3.1 Word Problems: Rates, Ratios, Proportions & Percentages

3.1 Word Problems: Rates, Ratios, Proportions & Percentages

The Arithmetic Reasoning (AR) subtest on the Armed Forces Classification Test (AFCT) evaluates your ability to analyze practical word problems, translate verbal statements into mathematical expressions, and solve them efficiently under strict time constraints. A substantial portion of the AR subtest consists of problems involving rates, ratios, proportions, and percentages.

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1. Ratios and Ratio Scaling

A ratio is a comparison of two quantities by division. Ratios can be written in three forms: as a fraction ($\frac{a}{b}$), with a colon ($a:b$), or using the word "to" ($a\text{ to }b$).

Part-to-Part vs. Part-to-Whole Ratios

Understanding whether a ratio represents a part-to-part relationship or a part-to-whole relationship is essential:

• Part-to-Part: Compares one subgroup to another subgroup. For example, if a unit has 15 officers and 45 enlisted personnel, the ratio of officers to enlisted is $15:45$, which simplifies to $1:3$.
• Part-to-Whole: Compares a subgroup to the total. In the same unit, the total personnel is $15 + 45 = 60$. The ratio of officers to total personnel is $15:60 = 1:4$.

The Ratio Box / Total Parts Method

When a total quantity must be divided according to a given ratio $a : b : c$:

1. Add the ratio terms together to find the total number of equal parts: $\text{Total Parts} = a + b + c$.
2. Divide the total quantity by the total number of parts to determine the value of one part: $\text{Value per Part} = \frac{\text{Total Quantity}}{\text{Total Parts}}$.
3. Multiply each ratio term by the value of one part to find the specific quantity for each category.

Step-by-Step Worked Example 1: A tactical supply unit receives a shipment of 450 cases of ammunition distributed among caliber types Alpha, Bravo, and Charlie in the ratio $2 : 4 : 3$. How many cases of Bravo ammunition were received?

• Step 1: Calculate total ratio parts: $2 + 4 + 3 = 9\text{ parts}$.
• Step 2: Find value per part: $\frac{450\text{ cases}}{9\text{ parts}} = 50\text{ cases per part}$.
• Step 3: Calculate Bravo cases ($4\text{ parts}$): $4 \times 50 = 200\text{ cases}$.

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2. Rates, Unit Rates & Dimensional Analysis

A rate is a ratio that compares two quantities measured in different units (such as miles per hour, dollars per gallon, or fuel consumption per engine hour). A unit rate is a rate expressed with a denominator of 1.

Dimensional Analysis (Unit Conversion)

To convert rates across different measurement units, multiply by conversion factors structured as fractions equal to 1, ensuring unwanted units cancel out diagonally.

$\text{Target Rate} = \text{Given Rate} \times \left( \frac{\text{Unit } B}{\text{Unit } A} \right) \times \left( \frac{\text{Unit } C}{\text{Unit } B} \right)$

Step-by-Step Worked Example 2: A generator consumes diesel fuel at a rate of 12 quarts per hour. If fuel costs $4.00 per gallon, what is the operational fuel cost of running the generator for an 8-hour shift?

• Step 1: Find total quarts consumed in 8 hours: $12\text{ quarts/hr} \times 8\text{ hrs} = 96\text{ quarts}$.
• Step 2: Convert quarts to gallons ($4\text{ quarts} = 1\text{ gallon}$): $\frac{96\text{ quarts}}{4\text{ quarts/gal}} = 24\text{ gallons}$.
• Step 3: Multiply by cost per gallon: $24\text{ gallons} \times $4.00 = $96.00$.

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3. Direct vs. Inverse Proportions

A proportion is an equation stating that two ratios are equal: $\frac{a}{b} = \frac{c}{d}$. Proportions fall into two distinct categories: direct and inverse.

Direct Proportions ($y = kx$)

In a direct proportion, as one variable increases, the other variable increases at the same multiplier rate. The ratio $\frac{y}{x} = k$ remains constant.

$\frac{a_1}{b_1} = \frac{a_2}{b_2} \implies a_1 \cdot b_2 = b_1 \cdot a_2$

Inverse Proportions ($x \cdot y = k$)

In an inverse proportion, as one variable increases, the other variable decreases proportionally. The product of the two variables $x \cdot y = k$ remains constant. Common AFCT inverse proportion scenarios include workers vs. completion time or travel speed vs. travel time.

$a_1 \cdot b_1 = a_2 \cdot b_2$

Step-by-Step Worked Example 3 (Inverse Proportion): If 8 soldiers can dig a trench system in 15 hours, how many hours will it take 12 soldiers working at the exact same rate to dig the identical trench system?

• Step 1: Identify relationship: More soldiers means fewer hours (inverse proportion).
• Step 2: Set up product equality: $\text{Soldiers}_1 \times \text{Hours}_1 = \text{Soldiers}_2 \times \text{Hours}_2$.
• Step 3: Substitute known values: $8 \times 15 = 12 \times H_2$.
• Step 4: Solve for $H_2$: $120 = 12 \cdot H_2 \implies H_2 = 10\text{ hours}$.

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4. Percentages, Percent Change & Multi-Step Discounts

A percentage is a fraction expressed out of 100. The foundational percentage relationship is:

$\text{Part} = \text{Percent (in decimal form)} \times \text{Whole}$

Percent Increase and Decrease

When calculating percentage increase or decrease, the change is always evaluated relative to the original starting amount:

$\text{Percent Change} = \frac{|\text{New Value} - \text{Original Value}|}{\text{Original Value}} \times 100\%$

AFCT Common Pitfall: Never divide the change by the new value! Always divide by the starting (original) value.

Successive (Multi-Step) Percentages

When multiple percentage changes occur sequentially (e.g., a $20%$ discount followed by an additional $10%$ discount), do not add the percentages together.

Step-by-Step Worked Example 4: A piece of tactical equipment originally priced at $500 is placed on sale for $20%$ off. A military personnel discount reduces the sale price by an additional $10%$. What is the final price?

• Step 1: Calculate price after first discount ($20%$ off means paying $80%$): $$500 \times 0.80 = $400$.
• Step 2: Apply second discount to the new price ($10%$ off of $400 means paying $90%$): $$400 \times 0.90 = $360$.
• Step 3: Note total discount is $$500 - $360 = $140$, which is $\frac{140}{500} = 28%$, not $30%$.

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5. AFCT Word Problem Translation Strategy

To translate word problems quickly into solvable mathematical equations, use this translation matrix: