GED Math Quiz: Word Problems

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.

Question 1

1. $9m < 60$ (correct answer)
2. $60 < 9m$
3. $9m + 60 < 0$
4. $\dfrac{9}{m} < 60$

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 $$m$ $months, the total cost is$ $9m$ $dollars. Plan B costs a flat$60 per year regardless of usage. To find when Plan A is cheaper than Plan B, you set up the inequality: Plan A cost < Plan B cost, which gives you $9m < 60$. Let's verify this makes sense: if you use the service for 6 months, Plan A costs $9 \times 6 = 54$ dollars, while Plan B costs $60. Since$ $54 < 60$ $, Plan A is indeed cheaper, confirming our inequality is correct. Looking at the wrong answers: Choice B ($$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 2

1. $15c + 7g \ge 600$ (correct answer)
2. $15c + 7g \le 600$
3. $15 + 7 \ge 600(c + g)$
4. $15cg + 7 = 600$

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 $$c$ $candles generate$ $15c$ $dollars. Greeting cards cost$7 each, so $g$ cards generate $7g$ dollars. The total revenue is $15c + 7g$. Since the class "must raise at least $600," they need their total revenue to be greater than or equal to$600. This gives us $15c + 7g \ge 600$, which is answer choice A. The "greater than or equal to" symbol (≥) correctly captures "at least $600." Let's examine why the other choices fail: Answer B uses $$15c + 7g \le 600$$, which would mean they want to raise $600 or less—the opposite of their goal. Answer C rearranges the variables incorrectly, putting the unit prices outside the inequality and multiplying$600 by the quantities, which doesn't match the revenue calculation. Answer D uses $$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 3

1. $50e + 280 = 1080$ (correct answer)
2. $280e + 50 = 1080$
3. $50 + 280e = 1080$
4. $50e - 280 = 1080$

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 $$e$ $employees, and each gets a$50 gift card, then the total spent on gift cards is $50e$ (50 dollars times the number of employees). The correct equation is $50e + 280 = 1080$ (choice A). This represents: (cost per employee × number of employees) + (coffee machine cost) = (total spent). Let's examine why the other choices are wrong: Choice B ($280e + 50 = 1080$) incorrectly suggests Carmen buys $e$ coffee machines at $280 each, plus something costing$50. This misinterprets what the variables represent. Choice C ($50 + 280e = 1080$) implies Carmen spends only $50 total on gift cards regardless of employee count, then buys$ $e$ $ coffee machines. This contradicts the problem stating she buys one coffee machine. Choice D ($$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 4

1. $85 + 12g$ (correct answer)
2. $85g + 12$
3. $(85 + 12)g$
4. $85 \div 12g$

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 $$g$ $guests, she pays$12 for each guest's meal, giving us $12g$ for the total meal costs. Her complete expenses are the fixed hall cost plus the variable meal costs: $85 + 12g$. Let's examine why each answer choice is right or wrong: A) $85 + 12g$ correctly separates the fixed cost ($85) from the variable cost ($ $12g$ $). This matches our reasoning above. B) $$85g + 12$$ incorrectly makes the hall rental fee variable by multiplying it by $$g$$, suggesting Maria pays $85 per guest for the hall, which doesn't match the problem. C) $(85 + 12)g$ treats both costs as per-guest expenses, meaning Maria would pay $97 per guest with no fixed costs. This ignores that the hall fee is a one-time charge. D) $$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 5

1. $C = 45 + 10g$ (correct answer)
2. $C = 10 + 45g$
3. $C = 45(10g)$
4. $C = 45 - 10g$

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 $$g$ $represents gigabytes above the included 4 GB, the total cost is the base fee plus$ $10g$ $for the extra data. This gives us$ $C = 45 + 10g$ $, which is choice A. Let's examine why the other options are incorrect. Choice B ($$C = 10 + 45g$$) reverses the fixed and variable components—it treats the base cost as $10 and charges$45 per extra gigabyte, which doesn't match the problem. Choice C ($$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 6

1. $W = 45 + 12t$ (correct answer)
2. $W = 12 - 45t$
3. $W = 45t + 12$
4. $W = 12t - 45$

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 $t$ 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 $t$ minutes, that's $12t$ liters of additional water. So the total is $W = 45 + 12t$. Choice A ($W = 45 + 12t$) correctly represents this situation with the initial amount (45) plus the rate of change (12) multiplied by time. Choice B ($W = 12 - 45t$) 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 ($W = 45t + 12$) 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 ($W = 12t - 45$) 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 $t$). The general form is: Final Amount = Initial Amount + (Rate × Time).

Question 7

1. $d = 70 + 28g$ (correct answer)
2. $d = 28 + 70g$
3. $d = 70 - 28g$
4. $d = 70g - 28$

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 $g$ you add, the car can travel an additional 28 miles. This gives you $28g$ 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: $d = 70 + 28g$. This matches choice A. Choice B ($d = 28 + 70g$) 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 ($d = 70 - 28g$) 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 ($d = 70g - 28$) 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 8

1. $P = 14h + 25$ (correct answer)
2. $P = 14(h + 25)$
3. $P = 25h + 14$
4. $P = 14h - 25$

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 $$14h$ $where$ $h$ $is hours worked. Additionally, he receives a flat $25 bonus when he works at least 20 hours. Since the problem states $$h \ge 20$, we know he always qualifies for this bonus. His total pay is the sum of these two components:$P = 14h + 25$$. Looking at the wrong answers: Option B, $P = 14(h + 25)$, incorrectly multiplies both the hours and bonus by the hourly rate. This would mean Nick earns $14 per hour plus an additional$350 bonus ($14 \times 25$), which doesn't match the problem. Option C, $P = 25h + 14$, reverses the coefficients, suggesting he earns $25 per hour with a$14 bonus. Option D, $P = 14h - 25$, subtracts the bonus instead of adding it, which makes no sense since bonuses increase pay. The correct answer is A: $P = 14h + 25$. 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 9

1. $28 + pn = 96$
2. $28 + pn - 96 = 0$
3. $96 - 28 = pn$ (correct answer)
4. $28p + n = 96$

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 $96 - 28 = 68$ more ounces. Since she'll buy $n$ packages with $p$ ounces each, the packages will provide $pn$ ounces total. Therefore, $pn$ must equal the amount she still needs: $96 - 28 = pn$. This matches answer choice C. Let's examine why the other options are incorrect. Answer A ($28 + pn = 96$) 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 ($28 + pn - 96 = 0$) is also mathematically equivalent but unnecessarily complicated—you wouldn't typically set up the initial equation this way. Answer D ($28p + n = 96$) contains a critical error: it multiplies 28 by $p$ 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 10

1. $r + 3r = \frac{1}{6}$
2. $\frac{1}{r} + \frac{1}{3r} = 6$
3. $(r + 3r) \times 6 = 1$ (correct answer)
4. $6r + 18r = 1$

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 11

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 12

1. $8n - 3n - 150 \geq 200$ and $n \geq 40$
2. $8n - 3n + 150 \geq 200$ and $n \geq 40$
3. $8n - (3n + 150) \geq 200$ and $n \geq 40$ (correct answer)
4. $5n - 150 \geq 200$ and $n \leq 40$

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 13

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. Choice A ($74) results from using r = 5.5 (incorrect calculation). Choice C ($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 14

1. $\frac{45 + 60}{d} = 3.5$
2. $\frac{d}{45} - \frac{d}{60} = 3.5$
3. $45d + 60d = 3.5$
4. $\frac{d}{45} + \frac{d}{60} = 3.5$ (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 $\frac{d}{45}$ hours. On the return trip, it travels the same distance d at 60 mph, taking $\frac{d}{60}$ hours. Since the total trip takes 3.5 hours, you add these times: $\frac{d}{45} + \frac{d}{60} = 3.5$. Answer D correctly represents this situation by adding the time for each leg of the journey. Answer A is incorrect because $\frac{45 + 60}{d}$ 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 ($\frac{d}{45} - \frac{d}{60}$), which would give you the difference between travel times, not the total time. Answer C incorrectly multiplies distance by speed ($45d + 60d$), 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 15

1. $12(s + 150) + 8s = 2880$ (correct answer)
2. $12s + 8(s + 150) = 2880$
3. $12(150) + 8s = 2880$
4. $12s + 8s + 150 = 2880$

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 16

1. $2w + 2(2w + 4) = 320$
2. $8[2w + 2(2w + 4)] = 320$ (correct answer)
3. $8[w(2w + 4)] = 320$
4. $8[w + (2w + 4)] = 320$

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 $$2w + 2(2w + 4) = 320$ $, which calculates the perimeter in feet but ignores the $8 cost per foot—this would only work if fencing were $1 per foot. Choice C uses $$8[w(2w + 4)] = 320$, 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 17

1. $F = \dfrac{9}{5}C + 32$ (correct answer)
2. $F = \dfrac{5}{9}C + 32$
3. $F = \dfrac{9}{5}(C - 32)$
4. $F = \dfrac{5}{9}(C - 32)$

Explanation: The standard conversion is $F = \frac{9}{5}C + 32$, 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.

Question 18

1. $T = 235 + d$ (correct answer)
2. $T = d - 235$
3. $T = 235d$
4. $T = \dfrac{d}{235}$

Explanation: Monday–Thursday sales sum to $40 + 52 + 63 + 80 = 235$. Total cupcakes equal that constant plus Friday's sales $d$: $T = 235 + d$. B. Subtracts rather than adds. C. Multiplies, creating far too large a total. D. Divides, not a realistic aggregation method.

Question 19

1. $C = 25 + 0.08t$ (correct answer)
2. $C = 25t + 0.08$
3. $C = 25(0.08)t$
4. $C = 25 - 0.08t$

Explanation: This is a linear equation problem where you need to identify fixed costs versus variable costs. When you see word problems about monthly charges, think about what stays the same each month versus what changes based on usage. The phone company has two types of charges: a base fee of $25 that you pay regardless of usage, and an additional $0.08 for each text message sent. Since $$t$ $represents the number of text messages, the variable cost is$ $0.08t$ $(the per-message rate times the number of messages). Your total monthly charge$ $C$ $equals the fixed base fee plus the variable texting charges:$ $C = 25 + 0.08t$ $. Looking at the wrong answers: Choice B ($$C = 25t + 0.08$$) incorrectly multiplies the base fee by the number of texts and treats the per-message rate as a fixed cost - this reverses the relationship. Choice C ($$C = 25(0.08)t$$) multiplies all three values together, which would give you $$C = 2t$$, completely ignoring the structure of base fee plus variable cost. Choice D ($$C = 25 - 0.08t$$) subtracts the texting charges from the base fee, which would mean sending more texts somehow reduces your bill. The correct answer is A: $$C = 25 + 0.08t$$. Study tip: In cost problems, identify what's fixed (stays the same) and what's variable (depends on usage). Fixed costs are added as constants, while variable costs are the rate multiplied by the quantity used.

Question 20

1. $3n + 2p$ (correct answer)
2. $(3 + 2)(n + p)$
3. $3np + 2$
4. $n + p$

Explanation: When you encounter word problems involving multiple items with different costs, you need to carefully translate the language into mathematical expressions by identifying quantities and unit costs. Javier bought 3 notebooks at $n$ dollars each and 2 pens at $p$ dollars each. To find the total cost, you multiply each quantity by its unit price, then add: $(3 \times n) + (2 \times p) = 3n + 2p$. This gives you the correct expression in choice A. Let's examine why the other options are incorrect. Choice B, $(3 + 2)(n + p)$, represents buying 5 items where each item costs $(n + p)$ dollars. This would mean each notebook and pen costs the same amount, which contradicts the problem. Choice C, $3np + 2$, incorrectly multiplies the number of notebooks by both unit prices, then adds 2 (not $2p$). This makes no mathematical sense in the context. Choice D, $n + p$, only accounts for buying one notebook and one pen, completely ignoring the quantities specified. The key trap here is confusing addition with multiplication. Remember that when you buy multiple items, you multiply quantity times unit price for each type of item, then add those products together. Watch out for expressions that incorrectly combine quantities and prices, or that treat different items as having the same cost when they don't.