Pre-Algebra Quiz: Representing Constraints

10 questions · exam conditions

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Representing ConstraintsQuestion 1 of 10

A rectangular garden must have a perimeter of exactly 48 feet, and its length must be at least 4 feet longer than its width. If $w$ represents the width in feet, which system correctly represents both constraints?

2w + 2l = 48

and

l ≥ w + 4

2w + 2(w + 4) = 48

and

w > 0

2w + 2(w + 4) = 48

and

w + 4 ≥ w

w + (w + 4) = 48

and

w > 0

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Pre-Algebra Quiz

Pre-Algebra Quiz: Representing Constraints

Practice Representing Constraints in Pre-Algebra 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 Representing Constraints, giving you a quick way to practice the rules, question types, and explanations that matter most for Pre-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.

All questions

Question 1

$2w + 2l = 48$ and $l ≥ w + 4$
$2w + 2(w + 4) = 48$ and $w > 0$ (correct answer)
$2w + 2(w + 4) = 48$ and $w + 4 ≥ w$
$w + (w + 4) = 48$ and $w > 0$

Explanation: When you encounter word problems involving rectangles with constraints, you need to translate each condition into a mathematical equation or inequality, then determine how the variables relate to each other. Let's break down the constraints: the perimeter must be exactly 48 feet, and the length must be at least 4 feet longer than the width. Since we're told

w

represents width, we can express length as

w + 4

(assuming the minimum case where length is exactly 4 feet longer). For the perimeter constraint, we use the rectangle formula:

P = 2w + 2l

. Substituting our expression for length:

2w + 2(w + 4) = 48

. This gives us exactly 48 feet, as required. We also need

w > 0

since width must be positive for a real garden. Choice A uses two separate variables (

w

and

l

) without establishing their relationship, making it incomplete. Choice C includes the constraint

w + 4 ≥ w

, which is always true regardless of

w

's value, so it provides no useful restriction. Choice D incorrectly calculates perimeter as

w + (w + 4) = 48

, which represents the semi-perimeter (half the actual perimeter) rather than the full perimeter. Choice B correctly captures both constraints: the perimeter equation with the length-width relationship built in, plus the necessary condition that width must be positive. Study tip: In constraint problems, always substitute related variables into each other early. This reduces the number of unknowns and often reveals whether you've set up the relationships correctly.

Question 2

A delivery truck's fuel efficiency decreases as its load increases. When empty, the truck gets 18 mpg, but loses 0.5 mpg for every 1,000 pounds of cargo. If the truck has a 25-gallon tank and must travel 300 miles, which inequality represents the constraint on the cargo weight $w$ (in thousands of pounds)?

$25(18 - 0.5w) ≥ 300$ (correct answer)
$25(18 + 0.5w) ≥ 300$
$300 ≤ 25(18 - 0.5w)$
$300 ≥ 25(18 + 0.5w)$

Explanation: Fuel efficiency = 18 - 0.5w mpg. Maximum distance = tank capacity × efficiency = 25(18 - 0.5w) miles. For the 300-mile trip: 25(18 - 0.5w) ≥ 300. Choice B incorrectly adds 0.5w (efficiency increasing with weight). Choice C is mathematically equivalent to A but less conventional. Choice D has the inequality reversed and incorrectly shows efficiency increasing.

Question 3

A movie theater charges a membership fee plus an additional cost per movie ticket. Sarah paid $45 for her membership and 3 movie tickets, while Jake paid $65 for his membership and 7 movie tickets. Which equation correctly represents the constraint that relates the total cost $ $C$ $to the number of tickets$ $t$ $ purchased?

$C = 5t + 30$ (correct answer)
$C = 5t + 15$
$C = 15t + 5$
$C = 30t + 5$

Explanation: First, find the cost per ticket:

(65-45)÷(7-3) = 20÷4 = 5

dollars per ticket. Then find the membership fee using Sarah's data:

45 = 5(3) + membership

, so membership = $30. Therefore $C = 5t + 30

. Choice B uses the wrong membership fee ($15 instead of

30). Choice C switches the coefficient and constant. Choice D uses the wrong per-ticket cost ($30 instead of $5).

Question 4

A cell phone plan charges $25 per month plus $0.10 per text message over 500 texts. Emma's monthly bill cannot exceed $40. If she has already sent 300 text messages this month, which inequality represents the constraint on the number of additional texts $ $x$ $ she can send?

$25 + 0.10(x - 200) ≤ 40$ when $x ≥ 200$
$25 + 0.10(x + 300 - 500) ≤ 40$ when $x + 300 ≥ 500$ (correct answer)
$25 + 0.10x ≤ 40$
$25 + 0.10(300 + x) ≤ 40$

Explanation: Total texts will be 300 + x. Charges apply only when total > 500, so when x + 300 > 500 or x > 200. The constraint becomes: 25 + 0.10(total texts - 500) ≤ 40, which is 25 + 0.10(x + 300 - 500) ≤ 40. Choice A uses x - 200 instead of accounting for the 300 already sent. Choice C ignores the 500-text threshold. Choice D charges for all texts sent.

Question 5

A bakery makes profit by selling cupcakes for $3 each and cookies for $1.50 each. The bakery has enough ingredients to make at most 200 items total and wants to earn at least $450 in revenue. If $ $c$ $represents the number of cupcakes and$ $k$ $ represents the number of cookies, which pair of inequalities correctly represents these constraints?

$c + k ≤ 200$ and $3c + 1.5k = 450$
$c + k ≥ 200$ and $3c + 1.5k ≤ 450$
$c + k = 200$ and $3c + 1.5k ≥ 450$
$c + k ≤ 200$ and $3c + 1.5k ≥ 450$ (correct answer)

Explanation: When you encounter word problems involving constraints, you need to translate each condition into mathematical language using inequalities or equations based on the specific wording. Let's break down each constraint. The bakery can make "at most 200 items total" - this means the total number of cupcakes and cookies cannot exceed 200. Since

c

represents cupcakes and

k

represents cookies, this translates to

c + k ≤ 200

. For revenue, the bakery "wants to earn at least $450." Each cupcake sells for $3 and each cookie for $1.50, so total revenue is $

3c + 1.5k

. "At least $450" means the revenue must be $450 or more, giving us $

3c + 1.5k ≥ 450$$. Choice D correctly captures both constraints:

c + k ≤ 200

and

3c + 1.5k ≥ 450

. Choice A uses an equals sign for revenue (

3c + 1.5k = 450

), which means the bakery wants exactly $450, not "at least

450." Choice B flips both inequalities - it suggests the bakery needs at least 200 items ($$c + k \geq 200$$) and wants at most

450 in revenue (

3c + 1.5k ≤ 450

), which contradicts the problem. Choice C uses an equals sign for the item constraint, meaning exactly 200 items must be made rather than "at most 200." Remember: "at most" translates to ≤, "at least" translates to ≥, and "exactly" translates to =. Pay careful attention to these key phrases when setting up constraint problems.

Question 6

A swimming pool is being filled by two hoses. Hose A fills at 12 gallons per minute, and Hose B fills at 8 gallons per minute. Due to water pressure limitations, Hose A can run for at most twice as long as Hose B. If the pool needs exactly 640 gallons and $t$ represents the time Hose B runs (in minutes), which equation represents the constraint?

$12(2t) + 8(2t) = 640$
$12t + 8(2t) = 640$
$12t + 8t = 640$
$12(2t) + 8t = 640$ (correct answer)

Explanation: When you encounter rate problems with multiple constraints, carefully define your variables and translate each piece of information into mathematical expressions. Here, you need to track two hoses with different rates and a time limitation. Let's define

t

as the time Hose B runs. Since Hose A can run "at most twice as long" as Hose B, and we need exactly 640 gallons, Hose A must run for the maximum allowed time:

2t

minutes. Now calculate the water from each hose:

Hose A: $12 \text{ gallons/minute} \times 2t \text{ minutes} = 12(2t) \text{ gallons}$
Hose B: $8 \text{ gallons/minute} \times t \text{ minutes} = 8t \text{ gallons}$

The total must equal 640 gallons:

12(2t) + 8t = 640

Looking at the wrong answers: Choice A assumes both hoses run for

2t

minutes, ignoring that only Hose A has this time constraint. Choice B incorrectly gives Hose B the longer time (

2t

) while Hose A gets the shorter time (

t

). Choice C assumes both hoses run for the same time

t

, completely missing the "at most twice as long" constraint. The correct answer is D:

12(2t) + 8t = 640

. Study tip: In constraint problems, pay close attention to phrases like "at most," "at least," or "exactly." These determine whether you use the maximum, minimum, or exact value allowed. Also, always check that your variable assignments match what the problem defines—here,

t

specifically represents Hose B's time, not the total time.

Question 7

A company produces two types of widgets: Type X takes 3 hours to make and uses 2 pounds of material, while Type Y takes 2 hours to make and uses 4 pounds of material. The company has 120 hours of labor and 150 pounds of material available this week. If $x$ and $y$ represent the numbers of Type X and Type Y widgets respectively, which inequality correctly represents the material constraint?

$2x + 4y ≥ 150$
$3x + 2y ≤ 150$
$2x + 4y ≤ 150$ (correct answer)
$4x + 2y ≤ 150$

Explanation: When you encounter word problems involving constraints and limitations, you're working with inequalities that represent real-world boundaries. The key is to translate the given information into mathematical expressions that show what's possible within those limits. Let's focus on the material constraint. Type X widgets use 2 pounds of material each, and Type Y widgets use 4 pounds each. If you make

x

Type X widgets and

y

Type Y widgets, your total material usage is

2x + 4y

pounds. Since the company only has 150 pounds available, the total material used must be less than or equal to what's available:

2x + 4y ≤ 150

. Choice A (

2x + 4y ≥ 150

) uses the wrong inequality direction, suggesting you must use at least 150 pounds, which doesn't make sense for a limitation. Choice B (

3x + 2y ≤ 150

) incorrectly uses the time values (3 and 2 hours) instead of material amounts. Choice D (

4x + 2y ≤ 150

) swaps the material requirements, incorrectly assigning 4 pounds to Type X and 2 pounds to Type Y. The correct answer is C:

2x + 4y ≤ 150

. Study tip: In constraint problems, always match the numbers to what they actually represent (time, materials, cost, etc.) and remember that limitations create "less than or equal to" inequalities, while minimum requirements create "greater than or equal to" inequalities.

Question 8

A elevator has a weight limit and currently contains 4 adults averaging 160 pounds each and some children averaging 80 pounds each. The elevator can safely hold a maximum of 2,000 pounds total. If $c$ represents the number of children currently in the elevator, which inequality represents the constraint on the number of additional adults $a$ that can enter?

$4 + c + a ≤ 2000$
$160 + 80c + 160a ≤ 2000$
$640 + 80c + 160a ≤ 2000$ (correct answer)
$640 + 80c + 160a ≥ 2000$

Explanation: When you encounter inequality word problems, the key is systematically translating each piece of information into mathematical terms, then combining them to represent the constraint. Let's build this inequality step by step. The elevator currently has 4 adults at 160 pounds each, which gives us

4 \times 160 = 640

pounds from adults already inside. We also have

c

children at 80 pounds each, contributing

80c

pounds. Finally,

a

additional adults will add

160a

pounds. The total weight is

640 + 80c + 160a

, and this must be less than or equal to the 2,000-pound limit. Answer choice C correctly captures this:

640 + 80c + 160a ≤ 2000

. Answer choice A treats each person as weighing 1 pound instead of using their actual weights—this completely ignores the weight information given in the problem. Answer choice B makes a critical error by using 160 for the current adults instead of

4 \times 160 = 640

, essentially treating the current load as just one adult rather than four. Answer choice D uses the correct weight calculation but has the inequality flipped—it suggests the total weight should be greater than or equal to 2,000 pounds, which would mean the elevator is overloaded. When solving inequality word problems, always identify what's already present, what's being added, and what the constraint is. Calculate existing quantities completely before setting up your inequality, and double-check that your inequality symbol points in the logical direction for the constraint.

Question 9

A water tank is being filled at a rate of 15 gallons per minute and drained at a rate of 8 gallons per minute simultaneously. The tank starts with 120 gallons and must maintain at least 50 gallons at all times for safety reasons. Which inequality represents the time constraint $t$ (in minutes) for how long this process can continue?

$120 + 7t ≥ 50$ (correct answer)
$120 - 7t ≥ 50$
$120 + 15t - 8t ≥ 50$
$120 - 15t + 8t ≥ 50$

Explanation: Net rate = 15 - 8 = 7 gallons per minute (filling faster than draining). After t minutes: amount = 120 + 7t. The constraint is this amount ≥ 50, so 120 + 7t ≥ 50. Choice B incorrectly assumes the tank is losing water overall. Choices C and D are mathematically equivalent to A and B respectively, but don't simplify the net rate, making them less clear representations of the constraint.

Question 10

A parking garage has a maximum weight capacity. The garage currently contains 12 cars averaging 3,200 pounds each and 8 trucks averaging 7,500 pounds each. If the garage can hold a maximum total weight of 150,000 pounds, which inequality represents the constraint for the additional weight $w$ (in pounds) that can still be added?

$w ≤ 150,000 - 98,400$ (correct answer)
$w ≤ 150,000 - 38,400$
$w ≤ 98,400 - 150,000$
$w ≤ 150,000 + 98,400$

Explanation: Current weight = 12(3,200) + 8(7,500) = 38,400 + 60,000 = 98,400 pounds. The constraint is: current weight + additional weight ≤ maximum capacity, so 98,400 + w ≤ 150,000, which gives w ≤ 150,000 - 98,400. Choice B incorrectly uses only the car weight (38,400). Choice C reverses the subtraction, giving a negative result. Choice D adds instead of subtracts the current weight.