding periods per year} = m \\[3ex] time = t \\[3ex] amount = A \\[5ex] \underline{\text{Farhad: Compound Interest}} \\[3ex] P = \$1500 \\[3ex] r = 4\% = \dfrac{4}{100} = 0.04 \\[5ex] t = 2\;years \\[3ex] m = 1 \quad\dots\text{compounded once per year} \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = 1500 * \left(1 + \dfrac{0.04}{1}\right)^{1 * 2} \\[5ex] A = 1500(1.04)^2 \\[3ex] A = \$1622.40 \\[5ex] \underline{\text{Gulsan: Simple Interest}} \\[3ex] P = \$1500 \\[3ex] r = x\% = \dfrac{x}{100} = 0.01x \\[5ex] t = 2\;years \\[3ex] A = P(1 + rt) \\[3ex] A = 1500[1 + 2(0.01x)] \\[3ex] A = 1500(1 + 0.02x) \\[5ex] \text{The two amounts are the same after 2 years} \\[3ex] A = A \implies \\[3ex] 1500(1 + 0.02x) = 1622.4 \\[3ex] 1 + 0.02x = \dfrac{1622.4}{1500} \\[5ex] 0.02x = 1.0816 - 1 \\[3ex] x = \dfrac{0.0816}{0.02} \\[5ex] x = 4.08 \\[3ex] x\% = 4.08\% $
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