7.2 Classic Math Word Problems: Age, Work & Motion

Classic Math Word Problems: Age, Work & Motion

NAPOLCOM publishes Problem-Solving Skills within Quantitative Reasoning. Word problems are a practical way to train those skills, although the official current coverage does not specify a fixed mix of problem archetypes. Rather than evaluating rote calculation in isolation, word problems test a candidate's ability to extract mathematical structures from complex English prose, organize interrelated data points into coherent models, and execute multi-step solutions under rigorous time constraints. Master test-takers do not treat word problems as separate puzzles; they recognize that nearly all civil service math items belong to four classic, standardized mathematical archetypes: Age Problems, Work Problems, Uniform Motion Problems, and Mixture Problems.

Independent Preparation Notice: This study module is independently developed by OpenExamPrep to assist prospective applicants in mastering the quantitative reasoning concepts required for the examination. OpenExamPrep is an independent educational publisher and is not affiliated with, endorsed by, or partnered with the National Police Commission (NAPOLCOM) or the Philippine National Police (PNP).

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1. The Word Problem Translation Protocol: From Narrative to Numerical Model

To solve multi-step word problems efficiently within the overall examination limit, candidates should adopt a structured 4-phase translation protocol:

1. Phase 1: Variable Declaration with Temporal Reference: Identify precisely what unknown quantity the question asks for. Declare a single variable (x) representing the baseline entity, explicitly anchoring it in time (e.g., "Let x = Patrolman's present age in years").
2. Phase 2: Structured Tabular Modeling: Never attempt to hold multi-variable relationships in working memory under exam stress. Construct a structured 2-column or 3-column reference table (e.g., Past-Present-Future grid for Age; Rate-Time-Work table for Work; Distance-Speed-Time matrix for Motion).
3. Phase 3: Relational Equation Formulation: Translate the narrative sentence connecting the quantities into an algebraic equality. Verify that all measurement units (hours vs. minutes, kilometers vs. meters) are standardized before equating terms.
4. Phase 4: Contextual Reality Verification: After solving the algebraic equation, re-read the final sentence of the prompt. Verify whether the question asks for the solved variable x, or for a secondary quantity (e.g., the partner's age, the return travel time, or the combined total cost). Ensure answers are physically realistic (e.g., ages and speeds cannot be negative numbers).

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2. Age Problems: The Chronological Grid Strategy

Age problems assess a candidate's ability to track variable relationships across shifting temporal frames. They are governed by two universal principles:

• Uniform Temporal Shift: If N years elapse, every person's age increases by exactly +N years (into the future) or decreases by -N years (into the past).
• Invariance of Age Differences: The numerical difference between two people's ages remains constant throughout their lives. If Officer A is 8 years older than Officer B today, Officer A was 8 years older 20 years ago and will be 8 years older 30 years from now.

The Chronological Table Architecture

Whenever an age problem mentions past or future conditions, immediately draw a 3-column chronological grid on your scratchpad:

Worked Example 1: Multiplier Relationship with Future Frame

• Narrative: A Police Captain is currently 4 times as old as a newly appointed Patrolman. In 16 years, the Captain will be only twice as old as the Patrolman. What are their current ages?
• Step 1: Declare baseline variable:
  • Let the Patrolman's present age = x.
  • The Captain's present age = 4x.
• Step 2: Model the future frame (+16 years):
  • In 16 years, Patrolman's age = x + 16.
  • In 16 years, Captain's age = 4x + 16.
• Step 3: Translate narrative equality:
  • Captain's future age = 2 × (Patrolman's future age)
  • 4x + 16 = 2(x + 16)
• Step 4: Solve algebraically:
  • 4x + 16 = 2x + 32
  • 4x - 2x = 32 - 16 ⟹ 2x = 16 ⟹ x = 8
• Conclusion: The Patrolman's present age is 8 years old (or baseline junior age), and the Captain's present age is 4(8) = 32 years old.
• Verification: In 16 years, Patrolman will be 8 + 16 = 24, and Captain will be 32 + 16 = 48. Since 48 = 2 × 24, the solution satisfies all stated conditions.

Worked Example 2: Dual Temporal Frames (Past and Future Equations)

• Narrative: Seven years ago, an experienced investigator was five times as old as his junior partner. Five years from now, the investigator will be three times as old as his junior partner. What are their present ages?
• Setup:
  • Let J = Junior partner's present age; let I = Investigator's present age.
• Condition 1 (7 years ago):
  • I - 7 = 5(J - 7) ⟹ I - 7 = 5J - 35 ⟹ I = 5J - 28
• Condition 2 (5 years from now):
  • I + 5 = 3(J + 5) ⟹ I + 5 = 3J + 15 ⟹ I = 3J + 10
• Equate both expressions for I:
  • 5J - 28 = 3J + 10
  • 2J = 38 ⟹ J = 19 years old
• Solve for I:
  • I = 3(19) + 10 = 57 + 10 = 67 years old
• Verification: 7 years ago: J = 12, I = 60 (60 = 5 × 12). In 5 years: J = 24, I = 72 (72 = 3 × 24). Both conditions hold.

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3. Work Problems: The Reciprocal Rate Principle

Work problems evaluate tasks completed by individuals or machines working at constant productivity rates. The fundamental governing relationship is:

Work Done = Productivity Rate × Time (W = R × T)

The Reciprocal Rate Rule

When a problem states that an individual can complete an entire project alone in T hours, the entire job is defined as 1 whole unit (W = 1). Consequently, the individual's hourly productivity rate is the mathematical reciprocal of their completion time:

Rate = 1 / (Time to complete 1 job alone)

• If Detective A completes a dossier audit in 6 hours, Detective A's rate is 1/6 of the audit per hour.
• If Detective B completes the same audit in 4 hours, Detective B's rate is 1/4 of the audit per hour.

Collaborative Work (Joint Rates)

When multiple entities work together simultaneously toward the same goal, their individual work rates are strictly additive:

• Combined Rate = Rate_A + Rate_B = 1/A + 1/B = (A + B) / (A × B)
• Time to Complete Together (T_together) = 1 / Combined Rate = (A × B) / (A + B)

The Product-Over-Sum Shortcut: For two workers collaborating, their combined completion time is simply the product of their individual times divided by the sum of their individual times: T = (A × B) / (A + B).

Scenario A: Segmented Work (Workers Entering or Leaving Midway)

In complex civil service items, workers rarely work together from start to finish. One worker starts alone, or one departs before completion. Such problems are modeled by summing the fractional portions of work completed during each segment:

Work Segment 1 + Work Segment 2 = 1 (Whole Job)

• Problem: Technician Santos can process a station's backlog of forensic evidence in 10 hours alone, while Technician Cruz can complete it in 15 hours alone. Technician Santos works alone for 4 hours, after which Technician Cruz joins him to finish the remainder together. What is the total elapsed time required to complete the backlog?
• Step 1: Define individual rates:
  • Rate_S = 1/10 backlog/hour
  • Rate_C = 1/15 backlog/hour
• Step 2: Calculate work accomplished in Segment 1:
  • Santos works alone for 4 hours: W₁ = 4 × (1/10) = 4/10 = 2/5 of the backlog
• Step 3: Determine remaining work for Segment 2:
  • W_remaining = 1 - 2/5 = 3/5 of the backlog
• Step 4: Calculate combined rate for Segment 2:
  • Combined Rate = 1/10 + 1/15 = (3 + 2)/30 = 5/30 = 1/6 backlog/hour
• Step 5: Calculate time required for Segment 2:
  • t₂ = W_remaining / Combined Rate = (3/5) / (1/6) = (3/5) × 6 = 18/5 = 3.6 hours = 3 hours 36 minutes
• Step 6: Compute total elapsed time:
  • Total Time = t₁ + t₂ = 4 hours + 3.6 hours = 7.6 hours = 7 hours 36 minutes

Scenario B: Opposing Rates (Pipes, Reservoirs & Drainage Valves)

Inlet pipes fill a reservoir (+ rate), while outlet or drainage pipes empty it (- rate):

Net Rate = Sum of Inlet Rates - Sum of Drain Rates

• Problem: An emergency auxiliary water storage tank at a police barracks is fed by two inlet pipes and has one bottom drain valve. Inlet Pipe 1 can fill the empty tank in 4 hours. Inlet Pipe 2 can fill it in 6 hours. The drain valve can empty a completely full tank in 12 hours. If the tank is initially empty and all three valves are opened simultaneously, in how many hours will the tank be filled to capacity?
• Formulate Net Rate:
  • Net Rate = 1/4 + 1/6 - 1/12
  • Net Rate = 3/12 + 2/12 - 1/12 = 4/12 = 1/3 tank per hour
• Compute Time to Fill:
  • T = 1 / Net Rate = 1 / (1/3) = 3 hours

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4. Uniform Motion Problems: Relative Speed, Pursuit & Harmonic Average Speed

Uniform motion problems describe entities moving at constant cruising speeds. The foundational kinematic relationship is:

Distance = Speed × Time (d = vt) Speed = d / t, and Time = d / v

Archetype 1: Motion in Opposite Directions / Converging Motion (Closing Distance)

When two vehicles travel toward each other from two distant points, or depart from the same point in opposite directions, the distance between them changes at a combined relative speed:

v_relative = v₁ + v₂ Time to Meet / Total Separation Time = d_total / (v₁ + v₂)

• Problem: Station Alpha and Station Bravo are located 240 kilometers apart along a straight national highway. At 07:00 AM, Patrol Car 1 departs Station Alpha heading toward Station Bravo at an average speed of 55 km/h. At the exact same moment, Patrol Car 2 departs Station Bravo heading toward Station Alpha at an average speed of 65 km/h. At what time will the two patrol cars meet?
• Calculate Relative Closing Speed:
  • v_rel = 55 + 65 = 120 km/h
• Calculate Travel Time to Meeting Point:
  • t = d_total / v_rel = 240 km / 120 km/h = 2.0 hours
• Determine Clock Time:
  • 07:00 AM + 2 hours = 09:00 AM
• Distance Traveled by Car 1: 55 km/h × 2 h = 110 km from Station Alpha.

Archetype 2: Same-Direction Pursuit (Overtaking Problems)

In a pursuit problem, a faster vehicle attempts to catch up to a slower vehicle that departed earlier or from an advanced position. The distance between the vehicles closes at the difference of their speeds:

v_relative = v_faster - v_slower Time to Overtake = Head Start Distance / (v_faster - v_slower)

• Problem: A stolen vehicle passes an expressway checkpoint traveling at a steady speed of 72 km/h. Exactly 20 minutes (1/3 hour) later, a highway patrol interceptor departs the same checkpoint in pursuit at a steady speed of 108 km/h. How long will the interceptor take to overtake the suspect vehicle, and how far from the checkpoint will the interception occur?
• Step 1: Calculate the head start distance:
  • d_head start = 72 km/h × (1/3) hour = 24 kilometers
• Step 2: Calculate relative closing speed:
  • v_rel = 108 - 72 = 36 km/h
• Step 3: Calculate time to overtake:
  • t_overtake = 24 km / 36 km/h = 2/3 hour = (2/3 × 60) = 40 minutes
• Step 4: Calculate total distance from checkpoint:
  • d = 108 km/h × (2/3) hour = 72 kilometers

Archetype 3: Round-Trip Average Speed & The Harmonic Mean

The single most pervasive distractor on the NAPOLCOM quantitative section is the arithmetic average trap in round-trip speed questions.

The Cardinal Motion Rule: If an entity travels a fixed distance D at speed v₁ and returns over the exact same distance D at speed v₂, the average speed for the round trip is NEVER the arithmetic mean (v₁ + v₂) / 2. It is strictly the Harmonic Mean of the two speeds:

v_avg = Total Distance / Total Time = (2 · v₁ · v₂) / (v₁ + v₂)

Why the Arithmetic Mean Fails

Average speed is defined as total distance divided by total elapsed time. Because the traveler covers the same distance at different speeds, more time is spent traveling at the slower speed than at the faster speed. Consequently, the slower speed exerts a heavier temporal weight on the overall journey.

• Demonstration: A patrol car travels from headquarters to an outlying rural post at 40 km/h and returns over the exact same route at 60 km/h.
  • The Arithmetic Trap: (40 + 60)/2 = 50 km/h (INCORRECT).
  • The Harmonic Calculation: v_avg = (2 × 40 × 60) / (40 + 60) = 4,800 / 100 = 48 km/h
  • Step-by-Step Distance Proof: Assume the one-way distance is 120 km.
    • Outbound time: 120 / 40 = 3 hours.
    • Return time: 120 / 60 = 2 hours.
    • Total distance: 120 + 120 = 240 km.
    • Total time: 3 + 2 = 5 hours.
    • Average speed: 240 / 5 = 48 km/h.

Archetype 4: River Currents and Wind Vectors

When traveling in a medium that is itself moving (water currents or atmospheric wind):

• Downstream / Tailwind (Moving With the Current):
  • v_downstream = v_still + v_current
• Upstream / Headwind (Moving Against the Current):
  • v_upstream = v_still - v_current
• Isolating Still-Water Speed and Current Speed:
  • v_still = (v_downstream + v_upstream) / 2
  • v_current = (v_downstream - v_upstream) / 2

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5. Mixture & Concentration Problems: The Weighted Average & Alligation Framework

Mixture problems involve combining two or more ingredients with different concentrations, purities, or unit costs to achieve a desired composite concentration.

The Fundamental Solute Conservation Formula

The total quantity of active solute (pure alcohol, acid, salt, or specific chemical reagent) before mixing must equal the total quantity of active solute in the final composite mixture:

(Concentration₁ × Volume₁) + (Concentration₂ × Volume₂) = Concentration_final × (Volume₁ + Volume₂) C₁V₁ + C₂V₂ = C_f(V₁ + V₂)

Worked Example 1: Diluting Solutions with Pure Solute

• Problem: A police forensics laboratory technician has 40 liters of a 25% disinfectant solution. How many liters of pure disinfectant (100% concentration) must be added to produce a 40% disinfectant solution?
• Step 1: Set up the algebraic balance:
  • Let x = liters of pure (100%) disinfectant added.
  • Amount of disinfectant in 25% solution: 0.25 × 40 = 10 liters.
  • Amount of disinfectant in added solution: 1.00 × x = x liters.
  • Total final volume: (40 + x) liters.
  • Target concentration: 40% = 0.40.
• Step 2: Formulate equation:
  • 10 + x = 0.40(40 + x)
• Step 3: Solve for x:
  • 10 + x = 16 + 0.40x
  • x - 0.40x = 16 - 10 ⟹ 0.60x = 6
  • x = 6 / 0.60 = 10 liters
• Verification: Total solute = 10 + 10 = 20 liters. Total volume = 40 + 10 = 50 liters. Final concentration = 20 / 50 = 0.40 = 40%.

The Alligation Alternate Method for Rapid Mixing Ratios

When a problem asks for the ratio of parts needed to blend two known concentrations to achieve a middle target concentration, the Alligation Matrix eliminates the need for algebraic equations:

• Higher Concentration (C_H): Place at upper left
• Lower Concentration (C_L): Place at lower left
• Target Mean Concentration (C_M): Place in the center
• Parts of Higher = |C_M - C_L|
• Parts of Lower = |C_H - C_M|
• Ratio of Higher to Lower = (C_M - C_L) : (C_H - C_M)

Example: In what ratio must a 70% alcohol solution and a 20% alcohol solution be mixed to produce a 50% alcohol solution?

• Higher: 70%; Lower: 20%; Target: 50%.
• Parts of 70% solution = |50 - 20| = 30 parts.
• Parts of 20% solution = |70 - 50| = 20 parts.
• Ratio of 70% solution to 20% solution is 30 : 20 = 3 : 2.