5.1 Motion, Work Rate & Fluid Flow Word Problems

Armed Forces of the Philippines AFP

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5.1 Motion, Work Rate & Fluid Flow Word Problems 5.2 Practical Arithmetic: Simple Interest, Percentage Change, Profit/Loss & Age Problems 5.3 Geometry Fundamentals: Perimeter, Area, Volume, Pythagorean Theorem & Angles 5.4 Number Series, Letter Series, Averages & Data Interpretation

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5.1 Motion, Work Rate & Fluid Flow Word Problems

Key Takeaways Explain This Topic AI

Uniform motion problems pivot on Distance = Rate × Time (D = R · T); approaching objects sum their speeds (R_rel = R1 + R2), whereas overtaking objects subtract their speeds (R_rel = |R1 - R2|).

Average speed over equal round-trip distances is governed by the harmonic mean (2 · R1 · R2 / (R1 + R2)); applying the arithmetic mean ((R1 + R2) / 2) is a standard test trap that severely overestimates average velocity.

Fluid motion and aerodynamic drift modify ground speed additively: downstream/tailwind speed equals vehicle speed plus medium speed (R + C), whereas upstream/headwind speed equals vehicle speed minus medium speed (R - C).

Collaborative work rate operates inversely to completion time (Rate = 1 / T); combined multi-worker efficiency equals the sum of individual rates (1 / T_total = 1 / T1 + 1 / T2 + ... + 1 / Tn).

Fluid reservoir and cistern problems model inflows as positive production rates (+1 / T_in) and leakage or drainage conduits as negative consumption rates (-1 / T_drain).

Last updated: September 2026

5.1 Motion, Work Rate & Fluid Flow Word Problems

Core Principle: In military field operations, calculating logistical transit intervals, coordinating multi-column tactical movements, scheduling naval interdictions, and projecting combat engineering completion timelines require absolute mathematical accuracy under pressure. On the AFP Service Aptitude Test (AFPSAT) , calculators are strictly barred. Officer candidates must rapidly construct linear algebraic models and execute mental arithmetic using structured rate-time templates.

1. Uniform Motion Fundamentals ( D = R × T D = R \times T D = R × T )

All classical motion problems on the AFPSAT evaluate the relationship between three core variables: Distance ( D D D ) , Rate or Speed ( R R R ) , and Elapsed Time ( T T T ) . Distance = Rate × Time ( D = R ⋅ T ) \text{Distance} = \text{Rate} \times \text{Time} \quad (D = R \cdot T) Distance = Rate × Time ( D = R ⋅ T ) Rate = Distance Time ( R = D T ) Time = Distance Rate ( T = D R ) \text{Rate} = \frac{\text{Distance}}{\text{Time}} \quad \left(R = \frac{D}{T}\right) \qquad \text{Time} = \frac{\text{Distance}}{\text{Rate}} \quad \left(T = \frac{D}{R}\right) Rate = Time Distance ​ ( R = T D ​ ) Time = Rate Distance ​ ( T = R D ​ )

Unit Consistency and Dimensional Alignment

A frequent source of test errors is mismatched physical units. Examinees must verify dimensional alignment prior to calculating:

Speed in kilometers per hour (km/h) with time in hours ( h h h ) and distance in kilometers ( k m km km ).

Speed in meters per second (m/s) with time in seconds ( s s s ) and distance in meters ( m m m ). Conversion Factor: 1 km/h = 1 , 000 m 3 , 600 s = 5 18 m/s ⟺ 1 m/s = 18 5 km/h = 3.6 km/h \text{Conversion Factor: } 1 \text{ km/h} = \frac{1,000 \text{ m}}{3,600 \text{ s}} = \frac{5}{18} \text{ m/s} \quad \Longleftrightarrow \quad 1 \text{ m/s} = \frac{18}{5} \text{ km/h} = 3.6 \text{ km/h} Conversion Factor: 1 km/h = 3 , 600 s 1 , 000 m ​ = 18 5 ​ m/s ⟺ 1 m/s = 5 18 ​ km/h = 3.6 km/h

Naval Velocity: In maritime interdiction scenarios involving the Philippine Navy or Coast Guard, speed is expressed in knots (1 knot = 1 nautical mile per hour = 1.852 km/h). +--------------------------------------------------------------------------+

• | RATE-TIME-DISTANCE TABULAR SETUP |
• +---------------------+-------------------+----------------+---------------+
• | Entity / Scenario | Rate (R) | Time (T) | Distance (D) |
• +---------------------+-------------------+----------------+---------------+
• | Element A (Lead) | R_A | T_A | D_A = R_A·T_A |
• | Element B (Chase) | R_B | T_B | D_B = R_B·T_B |
• +---------------------+-------------------+----------------+---------------+
• | System Condition | Closing: R_B - R_A| Elapsed Time | D_B - D_A = 0 |
• | | Opposing: R_A+R_B | Meeting Time | D_A+D_B = Tot |
• +---------------------+-------------------+----------------+---------------+
• 2. Core Motion Problem Archetypes

Every motion word problem on military entrance examinations falls into one of four structural archetypes. Identifying the archetype immediately dictates the setup equation.

Archetype A: Opposite Direction / Converging Motion (Approaching Objects)

When two bodies move directly toward each other from a fixed initial separation distance ( D total D_{\text{total}} D total ​ ), or depart simultaneously in opposite directions from a single central point:

Their effective closing rate is the sum of their individual speeds: R relative = R 1 + R 2 R_{\text{relative}} = R_1 + R_2 R relative ​ = R 1 ​ + R 2 ​ .

The total distance traversed collectively equals the sum of the individual distances: D total = D 1 + D 2 = ( R 1 ⋅ T ) + ( R 2 ⋅ T ) = ( R 1 + R 2 ) ⋅ T D_{\text{total}} = D_1 + D_2 = (R_1 \cdot T) + (R_2 \cdot T) = (R_1 + R_2) \cdot T D total ​ = D 1 ​ + D 2 ​ = ( R 1 ​ ⋅ T ) + ( R 2 ​ ⋅ T ) = ( R 1 ​ + R 2 ​ ) ⋅ T .

The time elapsed until convergence is: T meet = D total R 1 + R 2 T_{\text{meet}} = \frac{D_{\text{total}}}{R_1 + R_2} T meet ​ = R 1 ​ + R 2 ​ D total ​ ​

Worked Tactical Scenario: A Philippine Army logistics convoy departs Camp O'Donnell traveling south along the expressway toward Fort Magsaysay at an average sustained speed of 45 km/h 45 \text{ km/h} 45 km/h . At the identical moment, an armored reconnaissance patrol leaves Fort Magsaysay heading north toward Camp O'Donnell at 55 km/h 55 \text{ km/h} 55 km/h . The two military installations are separated by a highway distance of 250 km 250 \text{ km} 250 km . How long after departure do the two military units meet, and how far has the logistics convoy traveled from Camp O'Donnell?

Isolate Knowns: R 1 = 45 km/h R_1 = 45 \text{ km/h} R 1 ​ = 45 km/h , R 2 = 55 km/h R_2 = 55 \text{ km/h} R 2 ​ = 55 km/h , D total = 250 km D_{\text{total}} = 250 \text{ km} D total ​ = 250 km .

Compute Relative Speed: R rel = 45 + 55 = 100 km/h R_{\text{rel}} = 45 + 55 = 100 \text{ km/h} R rel ​ = 45 + 55 = 100 km/h .

Solve for Meeting Time: T = 250 km 100 km/h = 2.5 hours T = \frac{250 \text{ km}}{100 \text{ km/h}} = 2.5 \text{ hours} T = 100 km/h 250 km ​ = 2.5 hours (2 hours and 30 minutes).

Calculate Convoy Distance: D convoy = R 1 ⋅ T = 45 km/h × 2.5 h = 112.5 km D_{\text{convoy}} = R_1 \cdot T = 45 \text{ km/h} \times 2.5 \text{ h} = 112.5 \text{ km} D convoy ​ = R 1 ​ ⋅ T = 45 km/h × 2.5 h = 112.5 km from Camp O'Donnell.

Archetype B: Same Direction Motion (Catch-Up & Tactical Interception)

In an overtaking or interception problem, a faster pursuit craft pursues a slower vehicle that has already departed or holds an initial positional lead ( D lead D_{\text{lead}} D lead ​ ):

The net closing speed is the absolute difference between their rates: R closing = R pursuit − R target R_{\text{closing}} = R_{\text{pursuit}} - R_{\text{target}} R closing ​ = R pursuit ​ − R target ​ , where R pursuit > R target R_{\text{pursuit}} > R_{\text{target}} R pursuit ​ > R target ​ .

If the target has a head start of t headstart t_{\text{headstart}} t headstart ​ hours, the initial gap is D lead = R target × t headstart D_{\text{lead}} = R_{\text{target}} \times t_{\text{headstart}} D lead ​ = R target ​ × t headstart ​ .

The time required for the pursuit unit to eliminate the lead distance is: T intercept = D lead R pursuit − R target T_{\text{intercept}} = \frac{D_{\text{lead}}}{R_{\text{pursuit}} - R_{\text{target}}} T intercept ​ = R pursuit ​ − R target ​ D lead ​ ​

Worked Naval Interdiction Scenario: A suspicious, unflagged maritime vessel is detected fleeing a coastal anchorage in the Sulu Sea at a speed of 12 knots 12 \text{ knots} 12 knots . Two hours later, a Philippine Navy Littoral Combat Vessel departs the same base in direct pursuit along the identical maritime corridor at 20 knots 20 \text{ knots} 20 knots . How many hours will the naval vessel take to overtake the target, and at what distance from the base will the interdiction occur?

Calculate Target's Lead Distance: D lead = 12 knots × 2 hours = 24 nautical miles (NM) D_{\text{lead}} = 12 \text{ knots} \times 2 \text{ hours} = 24 \text{ nautical miles (NM)} D lead ​ = 12 knots × 2 hours = 24 nautical miles (NM) .

Compute Closing Rate: R closing = 20 − 12 = 8 knots R_{\text{closing}} = 20 - 12 = 8 \text{ knots} R closing ​ = 20 − 12 = 8 knots .

Solve Intercept Time: T intercept = 24 NM 8 knots = 3 hours T_{\text{intercept}} = \frac{24 \text{ NM}}{8 \text{ knots}} = 3 \text{ hours} T intercept ​ = 8 knots 24 NM ​ = 3 hours .

Determine Interception Range from Base: D = R pursuit × T intercept = 20 knots × 3 hours = 60 nautical miles D = R_{\text{pursuit}} \times T_{\text{intercept}} = 20 \text{ knots} \times 3 \text{ hours} = 60 \text{ nautical miles} D = R pursuit ​ × T intercept ​ = 20 knots × 3 hours = 60 nautical miles . Cross-Verification: The target traveled for 2 + 3 = 5 hours 2 + 3 = 5 \text{ hours} 2 + 3 = 5 hours at 12 knots 12 \text{ knots} 12 knots : 12 × 5 = 60 NM 12 \times 5 = 60 \text{ NM} 12 × 5 = 60 NM . The positions coincide exactly.

Archetype C: The Average Speed Trap and Harmonic Mean

A notorious trap on standardized military aptitude examinations is calculating average speed for a round trip over identical outbound and inbound distances.

The Trap: If a vehicle travels from Point A to Point B at R 1 R_1 R 1 ​ and returns along the identical route from Point B to Point A at R 2 R_2 R 2 ​ , the average speed is never the simple arithmetic mean R 1 + R 2 2 \frac{R_1 + R_2}{2} 2 R 1 ​ + R 2 ​ ​ . Average Speed = Total Distance Total Time \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} Average Speed = Total Time Total Distance ​

Because the vehicle travels at a lower speed over one leg, it spends significantly more time moving at that slower speed. Time acts as a weighting factor that pulls the true average speed toward the lower velocity.

Algebraic Derivation of the Harmonic Mean:

Let d d d represent the one-way distance between endpoints.

Total round-trip distance = d + d = 2 d = d + d = 2d = d + d = 2 d .

Outbound transit time: T 1 = d R 1 T_1 = \frac{d}{R_1} T 1 ​ = R 1 ​ d ​ .

Inbound transit time: T 2 = d R 2 T_2 = \frac{d}{R_2} T 2 ​ = R 2 ​ d ​ .

Total round-trip time: T total = T 1 + T 2 = d R 1 + d R 2 = d ( R 1 + R 2 R 1 R 2 ) T_{\text{total}} = T_1 + T_2 = \frac{d}{R_1} + \frac{d}{R_2} = d \left(\frac{R_1 + R_2}{R_1 R_2}\right) T total ​ = T 1 ​ + T 2 ​ = R 1 ​ d ​ + R 2 ​ d ​ = d ( R 1 ​ R 2 ​ R 1 ​ + R 2 ​ ​ ) .

Substituting into the average speed definition: Average Speed = 2 d d ( R 1 + R 2 R 1 R 2 ) = 2 ⋅ R 1 ⋅ R 2 R 1 + R 2 \text{Average Speed} = \frac{2d}{d \left(\frac{R_1 + R_2}{R_1 R_2}\right)} = \frac{2 \cdot R_1 \cdot R_2}{R_1 + R_2} Average Speed = d ( R 1 ​ R 2 ​ R 1 ​ + R 2 ​ ​ ) 2 d ​ = R 1 ​ + R 2 ​ 2 ⋅ R 1 ​ ⋅ R 2 ​ ​

Numerical Proof: A military dispatch motorcycle travels outbound through an urban zone at 30 km/h 30 \text{ km/h} 30 km/h and returns along the identical road during off-peak hours at 60 km/h 60 \text{ km/h} 60 km/h .

Arithmetic Mean (WRONG): 30 + 60 2 = 45 km/h \frac{30 + 60}{2} = 45 \text{ km/h} 2 30 + 60 ​ = 45 km/h .

Harmonic Mean (CORRECT): 2 × 30 × 60 30 + 60 = 3 , 600 90 = 40 km/h \frac{2 \times 30 \times 60}{30 + 60} = \frac{3,600}{90} = 40 \text{ km/h} 30 + 60 2 × 30 × 60 ​ = 90 3 , 600 ​ = 40 km/h .

Verification via arbitrary distance ( d = 60 km d = 60 \text{ km} d = 60 km ): Outbound time = 60 / 30 = 2 h = 60 / 30 = 2 \text{ h} = 60/30 = 2 h . Inbound time = 60 / 60 = 1 h = 60 / 60 = 1 \text{ h} = 60/60 = 1 h . Total distance = 120 km = 120 \text{ km} = 120 km . Total time = 3 h = 3 \text{ h} = 3 h . Average speed = 120 / 3 = 40 km/h = 120 / 3 = 40 \text{ km/h} = 120/3 = 40 km/h .

Archetype D: River Currents and Headwind/Tailwind Problems

When a vehicle operates within a moving medium (water current for boats, wind for aircraft), the medium directly modifies the craft's effective ground speed ( R effective R_{\text{effective}} R effective ​ ):

Let R v = craft’s velocity in still water or calm air R_v = \text{craft's velocity in still water or calm air} R v ​ = craft’s velocity in still water or calm air .

Let R m = velocity of the moving fluid medium (current or wind) R_m = \text{velocity of the moving fluid medium (current or wind)} R m ​ = velocity of the moving fluid medium (current or wind) . Downstream / With Tailwind: R effective = R v + R m \text{Downstream / With Tailwind: } R_{\text{effective}} = R_v + R_m Downstream / With Tailwind: R effective ​ = R v ​ + R m ​ Upstream / Against Headwind: R effective = R v − R m \text{Upstream / Against Headwind: } R_{\text{effective}} = R_v - R_m Upstream / Against Headwind: R effective ​ = R v ​ − R m ​

Adding and subtracting these two fundamental equations yields direct formulas to isolate craft speed and current speed: R v = R down + R up 2 R m = R down − R up 2 R_v = \frac{R_{\text{down}} + R_{\text{up}}}{2} \qquad R_m = \frac{R_{\text{down}} - R_{\text{up}}}{2} R v ​ = 2 R down ​ + R up ​ ​ R m ​ = 2 R down ​ − R up ​ ​

Worked Aerial Navigation Example: A Philippine Air Force C-295 transport aircraft flies a 720 km 720 \text{ km} 720 km route from Villamor Air Base to Palawan in 2 hours 2 \text{ hours} 2 hours assisted by a tailwind. On the return flight along the exact reverse heading, facing an identical opposing headwind, the flight takes 2.4 hours 2.4 \text{ hours} 2.4 hours . What is the aircraft's airspeed in calm conditions, and what is the velocity of the prevailing wind?

Compute Effective Speeds:

Tailwind Ground Speed: R down = 720 km 2 h = 360 km/h R_{\text{down}} = \frac{720 \text{ km}}{2 \text{ h}} = 360 \text{ km/h} R down ​ = 2 h 720 km ​ = 360 km/h .

Headwind Ground Speed: R up = 720 km 2.4 h = 300 km/h R_{\text{up}} = \frac{720 \text{ km}}{2.4 \text{ h}} = 300 \text{ km/h} R up ​ = 2.4 h 720 km ​ = 300 km/h .

Isolate Calm Airspeed ( R v R_v R v ​ ): R v = 360 + 300 2 = 660 2 = 330 km/h R_v = \frac{360 + 300}{2} = \frac{660}{2} = 330 \text{ km/h} R v ​ = 2 360 + 300 ​ = 2 660 ​ = 330 km/h

Isolate Wind Velocity ( R m R_m R m ​ ): R m = 360 − 300 2 = 60 2 = 30 km/h R_m = \frac{360 - 300}{2} = \frac{60}{2} = 30 \text{ km/h} R m ​ = 2 360 − 300 ​ = 2 60 ​ = 30 km/h

3. Work Rate Problems and Collaborative Production

Work problems apply the uniform motion equation to tasks and civil-military engineering projects. Rather than traversing a spatial distance ( D D D ), workers generate units of finished work ( W W W ). Work Done = Rate × Time ( W = R ⋅ T ) \text{Work Done} = \text{Rate} \times \text{Time} \quad (W = R \cdot T) Work Done = Rate × Time ( W = R ⋅ T )

The Reciprocal Rule of Work

If a single military engineering unit can complete an entire project (defined as W = 1 W = 1 W = 1 complete job) in T T T days working alone, its daily work rate is the reciprocal of time: R = 1 T of the total job completed per day R = \frac{1}{T} \quad \text{of the total job completed per day} R = T 1 ​ of the total job completed per day

Combined Collaborative Rate Formula

When multiple entities work concurrently without interfering with one another, their production rates are additive: R combined = R 1 + R 2 + R 3 + ⋯ + R n = 1 T 1 + 1 T 2 + 1 T 3 + ⋯ + 1 T n R_{\text{combined}} = R_1 + R_2 + R_3 + \dots + R_n = \frac{1}{T_1} + \frac{1}{T_2} + \frac{1}{T_3} + \dots + \frac{1}{T_n} R combined ​ = R 1 ​ + R 2 ​ + R 3 ​ + ⋯ + R n ​ = T 1 ​ 1 ​ + T 2 ​ 1 ​ + T 3 ​ 1 ​ + ⋯ + T n ​ 1 ​

For two workers with individual completion times T 1 T_1 T 1 ​ and T 2 T_2 T 2 ​ : 1 T together = 1 T 1 + 1 T 2 = T 1 + T 2 T 1 ⋅ T 2 ⟹ T together = T 1 ⋅ T 2 T 1 + T 2 ("Product over Sum" Rule) \frac{1}{T_{\text{together}}} = \frac{1}{T_1} + \frac{1}{T_2} = \frac{T_1 + T_2}{T_1 \cdot T_2} \quad \implies \quad T_{\text{together}} = \frac{T_1 \cdot T_2}{T_1 + T_2} \quad \text{("Product over Sum" Rule)} T together ​ 1 ​ = T 1 ​ 1 ​ + T 2 ​ 1 ​ = T 1 ​ ⋅ T 2 ​ T 1 ​ + T 2 ​ ​ ⟹ T together ​ = T 1 ​ + T 2 ​ T 1 ​ ⋅ T 2 ​ ​ ("Product over Sum" Rule)

Workers Joining or Leaving Midway Through Tasks

When participants work for varying time intervals, set up an equation where the sum of the fractional work completed equals 1 1 1 : ∑ ( R i ⋅ t i ) = ( R 1 ⋅ t 1 ) + ( R 2 ⋅ t 2 ) + ⋯ + ( R n ⋅ t n ) = 1 \sum (R_i \cdot t_i) = (R_1 \cdot t_1) + (R_2 \cdot t_2) + \dots + (R_n \cdot t_n) = 1 ∑ ( R i ​ ⋅ t i ​ ) = ( R 1 ​ ⋅ t 1 ​ ) + ( R 2 ​ ⋅ t 2 ​ ) + ⋯ + ( R n ​ ⋅ t n ​ ) = 1

Worked Engineering Scenario: Detachment Alpha can construct a tactical perimeter trench in 12 days 12 \text{ days} 12 days . Detachment Bravo can construct the same trench in 18 days 18 \text{ days} 18 days . Detachment Alpha works on the fortification alone for 4 days 4 \text{ days} 4 days , after which Detachment Bravo is ordered to assist. How many additional days will both detachments take working together to finish the trench?

Identify Individual Rates: R A = 1 12 R_A = \frac{1}{12} R A ​ = 12 1 ​ job/day, R B = 1 18 R_B = \frac{1}{18} R B ​ = 18 1 ​ job/day.

Calculate Work Accomplished Solo by Alpha: W initial = R A × 4 = 4 12 = 1 3 of the trench completed W_{\text{initial}} = R_A \times 4 = \frac{4}{12} = \frac{1}{3} \text{ of the trench completed} W initial ​ = R A ​ × 4 = 12 4 ​ = 3 1 ​ of the trench completed

Determine Remaining Balance: W remaining = 1 − 1 3 = 2 3 of the trench W_{\text{remaining}} = 1 - \frac{1}{3} = \frac{2}{3} \text{ of the trench} W remaining ​ = 1 − 3 1 ​ = 3 2 ​ of the trench

Compute Combined Daily Rate: R combined = 1 12 + 1 18 = 3 36 + 2 36 = 5 36 job/day R_{\text{combined}} = \frac{1}{12} + \frac{1}{18} = \frac{3}{36} + \frac{2}{36} = \frac{5}{36} \text{ job/day} R combined ​ = 12 1 ​ + 18 1 ​ = 36 3 ​ + 36 2 ​ = 36 5 ​ job/day

Solve for Collaborative Time ( t t t ): t = W remaining R combined = 2 / 3 5 / 36 = 2 3 × 36 5 = 24 5 = 4.8 days t = \frac{W_{\text{remaining}}}{R_{\text{combined}}} = \frac{2/3}{5/36} = \frac{2}{3} \times \frac{36}{5} = \frac{24}{5} = 4.8 \text{ days} t = R combined ​ W remaining ​ ​ = 5/36 2/3 ​ = 3 2 ​ × 5 36 ​ = 5 24 ​ = 4.8 days Both detachments require an additional 4.8 days 4.8 \text{ days} 4.8 days (4 days, 19 hours, and 12 minutes) to complete the project.

4. Fluid Flow: Pipes, Cisterns & Reservoirs

Fluid flow problems model storage reservoirs, fuel bladders, and naval ballast systems. They follow the exact logic of work problems with one critical distinction: flow directions have opposing algebraic signs .

Inlet Conduits (Filling): Add volume to the system; treated as positive rates ( + R i = + 1 T fill +R_i = +\frac{1}{T_{\text{fill}}} + R i ​ = + T fill ​ 1 ​ ).

Drainage Conduits / Leaks (Emptying): Remove volume from the system; treated as negative rates ( − R d = − 1 T drain -R_d = -\frac{1}{T_{\text{drain}}} − R d ​ = − T drain ​ 1 ​ ). Net Rate = ∑ R inlets − ∑ R outlets = ( 1 T in , 1 + 1 T in , 2 ) − ( 1 T out , 1 + 1 T leak ) \text{Net Rate} = \sum R_{\text{inlets}} - \sum R_{\text{outlets}} = \left(\frac{1}{T_{\text{in},1}} + \frac{1}{T_{\text{in},2}}\right) - \left(\frac{1}{T_{\text{out},1}} + \frac{1}{T_{\text{leak}}}\right) Net Rate = ∑ R inlets ​ − ∑ R outlets ​ = ( T in , 1 ​ 1 ​ + T in , 2 ​ 1 ​ ) − ( T out , 1 ​ 1 ​ + T leak ​ 1 ​ )

Worked Fluid Mechanics Example: An expeditionary field hospital maintains a 12,000-liter potable water bladder. Main Supply Line 1 can fill the empty bladder in 6 hours 6 \text{ hours} 6 hours . Secondary Line 2 can fill it in 8 hours 8 \text{ hours} 8 hours . An automated purification distribution valve empties a full bladder in 12 hours 12 \text{ hours} 12 hours . If both supply lines and the distribution valve are opened simultaneously when the bladder is completely empty, how many hours will it take to fill the bladder completely?

Establish Individual Rates: R 1 = + 1 6 R_1 = +\frac{1}{6} R 1 ​ = + 6 1 ​ , R 2 = + 1 8 R_2 = +\frac{1}{8} R 2 ​ = + 8 1 ​ , R drain = − 1 12 R_{\text{drain}} = -\frac{1}{12} R drain ​ = − 12 1 ​ .

Calculate Net Rate using Least Common Denominator (LCD = 24): R net = 1 6 + 1 8 − 1 12 = 4 24 + 3 24 − 2 24 = 5 24 bladder per hour R_{\text{net}} = \frac{1}{6} + \frac{1}{8} - \frac{1}{12} = \frac{4}{24} + \frac{3}{24} - \frac{2}{24} = \frac{5}{24} \text{ bladder per hour} R net ​ = 6 1 ​ + 8 1 ​ − 12 1 ​ = 24 4 ​ + 24 3 ​ − 24 2 ​ = 24 5 ​ bladder per hour

Invert Net Rate to Find Total Time: T = 1 R net = 24 5 = 4.8 hours = 4 hours and 48 minutes T = \frac{1}{R_{\text{net}}} = \frac{24}{5} = 4.8 \text{ hours} = 4 \text{ hours and } 48 \text{ minutes} T = R net ​ 1 ​ = 5 24 ​ = 4.8 hours = 4 hours and 48 minutes

Comprehensive Word Problem Archetype Matrix

The following master reference table summarizes the governing equations, algebraic setups, and military operational contexts tested on the AFPSAT:

Problem Archetype Primary Unknowns Governing Formula Key Operational Setup Representative Military Scenario

Converging Motion (Opposite Directions) Time to meet ( T T T ), meeting point distance ( D D D ) T = D initial R 1 + R 2 T = \frac{D_{\text{initial}}}{R_1 + R_2} T = R 1 ​ + R 2 ​ D initial ​ ​ Add individual rates: R rel = R 1 + R 2 R_{\text{rel}} = R_1 + R_2 R rel ​ = R 1 ​ + R 2 ​ ; total distance D = D 1 + D 2 D = D_1 + D_2 D = D 1 ​ + D 2 ​ . Two tactical convoys moving toward each other along the MacArthur Highway from separated regional commands.

Overtaking Motion (Same Direction) Intercept time ( T T T ), distance traveled before catch-up T = D lead R chase − R target T = \frac{D_{\text{lead}}}{R_{\text{chase}} - R_{\text{target}}} T = R chase ​ − R target ​ D lead ​ ​ Subtract individual rates: R closing = R chase − R target R_{\text{closing}} = R_{\text{chase}} - R_{\text{target}} R closing ​ = R chase ​ − R target ​ ; equate total distances at intercept. Philippine Navy fast attack craft pursuing an unidentified maritime smuggling vessel fleeing toward territorial limits.

Round-Trip Motion (Equal Distances) Mean system velocity over complete cycle R avg = 2 ⋅ R 1 ⋅ R 2 R 1 + R 2 R_{\text{avg}} = \frac{2 \cdot R_1 \cdot R_2}{R_1 + R_2} R avg ​ = R 1 ​ + R 2 ​ 2 ⋅ R 1 ​ ⋅ R 2 ​ ​ Harmonic mean required; never calculate the arithmetic average R 1 + R 2 2 \frac{R_1 + R_2}{2} 2 R 1 ​ + R 2 ​ ​ . Armored personnel carrier conducting reconnaissance to a mountain outpost uphill and returning downhill.

Fluid/Wind Drift (Current / Headwind) Craft speed in calm medium ( R v R_v R v ​ ), current/wind velocity ( R m R_m R m ​ ) R down = R v + R m R_{\text{down}} = R_v + R_m R down ​ = R v ​ + R m ​ ; R up = R v − R m R_{\text{up}} = R_v - R_m R up ​ = R v ​ − R m ​ R v = R down + R up 2 R_v = \frac{R_{\text{down}} + R_{\text{up}}}{2} R v ​ = 2 R down ​ + R up ​ ​ ; R m = R down − R up 2 R_m = \frac{R_{\text{down}} - R_{\text{up}}}{2} R m ​ = 2 R down ​ − R up ​ ​ PAF maritime patrol aircraft executing aerial reconnaissance with prevailing monsoon winds.

Collaborative Work (Concurrent Effort) Combined duration ( T T T ), individual work fraction 1 T = 1 T 1 + 1 T 2 + ⋯ + 1 T n \frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2} + \dots + \frac{1}{T_n} T 1 ​ = T 1 ​ 1 ​ + T 2 ​ 1 ​ + ⋯ + T n ​ 1 ​ Sum individual reciprocal rates; for 2 workers use Product over Sum: T = T 1 T 2 T 1 + T 2 T = \frac{T_1 T_2}{T_1 + T_2} T = T 1 ​ + T 2 ​ T 1 ​ T 2 ​ ​ . Combat engineers erecting pre-fabricated modular bridging structures across river barriers.

Staggered Work (Departing/Joining) Remaining duration, total project calendar days ∑ ( R i ⋅ t i ) = 1.0 \sum (R_i \cdot t_i) = 1.0 ∑ ( R i ​ ⋅ t i ​ ) = 1.0 Calculate completed fraction: W 1 = R 1 ⋅ t 1 W_1 = R_1 \cdot t_1 W 1 ​ = R 1 ​ ⋅ t 1 ​ ; remaining work is 1 − W 1 1 - W_1 1 − W 1 ​ ; divide by active rates. Runway damage repair teams where heavy equipment operators arrive on site hours after hand-clearing teams.

Cisterns & Reservoirs (Inflows & Outflows) Net fill/drain time, net flow rate R net = ∑ R in − ∑ R out R_{\text{net}} = \sum R_{\text{in}} - \sum R_{\text{out}} R net ​ = ∑ R in ​ − ∑ R out ​ Treat fill conduits as positive ( + ) (+) ( + ) and drainage conduits as negative ( − ) (-) ( − ) . Invert net rate for time. Quartermaster logistics teams managing forward operating base aviation fuel and bulk water bladders.