7.1 Multi-Step Real-World Word Problems

Translating Words to Math

The Specialized High Schools Admissions Test (SHSAT) Math section does not permit the use of a calculator. This constraint means that word problems are designed to test your logical reasoning and ability to construct mathematical models rather than your capacity to perform tedious, large-number computations. To succeed, you must master the art of translation: turning English sentences into algebraic expressions and equations.

The first step in any word problem is to identify the unknown quantity and assign it a variable, usually $x$. The second step is to translate verbs and prepositions into mathematical operations. The table below outlines the most common English terms and their algebraic translations:

Consecutive Integers

Consecutive integers are whole numbers that follow each other in order without gaps, such as $5, 6, 7$. In algebra, consecutive integers are represented as $x, x+1, x+2, \dots$. Consecutive even integers (like $4, 6, 8$) or consecutive odd integers (like $5, 7, 9$) both increase by 2. Therefore, they are represented as $x, x+2, x+4, \dots$ regardless of whether they are even or odd. A common trap is representing consecutive odd integers as $x, x+1, x+3$ — remember that the difference between any two consecutive odd numbers is always 2.

Let's look at a worked example: The sum of three consecutive odd integers is 57. What is the value of the largest of these integers? First, define the variables: let the three consecutive odd integers be $x$, $x+2$, and $x+4$. Second, set up the equation: $x + (x + 2) + (x + 4) = 57$ Combine like terms: $3x + 6 = 57$ Subtract 6 from both sides: $3x = 51$ Divide by 3: $x = 17$ The question asks for the largest integer, which is $x+4$. Therefore, the largest integer is $17 + 4 = 21$.

Identifying the "Target Variable" Trap

One of the most frequent traps on the SHSAT is solving for the variable $x$ and immediately selecting it as the answer, only to realize too late that the question asked for a different quantity. For example, if the question asks for "the value of the largest of these integers," and you defined $x$ as the smallest integer, writing down $x$ as your final answer results in an incorrect response. Always reread the final line of the question before bubbling in your answer to ensure you are answering the exact question asked.

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Rates and Unit Rates

A rate is a comparison of two quantities with different units, such as miles per hour, dollars per hour, or words per minute. A unit rate is a rate where the second quantity is exactly 1 unit. For example, if you run 12 miles in 2 hours, your rate is 6 miles per 1 hour, or 6 mph.

On the SHSAT, rate problems often require you to find a unit rate first and then use it to solve a multi-step problem. Because you cannot use a calculator, look for common factors that allow for simple division and multiplication.

Consider this scenario: A commercial printer can produce 360 color brochures in 18 minutes. At this rate, how many brochures can the printer produce in 2.5 hours? First, calculate the unit rate (brochures per minute): $\text{Unit Rate} = \frac{360 \text{ brochures}}{18 \text{ minutes}} = 20 \text{ brochures per minute}$ Second, convert the target time into the same units as the rate. Since the rate is per minute, convert 2.5 hours into minutes: $2.5 \text{ hours} \times 60 \text{ minutes per hour} = 150 \text{ minutes}$ Third, multiply the unit rate by the total minutes: $20 \text{ brochures per minute} \times 150 \text{ minutes} = 3,000 \text{ brochures}$

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Currency and Coin Problems

Currency problems require you to distinguish between the quantity of the items and the value of the items. For example, having 10 dimes is different from having $1.00.

To solve coin problems, set up two distinct equations:

1. A quantity equation: tracking the number of coins.
2. A value equation: tracking the total monetary worth.

Let's solve a typical problem: A jar contains 45 coins, consisting only of nickels and dimes. If the total value of the coins is $3.50, how many dimes are in the jar? Let $n$ be the number of nickels, and $d$ be the number of dimes. Set up the quantity equation: $n + d = 45$ Set up the value equation (using cents to avoid working with decimals): $5n + 10d = 350$ (since a nickel is 5 cents, a dime is 10 cents, and $3.50 is 350 cents) Now, solve the system of equations. Express $n$ in terms of $d$ from the first equation: $n = 45 - d$ Substitute this into the value equation: $5(45 - d) + 10d = 350$ Distribute the 5: $225 - 5d + 10d = 350$ Combine like terms: $225 + 5d = 350$ Subtract 225 from both sides: $5d = 125$ Divide by 5: $d = 25$ There are 25 dimes in the jar (and $45 - 25 = 20$ nickels). Let's verify: 20 nickels are worth $1.00, and 25 dimes are worth $2.50. The sum is $3.50, which matches the problem description.

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Mixture Problems

Mixture problems involve combining two or more substances with different concentrations (e.g., acid solutions, salt water, or food mixes) to create a new mixture with a target concentration. The key to solving mixture problems is to track the amount of the pure substance (e.g., pure acid, pure salt, or pure ingredient) because the total amount of pure substance remains constant before and after mixing.

The basic formula for the amount of pure substance is: $\text{Pure Substance} = \text{Total Volume/Weight} \times \text{Concentration Percent}$

Let's analyze a standard mixture problem: A chemist has 12 ounces of a 50% acid solution. How many ounces of pure water must be added to dilute the solution to a 30% acid solution? Let $w$ be the number of ounces of pure water to be added. Understand that pure water contains 0% acid. Set up an equation tracking the total amount of pure acid: $\text{Pure acid in original solution} + \text{Pure acid in water added} = \text{Pure acid in final solution}$ $(0.50 \times 12) + (0 \times w) = 0.30 \times (12 + w)$ Simplify the terms: $6 + 0 = 3.6 + 0.30w$ Subtract 3.6 from both sides: $2.4 = 0.30w$ To solve for $w$, divide both sides by 0.30: $w = \frac{2.4}{0.30} = \frac{24}{3} = 8$ The chemist must add 8 ounces of pure water.

Key Concept: Concentration and Total Quantity

In any mixture problem, the total weight or volume of the final mixture is always the sum of the weights or volumes of the components. The percentage concentration, however, is a weighted average and will always lie strictly between the concentrations of the original components. For instance, if you mix a 10% salt solution with a 30% salt solution, the final concentration must be greater than 10% and less than 30%. If your calculated answer lies outside this range, you have made an algebraic error.

Visualizing Mixture Setups

To visualize a mixture problem, you can sketch boxes representing each container:

• Container 1: 12 oz @ 50% concentration $\rightarrow$ holds $12 \times 0.50 = 6$ oz of pure acid.
• Container 2: $w$ oz @ 0% concentration $\rightarrow$ holds $w \times 0 = 0$ oz of pure acid.
• Final Container: $(12 + w)$ oz @ 30% concentration $\rightarrow$ holds $(12 + w) \times 0.30$ oz of pure acid.

By equating the sum of the ingredients' pure substance to the final mixture's pure substance, you build a solid algebraic foundation that prevents conceptual errors.