5.2 Practical Arithmetic: Simple Interest, Percentage Change, Profit/Loss & Age Problems

5.2 Practical Arithmetic: Simple Interest, Percentage Change, Profit/Loss & Age Problems

Context: Commissioned officers in the Armed Forces of the Philippines bear significant administrative and fiduciary responsibilities. Officers regularly oversee unit operational budgets, manage commissary and quartermaster procurement, process AFP Savings and Loan Association (AFPSLAI) and Armed Forces Mutual Benefit Association, Inc. (AFP-MBAI) member financial transactions, and blend critical resources such as military aviation fuel and medical antiseptic supplies. The AFPSAT tests these practical arithmetic competencies extensively.

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1. Financial Mathematics: Simple Interest ($I = P \cdot r \cdot t$)

Use the interest model stated in the question. Simple interest is calculated solely on the original principal amount deposited or borrowed for the entire duration of the transaction.

The Fundamental Interest Formula

$$I = P \cdot r \cdot t$$

Where:

• $I$ = Total interest accrued or paid (in Pesos).
• $P$ = Principal sum of money originally invested or borrowed.
• $r$ = Annual interest rate expressed as a decimal or fraction (e.g., $6\% = 0.06 = \frac{6}{100} = \frac{3}{50}$).
• $t$ = Time span expressed strictly in years.

Time Conversions and Adjustments

Examinees must convert non-annual timeframes into fractional years before inserting them into the formula:

• Time in Months ($m$): $t = \frac{m}{12}$
• Time in Days ($d$):
  • Ordinary interest (when a 360-day year is specified): $t = \frac{d}{360}$
  • Exact simple interest (when a 365-day year is specified): $t = \frac{d}{365}$

Total Accumulated Maturity Amount ($A$)

The total amount owed or collected at the conclusion of the term is the sum of the principal and accrued interest:

$$A = P + I = P + (P \cdot r \cdot t) = P(1 + r \cdot t)$$

Algebraic Manipulations to Isolate Specific Variables

$$P = \frac{I}{r \cdot t} \qquad r = \frac{I}{P \cdot t} \qquad t = \frac{I}{P \cdot r}$$

Hypothetical Simple-Interest Loan: An officer borrows ₱150,000 at an annual simple interest rate of $8\%$. The loan term is set for $18 \text{ months}$. What is the total interest accrued, and what is the total maturity amount that must be repaid?

1. Convert Time to Years: $t = \frac{18 \text{ months}}{12 \text{ months/year}} = 1.5 \text{ years}$.
2. Convert Rate to Decimal: $r = 8\% = 0.08 = \frac{8}{100}$.
3. Compute Simple Interest: $$I = 150,000 \times 0.08 \times 1.5 = 150,000 \times 0.12 = 18,000$$ Total interest accrued is ₱18,000.
4. Compute Total Maturity Amount: $$A = P + I = 150,000 + 18,000 = 168,000$$ Total maturity amount is ₱168,000. Equal monthly payment under this example’s add-on-interest assumption (with no fees): $\frac{168,000}{18} =$ ₱9,333.33 per month.

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2. Commercial Mathematics: Cost, Selling Price, Profit & Loss

Quartermaster supply and commissary operations require precise accounting of acquisition costs, markups, discounts, and realized profit margins.

Key Commercial Definitions

• Cost Price (CP): The baseline purchase or production expenditure incurred to obtain an item.
• Selling Price (SP): The actual gross revenue collected from selling the item.
• Profit (Gain): Realized when $\text{SP} > \text{CP}$: $\text{Profit} = \text{SP} - \text{CP}$.
• Loss: Incurred when $\text{CP} > \text{SP}$: $\text{Loss} = \text{CP} - \text{SP}$.

Choose the percentage base: The profit and loss percentages below are measured relative to Cost Price (CP). A profit margin is a different ratio, profit divided by Selling Price (SP). Follow the base named in the question.

$$\text{Profit } \% = \left(\frac{\text{Profit}}{\text{CP}}\right) \times 100\% = \left(\frac{\text{SP} - \text{CP}}{\text{CP}}\right) \times 100\%$$ $$\text{Loss } \% = \left(\frac{\text{Loss}}{\text{CP}}\right) \times 100\% = \left(\frac{\text{CP} - \text{SP}}{\text{CP}}\right) \times 100\%$$

Direct Relations Between Selling Price and Cost Price

Here p and l are the numerical percentages: p = 20 means a profit of 20% on cost, and l = 20 means a loss of 20% on cost. Divide each by 100 before using it as a decimal rate.

$$\text{When selling at a profit of } p\%: \quad \text{SP} = \text{CP} \times \left(1 + \frac{p}{100}\right) = \text{CP} \times \left(\frac{100 + p}{100}\right)$$ $$\text{When selling at a loss of } l\%: \quad \text{SP} = \text{CP} \times \left(1 - \frac{l}{100}\right) = \text{CP} \times \left(\frac{100 - l}{100}\right)$$ $$\text{Solving for Cost Price from Selling Price: } \quad \text{CP} = \frac{\text{SP}}{1 + p/100} \quad \text{or} \quad \text{CP} = \frac{\text{SP}}{1 - l/100}$$

Markup, Marked Price & Markdown (Discounts)

Use d, d₁, and d₂ as numerical discount percentages, so d = 20 represents a 20% discount.

• Marked Price / List Price (MP): The initial retail sticker price displayed on goods.
• Discount (Markdown): A reduction granted from the Marked Price: $\text{Discount} = \text{MP} - \text{SP}$.

$$\text{Discount } \% = \left(\frac{\text{Discount}}{\text{MP}}\right) \times 100\% \quad \implies \quad \text{SP} = \text{MP} \times \left(1 - \frac{d}{100}\right)$$

Consecutive (Successive) Discounts

When an item is subjected to multiple successive discounts (e.g., a $20\%$ commercial markdown followed by a $10\%$ military personnel discount), the second discount applies only to the reduced intermediate price, not the original list price.

$$\text{Final Price} = \text{MP} \times \left(1 - \frac{d_1}{100}\right) \times \left(1 - \frac{d_2}{100}\right)$$

Single Equivalent Discount Formula ($D_{\text{eff}}$):

$$D_{\text{eff}} = 100\left[1 - \left(1 - \frac{d_1}{100}\right)\left(1 - \frac{d_2}{100}\right)\right] = d_1 + d_2 - \frac{d_1 \cdot d_2}{100}$$

Worked Successive Discount Example: A tactical equipment supply store lists a modular field pack for 4,000. The store advertises a storewide sale of $20\%$, and an active-duty military officer receives an additional $10\%$ armed forces discount at checkout. What is the single equivalent discount rate, and what is the final price paid?

1. Calculate Single Equivalent Discount: $$D_{\text{eff}} = 20 + 10 - \frac{20 \times 10}{100} = 30 - 2 = 28$$ The equivalent discount is 28%. (Note: The naive sum $20\% + 10\% = 30\%$ is completely incorrect).
2. Calculate Final Price Paid: $$\text{Final Price} = 4,000 \times (1 - 0.28) = 4,000 \times 0.72 = 2,880$$ Final price paid is ₱2,880. Step-by-step verification: After first discount (20% off ₱4,000): $4,000 - 800 =$ ₱3,200. After second discount (10% off ₱3,200): $3,200 - 320 =$ ₱2,880.

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3. Age Word Problems: Systematic Tabular Modeling

Age problems test an examinee's ability to translate temporal relationships into systems of linear equations. The defining mathematical invariant in all age problems is:

Invariant Principle: The chronological difference in age between two people remains strictly constant across all points in time.

The Standard Past-Present-Future Matrix

To eliminate confusion with past and future offsets, always construct a three-column grid:

+--------------------------------------------------------------------------+
|                       AGE TRANSLATION MATRIX SETUP                       |
+---------------------+-------------------+----------------+---------------+
| Person              | Past (- Y years)  | Present (Now)  | Future (+ Z y)|
+---------------------+-------------------+----------------+---------------+
| Senior Person (A)   | A - Y             | A              | A + Z         |
| Junior Person (B)   | B - Y             | B              | B + Z         |
+---------------------+-------------------+----------------+---------------+

Translating Verbal Clues to Algebraic Expressions

• "Ten years ago, Colonel A was three times as old as Captain B": $$(A - 10) = 3 \cdot (B - 10)$$
• "In six years, Colonel A will be twice as old as Captain B will be then": $$(A + 6) = 2 \cdot (B + 6)$$

Step-by-Step Algebraic Solution:

1. Expand Equation 1: $A - 10 = 3B - 30 \implies A = 3B - 20$.
2. Expand Equation 2: $A + 6 = 2B + 12 \implies A = 2B + 6$.
3. Equate Expressions for A: $$3B - 20 = 2B + 6 \implies B = 26$$
4. Solve for Senior Person A: $$A = 2(26) + 6 = 52 + 6 = 58$$
5. Verify Against Problem Constraints: Ten years ago, $A$ was $48$ and $B$ was $16$ ($48 = 3 \times 16$, holds). In six years, $A$ will be $64$ and $B$ will be $32$ ($64 = 2 \times 32$, holds). Colonel A is currently 58 years old, and Captain B is 26 years old.

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4. Mixture and Solution Problems: The Alligation Method

Mixture problems require balancing quantities and concentrations when combining two or more solutions of different potencies, using an idealized model in which volumes add and the same substance is conserved. This arithmetic model does not establish a medically effective concentration or imply that fuel octane blends linearly.

The Fundamental Conservation of Solute Equation

$$\text{Total Active Solute} = \text{Solute from Part 1} + \text{Solute from Part 2}$$ $$Q_{\text{final}} \cdot C_{\text{final}} = (Q_1 \cdot C_1) + (Q_2 \cdot C_2)$$

Where $Q_1 + Q_2 = Q_{\text{final}}$, $Q$ represents liquid volume, and $C$ represents concentration percentage.

The Alligation Cross Shortcut

When an exam question asks for the ratio of quantities ($Q_1 : Q_2$) needed to achieve a target mixture concentration ($C_m$) from a higher concentration ($C_H$) and a lower concentration ($C_L$):

   Higher Strength (C_H)           (C_m - C_L)  [Parts of Higher]
                     \           /
                      Mean (C_m)
                     /           \
   Lower Strength (C_L)            (C_H - C_m)  [Parts of Lower]

$$\frac{\text{Quantity of Higher } (Q_H)}{\text{Quantity of Lower } (Q_L)} = \frac{C_m - C_L}{C_H - C_m}$$

Idealized Concentration Exercise: A laboratory has $60 \text{ liters}$ of an antiseptic solution with a concentration of $20\%$ isopropyl alcohol. The exercise asks to increase the strength of the solution to $35\%$ by adding a stock solution containing $70\%$ alcohol. How many liters of the $70\%$ stock solution must be blended in?

1. Define Knowns and Variables:
   • $Q_1 = 60 \text{ L}$, $C_1 = 20\% = 0.20$.
   • $Q_2 = x \text{ L}$ (unknown volume of stock), $C_2 = 70\% = 0.70$.
   • $Q_{\text{final}} = 60 + x$, $C_{\text{final}} = 35\% = 0.35$.
2. Set Up the Solute Conservation Equation: $$(60 \times 0.20) + (x \times 0.70) = (60 + x) \times 0.35$$
3. Expand and Collect Terms: $$12 + 0.70x = 21 + 0.35x$$ $$0.70x - 0.35x = 21 - 12$$ $$0.35x = 9$$
4. Divide to Solve for $x$: $$x = \frac{9}{0.35} = \frac{900}{35} = \frac{180}{7} \approx 25.71 \text{ liters}$$

Alligation Cross Check: Ratio of $Q_{70} : Q_{20} = (35 - 20) : (70 - 35) = 15 : 35 = 3 : 7$. Since $Q_{20} = 60 \text{ L}$, $Q_{70} = 60 \times \frac{3}{7} = \frac{180}{7} \text{ liters}$. Both methods yield identical results.

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Comprehensive Practical Arithmetic Reference Matrix

The following table serves as a comprehensive formula and translation guide for practical arithmetic problems: