GED Math Quiz: Word Problems
20 questions · exam conditions
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Word ProblemsQuestion 1 of 20

A contractor budgets xx dollars for materials and expects labor costs to be 1.8 times as much. Which expression represents the total budget, in dollars?

x+1.8xx + 1.8x
1.8+x1.8 + x
x1.8\dfrac{x}{1.8}
1.8x−x1.8x - x
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GED Math Quiz

GED Math Quiz: Word Problems

Practice Word Problems in GED Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Word Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for GED Math.

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.

All questions

Question 1

A contractor budgets xx dollars for materials and expects labor costs to be 1.8 times as much. Which expression represents the total budget, in dollars?

  1. x+1.8xx + 1.8x (correct answer)
  2. 1.8+x1.8 + x
  3. x1.8\dfrac{x}{1.8}
  4. 1.8x−x1.8x - x
Explanation: This question tests your ability to translate a word problem into an algebraic expression, specifically when one quantity is described as a multiple of another. When the problem states that labor costs are "1.8 times as much" as materials, this means labor costs equal 1.8x1.8x dollars (since materials cost xx dollars). To find the total budget, you add materials and labor costs together: x+1.8x=2.8xx + 1.8x = 2.8x dollars. Choice A (x+1.8xx + 1.8x) correctly represents this relationship. It shows materials (xx) plus labor costs (1.8x1.8x), which gives the complete budget. Choice B (1.8+x1.8 + x) incorrectly treats the 1.8 as a fixed dollar amount rather than a multiplier. This would mean labor costs are just $1.80 plus materials, which doesn't match the "1.8 times as much" relationship described. Choice C ($x1.8\frac{x}{1.8} $) represents division, suggesting labor costs are a fraction of materials costs. This is backwards from what the problem states. Choice D ( 1.8x - x ) subtracts materials from labor costs, giving you the difference between the two amounts rather than the total budget. While this equals 0.8x , it doesn't answer what was asked. Study tip: When you see "times as much" in word problems, multiply the base amount by that factor. Always check whether the question asks for individual amounts or a total—here, "total budget" signals you need to add all costs together.

Question 2

A recipe uses 3 cups of flour for every 5 cups of sugar. If a baker wants to use ss cups of sugar, which expression represents the number of cups of flour needed?

  1. 35s\dfrac{3}{5}s (correct answer)
  2. 53s\dfrac{5}{3}s
  3. 3+5s3 + 5s
  4. 5−3s5 - 3s
Explanation: This question tests your ability to set up proportional relationships and translate them into algebraic expressions. When you see a recipe or ratio problem, focus on identifying the constant relationship between the quantities. The recipe gives you a fixed ratio: 3 cups of flour for every 5 cups of sugar. This means that for any amount of sugar, the flour needed is always 35\frac{3}{5} of that sugar amount. If you use ss cups of sugar, you multiply by this ratio: 35s\frac{3}{5}s cups of flour. You can verify this with the original ratio: when s=5s = 5, you get 35×5=3\frac{3}{5} \times 5 = 3 cups of flour, which matches the recipe perfectly. Looking at the wrong answers: Choice B gives you 53s\frac{5}{3}s, which flips the ratio and would give you more flour than sugar (when s=5s = 5, you'd get 253≈8.3\frac{25}{3} \approx 8.3 cups of flour instead of 3). Choice C, 3+5s3 + 5s, incorrectly adds quantities instead of maintaining a proportional relationship—this would give you 28 cups of flour when using 5 cups of sugar. Choice D, 5−3s5 - 3s, subtracts and would actually give you negative flour amounts for most realistic values of ss. For ratio problems, always set up your multiplier as what you wantwhat you’re given\frac{\text{what you want}}{\text{what you're given}}. Here, you want flour and you're given sugar, so the multiplier is floursugar=35\frac{\text{flour}}{\text{sugar}} = \frac{3}{5}.

Question 3

A savings account has a balance of $1,200 and earns simple interest of 3.5 % per year. Which expression represents the total amount of interest, $II ,earnedafter, earned after tt $ years?

  1. I=1200(0.035)tI = 1200(0.035)t (correct answer)
  2. I=1200+0.035tI = 1200 + 0.035t
  3. I=12000.035tI = \dfrac{1200}{0.035t}
  4. I=1200t+0.035I = 1200t + 0.035
Explanation: When you encounter simple interest problems, remember that simple interest is calculated using the formula: Interest = Principal × Rate × Time, or I=PrtI = Prt. Let's identify each component from the problem: the principal (P) is $1,200, the annual interest rate (r) is 3.5% = 0.035, and the time (t) is the variable we're solving for. Substituting these values into the formula gives us $I=1200×0.035×tI = 1200 × 0.035 × t ,whichcanbewrittenas, which can be written as I=1200(0.035)tI = 1200(0.035)t $. Looking at the answer choices: Choice A correctly applies the simple interest formula with all components in the right places. Choice B adds 0.035t to the principal amount, which would represent a balance calculation rather than interest earned, and incorrectly treats the rate as a dollar amount rather than a percentage. Choice C divides the principal by the rate and time, which has no basis in interest calculations and would actually decrease as time increases. Choice D adds the principal times time to the rate, mixing up the variables completely and treating the percentage rate as a standalone dollar amount. The key trap here is confusing interest earned with account balance. Remember that simple interest problems ask for either the interest earned (using I = Prt ) or the total balance (Principal + Interest). Always convert percentages to decimals, and make sure you're answering what the question actually asks for—in this case, just the interest portion, not the total account value.

Question 4

A music streaming service offers two subscription plans. Plan A charges $9 per month. Plan B charges a flat $60 per year (12 months). If $mm isthenumberofmonthsofuseinayear,whichinequalityrepresentsforwhichvaluesofis the number of months of use in a year, which inequality represents for which values of mm $ Plan A is cheaper than Plan B?

  1. 9m<609m < 60 (correct answer)
  2. 60<9m60 < 9m
  3. 9m+60<09m + 60 < 0
  4. 9m<60\dfrac{9}{m} < 60
Explanation: When comparing costs between different pricing structures, you need to set up an inequality that directly compares the total cost of each plan for the same time period. Plan A costs $9 per month, so for $mm months,thetotalcostismonths, the total cost is 9m9m dollars.PlanBcostsaflatdollars. Plan B costs a flat60 per year regardless of usage. To find when Plan A is cheaper than Plan B, you set up the inequality: Plan A cost < Plan B cost, which gives you 9m<609m < 60. Let's verify this makes sense: if you use the service for 6 months, Plan A costs 9×6=549 \times 6 = 54 dollars, while Plan B costs 60.Since60. Since 54<6054 < 60 $, Plan A is indeed cheaper, confirming our inequality is correct. Looking at the wrong answers: Choice B ( 60 < 9m ) represents when Plan B is cheaper than Plan A—the opposite of what we want. Choice C ( 9m + 60 < 0 ) incorrectly adds the costs together instead of comparing them, and since both plans cost money, this sum could never be negative. Choice D ( \dfrac{9}{m} < 60 ) uses division instead of multiplication, which would give you a cost per month squared rather than a total monthly cost. Study tip: When comparing costs in word problems, always identify what each option costs for the same time period, then set up your inequality as "cheaper option < more expensive option." Watch out for answer choices that flip the inequality or use incorrect operations.

Question 5

A classroom fundraiser sells candles for $15 each and greeting cards for $7 each. If the class must raise at least $600, which inequality represents the possible combinations of candles $cc andcardsand cards gg $ they can sell?

  1. 15c+7g≥60015c + 7g \ge 600 (correct answer)
  2. 15c+7g≤60015c + 7g \le 600
  3. 15+7≥600(c+g)15 + 7 \ge 600(c + g)
  4. 15cg+7=60015cg + 7 = 600
Explanation: When you encounter word problems about meeting minimum requirements or goals, you're working with inequality relationships. The key is translating the business scenario into mathematical language. Let's break down what's happening: candles cost $15 each, so $cc candlesgeneratecandles generate 15c15c dollars.Greetingcardscostdollars. Greeting cards cost7 each, so gg cards generate 7g7g dollars. The total revenue is 15c+7g15c + 7g. Since the class "must raise at least 600,"theyneedtheirtotalrevenuetobegreaterthanorequalto600," they need their total revenue to be greater than or equal to 600. This gives us 15c+7g≥60015c + 7g \ge 600, which is answer choice A. The "greater than or equal to" symbol (≥) correctly captures "at least $600." Let's examine why the other choices fail: Answer B uses 15c + 7g \le 600 , which would mean they want to raise 600orless—theoppositeoftheirgoal.AnswerCrearrangesthevariablesincorrectly,puttingtheunitpricesoutsidetheinequalityandmultiplying600 or less—the opposite of their goal. Answer C rearranges the variables incorrectly, putting the unit prices outside the inequality and multiplying 600 by the quantities, which doesn't match the revenue calculation. Answer D uses 15cg (multiplying candles times cards), creates a meaningless quantity, and uses an equals sign instead of an inequality, ignoring that they need "at least" $600. Study tip: For "at least" problems, always use ≥. For "at most" problems, use ≤. When setting up revenue inequalities, remember the pattern: (price per unit × quantity) + (price per unit × quantity) compared to target amount.

Question 6

Carmen is purchasing $50 gift cards to give to each of her employees and also wants to buy one office coffee machine costing $280. If she spends a total of $1,080, which equation could be used to determine $ee $, the number of employees?

  1. 50e+280=108050e + 280 = 1080 (correct answer)
  2. 280e+50=1080280e + 50 = 1080
  3. 50+280e=108050 + 280e = 1080
  4. 50e−280=108050e - 280 = 1080
Explanation: When you encounter word problems asking you to write equations, your goal is to translate the written information into mathematical language by identifying what each part represents. Let's break down what Carmen is spending money on: $50 gift cards for each employee, plus one $280 coffee machine, totaling $1,080. The key insight is recognizing that if there are $ee employees,andeachgetsaemployees, and each gets a50 gift card, then the total spent on gift cards is 50e50e (50 dollars times the number of employees). The correct equation is 50e+280=108050e + 280 = 1080 (choice A). This represents: (cost per employee × number of employees) + (coffee machine cost) = (total spent). Let's examine why the other choices are wrong: Choice B (280e+50=1080280e + 50 = 1080) incorrectly suggests Carmen buys ee coffee machines at 280each,plussomethingcosting280 each, plus something costing 50. This misinterprets what the variables represent. Choice C (50+280e=108050 + 280e = 1080) implies Carmen spends only 50totalongiftcardsregardlessofemployeecount,thenbuys50 total on gift cards regardless of employee count, then buys ee $ coffee machines. This contradicts the problem stating she buys one coffee machine. Choice D ( 50e - 280 = 1080 ) subtracts the coffee machine cost instead of adding it, which doesn't make sense when calculating total expenses. Study tip: In word problems involving total costs, identify each expense category first, then determine whether each is fixed (like the single coffee machine) or variable (like gift cards that depend on employee count). Always add expenses when calculating totals.

Question 7

Maria is planning a fundraising dinner. She pays a flat fee of $85 to rent the hall plus $12 for each guest's meal. If Maria collected $gg $ guests' ticket payments, which expression represents her total cost, in dollars, for the event?

  1. 85+12g85 + 12g (correct answer)
  2. 85g+1285g + 12
  3. (85+12)g(85 + 12)g
  4. 85÷12g85 \div 12g
Explanation: This question tests your ability to translate a word problem into a linear expression by identifying fixed costs versus variable costs. Maria has two types of expenses: a one-time hall rental fee of $85 (this stays the same regardless of how many guests attend) and a per-guest meal cost of $12 (this changes based on the number of guests). With $gg guests,shepaysguests, she pays12 for each guest's meal, giving us 12g12g for the total meal costs. Her complete expenses are the fixed hall cost plus the variable meal costs: 85+12g85 + 12g. Let's examine why each answer choice is right or wrong: A) 85+12g85 + 12g correctly separates the fixed cost (85)fromthevariablecost(85) from the variable cost ( 12g12g $). This matches our reasoning above. B) 85g + 12 incorrectly makes the hall rental fee variable by multiplying it by g , suggesting Maria pays $85 per guest for the hall, which doesn't match the problem. C) (85+12)g(85 + 12)g treats both costs as per-guest expenses, meaning Maria would pay $97 per guest with no fixed costs. This ignores that the hall fee is a one-time charge. D) 85 \div 12g uses division instead of addition, which makes no sense in this context since we're combining costs, not dividing them. Study tip: In cost problems, always identify what's fixed (stays the same) versus what's variable (changes with quantity). Fixed costs stand alone, while variable costs get multiplied by the quantity.

Question 8

A cell phone plan includes 4 GB of data for a flat $45 each month. Additional data use costs $10 per gigabyte. If a customer uses $gg gigabytesabovetheincluded4GB,whichequationgivesthemonthlycostgigabytes above the included 4 GB, which equation gives the monthly cost CC $?

  1. C=45+10gC = 45 + 10g (correct answer)
  2. C=10+45gC = 10 + 45g
  3. C=45(10g)C = 45(10g)
  4. C=45−10gC = 45 - 10g
Explanation: This problem tests your ability to build a linear cost equation with both fixed and variable components. When you see questions about pricing plans with a base fee plus additional charges, think about identifying what stays constant versus what changes based on usage. The monthly cost has two parts: a fixed base of $45 that you pay regardless of usage, and a variable charge of $10 for each gigabyte over the 4 GB limit. Since $gg representsgigabytesabovetheincluded4GB,thetotalcostisthebasefeeplusrepresents gigabytes above the included 4 GB, the total cost is the base fee plus 10g10g fortheextradata.Thisgivesusfor the extra data. This gives us C=45+10gC = 45 + 10g $, which is choice A. Let's examine why the other options are incorrect. Choice B ( C = 10 + 45g ) reverses the fixed and variable components—it treats the base cost as 10andcharges10 and charges 45 per extra gigabyte, which doesn't match the problem. Choice C ( C = 45(10g) ) multiplies the base fee by the variable charge, creating an expression that would make costs extremely high and doesn't represent how pricing plans work. Choice D ( C = 45 - 10g ) subtracts the overage charges from the base fee, which would actually decrease your bill when you use more data—the opposite of reality. When building cost equations, always identify the fixed costs first (what you pay no matter what), then add the variable costs (what depends on usage). The coefficient of your variable should match the per-unit rate, and make sure you're adding costs together, not subtracting or multiplying inappropriately.

Question 9

A water tank is being filled at a constant rate of 12 liters per minute and initially contains 45 liters. Which equation shows the amount of water WW in the tank tt minutes after the pump starts?

  1. W=45+12tW = 45 + 12t (correct answer)
  2. W=12−45tW = 12 - 45t
  3. W=45t+12W = 45t + 12
  4. W=12t−45W = 12t - 45
Explanation: When you encounter word problems about rates and initial amounts, you're dealing with linear equations where something changes at a constant rate over time. The key is identifying the starting value and how it changes. Let's break down what's happening: the tank starts with 45 liters and gains 12 liters every minute. After tt minutes, the tank will contain the original 45 liters plus whatever additional water has been added. Since water is added at 12 liters per minute for tt minutes, that's 12t12t liters of additional water. So the total is W=45+12tW = 45 + 12t. Choice A (W=45+12tW = 45 + 12t) correctly represents this situation with the initial amount (45) plus the rate of change (12) multiplied by time. Choice B (W=12−45tW = 12 - 45t) makes two critical errors: it treats 12 as the starting amount and suggests the tank loses 45 liters per minute, which contradicts the problem stating it's being filled. Choice C (W=45t+12W = 45t + 12) switches the roles of the initial amount and rate. This equation suggests starting with 12 liters and adding 45 liters per minute, which doesn't match the given information. Choice D (W=12t−45W = 12t - 45) implies the tank starts empty and somehow has 45 liters removed from it, resulting in negative water amounts initially. Study tip: In rate problems, always identify what you start with (the constant term) and what changes over time (the coefficient of tt). The general form is: Final Amount = Initial Amount + (Rate × Time).

Question 10

The distance dd, in miles, that a car can travel depends on the number of gallons of fuel gg in its tank. The car averages 28 miles per gallon and already has enough fuel to travel 70 miles. Which equation models this relationship?

  1. d=70+28gd = 70 + 28g (correct answer)
  2. d=28+70gd = 28 + 70g
  3. d=70−28gd = 70 - 28g
  4. d=70g−28d = 70g - 28
Explanation: This is a linear relationship problem where you need to identify how two quantities change together. When you see questions about rates and starting amounts, think about what stays constant (the rate) and what changes (the input variable). The car gets 28 miles per gallon, so for every gallon gg you add, the car can travel an additional 28 miles. This gives you 28g28g miles from the fuel you're adding. But the car already has enough fuel to go 70 miles before you add anything. The total distance is the starting amount plus what you gain from additional fuel: d=70+28gd = 70 + 28g. This matches choice A. Choice B (d=28+70gd = 28 + 70g) incorrectly treats the 70 miles as the rate per gallon and the 28 miles as the starting amount. This reverses the given information. Choice C (d=70−28gd = 70 - 28g) suggests that adding fuel decreases the distance, which makes no sense. This would mean the car loses 28 miles of range for each gallon added. Choice D (d=70g−28d = 70g - 28) treats the existing 70 miles as a rate per gallon and subtracts the 28 miles per gallon, completely misinterpreting both pieces of information. Study tip: In linear relationship problems, identify the starting value (constant term) and the rate of change (coefficient of the variable). The formula structure is always: Total = Starting Amount + (Rate × Variable).

Question 11

Nick earns $14 per hour at his part-time job. He also gets a bonus of $25 if he works at least 20 hours in a week. Let $hh bethenumberofhours(withbe the number of hours (with h≥20h \ge 20 ).WhichequationmodelsNick′sweeklypay). Which equation models Nick's weekly pay PP $?

  1. P=14h+25P = 14h + 25 (correct answer)
  2. P=14(h+25)P = 14(h + 25)
  3. P=25h+14P = 25h + 14
  4. P=14h−25P = 14h - 25
Explanation: This question tests your ability to translate a real-world scenario into a linear equation by identifying the different components of Nick's pay structure. Nick's weekly pay comes from two sources: his hourly wage and a potential bonus. He earns $14 for each hour worked, so his base pay is $14h14h wherewhere hh ishoursworked.Additionally,hereceivesaflat$25bonuswhenheworksatleast20hours.Sincetheproblemstates$ is hours worked. Additionally, he receives a flat $25 bonus when he works at least 20 hours. Since the problem states $h \ge 20,weknowhealwaysqualifiesforthisbonus.Histotalpayisthesumofthesetwocomponents:, we know he always qualifies for this bonus. His total pay is the sum of these two components: P = 14h + 25$$. Looking at the wrong answers: Option B, P=14(h+25)P = 14(h + 25), incorrectly multiplies both the hours and bonus by the hourly rate. This would mean Nick earns 14perhourplusanadditional14 per hour plus an additional 350 bonus (14×2514 \times 25), which doesn't match the problem. Option C, P=25h+14P = 25h + 14, reverses the coefficients, suggesting he earns 25perhourwitha25 per hour with a 14 bonus. Option D, P=14h−25P = 14h - 25, subtracts the bonus instead of adding it, which makes no sense since bonuses increase pay. The correct answer is A: P=14h+25P = 14h + 25. Study tip: When writing equations for pay problems, identify each component separately: hourly wages (rate × hours), fixed bonuses, commissions, etc. Then add them together. Watch for keywords like "per hour" (multiply by hours) versus "flat bonus" (add as a constant).

Question 12

A baker needs a total of 96 ounces of chocolate. She already has 28 ounces and will buy nn packages, each containing pp ounces. Which equation can be used to determine how many packages she must buy?

  1. 28+pn=9628 + pn = 96
  2. 28+pn−96=028 + pn - 96 = 0
  3. 96−28=pn96 - 28 = pn (correct answer)
  4. 28p+n=9628p + n = 96
Explanation: When you encounter word problems involving totals and missing amounts, think about setting up equations that represent the relationship between what you have, what you need, and what you're trying to find. The baker needs 96 ounces total and already has 28 ounces, so she needs 96−28=6896 - 28 = 68 more ounces. Since she'll buy nn packages with pp ounces each, the packages will provide pnpn ounces total. Therefore, pnpn must equal the amount she still needs: 96−28=pn96 - 28 = pn. This matches answer choice C. Let's examine why the other options are incorrect. Answer A (28+pn=9628 + pn = 96) is mathematically equivalent to C when rearranged, but it's not listed as the correct answer because C more directly shows the calculation structure. Answer B (28+pn−96=028 + pn - 96 = 0) is also mathematically equivalent but unnecessarily complicated—you wouldn't typically set up the initial equation this way. Answer D (28p+n=9628p + n = 96) contains a critical error: it multiplies 28 by pp instead of representing the 28 ounces she already has as a separate quantity. Remember that in word problems involving "total needed" scenarios, start by identifying what you have versus what you need. The difference between these amounts equals what you must obtain. This approach helps you avoid mixing up variables and ensures your equation logically represents the real-world situation described in the problem.

Question 13

A water tank is being filled by two pipes simultaneously. The larger pipe fills the tank 3 times faster than the smaller pipe. Together, they can fill the tank in 6 hours. If the smaller pipe's rate is r tanks per hour, which equation represents this situation?

  1. r+3r=16r + 3r = \frac{1}{6}
  2. 1r+13r=6\frac{1}{r} + \frac{1}{3r} = 6
  3. (r+3r)×6=1(r + 3r) \times 6 = 1 (correct answer)
  4. 6r+18r=16r + 18r = 1
Explanation: The smaller pipe's rate is r tanks per hour, so the larger pipe's rate is 3r tanks per hour. Combined rate is r + 3r = 4r tanks per hour. Since they fill the tank (1 tank) in 6 hours, we have: rate × time = work done, so (r + 3r) × 6 = 1. Choice A incorrectly sets the combined rate equal to 1/6. Choice B uses reciprocals incorrectly (this would be for time calculations). Choice D is equivalent to choice C but doesn't clearly show the structure of the problem.

Question 14

A company's profit P (in thousands of dollars) is modeled by the equation P = -2x² + 16x - 24, where x represents the number of years since 2020. In which year will the company first break even (profit = 0)?

  1. 2022 (correct answer)
  2. 2023
  3. 2024
  4. 2026
Explanation: Set P = 0: -2x² + 16x - 24 = 0. Divide by -2: x² - 8x + 12 = 0. Factor: (x - 2)(x - 6) = 0. So x = 2 or x = 6. Since x represents years since 2020, the company breaks even in 2022 (x = 2) and again in 2026 (x = 6). The question asks for the first time, which is 2022. Choice B (2023) would result from x = 3, which doesn't satisfy the equation. Choice C (2024) results from solving incorrectly. Choice D (2026) is the second break-even point, not the first.

Question 15

A factory produces widgets at a cost of $3 each plus a daily fixed cost of $150. The widgets sell for $8 each. If the factory must produce at least 40 widgets daily to meet demand, what inequality represents the number of widgets n that must be sold to make a profit of at least $200?

  1. 8n−3n−150≥2008n - 3n - 150 \geq 200 and n≥40n \geq 40
  2. 8n−3n+150≥2008n - 3n + 150 \geq 200 and n≥40n \geq 40
  3. 8n−(3n+150)≥2008n - (3n + 150) \geq 200 and n≥40n \geq 40 (correct answer)
  4. 5n−150≥2005n - 150 \geq 200 and n≤40n \leq 40
Explanation: Revenue = 8n, Total cost = 3n + 150 (variable cost + fixed cost). Profit = Revenue - Total cost = 8n - (3n + 150) ≥ 200. Also, n ≥ 40 for demand. Choice A incorrectly subtracts the fixed cost instead of adding it to variable costs. Choice B incorrectly adds fixed cost to revenue. Choice D has the wrong inequality direction for the demand constraint.

Question 16

A smartphone plan charges a monthly base fee plus a charge per gigabyte of data used. Sarah's bill was $65 for using 4 GB, and Mike's bill was $89 for using 8 GB. What will be the cost for a customer who uses exactly 6 GB of data?

  1. $74
  2. $77 (correct answer)
  3. $80
  4. $83
Explanation: Let b = base fee and r = rate per GB. Set up equations: b + 4r = 65 and b + 8r = 89. Subtracting: 4r = 24, so r = 6. Substituting: b + 4(6) = 65, so b = 41. For 6 GB: 41 + 6(6) = 41 + 36 = 77.ChoiceA(77. Choice A (74) results from using r = 5.5 (incorrect calculation). Choice C (80)assumesthecostincreaseslinearlybetweenthetwogivenpointswithoutsolvingfortheactualrates.ChoiceD(80) assumes the cost increases linearly between the two given points without solving for the actual rates. Choice D (83) results from incorrectly calculating b = 47.

Question 17

A delivery truck travels from warehouse A to warehouse B at 45 mph, then returns at 60 mph due to lighter traffic. If the total trip takes 3.5 hours, what equation represents this situation where d is the distance between warehouses?

  1. 45+60d=3.5\frac{45 + 60}{d} = 3.5
  2. d45−d60=3.5\frac{d}{45} - \frac{d}{60} = 3.5
  3. 45d+60d=3.545d + 60d = 3.5
  4. d45+d60=3.5\frac{d}{45} + \frac{d}{60} = 3.5 (correct answer)
Explanation: When you encounter distance, rate, and time problems involving round trips, remember the fundamental relationship: Time = Distance ÷ Rate. You need to set up equations for each leg of the journey and add the times together. For this problem, the truck travels distance d at 45 mph to warehouse B, taking d45\frac{d}{45} hours. On the return trip, it travels the same distance d at 60 mph, taking d60\frac{d}{60} hours. Since the total trip takes 3.5 hours, you add these times: d45+d60=3.5\frac{d}{45} + \frac{d}{60} = 3.5. Answer D correctly represents this situation by adding the time for each leg of the journey. Answer A is incorrect because 45+60d\frac{45 + 60}{d} puts the speeds in the numerator and distance in the denominator, which doesn't follow the Time = Distance ÷ Rate formula. This would give you an incorrect unit (speed per distance rather than time). Answer B is wrong because it subtracts the return time from the initial time (d45−d60\frac{d}{45} - \frac{d}{60}), which would give you the difference between travel times, not the total time. Answer C incorrectly multiplies distance by speed (45d+60d45d + 60d), which gives you distance squared rather than time. This completely misapplies the distance-rate-time relationship. Study tip: For round-trip problems, always identify each leg separately, write the time equation for each leg using Time = Distance ÷ Rate, then add or subtract the times based on what the problem asks for. Watch out for answer choices that flip the distance-rate relationship or use the wrong operation.

Question 18

A movie theater sells adult tickets for $12 and student tickets for $8. On Friday night, they sold 150 more adult tickets than student tickets and earned $2,880 in total revenue. Which equation correctly represents this situation if s represents the number of student tickets sold?

  1. 12(s+150)+8s=288012(s + 150) + 8s = 2880 (correct answer)
  2. 12s+8(s+150)=288012s + 8(s + 150) = 2880
  3. 12(150)+8s=288012(150) + 8s = 2880
  4. 12s+8s+150=288012s + 8s + 150 = 2880
Explanation: Let s = student tickets. Since they sold 150 more adult tickets than student tickets, adult tickets = s + 150. Revenue = (adult price × adult tickets) + (student price × student tickets) = 12(s + 150) + 8s = 2880. Choice B incorrectly assigns the higher price to student tickets and adds 150 to student tickets instead of adult. Choice C assumes exactly 150 adult tickets were sold. Choice D treats 150 as additional revenue instead of additional tickets.

Question 19

A rectangular garden's length is 4 feet more than twice its width. If fencing costs $8 per foot and the total cost to fence the garden is $320, which equation can be used to find the width w of the garden?

  1. 2w+2(2w+4)=3202w + 2(2w + 4) = 320
  2. 8[2w+2(2w+4)]=3208[2w + 2(2w + 4)] = 320 (correct answer)
  3. 8[w(2w+4)]=3208[w(2w + 4)] = 320
  4. 8[w+(2w+4)]=3208[w + (2w + 4)] = 320
Explanation: When you see a word problem involving perimeter and cost, you need to work systematically through two key relationships: the geometric constraint and the cost calculation. First, establish the dimensions. The length is "4 feet more than twice the width," so if width = w, then length = 2w + 4. The perimeter of a rectangle is 2(length + width), which gives us 2w + 2(2w + 4). Next, apply the cost constraint. Since fencing costs $8 per foot and the total cost is $320, you multiply the perimeter by the cost per foot: 8 × [2w + 2(2w + 4)] = 320. This matches answer choice B. Looking at the wrong answers: Choice A gives you $2w+2(2w+4)=3202w + 2(2w + 4) = 320 ,whichcalculatestheperimeterinfeetbutignoresthe$8costperfoot—thiswouldonlyworkiffencingwere$1perfoot.ChoiceCuses$, which calculates the perimeter in feet but ignores the $8 cost per foot—this would only work if fencing were $1 per foot. Choice C uses $8[w(2w + 4)] = 320,whichmultipliescostbyarea(length×width)insteadofperimeter—acommonmix−up.ChoiceDgivesyou, which multiplies cost by area (length × width) instead of perimeter—a common mix-up. Choice D gives you 8[w + (2w + 4)] = 320$$, which represents the cost of fencing just the length plus width, forgetting that you need to fence all four sides of the rectangle. For word problems involving perimeter and cost, always remember the two-step process: first find the perimeter formula using the given relationships, then multiply by the unit cost. Don't confuse perimeter (distance around) with area (space inside).

Question 20

Temperature Conversion Graph

Based on the graph shown, which linear equation best models the relationship between temperature in degrees Celsius CC and degrees Fahrenheit FF?

  1. F=95C+32F = \dfrac{9}{5}C + 32 (correct answer)
  2. F=59C+32F = \dfrac{5}{9}C + 32
  3. F=95(C−32)F = \dfrac{9}{5}(C - 32)
  4. F=59(C−32)F = \dfrac{5}{9}(C - 32)
Explanation: The standard conversion is F=95C+32F = \frac{9}{5}C + 32, matching slope and intercept of the plotted line. B. Uses reciprocal slope. C. Subtracts 32 before multiplying, altering intercept. D. Both slope and intercept are incorrect for the graph.