14.3 Word Problems & Applications

14.3 Word Problems & Applications

Quick Answer: For PERT word problems, define a clear variable, translate English into algebraic expressions, write one equation or inequality that models the whole situation, solve with your linear/quadratic toolkit, and check that the result is realistic. Classic setups include rate ($d = rt$), percent ($\text{part} = \text{percent} \times \text{whole}$), and simple mixture or cost combinations. The algebra is rarely harder than what you already practice—the challenge is accurate translation.

PERT Mathematics expects you to apply algebra outside bare equations. Items may describe a trip, a discount, a mixture of solutions, or a geometric constraint. This section focuses on a repeatable translation process and on the problem types that appear at college-readiness level.

Universal Translation Framework

Keyword → symbol map

Order trap: “7 less than a number” is $x - 7$, not $7 - x$. “7 less a number” is ambiguous in casual speech—on tests, “less than” almost always means subtract from the number mentioned second.

Worked Example 1: Linear “consecutive” style setup

Problem: The sum of two consecutive even integers is 46. Find the integers.

Step 1 — Define. Let $x$ = the first even integer. Then the next consecutive even integer is $x + 2$.

Step 2 — Equation. $x + (x + 2) = 46$.

Step 3 — Solve. $2x + 2 = 46 \Rightarrow 2x = 44 \Rightarrow x = 22$. The integers are 22 and 24.

Step 4 — Check. $22 + 24 = 46$ ✓; both even and consecutive even ✓.

Worked Example 2: Age / comparison wording

Problem: Maya is 4 years older than twice her brother’s age. Together they are 28 years old. How old is the brother?

Define: Let $b$ = brother’s age. Maya’s age = $2b + 4$.

Equation: $b + (2b + 4) = 28 \Rightarrow 3b + 4 = 28 \Rightarrow 3b = 24 \Rightarrow b = 8$.

Check: Maya is $2(8) + 4 = 20$; $8 + 20 = 28$ ✓.

Rate Problems: Distance, Work, and Simple Speed

Core formula:

$\text{distance} = \text{rate} \times \text{time} \quad (d = rt)$

Rearrangements: $r = \frac{d}{t}$, $t = \frac{d}{r}$.

Worked Example 3: Distance–rate–time

Problem: A car travels 150 miles in the same time a train travels 210 miles. The train’s speed is 20 mph faster than the car’s. Find both speeds.

Define: Let $r$ = car speed (mph). Train speed = $r + 20$.

Equal times: Time car = $\frac{150}{r}$; time train = $\frac{210}{r + 20}$. Set equal:

$\frac{150}{r} = \frac{210}{r + 20}$

Cross-multiply: $150(r + 20) = 210r$

$150r + 3000 = 210r$

$3000 = 60r$

$r = 50$

Car 50 mph; train 70 mph.

Check: Times $\frac{150}{50} = 3$ h and $\frac{210}{70} = 3$ h ✓.

Worked Example 4: Simple work rate

Problem: Pipe A fills a tank in 6 hours; pipe B fills it in 3 hours. How long do they take working together?

Rates: A does $\frac{1}{6}$ tank/hour; B does $\frac{1}{3}$ tank/hour.

Together: $\frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$ tank/hour.

Time for 1 tank: 2 hours.

Percent Problems

$\text{part} = \frac{\text{percent}}{100} \times \text{whole}$

Also: percent change $= \frac{\text{new} - \text{old}}{\text{old}} \times 100%$.

Worked Example 5: Discount

Problem: A jacket costs $80. During a sale it is 25% off. What is the sale price?

$\text{Sale price} = 80\left(1 - 0.25\right) = 80 \times 0.75 = 60$

Alternatively, discount amount $0.25 \times 80 = 20$; $80 - 20 = 60$.

Worked Example 6: “What percent”

Problem: 18 is what percent of 45?

$\frac{18}{45} = 0.4 = 40\%$

Worked Example 7: Percent mixture of a whole

Problem: 30% of a number is 24. What is the number?

$0.30x = 24 \Rightarrow x = \frac{24}{0.30} = 80$

Simple Mixture Problems

Track the amount of pure ingredient in each part. If you mix $A$ liters at $p%$ with $B$ liters at $q%$, the pure amounts add:

$A \cdot \frac{p}{100} + B \cdot \frac{q}{100} = (A + B) \cdot \frac{r}{100}$

where $r%$ is the resulting concentration.

Worked Example 8: Mixture

Problem: How many liters of a 20% acid solution should be mixed with 10 liters of a 50% acid solution to get a 30% solution?

Define: Let $x$ = liters of 20% solution.

Pure acid equation:

$0.20x + 0.50(10) = 0.30(x + 10)$

$0.20x + 5 = 0.30x + 3$

$5 - 3 = 0.30x - 0.20x$

$2 = 0.10x$

$x = 20$

Check: Pure acid: $0.20(20) + 5 = 4 + 5 = 9$. Total volume 30 L; $9/30 = 0.30 = 30%$ ✓.

Geometric and Quadratic Applications

Some PERT items wrap a quadratic in a rectangle or area story.

Worked Example 9: Area leading to a quadratic

Problem: The length of a rectangle is 3 meters more than the width. The area is 40 square meters. Find the width.

Define: Width $w$; length $w + 3$.

Equation: $w(w + 3) = 40 \Rightarrow w^2 + 3w - 40 = 0$.

Factor: $(w + 8)(w - 5) = 0 \Rightarrow w = -8$ or $w = 5$.

Reject $w = -8$ (length cannot be negative). Width = 5 m; length = 8 m.

Check: $5 \times 8 = 40$ ✓.

This is where quadratic solving from Section 14.1 meets applications: algebra may produce an extra root that the story forbids.

Inequalities in Word Problems

Translate “at least,” “at most,” and “minimum/maximum” into $\ge$ or $\le$. Solve the inequality; remember to reverse the inequality sign if you multiply or divide by a negative number.

Worked Example 10: Budget inequality

Problem: Tickets cost $12 each. You have at most $100. What is the maximum number of whole tickets you can buy?

$12n \le 100 \Rightarrow n \le \frac{100}{12} \approx 8.33$

Maximum whole tickets: $n = 8$.
Check: $12 \times 8 = 96 \le 100$; $12 \times 9 = 108 > 100$.

Reasonableness Checks (Do Not Skip)

Quick validation habits

1. Substitute your answer back into the English sentence, not only the equation.
2. Estimate first: 25% of 80 should be near 20—if you get 200, a decimal place is wrong.
3. Units: mph times hours gives miles; mixing liters with liters; dollars with dollars.
4. Discard algebraic solutions that violate constraints you noted at setup.

Putting It Together: Full Translation Drill

Problem: A student scored 78, 84, and 90 on three tests. What score is needed on a fourth test so that the average is at least 85?

Define: Let $x$ = fourth score.

Inequality: $\frac{78 + 84 + 90 + x}{4} \ge 85$

$\frac{252 + x}{4} \ge 85$

$252 + x \ge 340$

$x \ge 88$

Answer: At least 88 on the fourth test.

Check: Average of 78, 84, 90, 88: $\frac{340}{4} = 85$ exactly ✓.

PERT Strategy Summary Table

Word problems reward slow setup and fast algebra. On an adaptive placement test, a correct equation almost always yields a correct answer if you finish the solve carefully—and a wrong equation almost never recovers. Spend your careful seconds on the translation, not on second-guessing after a clean check.