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

Key Takeaways

  • Simple interest operates strictly on initial principal: Interest = Principal × annual rate × time (I = P · r · t); time must always be converted to fractional years when given in months (m / 12) or days (d / 360).
  • Profit percentage and loss percentage are calculated with Cost Price (CP) as the mandatory denominator: Profit % = (Profit / CP) × 100%; calculating margins on selling price is an accounting error on the AFPSAT.
  • Successive or consecutive percentage discounts cannot be added directly; two successive discounts of d1% and d2% produce a net effective discount of D_eff = d1 + d2 - (d1 · d2) / 100.
  • Age word problems require a structured Past-Present-Future matrix, translating verbal clues into linear equations while maintaining constant age differentials between individuals over time.
  • Mixture and solution problems apply the alligation principle of weighted concentrations: Quantity1 · Concentration1 + Quantity2 · Concentration2 = Total Quantity · Final Concentration.
Last updated: September 2026

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.


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

On the AFPSAT, all interest problems evaluate Simple Interest unless explicitly designated as compound interest. 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=PrtI = 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 / Banker's Interest (Standard AFPSAT convention): $t = \frac{d}{360}$
    • Exact Interest: $t = \frac{d}{365}$

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

Algebraic Manipulations to Isolate Specific Variables

P=Irtr=IPtt=IPrP = \frac{I}{r \cdot t} \qquad r = \frac{I}{P \cdot t} \qquad t = \frac{I}{P \cdot r}

Worked AFPSLAI Loan Calculation: A first lieutenant secures a salary loan of 150,000 from AFPSLAI 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=150,000×0.08×1.5=150,000×0.12=18,000I = 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=168,000A = P + I = 150,000 + 18,000 = 168,000 Total maturity amount is ₱168,000. Monthly Amortization: $\frac{168,000}{18} =$ ₱9,333.33 per month.

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 $\text{CP} > \text{SP}$: $\text{Loss} = \text{CP} - \text{SP}$.

Golden Rule of AFPSAT Commercial Math: Profit Percentage and Loss Percentage are always calculated relative to Cost Price (CP), never Selling Price, unless explicitly instructed.

Profit %=(ProfitCP)×100%=(SPCPCP)×100%\text{Profit } \% = \left(\frac{\text{Profit}}{\text{CP}}\right) \times 100\% = \left(\frac{\text{SP} - \text{CP}}{\text{CP}}\right) \times 100\%\nLoss %=(LossCP)×100%=(CPSPCP)×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

When selling at a profit of p%:SP=CP×(1+p)=CP×(100+p100)\text{When selling at a profit of } p\%: \quad \text{SP} = \text{CP} \times (1 + p) = \text{CP} \times \left(\frac{100 + p}{100}\right) When selling at a loss of l%:SP=CP×(1l)=CP×(100l100)\text{When selling at a loss of } l\%: \quad \text{SP} = \text{CP} \times (1 - l) = \text{CP} \times \left(\frac{100 - l}{100}\right) Solving for Cost Price from Selling Price: CP=SP1+porCP=SP1l\text{Solving for Cost Price from Selling Price: } \quad \text{CP} = \frac{\text{SP}}{1 + p} \quad \text{or} \quad \text{CP} = \frac{\text{SP}}{1 - l}

Markup, Marked Price & Markdown (Discounts)

  • Marked Price / List Price (MP): The initial retail sticker price displayed on goods.
  • Discount (Markdown): A reduction granted from the Marked Price: $\text{Discount} = \text{MP} - \text{SP}$. Discount %=(DiscountMP)×100%    SP=MP×(1d)\text{Discount } \% = \left(\frac{\text{Discount}}{\text{MP}}\right) \times 100\% \quad \implies \quad \text{SP} = \text{MP} \times (1 - d)

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.

Final Price=MP×(1d1)×(1d2)\text{Final Price} = \text{MP} \times (1 - d_1) \times (1 - d_2)

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

Deff=1(1d1)(1d2)=d1+d2d1d2100D_{\text{eff}} = 1 - (1 - d_1)(1 - d_2) = 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: Deff=20%+10%20×10100%=30%2%=28%D_{\text{eff}} = 20\% + 10\% - \frac{20 \times 10}{100}\% = 30\% - 2\% = 28\% (Note: The naive sum $20% + 10% = 30%$ is completely incorrect).
  2. Calculate Final Price Paid: Final Price=4,000×(10.28)=4,000×0.72=2,880\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.

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": (A10)=3(B10)(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(B+6)(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: 3B20=2B+6    B=263B - 20 = 2B + 6 \implies B = 26
  4. Solve for Senior Person A: A=2(26)+6=52+6=58A = 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.

4. Mixture and Solution Problems: The Alligation Method

Mixture problems require balancing quantities and concentrations when combining two or more solutions of different potencies, such as combining alcohol solutions for field medical sterilization or blending fuels of differing octane ratings.

The Fundamental Conservation of Solute Equation

Total Active Solute=Solute from Part 1+Solute from Part 2\text{Total Active Solute} = \text{Solute from Part 1} + \text{Solute from Part 2} QfinalCfinal=(Q1C1)+(Q2C2)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]

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

Worked Chemical Blending Scenario: An AFP field hospital possesses $60 \text{ liters}$ of an antiseptic solution with a concentration of $20%$ isopropyl alcohol. The medical officer needs 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: (60×0.20)+(x×0.70)=(60+x)×0.35(60 \times 0.20) + (x \times 0.70) = (60 + x) \times 0.35
  3. Expand and Collect Terms: 12+0.70x=21+0.35x12 + 0.70x = 21 + 0.35x 0.70x0.35x=21120.70x - 0.35x = 21 - 12 0.35x=90.35x = 9
  4. Divide to Solve for $x$: x=90.35=90035=180725.71 litersx = \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.


Comprehensive Practical Arithmetic Reference Matrix

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

Mathematical DomainCore EquationVariable DefinitionsVerbal Clue TranslationComputational Shortcut / Rule
Simple Interest$I = P \cdot r \cdot t$<br>$A = P(1 + r \cdot t)$$P = \text{Principal}$, $r = \text{Annual Rate}$, $t = \text{Years}$, $A = \text{Maturity Sum}$"Accrued over 9 months" $\implies t = \frac{9}{12} = \frac{3}{4}$<br>"Ordinary interest over 120 days" $\implies t = \frac{120}{360} = \frac{1}{3}$Always check if question asks for Interest ($I$) alone or Total Repayment ($A$).
Profit Margin$\text{Profit} = \text{SP} - \text{CP}$<br>$\text{Profit } % = \frac{\text{Profit}}{\text{CP}} \cdot 100$$\text{CP} = \text{Cost Price}$, $\text{SP} = \text{Selling Price}$"A gain of $25%$ on procurement cost" $\implies \text{SP} = 1.25 \cdot \text{CP}$Base of percentage is strictly Cost Price (CP). Never divide profit by Selling Price.
Loss Margin$\text{Loss} = \text{CP} - \text{SP}$<br>$\text{Loss } % = \frac{\text{Loss}}{\text{CP}} \cdot 100$$\text{CP} = \text{Cost Price}$, $\text{SP} = \text{Selling Price}$"Sold at a loss of $15%$" $\implies \text{SP} = 0.85 \cdot \text{CP}$$\text{CP} = \frac{\text{SP}}{1 - l}$; buying price is always larger than selling price during a loss.
Successive Discounts$D_{\text{eff}} = d_1 + d_2 - \frac{d_1 d_2}{100}$$d_1, d_2 = \text{Discount percentages}$"A $20%$ discount followed by an additional $15%$" $\implies 20 + 15 - \frac{300}{100} = 32%$Effective discount is strictly less than the arithmetic sum of discounts ($32% < 35%$).
Age Problems$(A \pm k) = m \cdot (B \pm k)$$A, B = \text{Current ages}$, $k = \text{Years offset}$, $m = \text{Multiple}$"Five years ago" $\implies (\text{Age} - 5)$<br>"In eight years" $\implies (\text{Age} + 8)$The age gap between two individuals $(A - B)$ remains strictly constant through time.
Mixture / Alligation$Q_1 C_1 + Q_2 C_2 = Q_{\text{tot}} C_m$<br>$\frac{Q_1}{Q_2} = \frac{C_m - C_2}{C_1 - C_m}$$Q = \text{Volume/Weight}$, $C = \text{Concentration}$"Diluted with pure water" $\implies C = 0%$<br>"Added pure alcohol" $\implies C = 100%$Use Alligation Cross for rapid ratio determination without setting up full polynomials.
Test Your Knowledge

An AFP quartermaster procurement officer acquires tactical combat boots at a cost price of ₱2,400 per pair. At what list price must the boots be marked so that the commissary can offer a 20% promotional discount to active service members and still secure a profit of 25% on the original cost price?

A
B
C
D
Test Your Knowledge

A staff sergeant deposits ₱80,000 into an Armed Forces financial cooperative savings account earning simple interest. If the account generates ₱7,200 in total interest over an 18-month deployment period, what annual simple interest rate does the cooperative provide?

A
B
C
D
Test Your Knowledge

Major Alonzo is currently four times as old as his son, Cadet Alonzo. In 14 years, Major Alonzo will be exactly twice as old as his son will be at that time. What is Major Alonzo's current age?

A
B
C
D