5.3 Mathematical Word Problems & Quantitative Formulation

5.3 Mathematical Word Problems & Quantitative Formulation

Pure algebraic equations rarely present themselves neatly packaged in operational military environments. A staff officer does not encounter printed equations like $3x + 4y = 29000$; instead, the officer receives a radio dispatch reporting fuel consumption across a forward motorized patrol, troop age profiles across an infantry company, or chemical decontamination blends required for field water filtration. The true test of quantitative aptitude is formulation: the ability to deconstruct a verbal narrative, extract underlying numerical constraints, map verbal relationships into rigorous symbolic equations, and verify that solutions satisfy real-world operational realities.

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The 4-Phase Modeling Framework

Every quantitative word problem on the GAF Officer Cadet Written Examination can be methodically resolved using the 4-Phase Modeling Framework:

[Phase 1: DECODE]       --> Read narrative, define variables with units, extract boundary conditions
[Phase 2: MODEL]        --> Translate verbal syntax into formal equations or inequalities
[Phase 3: SOLVE]        --> Apply algebraic techniques cleanly to isolate target unknowns
[Phase 4: SANITY-CHECK] --> Validate solution against domain rules (non-negative, integer, physical logic)

The Verbal-to-Algebraic Translation Syntax

Word problems utilize predictable linguistic signposts that map directly onto mathematical operators:

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Archetype 1: Age Progression & Ratio Shift Problems

Age problems assess your ability to track quantities shifting across time. The essential mathematical principle is that time advances equally for all participants. If 6 years elapse, every individual's age increases by exactly 6 years.

Standard Formulation Protocol

1. Let the present age of the entities be the primary variables (e.g., $M$ and $S$).
2. Construct an Age Timeline Table:
   • Past ($n$ years ago): Age $= (\text{Present} - n)$
   • Present: Age $= \text{Present}$
   • Future ($m$ years hence): Age $= (\text{Present} + m)$
3. Formulate equations relating the past or future ages based on the given ratio or multiple.

Worked Example 1: Age Multiplier Shift

A senior warrant officer is currently twice as old as a new recruit. Ten years ago, the warrant officer was three times as old as the recruit was then. What are their present ages?

Phase 1: Decode

• Let recruit's present age $= r$ (in years).
• Warrant officer's present age $= 2r$ (in years).
• Past timeframe $= -10$ years.

Phase 2: Model Ten years ago:

• Recruit's age was $(r - 10)$.
• Warrant officer's age was $(2r - 10)$.
• Condition: Warrant officer's past age $= 3 \times$ recruit's past age.

$2r - 10 = 3(r - 10)$

Phase 3: Solve

$2r - 10 = 3r - 30$

$30 - 10 = 3r - 2r$

$r = 20$

• Recruit's present age: 20 years.
• Warrant officer's present age: $2 \times 20 = 40$ years.

Phase 4: Sanity-Check Ten years ago, the recruit was $20 - 10 = 10$ and the warrant officer was $40 - 10 = 30$. Since $30 = 3 \times 10$, the condition holds.

Worked Example 2: Dual Ratio Shift Across Past and Future

Three years ago, the ratio of a captain's age to a lieutenant's age was $4 : 3$. In five years, the ratio of their ages will become $5 : 4$. Determine their present ages.

Phase 1: Decode & Model using a common scaling parameter $x$ Let the ages three years ago be $4x$ and $3x$:

• Captain's age 3 years ago $= 4x$
• Lieutenant's age 3 years ago $= 3x$

Advance to the present (add 3 years):

• Captain's present age $= 4x + 3$
• Lieutenant's present age $= 3x + 3$

Advance to 5 years in the future (add 5 more years, total $+8$ years from baseline):

• Captain's future age $= 4x + 3 + 5 = 4x + 8$
• Lieutenant's future age $= 3x + 3 + 5 = 3x + 8$

Phase 2: Formulate the future ratio equation

$\frac{4x + 8}{3x + 8} = \frac{5}{4}$

Phase 3: Cross-multiply and Solve

$4(4x + 8) = 5(3x + 8)$

$16x + 32 = 15x + 40$

$16x - 15x = 40 - 32$

$x = 8$

Now calculate their present ages:

• Captain's present age: $4(8) + 3 = 32 + 3 = 35$ years.
• Lieutenant's present age: $3(8) + 3 = 24 + 3 = 27$ years.

Phase 4: Sanity-Check

• 3 years ago: Captain was 32, Lieutenant was 24. Ratio: $32 : 24 = 4 : 3$. Verified.
• In 5 years: Captain will be 40, Lieutenant will be 32. Ratio: $40 : 32 = 5 : 4$. Verified.

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Archetype 2: Positional Value & Reversed Two-Digit Numbers

In our base-10 positional numeral system, the face value of a digit is multiplied by its positional weight. For a two-digit integer with tens digit $t$ and units digit $u$:

$\text{Original Number: } N = 10t + u$ $\text{Reversing the digits yields: } N_{\text{rev}} = 10u + t$

where $t, u \in \{0, 1, 2, \dots, 9\}$ and $t \neq 0$.

The Fundamental Invariants of Digit Reversal

Two mathematical invariants govern all two-digit reversal problems:

1. Difference Invariant (Divisible by 9): $N - N_{\text{rev}} = (10t + u) - (10u + t) = 9(t - u)$ The difference between any two-digit number and its reverse is always divisible by 9, and dividing that difference by 9 immediately reveals the difference between the digits $(t - u)$.

2. Sum Invariant (Divisible by 11): $N + N_{\text{rev}} = (10t + u) + (10u + t) = 11(t + u)$ The sum of any two-digit number and its reverse is always divisible by 11, and dividing that sum by 11 reveals the sum of the digits $(t + u)$.

Worked Example 3: Reversal of Tactical Unit Identifier

A two-digit tactical callsign number has the property that the sum of its digits is 11. If the digits are reversed, the resulting number exceeds the original number by 45. Determine the original tactical callsign number.

Phase 1: Decode

• Let the tens digit be $t$ and units digit be $u$.
• Original number $N = 10t + u$.
• Reversed number $N_{\text{rev}} = 10u + t$.

Phase 2: Model

• Condition 1 (Sum of digits): $t + u = 11$
• Condition 2 (Reversal value shift): $N_{\text{rev}} - N = 45$

$(10u + t) - (10t + u) = 45$

$9(u - t) = 45 \implies u - t = 5$

Phase 3: Solve the simultaneous system

$\text{Equation (1): } u + t = 11$ $\text{Equation (2): } u - t = 5$

Add Equation (1) and Equation (2):

$(u + t) + (u - t) = 11 + 5$

$2u = 16 \implies u = 8$

Substitute $u = 8$ into Equation (1):

$t = 11 - 8 = 3$

The original number is $N = 10(3) + 8 = 38$.

Phase 4: Sanity-Check

• Sum of digits: $3 + 8 = 11$. Correct.
• Reversed number: $83$.
• Shift: $83 - 38 = 45$. Correct.

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Archetype 3: Mixture, Concentration & The Rule of Alligation

Mixture problems involve combining substances of differing concentrations (such as fuel blends, saline disinfectants, or metal alloys) to produce a compound of desired target concentration. All mixture problems rest upon the Conservation of Pure Substance:

$\text{Total Pure Solute} = q_1 c_1 + q_2 c_2 = (q_1 + q_2) c_m$

where $q_1, q_2$ are quantities and $c_1, c_2, c_m$ are respective percentage concentrations.

The Rule of Alligation (Cross Method)

The Rule of Alligation provides a rapid visual technique for determining the ratio in which two ingredients at concentrations $C_c$ (cheaper/lower) and $C_d$ (dearer/higher) must be blended to produce a mean concentration $C_m$:

  Higher Concentration (Cd)             Difference (Cm - Cc)
                              \       /
                               Mean (Cm)
                              /       \
  Lower Concentration (Cc)              Difference (Cd - Cm)

The ratio of the quantity of lower concentration ($Q_c$) to higher concentration ($Q_d$) is given by:

$\frac{Q_c}{Q_d} = \frac{C_d - C_m}{C_m - C_c}$

Worked Example 4: Workshop Coolant Blend

An Electrical and Mechanical Engineers (EME) vehicle workshop must prepare 50 liters of engine coolant containing 40% glycol by volume. Its store holds two stocks: a 70% glycol concentrate and a 20% glycol pre-mix. How many liters of each stock must be blended to produce the required 50 liters?

Phase 1: Decode

• Lower concentration $C_c = 20\%$
• Higher concentration $C_d = 70\%$
• Target mean concentration $C_m = 40\%$
• Total target volume $V = Q_c + Q_d = 50\text{ liters}$

Phase 2: Model via the Rule of Alligation

$\text{Numerator difference: } C_d - C_m = 70 - 40 = 30$ $\text{Denominator difference: } C_m - C_c = 40 - 20 = 20$

$\frac{Q_{20\%}}{Q_{70\%}} = \frac{30}{20} = \frac{3}{2}$

Phase 3: Solve for individual volumes The required volume ratio is $3 : 2$, giving a total of $3 + 2 = 5$ proportional parts.

• Value of 1 part: $\frac{50\text{ liters}}{5} = 10\text{ liters}$
• Volume of 20% solution required: $3 \times 10 = 30$ liters
• Volume of 70% solution required: $2 \times 10 = 20$ liters

Phase 4: Sanity-Check via Solute Balance

• Pure glycol from 20% solution: $30 \times 0.20 = 6\text{ liters}$
• Pure glycol from 70% solution: $20 \times 0.70 = 14\text{ liters}$
• Total pure glycol: $6 + 14 = 20\text{ liters}$
• Final concentration: $\frac{20\text{ liters}}{50\text{ liters}} = 0.40 = 40\%$. The blend is exact.

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Archetype 4: Multi-Denomination Currency & Resource Allocation

Currency and resource distribution problems model scenarios where two distinct items possess both a count (number of units) and a weighted value (monetary denomination or capacity).

Worked Example 5: Paymaster Cash Disbursement

A Ghana Armed Forces paymaster at Burma Camp disburses GH¢4,800 in operational allowances to a security detachment. The disbursement consists exclusively of GH¢50 and GH¢100 banknotes, totaling exactly 68 banknotes. How many notes of each denomination were disbursed?

Phase 1: Decode

• Let $x$ be the number of GH¢50 notes.
• Let $y$ be the number of GH¢100 notes.
• Total banknote count $= 68$.
• Total monetary value $= \text{GHS } 4,800$.

Phase 2: Model Formulate the count equation and value equation:

$\text{Count Equation: } x + y = 68$ $\text{Value Equation: } 50x + 100y = 4800$

Phase 3: Solve Simplify the value equation by dividing every term by 50:

$\frac{50x}{50} + \frac{100y}{50} = \frac{4800}{50} \implies x + 2y = 96$

Now subtract the count equation ($x + y = 68$) from this simplified value equation:

$(x + 2y) - (x + y) = 96 - 68$

$y = 28$

Back-substitute $y = 28$ into the count equation:

$x + 28 = 68 \implies x = 68 - 28 = 40$

The paymaster disbursed 40 notes of GH¢50 and 28 notes of GH¢100.

Phase 4: Sanity-Check

• Total notes: $40 + 28 = 68$ notes. Verified.
• Value of GH¢50 notes: $40 \times 50 = \text{GHS } 2,000$.
• Value of GH¢100 notes: $28 \times 100 = \text{GHS } 2,800$.
• Total cash: $2000 + 2800 = \text{GHS } 4,800$. Verified.

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Archetype 5: Fraction & Remainder Supply Depletion

A recurrent trap in word problems involves distinguishing between a fraction of the original total versus a fraction of the remaining balance.

Worked Example 6: Sequential Logistics Fuel Consumption

A peacekeeping supply convoy departs base with a full auxiliary fuel reserve. On Day 1 of the mission, the convoy consumes $\frac{1}{4}$ of the total initial reserve. On Day 2, tactical maneuvers consume $\frac{2}{5}$ of the remaining fuel. On Day 3, the convoy burns 360 liters, leaving exactly $\frac{1}{5}$ of the original full reserve in the tanks. What was the total initial fuel reserve in liters?

Phase 1: Decode

• Let the total initial fuel reserve be $F$ liters.

Phase 2: Model the step-by-step balance

1. Day 1 Consumption: $\text{Consumed}_1 = \frac{1}{4}F$ $\text{Remainder}_1 = F - \frac{1}{4}F = \frac{3}{4}F$

2. Day 2 Consumption (Fraction of Remainder): $\text{Consumed}_2 = \frac{2}{5} \times \left( \frac{3}{4}F \right) = \frac{6}{20}F = \frac{3}{10}F$ $\text{Remainder}_2 = \frac{3}{4}F - \frac{3}{10}F = \frac{15F - 6F}{20} = \frac{9}{20}F$

3. Day 3 Balance: On Day 3, 360 liters are consumed, and the remaining fuel equals $\frac{1}{5}F$:

$\frac{9}{20}F - 360 = \frac{1}{5}F$

Phase 3: Solve for $F$ Transpose terms to collect coefficients of $F$:

$\frac{9}{20}F - \frac{1}{5}F = 360$

Convert $\frac{1}{5}$ to twentieths ($\frac{4}{20}$):

$\frac{9}{20}F - \frac{4}{20}F = 360$

$\frac{5}{20}F = 360$

$\frac{1}{4}F = 360$

$F = 360 \times 4 = 1440\text{ liters}$

The convoy's initial fuel reserve was 1,440 liters.

Phase 4: Sanity-Check

• Initial: $1,440\text{ L}$.
• Day 1: Burns $\frac{1}{4}(1440) = 360\text{ L}$. Remainder $= 1,080\text{ L}$.
• Day 2: Burns $\frac{2}{5}(1080) = 432\text{ L}$. Remainder $= 1080 - 432 = 648\text{ L}$.
• Day 3: Burns $360\text{ L}$. Remainder $= 648 - 360 = 288\text{ L}$.
• Check against $\frac{1}{5}$ of initial: $\frac{1}{5}(1440) = 288\text{ L}$. The balance matches down to the single liter.