4.2 Ratios, Proportions, Rates, and Combined Work Problems

4.2 Ratios, Proportions, Rates, and Combined Work Problems

Core Principle: Rate and proportion problems constitute the mathematical backbone of military logistics, mission planning, and tactical engineering. Whether calculating vehicle convoy transit times, battalion staffing allocations, or fuel pipeline fill rates, solving these problems requires identifying whether the relationship is direct (both variables scale together) or inverse (one variable scales up while the other scales down).

Mastering algebraic setups for ratios, relative motion, harmonic mean speeds, and combined work rates ensures rapid, accurate solutions on the CAT-ASVAB Arithmetic Reasoning subtest.

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Direct vs. Inverse Proportions

The most fundamental decision when approaching a proportion word problem is determining whether the variables vary directly or inversely.

+---------------------------------------------------------------------------------------------------+
|                            DIRECT VS. INVERSE PROPORTION COMPARISON                               |
+----------------------+------------------------------------+---------------------------------------+
| Characteristic       | Direct Proportion                  | Inverse Proportion                    |
+----------------------+------------------------------------+---------------------------------------+
| Core Algebraic Form  | y / x = k  (or y = k·x)            | x · y = k                             |
| Dynamic Behavior     | As x increases, y increases        | As x increases, y decreases           |
| Proportion Setup     | x₁ / y₁ = x₂ / y₂                  | x₁ · y₁ = x₂ · y₂                     |
| Cross-Multiplication | Valid (x₁·y₂ = x₂·y₁)              | INVALID (do not cross-multiply!)      |
| Real-World Examples  | • Fuel used vs. Miles driven       | • Number of workers vs. Project days  |
|                      | • Map scale distance vs. Real km   | • Speed of travel vs. Transit time    |
|                      | • Supply cost vs. Quantity bought  | • Pipe flow rate vs. Tank fill hours  |
+----------------------+------------------------------------+---------------------------------------+

1. Direct Proportion Mechanics (Constant Ratio)

In a direct proportion, dividing one quantity by the other always yields a constant ratio ($k$).

Tactical Scenario: A logistics company discovers that 6 transport trucks consume 420 gallons of diesel fuel over a standardized supply route. How many gallons will 14 identical trucks consume over the same route?

$\frac{6 \text{ trucks}}{420 \text{ gallons}} = \frac{14 \text{ trucks}}{x \text{ gallons}}$

Cross-multiply to solve: $6x = 14 \times 420 \implies x = \frac{14 \times 420}{6} = 14 \times 70 = 980 \text{ gallons}$

Scaling Factor Shortcut: Calculate the unit consumption rate first: $\frac{420}{6} = 70 \text{ gallons/truck}$. Then multiply directly: $14 \times 70 = 980 \text{ gallons}$.

2. Inverse Proportion Mechanics (Constant Product)

In an inverse proportion, multiplying the two variables produces a constant total product ($k$).

Tactical Scenario: A combat engineer detachment of 8 soldiers can assemble an expeditionary bridge in 15 hours. If urgent operations require the bridge to be completed in 6 hours, how many total soldiers (working at the same rate) must be assigned to the task?

$\text{Total Labor Required} = 8 \text{ soldiers} \times 15 \text{ hours} = 120 \text{ soldier-hours}$

Set up the constant-product equation for the target time: $N \text{ soldiers} \times 6 \text{ hours} = 120 \text{ soldier-hours} \implies N = \frac{120}{6} = 20 \text{ soldiers}$

Trap Warning: Never cross-multiply inverse proportions! Setting up $\frac{8}{15} = \frac{N}{6}$ would yield $N = 3.2$ soldiers. Adding more time would require fewer soldiers, but completing the job faster requires more soldiers.

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Ratio Parts & Proportional Allocation

A ratio expresses the relative magnitude of two or more quantities. On the ASVAB, ratio problems frequently ask you to partition a total sum into unequal shares based on a given ratio.

+---------------------------------------------------------------------------------------------------+
|                                 THE 3-STEP RATIO PARTS ALGORITHM                                  |
+---------------------------------------------------------------------------------------------------+
| STEP 1: SUM THE RATIO PARTS                                                                       |
| Add all individual numbers in the ratio to find the total number of parts (N).                    |
| For ratio a : b : c, Total Parts = a + b + c.                                                     |
+---------------------------------------------------------------------------------------------------+
| STEP 2: FIND THE VALUE OF A SINGLE UNIT PART                                                      |
| Divide the total quantity by the total number of parts:                                           |
| Unit Value = (Total Quantity) / (Total Parts).                                                    |
+---------------------------------------------------------------------------------------------------+
| STEP 3: SCALE TO THE REQUESTED TARGET                                                             |
| Multiply the single unit value by the specific ratio term requested in the problem stem.          |
+---------------------------------------------------------------------------------------------------+

Worked Ratio Partition Example

A military task force of 720 personnel is composed of Infantry, Armor, and Artillery troops in the ratio of $5 : 3 : 1$. How many Armor personnel are in the task force?

1. Sum the parts: $5 + 3 + 1 = 9 \text{ total parts}$.
2. Find the value of 1 part: $\frac{720 \text{ personnel}}{9 \text{ parts}} = 80 \text{ personnel per part}$.
3. Calculate Armor component (3 parts): $3 \times 80 = 240 \text{ Armor personnel}$.
   • (Check: Infantry $= 5 \times 80 = 400$; Artillery $= 1 \times 80 = 80$; Total $= 400 + 240 + 80 = 720$).

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Uniform Motion Mechanics: Distance, Rate & Time

All kinematics word problems on the ASVAB derive from the fundamental rate equation:

$\text{Distance} = \text{Rate} \times \text{Time} \quad (D = R \cdot T)$ $\text{Rate} = \frac{\text{Distance}}{\text{Time}} \quad \left(R = \frac{D}{T}\right), \qquad \text{Time} = \frac{\text{Distance}}{\text{Rate}} \quad \left(T = \frac{D}{R}\right)$

+---------------------------------------------------------------------------------------------------+
|                              MULTI-BODY MOTION CONFIGURATIONS                                     |
+-----------------------+---------------------------------------+-----------------------------------+
| Configuration Type    | Relative Speed Formula                | Equation Setup                    |
+-----------------------+---------------------------------------+-----------------------------------+
| 1. Converging /       | Rates ADD together:                   | Time to Meet:                     |
|    Opposite Head-on   | R_relative = R₁ + R₂                  | T = Total Distance / (R₁ + R₂)    |
+-----------------------+---------------------------------------+-----------------------------------+
| 2. Pursuit /          | Rates SUBTRACT:                       | Time to Catch:                    |
|    Overtake / Catch-up| R_closing = R_fast - R_slow           | T = Lead Distance / (R_fast - R_s)|
+-----------------------+---------------------------------------+-----------------------------------+
| 3. Round-Trip         | Distance is identical for both legs:  | Average Speed (Harmonic Mean):    |
|    Equal Distance     | D₁ = D₂ = D                           | V_avg = (2 · v₁ · v₂) / (v₁ + v₂) |
+-----------------------+---------------------------------------+-----------------------------------+

1. Converging Motion (Closing the Gap)

Two naval vessels start 300 nautical miles apart and sail directly toward each other. Ship A travels at 18 knots (nautical miles per hour) and Ship B travels at 12 knots. How long until they pass each other? $R_{\text{combined}} = 18 + 12 = 30 \text{ knots}$ $T = \frac{300 \text{ NM}}{30 \text{ knots}} = 10 \text{ hours}$

2. Pursuit / Head-Start Motion (Overtaking)

A supply truck departs a base traveling at 40 mph. Two hours later, a military courier departs the same base along the identical route traveling at 60 mph. How many hours after departing will the courier overtake the truck?

1. Calculate the Truck's Lead Distance: $\text{Lead Distance} = 40 \text{ mph} \times 2 \text{ hours} = 80 \text{ miles}$
2. Calculate the Courier's Net Closing Rate: $R_{\text{closing}} = 60 \text{ mph} - 40 \text{ mph} = 20 \text{ mph}$
3. Calculate Time to Overtake: $T = \frac{80 \text{ miles}}{20 \text{ mph}} = 4 \text{ hours}$

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The Infamous Average Speed Trap (Harmonic Mean)

The single most common distractor trap on the Arithmetic Reasoning subtest occurs when calculating average speed for a round trip over equal distances.

The Trap: If a vehicle drives to a destination at $60 \text{ mph}$ and returns over the exact same route at $30 \text{ mph}$, examinees instinctively take the simple arithmetic mean: $\frac{60 + 30}{2} = 45 \text{ mph}$. THIS IS MATHEMATICALLY FALSE.

Why the Arithmetic Mean Fails

Average speed is defined strictly as total distance divided by total elapsed time:

$V_{\text{avg}} = \frac{D_{\text{total}}}{T_{\text{total}}}$

Because the vehicle travels at a slower speed ($30 \text{ mph}$) on the return trip, it spends twice as much time driving at the slower speed than at the faster speed. As a result, the slower speed exerts greater weight on the overall average.

Complete Mathematical Derivation

Let the one-way distance be $D = 180 \text{ miles}$:

• Outbound Time ($T_1$ at $60 \text{ mph}$): $T_1 = \frac{180}{60} = 3 \text{ hours}$.
• Return Time ($T_2$ at $30 \text{ mph}$): $T_2 = \frac{180}{30} = 6 \text{ hours}$.
• Total Distance ($D_{\text{total}}$): $180 + 180 = 360 \text{ miles}$.
• Total Time ($T_{\text{total}}$): $3 + 6 = 9 \text{ hours}$.
• True Average Speed: $V_{\text{avg}} = \frac{360 \text{ miles}}{9 \text{ hours}} = 40 \text{ mph}$

The Harmonic Mean Formula for Equal Distances

Whenever two legs of a journey cover the identical distance, use the harmonic mean shortcut:

$V_{\text{avg}} = \frac{2 \cdot v_1 \cdot v_2}{v_1 + v_2}$

$V_{\text{avg}} = \frac{2 \times 60 \times 30}{60 + 30} = \frac{3,600}{90} = 40 \text{ mph}$

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Combined Work-Rate Mechanics

Work-rate word problems describe individuals, machines, or pipes operating simultaneously to complete a task.

The Cardinal Rule of Work Problems

Never add completion times together! ALWAYS add rates of work per unit time.

If Worker A completes a job in $T_1$ hours, their hourly rate is $\frac{1}{T_1}$ of the job per hour. If Worker B completes the job in $T_2$ hours, their hourly rate is $\frac{1}{T_2}$.

$\text{Combined Hourly Rate} = \frac{1}{T_{\text{total}}} = \frac{1}{T_1} + \frac{1}{T_2}$

The "Product-over-Sum" Shortcut (Two Agents)

Solving $\frac{1}{T} = \frac{1}{A} + \frac{1}{B}$ algebraically produces the most powerful mental shortcut on the ASVAB:

$T_{\text{total}} = \frac{A \times B}{A + B}$

Example: Specialist Vance can inspect an engine in 6 hours. Specialist Torres can inspect the same engine in 3 hours. How long will the inspection take if they work together?

$T_{\text{total}} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2 \text{ hours}$

Three or More Workers

For three workers, determine the least common denominator (LCD) across all individual rates:

$\frac{1}{T_{\text{total}}} = \frac{1}{T_1} + \frac{1}{T_2} + \frac{1}{T_3}$

Opposing Work / Fluid Reservoir Problems

When one inlet pipe fills a tank while an outlet valve drains the tank, subtract the drain rate from the fill rate:

$\text{Net Rate} = \text{Fill Rate} - \text{Drain Rate} = \frac{1}{T_{\text{fill}}} - \frac{1}{T_{\text{drain}}}$