College Algebra Quiz: Translate Word Problems Into Expressions Equations

What this quiz covers

This quiz focuses on Translate Word Problems Into Expressions Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $(20 + w)(30 + w) = 1200$
2. $(20 + 2w)(30 + 2w) - 20 \cdot 30 = 1200$
3. $20 \cdot 30 + 2w(20 + 30) = 1200$
4. $(20 + 2w)(30 + 2w) = 1200$ (correct answer)

Explanation: When you encounter a problem about objects with uniform borders or frames, visualization is key. You need to think carefully about how the width affects the overall dimensions. The pool is 20 feet by 30 feet, and it's surrounded by a walkway of width $w$ feet on all sides. This means the walkway adds $w$ feet to each side of the pool. Since there are two sides in each direction (left and right, top and bottom), the total dimensions become $(20 + 2w)$ by $(30 + 2w)$. The total area is the area of this larger rectangle, which equals $(20 + 2w)(30 + 2w) = 1200$. This confirms answer choice D is correct. Let's examine why the other options fail. Choice A uses $(20 + w)(30 + w)$, which only adds the width $w$ once to each dimension instead of twice. This represents a common mistake of forgetting that the walkway extends on both sides. Choice B calculates $(20 + 2w)(30 + 2w) - 20 \cdot 30 = 1200$, which would give you just the area of the walkway itself, not the total area. Choice C uses $20 \cdot 30 + 2w(20 + 30) = 1200$, which incorrectly assumes the walkway is only along the perimeter rather than a full border. Strategy tip: When dealing with uniform borders or frames, always remember that the width gets added twice to each dimension—once for each side. Draw a quick sketch if needed to visualize how the border affects the total dimensions.

Question 2

1. $P = 500 \cdot 2^{t/3} - 50t$ (correct answer)
2. $P = 500 \cdot 2^{t/3} - 50$
3. $P = (500 - 50) \cdot 2^{t/3}$
4. $P = 500 \cdot (2^{1/3} - 50)^t$

Explanation: The population would grow to 500 · 2^(t/3) if there were no deaths. However, 50 bacteria die each hour, so after t hours, 50t bacteria have died in total. The net population is the growth minus total deaths: P = 500 · 2^(t/3) - 50t. Choice B only subtracts 50 bacteria total instead of 50 per hour. Choice C incorrectly reduces the initial population and then applies exponential growth. Choice D incorrectly combines the growth and death rates inside the exponential function.

Question 3

1. $A = x(120 - x)$
2. $A = x(60 - \frac{x}{2})$ (correct answer)
3. $A = x(120 - 2x)$
4. $A = \frac{x(120 - x)}{2}$

Explanation: Let the side parallel to the existing fence be x feet, and the perpendicular sides each be y feet. Since only three sides need fencing: x + 2y = 120, so y = (120 - x)/2 = 60 - x/2. The area is A = xy = x(60 - x/2). Choice A assumes all four sides need fencing. Choice C uses the wrong constraint equation. Choice D incorrectly divides the entire area expression by 2.

Question 4

1. $\begin{cases} 150d + 3m = 89 \\ 280d + 5m = 142 \end{cases}$
2. $\begin{cases} 3d + 150 = 89m \\ 5d + 280 = 142m \end{cases}$
3. $\begin{cases} 3d + 150m = 89 \\ 5d + 280m = 142 \end{cases}$ (correct answer)
4. $\begin{cases} 3(d + 150m) = 89 \\ 5(d + 280m) = 142 \end{cases}$

Explanation: When you encounter word problems involving multiple variables and different scenarios, you need to translate each situation into an equation where the total cost equals the sum of all charges. Here, each customer pays a daily rate ($d$ dollars per day) plus a mileage rate ($m$ dollars per mile). So the total cost formula is: (days × daily rate) + (miles × mileage rate) = total paid. For Customer A: 3 days × $d$ + 150 miles × $m$ = $89, giving us $3d + 150m = 89$ For Customer B: 5 days × $d$ + 280 miles × $m$ = $142, giving us $5d + 280m = 142$ This matches option C exactly. Option A incorrectly swaps the coefficients—it puts miles where days should be and days where miles should be ($150d + 3m = 89$). This would mean Customer A paid for 150 days and 3 miles, which contradicts the problem. Option B rearranges the terms incorrectly, setting $3d + 150 = 89m$. This suggests that the daily and mileage charges somehow equal a multiple of the mileage rate, which doesn't reflect the additive cost structure. Option D uses $3(d + 280m) = 142$, which would mean the daily rate gets multiplied by the number of days, but then you'd add the same daily rate to each mile driven—an illogical pricing structure. Study tip: In multi-variable word problems, always identify what each variable represents, then systematically substitute the given values. Write out the cost formula first: (quantity₁ × rate₁) + (quantity₂ × rate₂) = total cost.

Question 5

1. $250t = 25(t^2 + 4)$
2. $\frac{250}{t^2 + 4} = 25$
3. $\frac{250t}{t^2 + 4} = 25$ (correct answer)
4. $\frac{250t}{(t^2 + 4)^2} = 25$

Explanation: When you encounter a problem asking when a function equals a specific value, you're solving an equation where the function output equals that target value. This is a fundamental skill in working with mathematical models. To find when the drug concentration equals 25 mg/L, you need to set the concentration function equal to 25 and solve for $t$. Since $C(t) = \frac{250t}{t^2 + 4}$ represents the concentration at time $t$, the equation becomes $\frac{250t}{t^2 + 4} = 25$. This is exactly what choice C shows. Looking at the incorrect options: Choice A, $250t = 25(t^2 + 4)$, results from incorrectly moving the denominator $(t^2 + 4)$ to the right side without maintaining the fraction structure. Choice B, $\frac{250}{t^2 + 4} = 25$, omits the $t$ in the numerator entirely, which would represent a different function altogether. Choice D, $\frac{250t}{(t^2 + 4)^2} = 25$, incorrectly squares the denominator, which would actually represent the derivative of the original function rather than the function itself. Remember: when solving "when does a function equal a value" problems, always set the complete function expression equal to that value. Don't manipulate the function before setting up the equation—that's where algebra mistakes typically occur. Set up the equation first, then solve.

Question 6

1. $P = p(200 - 5p) - 8(200 - 5p) - 300$
2. $P = (p - 8)(200 - 5p) - 300$ (correct answer)
3. $P = p(200 - 5p) - 8p - 300$
4. $P = (200 - 5p)(p - 8 - 300)$

Explanation: Daily profit = Revenue - Total Costs. Revenue = price × quantity = p(200 - 5p). Total costs = variable costs + fixed costs = cost per unit × quantity + fixed costs = 8(200 - 5p) + 300. So P = p(200 - 5p) - [8(200 - 5p) + 300] = (200 - 5p)(p - 8) - 300. Choice A shows the expanded form but is equivalent to choice B. Choice C incorrectly calculates variable costs as 8p instead of 8(200 - 5p). Choice D incorrectly subtracts fixed costs from the per-unit profit margin instead of from total profit.

Question 7

1. $\frac{d}{12} + \frac{d}{18} + \frac{r}{12} = 1$ (correct answer)
2. $\frac{d}{12} + \frac{d}{18} = 1 - \frac{r}{12}$
3. $\frac{d}{12} + \frac{r}{18} = 1$
4. $12d + 18d + 12r = 1$

Explanation: The first crew works at rate 1/12 jobs per day, the second at rate 1/18 jobs per day. Working together for d days, they complete d/12 + d/18 of the job. Then the first crew works alone for r additional days, completing r/12 more of the job. The total work done must equal 1 complete job: d/12 + d/18 + r/12 = 1. Choice B incorrectly rearranges the equation. Choice C omits the work done by the second crew during the joint period. Choice D uses incorrect rates (should be fractions, not whole numbers).