4.2 Percentages, Profit and Loss, Simple and Compound Interest

Percentages, Profit and Loss, Simple and Compound Interest

Financial stewardship and logistical precision are paramount competencies for commissioned officers in the Ghana Armed Forces. Senior and junior officers routinely oversee quartermaster acquisitions, monitor bulk fuel contracts, manage defense infrastructure project disbursements, and audit unit operational accounts. The arithmetic of percentages, profit margins, discounts, and capital interest rates tests a candidate's ability to manipulate financial data rapidly and accurately.

On the Officer Cadet Written Examination, commercial mathematics questions are structured to penalize candidates who rely on brute-force decimal multiplication. Success requires mastering fraction-to-percentage equivalencies, net change multipliers, and rapid difference formulas.

Core Percentage Mechanics & Fractional Equivalents

A percentage is a dimensionless ratio expressing a fraction of $100$. The multiplier formulation allows immediate one-step calculation of increases and decreases:

• Percentage Increase by $r\%$: Multiply by $\left(1 + \frac{r}{100}\right)$.
• Percentage Decrease by $r\%$: Multiply by $\left(1 - \frac{r}{100}\right)$.

Essential Fraction-to-Percentage Conversion Matrix

Candidates must commit these high-frequency fractional benchmarks to memory to avoid manual division during the examination.

The Asymmetry of Percentage Changes

A critical trap in military budgeting arises when a quantity increases or decreases, and the analyst attempts to reverse the operation by applying the same percentage figure.

• If a baseline quantity $B$ is reduced by $x\%$, the new baseline becomes $B\left(1 - \frac{x}{100}\right)$.
• To restore this reduced value back to the original baseline $B$, the required percentage increase is: $\text{Required Restoration } \% = \left(\frac{x}{100 - x} \times 100\right)\%$
• Conversely, if a quantity is increased by $x\%$, returning to the original value requires a reduction of: $\text{Required Reduction } \% = \left(\frac{x}{100 + x} \times 100\right)\%$

Operational Example: A battalion's fuel ration is cut by $20\%$. To return to the full allocation, the remaining fuel supply must be increased not by $20\%$, but by $\frac{20}{100 - 20} \times 100\% = \frac{20}{80} \times 100\% = 25\%$.

Successive Percentage Changes & The Net Change Formula

When a value is altered by multiple percentage adjustments in sequence, the net effect is NOT the simple arithmetic sum of the percentages. Instead, each successive adjustment compounds upon the previously adjusted value.

The 2-Variable Net Formula

If a quantity undergoes two successive percentage changes of $a\%$ and $b\%$, the net effective percentage change is given by: $\text{Net Change } \% = \left(a + b + \frac{a \times b}{100}\right)\%$

Sign Convention Rules:

• An increase is entered as a positive value ($+a$, $+b$).
• A decrease or discount is entered as a negative value ($-a$, $-b$).

Exemplar Cases

1. Two Successive Increases ($+20\%$ and $+10\%$): $\text{Net } \% = 20 + 10 + \frac{20 \times 10}{100} = 30 + 2 = +32\%$
2. Two Successive Discounts ($-20\%$ and $-10\%$): $\text{Net } \% = -20 - 10 + \frac{(-20)(-10)}{100} = -30 + 2 = -28\%$ (A net reduction of $28\%$, never $30\%$).
3. An Increase Followed by an Equal Decrease ($+x\%$ and $-x\%$): $\text{Net } \% = x - x + \frac{x(-x)}{100} = -\frac{x^2}{100}\%$ Whenever a quantity is increased by $x\%$ and subsequently decreased by $x\%$, there is ALWAYS a net loss equal to $\frac{x^2}{100}\%$.

Chained Multiplier Method for Multi-Stage Scenarios

For three or more successive adjustments, express each step as a decimal multiplier and compute their product: $\text{Final Value} = \text{Initial Value} \times \left(1 + \frac{r_1}{100}\right) \left(1 + \frac{r_2}{100}\right) \cdots \left(1 + \frac{r_k}{100}\right)$

Logistical Scenario: An ammunition reserve of $50,000$ rounds expands by $10\%$ in Phase 1, contracts by $20\%$ in Phase 2 due to field exercises, and is reinforced by $15\%$ in Phase 3: $\text{Final Reserve} = 50,000 \times (1.10) \times (0.80) \times (1.15) = 50,000 \times 1.012 = 50,600\text{ rounds}$ Net effective change: $+1.2\%$.

Profit, Loss, Margin vs. Markup & Discount Mechanics

Commercial procurement questions revolve around the core economic concepts of Cost Price, Selling Price, and Marked Price.

COST PRICE (CP)  ---+--- [Add Markup] --->  MARKED PRICE (MP)
                    |                              |
               [Add Profit]                [Subtract Discount]
                    |                              |
                    +----------->  SELLING PRICE (SP)

Core Equations of Commercial Arithmetic

• Cost Price ($CP$): The net expenditure incurred to produce or acquire an asset.
• Selling Price ($SP$): The revenue realized upon disposal or sale of the asset.
• Profit ($P$): Realized when $SP > CP$, where $P = SP - CP$.
• Loss ($L$): Realized when $CP > SP$, where $L = CP - SP$.
• Percentage Profit: Always calculated on Cost Price unless specifically stated otherwise: $\text{Profit } \% = \left(\frac{SP - CP}{CP} \times 100\right)\%$
• Percentage Loss: Always calculated on Cost Price: $\text{Loss } \% = \left(\frac{CP - SP}{CP} \times 100\right)\%$

Markup vs. Margin: The Critical Distinction

Confusion between Markup and Margin causes extensive score loss on officer selection tests:

• Markup: The profit expressed as a percentage of the Cost Price ($CP$): $\text{Markup } \% = \frac{\text{Profit}}{CP} \times 100\%$
• Margin (Profit Margin): The profit expressed as a percentage of the Selling Price ($SP$): $\text{Margin } \% = \frac{\text{Profit}}{SP} \times 100\%$

Rapid Conversion Invariants

Let markup be $m = \frac{P}{CP}$ and margin be $g = \frac{P}{SP}$: $\text{Margin} = \frac{\text{Markup}}{1 + \text{Markup}}, \qquad \text{Markup} = \frac{\text{Margin}}{1 - \text{Margin}}$

Discounts and Successive Trade Discounts

Discounts are ALWAYS calculated on the Marked Price ($MP$) (also called the List Price or Catalogue Price): $SP = MP \times \left(1 - \frac{d}{100}\right)$

When a vendor offers successive trade discounts of $d_1\%$ and $d_2\%$, the single equivalent discount $D_{\text{eq}}$ is: $D_{\text{eq}} = \left(d_1 + d_2 - \frac{d_1 d_2}{100}\right)\%$

Procurement Problem: A defense supplier lists tactical body armor at GH¢4,000 per unit, offering successive trade discounts of $15\%$ and $10\%$. What is the net purchase price?

• $D_{\text{eq}} = 15 + 10 - \frac{15 \times 10}{100} = 25 - 1.5 = 23.5\%$.
• Net purchase price: $SP = 4000 \times (1 - 0.235) = 4000 \times 0.765 = \text{GHS } 3,060$.

Simple and Compound Interest Dynamics

Interest represents the time value of money, reflecting capital borrowing costs or investment accrual.

1. Simple Interest (SI)

Simple interest accrues uniformly over time, calculated strictly on the initial principal $P$: $I = \frac{P \times R \times T}{100}$ $\text{Total Accumulated Amount } A = P + I = P\left(1 + \frac{R \times T}{100}\right)$ where $P$ is principal, $R$ is annual interest rate in percent, and $T$ is time in years.

2. Compound Interest (CI)

Compound interest accrues on the principal plus all interest accumulated from preceding periods. The total compounded amount $A$ after $t$ years with compounding frequency $n$ times per year is: $A = P\left(1 + \frac{R}{100n}\right)^{nt}$ $\text{Compound Interest Earned } CI = A - P = P\left[\left(1 + \frac{R}{100n}\right)^{nt} - 1\right]$

• Compounded Annually ($n = 1$): $A = P\left(1 + \frac{R}{100}\right)^t$
• Compounded Semi-Annually ($n = 2$): Rate becomes $\frac{R}{2}$, periods become $2t$: $A = P\left(1 + \frac{R}{200}\right)^{2t}$
• Compounded Quarterly ($n = 4$): Rate becomes $\frac{R}{4}$, periods become $4t$: $A = P\left(1 + \frac{R}{400}\right)^{4t}$

3. Rapid Difference Formulas: CI vs. SI

Examiners frequently construct questions comparing 2-year and 3-year CI and SI on the same principal and rate. Deriving these long-hand takes several minutes; deploying algebraic shortcuts solves them in seconds.

The 2-Year Difference Identity

$CI_2 - SI_2 = P\left(\frac{R}{100}\right)^2$

The 3-Year Difference Identity

$CI_3 - SI_3 = P\left(\frac{R}{100}\right)^2 \left(\frac{300 + R}{100}\right) = 3(CI_2 - SI_2) + P\left(\frac{R}{100}\right)^3$

4. Capital Asset Depreciation (Declining-Balance Method)

Military equipment, vehicles, and generators lose value over operational life cycles. Under the declining-balance method, asset value depreciates at an annual percentage rate $r\%$: $V_t = V_0 \left(1 - \frac{r}{100}\right)^t$ where $V_0$ is the acquisition cost, $r$ is the depreciation rate, and $V_t$ is the book value after $t$ years.

Worked Defense Example: A tactical troop transport truck is procured for GH¢500,000. If its value depreciates at $20\%$ per annum, what is its salvage value after $3$ years? $V_3 = 500,000 \times (1 - 0.20)^3 = 500,000 \times (0.80)^3 = 500,000 \times 0.512 = \text{GHS } 256,000$