2.2 Word Problems

Introduction to Word Problems

Word problems are often the most intimidating part of the math section, as they require translating a real-world scenario into a mathematical equation. The most critical step in solving any word problem is defining your variables and deeply understanding the relationships described in the text. Taking a few seconds to write down what you know and what you are trying to find can prevent careless errors.

Translating English to Math

Before diving into specific types of word problems, it is crucial to recognize common phrases and their mathematical equivalents:

• Addition: sum, increased by, more than, total, older than.
• Subtraction: difference, decreased by, less than, younger than.
• Multiplication: product, times, twice, of (when dealing with fractions/percents).
• Division: quotient, ratio, divided by, half.
• Equals: is, will be, was, yields.

Age Problems

Age problems typically involve comparing the ages of two or more people at different points in time, such as the past, present, or future.

Strategy for Age Problems

1. Explicitly define variables for the present ages of the individuals.
2. Express their past or future ages by adding or subtracting the relevant number of years from the present age variables.
3. Set up an equation based on the relationship given in the problem.

Example: John is twice as old as his brother. In 5 years, the sum of their ages will be 40. How old is John now?

• Let the brother's present age be $x$.
• John's present age is $2x$ (since he is twice as old).
• In 5 years, the brother will be $x + 5$ and John will be $2x + 5$.
• The sum of their ages in 5 years is 40. Equation: $(x + 5) + (2x + 5) = 40$
• Combine like terms: $3x + 10 = 40$
• $3x = 30 \implies x = 10$
• The brother is 10. John's age is $2x = 20$. John is 20 years old.

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Work Problems

Work problems involve people or machines working together to complete a specific task. The key concept is to find the rate of work, which is the portion of the job completed in one unit of time.

Strategy for Work Problems

If someone can complete an entire job in $t$ hours, their work rate is $1/t$ of the job per hour. If two people work together simultaneously, their combined rate is the sum of their individual rates: $\frac{1}{t_1} + \frac{1}{t_2} = \frac{1}{T_{total}}$.

Example: Alice can paint a room in 4 hours. Bob can paint the same room in 6 hours. How long will it take them to paint the room if they work together?

• Alice's rate: $1/4$ of the room per hour.
• Bob's rate: $1/6$ of the room per hour.
• Combined rate: $\frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}$ of the room per hour.
• Let $T$ be the total time required. $\frac{1}{T} = \frac{5}{12} \implies T = \frac{12}{5} = 2.4$ hours.

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Mixture Problems

Mixture problems involve combining two solutions of different concentrations to achieve a new, desired concentration. A table is highly recommended to organize the amount, concentration, and total pure substance of each component.

Strategy for Mixture Problems

The core equation is: $\text{Amount}_1 \times \text{Concentration}_1 + \text{Amount}2 \times \text{Concentration}2 = \text{Amount}{total} \times \text{Concentration}{final}$

Example: How many liters of a 20% acid solution must be added to 10 liters of a 50% acid solution to get a 30% acid solution?

• Let $x$ be the volume of the 20% solution.
• Total volume will be $x + 10$.
• Pure acid from the 20% solution: $0.20x$
• Pure acid from the 50% solution: $0.50(10) = 5$
• Pure acid in the final mixture: $0.30(x + 10)$
• Set up the equation: $0.20x + 5 = 0.30(x + 10)$
• Distribute: $0.20x + 5 = 0.30x + 3$
• Solve for x: $2 = 0.10x \implies x = 20$
• 20 liters of the 20% solution must be added.

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Rate, Distance, and Time

The fundamental formula for motion problems is $d = r \times t$ (Distance = Rate × Time).

Strategy for Motion Problems

Depending on the specific problem, you may need to set distances equal to each other, add distances together, or relate the travel times.

Example: A train leaves a station traveling at 60 mph. Two hours later, an express train leaves the same station traveling in the same direction at 90 mph. How long will it take the express train to catch up to the first train?

• Let $t$ be the time the express train travels.
• The first train left two hours earlier, so it has been traveling for $t + 2$ hours.
• Distance of the first train: $60(t + 2)$
• Distance of the express train: $90t$
• Since they catch up to each other, their traveled distances must be equal: $60(t + 2) = 90t$
• Distribute: $60t + 120 = 90t$
• Solve for t: $120 = 30t \implies t = 4$
• It takes the express train 4 hours to catch up.

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Ratio and Proportion

A ratio is a comparison of two numbers (e.g., $a:b$ or $a/b$). A proportion is an algebraic equation stating that two ratios are equal: $\frac{a}{b} = \frac{c}{d}$.

Strategy for Proportions

To solve a proportion with an unknown, use cross-multiplication: if $\frac{a}{b} = \frac{c}{d}$, then $ad = bc$.

Example: A recipe calls for 2 cups of sugar for every 3 cups of flour. If you intend to use 9 cups of flour, how many cups of sugar do you need?

• Set up the proportion: $\frac{\text{Sugar}}{\text{Flour}} = \frac{2}{3} = \frac{x}{9}$
• Cross-multiply: $3x = 18$
• Divide by 3: $x = 6$
• You will need 6 cups of sugar.

Common Traps to Avoid

• Trap: Forgetting that if a person works twice as fast, their completion time is halved, not doubled.
• Trap: In mixture problems, always ensure that the total volume on the right side of the equation exactly equals the sum of the individual volumes on the left side.