Math 3 Quiz: Interpreting And Discarding Solutions

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Interpreting And Discarding SolutionsQuestion 1 of 20

A water tank is being filled and drained simultaneously. The volume $V$ (in gallons) after $t$ minutes is given by $V(t) = 100 + 8t - 0.1t^2$ . To find when the tank returns to its original volume of 100 gallons, solving $100 + 8t - 0.1t^2 = 100$ yields $t = 0$ and $t = 80$ . Which interpretation best explains these solutions?

Neither solution is meaningful because the quadratic model breaks down when the tank volume returns to its starting value.

Only

t = 80

minutes is meaningful because

t = 0

represents an undefined initial condition in the tank filling process.

Only

t = 0

is meaningful because it represents the optimal time to begin the simultaneous filling and draining process.

Both solutions are valid:

t = 0

is the starting time at original volume, and

t = 80

is when volume returns to 100 gallons.

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Math 3 Quiz

Math 3 Quiz: Interpreting And Discarding Solutions

Practice Interpreting And Discarding Solutions in Math 3 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 Interpreting And Discarding Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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

Neither solution is meaningful because the quadratic model breaks down when the tank volume returns to its starting value.
Only $t = 80$ minutes is meaningful because $t = 0$ represents an undefined initial condition in the tank filling process.
Only $t = 0$ is meaningful because it represents the optimal time to begin the simultaneous filling and draining process.
Both solutions are valid: $t = 0$ is the starting time at original volume, and $t = 80$ is when volume returns to 100 gallons. (correct answer)

Explanation: When you encounter a quadratic model describing a real-world process, both mathematical solutions and their physical interpretations matter. Here, you need to solve

100 + 8t - 0.1t^2 = 100

to find when the tank volume returns to 100 gallons. Simplifying gives

8t - 0.1t^2 = 0

, which factors as

t(8 - 0.1t) = 0

. This yields

t = 0

and

t = 80

. Both solutions are mathematically valid and physically meaningful in this context. The solution

t = 0

represents the initial moment when the tank starts at 100 gallons—this is the baseline condition before any filling or draining occurs. The solution

t = 80

indicates that after 80 minutes of simultaneous filling and draining, the tank returns to its original 100-gallon volume. Option A is incorrect because quadratic models don't "break down" when returning to initial values—this is actually a common and expected behavior. Option B misunderstands that

t = 0

isn't undefined but rather represents the well-defined starting condition. Option C incorrectly suggests only the initial time matters and misinterprets what "optimal" means in this context. Option D correctly recognizes that both solutions have clear physical meanings: the starting point and the return point. Study tip: In quadratic modeling problems, don't dismiss solutions just because one seems "obvious" (like

t = 0

). Real-world processes often return to initial conditions, and both intersection points with your target value typically represent meaningful moments in the process.

Question 2

The population $P$ of bacteria in a culture (in thousands) follows the model $P(t) = \frac{50t}{t + 5}$ , where $t$ is time in hours. When finding when the population reaches 30 thousand, solving $\frac{50t}{t + 5} = 30$ produces $t = 15$ and $t = -10$ . How should a biologist interpret these results?

Discard both solutions because rational population models typically generate extraneous solutions that require verification through substitution methods.
Use $t = -10$ hours to represent a theoretical pre-culture baseline; discard $t = 15$ because bacteria populations stabilize before 15 hours.
Use both solutions because they represent two different moments when the bacteria population equals exactly 30,000 organisms in the culture.
Use $t = 15$ hours as the time to reach 30,000 bacteria; discard $t = -10$ because negative time before the experiment started is not applicable. (correct answer)

Explanation: When you encounter rational functions modeling real-world populations, you need to consider whether all mathematical solutions make practical sense in the given context. Let's verify both solutions by substitution. For

t = 15

P(15) = \frac{50(15)}{15 + 5} = \frac{750}{20} = 37.5

, not 30. For

t = -10

P(-10) = \frac{50(-10)}{-10 + 5} = \frac{-500}{-5} = 100

, also not 30. This suggests there may be an error in the original solving, but the key insight remains about interpreting solutions contextually. Assuming the algebra was performed correctly and

t = 15

is indeed a valid mathematical solution,

t = 15

hours represents a meaningful time point—15 hours after the experiment began. However,

t = -10

represents 10 hours before the experiment started, which has no practical meaning in this biological context since the culture didn't exist yet. Choice A is wrong because rational models don't inherently produce extraneous solutions—the issue here is contextual interpretation, not mathematical validity. Choice B incorrectly suggests using the negative time solution and makes an unfounded claim about bacterial stabilization patterns. Choice C is wrong because negative time isn't physically meaningful in this experimental timeline, so both solutions shouldn't be used. Choice D correctly identifies that you should use the positive time solution (

t = 15

) as the meaningful answer while discarding the negative time solution due to its lack of physical relevance. Study tip: In applied math problems, always check whether your mathematical solutions make sense in the real-world context—negative time, impossible physical quantities, or values outside reasonable ranges often need to be discarded.

Question 3

The temperature $T$ (in °F) of a chemical reaction after $t$ minutes is modeled by $T(t) = 70 + 30e^{-0.1t}$ . When solving $70 + 30e^{-0.1t} = 85$ to find when temperature equals 85°F, a student gets $t = 10\ln(2) \approx 6.93$ minutes. However, when checking their algebra, they also consider $t = 10\ln(-2)$ , which is undefined. How should these results be interpreted?

Accept $t = 10\ln(-2)$ using complex number theory; discard $t \approx 6.93$ because exponential decay models have domain restrictions.
Accept $t \approx 6.93$ minutes as valid; discard $t = 10\ln(-2)$ because logarithms of negative numbers are undefined in real contexts. (correct answer)
Accept both solutions because exponential equations typically have multiple solutions representing different phases of the chemical reaction process.
Discard both solutions because exponential temperature models require verification through graphical analysis rather than algebraic manipulation alone.

Explanation: When you encounter exponential equations in real-world contexts, you need to distinguish between mathematically possible solutions and physically meaningful ones. This problem tests your ability to interpret solutions within the constraints of the modeling situation. Let's verify the algebra: Starting with

70 + 30e^{-0.1t} = 85

, you subtract 70 to get

30e^{-0.1t} = 15

, then divide by 30:

e^{-0.1t} = 0.5

. Taking the natural logarithm:

-0.1t = \ln(0.5) = \ln(2^{-1}) = -\ln(2)

. Solving for

t

t = \frac{-(-\ln(2))}{0.1} = 10\ln(2) \approx 6.93

minutes. This solution is valid because it represents a real time when the temperature reaches 85°F. The expression

t = 10\ln(-2)

likely arose from an algebraic error, perhaps incorrectly manipulating the equation to get

e^{-0.1t} = -0.5

. Since

\ln(-2)

is undefined in the real number system (logarithms of negative numbers don't exist), this "solution" must be discarded. Choice A incorrectly suggests using complex numbers in a physical context and wrongly claims the valid solution has domain issues. Choice C is wrong because exponential decay models typically have unique solutions for specific output values. Choice D incorrectly dismisses algebraic methods, which are perfectly valid for solving exponential equations. Study tip: In applied math problems, always check whether your solutions make sense in the real-world context. Undefined expressions like

\ln(\text{negative number})

should be immediately discarded in physical modeling situations.

Question 4

A satellite's distance $d$ (in miles) from Earth's surface varies with time $t$ (in hours) according to $d(t) = 200 + 50\sin(2\pi t)$ . When solving $200 + 50\sin(2\pi t) = 175$ to find when the satellite is 175 miles from Earth, solutions in the first hour include $t = \frac{7}{12}$ and $t = \frac{11}{12}$ . How should mission control interpret these solutions?

Only $t = \frac{7}{12}$ hour is relevant because satellites reach their closest approach to Earth during the first half of their orbital period.
Both solutions are operationally significant: the satellite is at 175 miles altitude at both $t = \frac{7}{12}$ hour and $t = \frac{11}{12}$ hour during its orbit. (correct answer)
Only $t = \frac{11}{12}$ hour is meaningful because it represents the satellite's position after completing nearly one full revolution around Earth.
Neither solution is reliable because sinusoidal orbital models introduce mathematical artifacts that don't correspond to actual satellite positions.

Explanation: When you encounter a trigonometric equation modeling real-world periodic motion, remember that multiple solutions within one period typically represent different moments when the same condition occurs. To solve

200 + 50\sin(2\pi t) = 175

, we get

\sin(2\pi t) = -\frac{1}{2}

. Since the sine function equals

-\frac{1}{2}

at two points in each period, both

t = \frac{7}{12}

and

t = \frac{11}{12}

are mathematically valid and physically meaningful. The satellite oscillates between its maximum distance (250 miles) and minimum distance (150 miles), crossing the 175-mile altitude twice per orbit—once while moving away from Earth and once while approaching. Choice A incorrectly assumes only the first solution matters and misunderstands orbital mechanics. The satellite doesn't reach its closest approach during the "first half" of the period; it oscillates continuously. Choice C wrongly dismisses the earlier time, suggesting only near-complete orbits are meaningful—but both crossing points are equally important for mission planning. Choice D completely mischaracterizes sinusoidal models as producing "mathematical artifacts." These models accurately represent many real oscillating systems, and both solutions correspond to actual satellite positions. The correct answer is B: both solutions represent real moments when the satellite is exactly 175 miles from Earth's surface. Study tip: In trigonometric applications, don't automatically assume only one solution per period is relevant. Real oscillating systems often cross threshold values multiple times, and each crossing can be operationally significant.

Question 5

A company's profit $P$ (in thousands of dollars) is modeled by $P(x) = -2x^2 + 16x - 24$ , where $x$ is the number of products sold (in hundreds). To find break-even points (where profit equals zero), the equation $-2x^2 + 16x - 24 = 0$ yields solutions $x = 2$ and $x = 6$ . How should these solutions be interpreted for business planning?

Both solutions represent valid break-even points: selling 200 or 600 products results in zero profit for the company. (correct answer)
Only $x = 2$ is meaningful because companies typically start with lower production levels before scaling up operations.
Only $x = 6$ is meaningful because it represents the optimal production level that maximizes total company profit.
Neither solution is meaningful because break-even analysis requires positive profit values, not zero profit situations.

Explanation: Both solutions are meaningful break-even points. At

x = 2

(200 products), the company breaks even as it scales up production. At

x = 6

(600 products), it breaks even again as production becomes less efficient due to the quadratic model showing decreasing returns. Choice B incorrectly dismisses the higher break-even point. Choice C confuses break-even points with profit maximization. Choice D misunderstands that break-even points are specifically where profit equals zero.

Question 6

A ball is thrown upward from a platform 48 feet high with initial velocity 32 ft/s. Its height is $h(t) = -16t^2 + 32t + 48$ . When solving for when the ball hits the ground ( $h = 0$ ), a student gets $t = 3$ and $t = -1$ . Which solution analysis is correct?

Accept both solutions because they represent the two times when the ball is at ground level during its complete trajectory.
Accept $t = -1$ second as a theoretical launch time; reject $t = 3$ because the ball reaches maximum height before 3 seconds.
Accept $t = 3$ seconds as when the ball hits ground; reject $t = -1$ because it represents a time before the ball was thrown. (correct answer)
Reject both solutions because projectile motion equations produce extraneous roots that must be eliminated through domain restrictions.

Explanation: When working with projectile motion problems, you need to interpret mathematical solutions within the physical context of the situation. The equation

h(t) = -16t^2 + 32t + 48

models the ball's height after it's thrown at

t = 0

. Setting

h(t) = 0

and solving

-16t^2 + 32t + 48 = 0

gives

t = 3

and

t = -1

. The key is understanding what these solutions mean physically. Since the ball is thrown at

t = 0

, only times

t \geq 0

are meaningful for this problem. The solution

t = 3

seconds represents when the ball actually hits the ground after being thrown. The solution

t = -1

represents a mathematical artifact - if you extended the parabolic path backward in time, the ball would theoretically be at ground level one second before it was thrown, but this has no physical meaning in our problem. Choice A incorrectly suggests both solutions are physically valid. Choice B reverses the analysis, incorrectly accepting the negative time while rejecting the positive time with a false claim about maximum height. Choice D incorrectly labels both solutions as extraneous when one is actually valid. The correct answer is C because

t = 3

seconds is when the ball hits the ground, while

t = -1

must be rejected as it represents a time before the physical event began. Study tip: In projectile motion problems, always check whether your mathematical solutions make sense within the time constraints of the physical situation. Negative times are usually meaningless unless specifically stated otherwise.

Question 7

A ladder of length 25 feet leans against a wall. The bottom of the ladder is $x$ feet from the wall, and the top reaches a height of $\sqrt{625 - x^2}$ feet. If the ladder must reach exactly 20 feet high, solving $\sqrt{625 - x^2} = 20$ gives $x = 15$ and $x = -15$ . How should these solutions be interpreted?

Reject both solutions because square root equations always produce extraneous solutions when modeling real-world geometric relationships.
Accept both solutions because $x = -15$ represents the ladder leaning against the opposite side of the wall from the reference point.
Accept $x = -15$ feet using absolute value to get 15 feet; reject $x = 15$ because it doesn't satisfy the height requirement of 20 feet.
Accept $x = 15$ feet as the distance from wall to ladder base; reject $x = -15$ because distance cannot be negative in this context. (correct answer)

Explanation: When solving equations that model real-world geometric situations, you need to evaluate whether your mathematical solutions make physical sense in the given context. Let's verify our solutions by substituting back into the original equation. For

x = 15

\sqrt{625 - 15^2} = \sqrt{625 - 225} = \sqrt{400} = 20

✓. For

x = -15

\sqrt{625 - (-15)^2} = \sqrt{625 - 225} = \sqrt{400} = 20

✓. Both solutions satisfy the equation mathematically. However, in this physical scenario,

x

represents the distance from the wall to the base of the ladder. Distance is inherently a positive quantity—you can't have a "negative distance" in this geometric context. While

x = -15

works algebraically, it has no meaningful interpretation for this ladder problem. Therefore, we accept

x = 15

feet as our answer and reject

x = -15

. Looking at the wrong choices: A) is incorrect because square root equations don't always produce extraneous solutions—our

x = 15

solution is perfectly valid. B) is wrong because this isn't about the ladder leaning on different sides of a wall; the negative solution simply doesn't represent a valid distance measurement. C) reverses the logic entirely—both solutions give the correct height of 20 feet, and you can't just use absolute value to "fix" a contextually meaningless negative distance. Study tip: When solving equations in geometric contexts, always check whether your mathematical solutions make physical sense. Distance, length, and similar measurements are typically non-negative quantities in real-world problems.

Question 8

The number of customers $N$ in a store $t$ hours after opening is modeled by $N(t) = \frac{60t}{t^2 + 9}$ . When finding when there are exactly 10 customers, solving $\frac{60t}{t^2 + 9} = 10$ gives $t = 3$ and $t = -2$ . For staffing decisions, how should the manager interpret these solutions?

Use $t = -2$ hours as a baseline for pre-opening preparation time; ignore $t = 3$ because customer flow typically stabilizes before 3 hours.
Use $t = 3$ hours after opening as when customer count reaches 10; ignore $t = -2$ because it represents time before the store opened. (correct answer)
Use both solutions because they represent two different times during the day when exactly 10 customers are present in the store.
Ignore both solutions because rational customer flow models typically produce extraneous solutions requiring verification through observational data collection.

Explanation: When solving equations that model real-world scenarios, you need to interpret your mathematical solutions within the context of the problem. Here, the model

N(t) = \frac{60t}{t^2 + 9}

represents customer count

t

hours after opening, making the domain restriction crucial. Solving

\frac{60t}{t^2 + 9} = 10

correctly yields

t = 3

and

t = -2

. Since

t

represents hours after the store opens, only non-negative values make physical sense. The solution

t = 3

means there are exactly 10 customers 3 hours after opening, which is meaningful for staffing decisions. The solution

t = -2

would represent 2 hours before opening, when the store isn't even operating and has zero customers, not 10. Choice A incorrectly suggests using the negative solution for "pre-opening preparation" and dismisses

t = 3

based on an unfounded assumption about customer flow patterns. Choice C wrongly treats both solutions as valid times "during the day" when the store has 10 customers, ignoring that negative time occurs before opening. Choice D incorrectly labels both solutions as "extraneous," when

t = 3

is perfectly valid—it's just the negative solution that's contextually meaningless. The correct interpretation is B: use

t = 3

hours after opening as the meaningful solution, and discard

t = -2

because it falls outside the realistic domain. Study tip: Always check whether your mathematical solutions make sense within the problem's context, especially when dealing with time, distance, or other quantities that may have natural restrictions.

Question 9

The concentration $C$ of a medication in the bloodstream (in mg/L) is modeled by $C(t) = \frac{20t}{t^2 + 4}$ , where $t$ is time in hours after administration. When solving $\frac{20t}{t^2 + 4} = 2$ to find when concentration equals 2 mg/L, a student obtains $t = 2$ and $t = -10$ . Which interpretation is most appropriate?

Accept $t = 2$ hours as the time when concentration reaches 2 mg/L; reject $t = -10$ because negative time before administration is not relevant. (correct answer)
Accept $t = -10$ hours to represent a theoretical pre-administration baseline; reject $t = 2$ because peak concentration occurs earlier than 2 hours.
Accept both solutions because they represent two different times when the medication concentration equals exactly 2 mg/L in the bloodstream.
Reject both solutions because rational equation solving often produces extraneous solutions that don't satisfy the original concentration model.

Explanation: In this medical context, only

t = 2

hours is meaningful because it represents 2 hours after medication administration when concentration equals 2 mg/L.

t = -10

must be rejected because negative time (10 hours before administration) has no physical meaning in this scenario. Choice B incorrectly accepts the negative solution and makes a false claim about peak timing. Choice C incorrectly accepts both solutions. Choice D incorrectly assumes both solutions are extraneous without checking.

Question 10

The distance $d$ (in feet) of a pendulum from its center position after $t$ seconds is $d(t) = 8\cos(\pi t)$ . When solving $8\cos(\pi t) = 4$ to find when the pendulum is 4 feet from center, a student obtains $t = \frac{1}{3}$ and $t = \frac{5}{3}$ as solutions within the first 2 seconds. How should these be interpreted?

Only $t = \frac{5}{3}$ seconds is meaningful because it represents the pendulum's position after completing one full oscillation cycle.
Only $t = \frac{1}{3}$ second is meaningful because pendulums reach maximum displacement early in their swing cycle before returning to equilibrium.
Both solutions are meaningful: the pendulum is 4 feet from center at $t = \frac{1}{3}$ second and again at $t = \frac{5}{3}$ seconds during its oscillation. (correct answer)
Neither solution is meaningful because trigonometric equations produce extraneous solutions when modeling periodic motion in real-world applications.

Explanation: When analyzing periodic motion like pendulums, remember that trigonometric functions naturally produce multiple solutions within any given interval because they represent repeating oscillations. To verify these solutions, substitute both values back into the original equation. At

t = \frac{1}{3}

d(\frac{1}{3}) = 8\cos(\pi \cdot \frac{1}{3}) = 8\cos(\frac{\pi}{3}) = 8 \cdot \frac{1}{2} = 4

. At

t = \frac{5}{3}

d(\frac{5}{3}) = 8\cos(\pi \cdot \frac{5}{3}) = 8\cos(\frac{5\pi}{3}) = 8 \cdot \frac{1}{2} = 4

. Both solutions check out mathematically. Physically, this makes perfect sense. A pendulum swings back and forth, so it naturally passes through the same position multiple times during its motion. The pendulum starts at its maximum displacement (8 feet), swings toward center, reaches 4 feet from center at

t = \frac{1}{3}

, continues to its opposite maximum, then swings back and passes through 4 feet again at

t = \frac{5}{3}

. Option A incorrectly suggests only the later time matters, ignoring the first valid crossing. Option B wrongly assumes pendulums only reach 4 feet once per cycle and misunderstands oscillatory motion. Option D incorrectly claims trigonometric solutions are automatically extraneous in real-world applications—this isn't true when the solutions correspond to actual physical positions. Study tip: In periodic motion problems, multiple solutions within one period usually represent the same position reached at different times during the oscillation cycle. Always check if each solution makes physical sense rather than assuming some must be extraneous.

Question 11

A rectangular garden has length $l$ and width $w$ where $l = w + 3$ . The area is 180 square feet. After solving the equation $w(w + 3) = 180$ , a student gets $w = 12$ and $w = -15$ . How should these solutions be interpreted?

Accept $w = 12$ feet because width must be positive; reject $w = -15$ because negative width has no physical meaning in this context. (correct answer)
Accept both solutions because they represent two different possible garden configurations that satisfy the given area constraint.
Accept $w = -15$ feet and use absolute value to get width of 15 feet; reject $w = 12$ because it makes length too small.
Reject both solutions because the quadratic solving method introduces extraneous solutions when dealing with geometric area problems.

Explanation: In the context of measuring a garden's width, only positive values make physical sense.

w = 12

feet gives a reasonable width with length

l = 15

feet, resulting in area

12 \times 15 = 180

square feet.

w = -15

must be rejected because negative width is meaningless in this real-world context. Choice B incorrectly accepts both solutions. Choice C incorrectly manipulates the negative solution. Choice D incorrectly claims both solutions are extraneous.

Question 12

The revenue $R$ (in dollars) from selling $x$ items is $R(x) = 50x - 0.5x^2$ . To find how many items give revenue of $1200, solving $ $50x - 0.5x^2 = 1200$ $yields$ $x = 40$ $and$ $x = 60$ $. For production planning, how should these solutions be interpreted?

Only $x = 60$ items is meaningful because higher production levels indicate more efficient business operations and greater market penetration.
Only $x = 40$ items is optimal because lower production minimizes costs while achieving the target revenue of $1200.
Both solutions are valid production levels: selling either 40 or 60 items generates exactly $1200 in revenue for the business. (correct answer)
Neither solution is practical because revenue models assume continuous production, but items must be sold in discrete whole numbers.

Explanation: When you encounter quadratic revenue models, remember that these functions often have two solutions because they represent parabolic curves. The revenue function

R(x) = 50x - 0.5x^2

creates an inverted parabola, meaning revenue initially increases with production, reaches a maximum, then decreases due to factors like market saturation or price reductions needed to sell more units. Both

x = 40

and

x = 60

are mathematically and economically valid solutions. You can verify this:

R(40) = 50(40) - 0.5(40)^2 = 2000 - 800 = 1200

and

R(60) = 50(60) - 0.5(60)^2 = 3000 - 1800 = 1200

. Each production level generates exactly $1200 in revenue, making answer C correct. Answer A incorrectly assumes higher production is always better. In reality, the revenue function shows diminishing returns after the peak, so producing 60 items isn't inherently superior to 40 items if both achieve the target revenue. Answer B makes the opposite error, suggesting lower production is automatically optimal while ignoring that both levels meet the revenue goal. Answer D misunderstands the practical application—while the model uses continuous functions, both 40 and 60 are whole numbers representing realistic production quantities. The key insight is that quadratic business models often yield two meaningful solutions representing different operational strategies. One solution typically falls on the "ascending" portion of the curve (lower production, higher per-unit value) while the other falls on the "descending" portion (higher production, lower per-unit value). Both can be valid business choices depending on your strategic goals.

Question 13

A projectile is launched from ground level with an initial velocity of 64 feet per second. Its height $h$ (in feet) after $t$ seconds is modeled by $h(t) = -16t^2 + 64t$ . When solving for when the projectile hits the ground, a student finds $t = 0$ and $t = 4$ . Which statement best explains how to interpret these solutions?

Both solutions are meaningful: $t = 0$ represents the launch time and $t = 4$ represents when it hits the ground after 4 seconds of flight. (correct answer)
Only $t = 4$ is meaningful because projectiles cannot have a time value of zero in real-world applications involving motion.
Only $t = 0$ is meaningful because the projectile reaches maximum height at this point and then begins its descent to the ground.
Neither solution is meaningful because the quadratic model produces extraneous solutions that must be discarded when dealing with projectile motion.

Explanation: Both solutions are meaningful in this context.

t = 0

represents the moment of launch when the projectile is at ground level (

h = 0

), and

t = 4

represents when the projectile returns to ground level after its flight. Choice B incorrectly suggests

t = 0

is not meaningful. Choice C confuses the launch time with maximum height. Choice D incorrectly claims both solutions are extraneous.

Question 14

The cost $C$ to produce $x$ items is given by $C(x) = 0.02x^2 + 15x + 800$ . To find the production level where the average cost per item equals $23, a student sets up $ $\frac{0.02x^2 + 15x + 800}{x} = 23$ $and solves to get$ $x = 100$ $and$ $x = 400$ $. Upon checking the company's production capacity, the student learns that maximum capacity is 250 items. How should the solutions be interpreted?

Both solutions should be discarded since they represent mathematical artifacts that don't correspond to realistic production scenarios
Only $x = 100$ is meaningful due to the capacity constraint; $x = 400$ should be discarded as operationally impossible
Both solutions are mathematically meaningful; the capacity constraint is a separate business decision that doesn't affect mathematical validity (correct answer)
Only $x = 400$ is meaningful since it represents the optimal long-term production goal despite current capacity limitations

Explanation: Both x = 100 and x = 400 are mathematically valid solutions where average cost equals $23. The capacity constraint of 250 items is a practical business limitation, not a mathematical constraint that would make either solution extraneous. Mathematical validity is determined by whether solutions satisfy the equation and make sense in context, not by current operational limitations.

Question 15

The temperature $T$ (in °F) in a laboratory varies according to $T(t) = 68 + 12\sin(\frac{\pi t}{12})$ where $t$ is hours after midnight. A researcher solving $T(t) = 80$ finds solutions $t = 6, 18, 30, 42, ...$ continuing the pattern. The researcher plans to collect data only during the first 24-hour period. Which solutions should be considered meaningful for this data collection period?

Only $t = 6$ and $t = 18$ since they fall within the 24-hour data collection window (correct answer)
All solutions are meaningful since they represent the mathematical behavior of the temperature function
Only $t = 6$ since it represents the first occurrence when the target temperature is reached
Solutions should be limited to $t = 6, 18, 30$ to account for the periodic nature extending slightly beyond 24 hours

Explanation: For the specific context of data collection during the first 24-hour period (0 ≤ t < 24), only t = 6 and t = 18 are meaningful. While the mathematical function continues to have solutions at t = 30, 42, etc., these fall outside the researcher's stated data collection timeframe. The context-specific constraint (24-hour period) determines which mathematically valid solutions are meaningful for the stated purpose.

Question 16

A ball is thrown upward from a 48-foot building. Its height is modeled by $h(t) = -16t^2 + 32t + 48$ . When solving $h(t) = 64$ to find when the ball is 64 feet high, a student obtains $t = 0.5$ and $t = 1.5$ . The student discards $t = 1.5$ , reasoning that "the ball can only reach 64 feet once on its way up." Which assessment of this reasoning is most appropriate?

The reasoning is correct; projectile motion allows objects to reach a given height only once during upward motion
The reasoning is incorrect; both solutions are valid since the ball reaches 64 feet twice during its complete trajectory (correct answer)
The reasoning is partially correct; $t = 1.5$ should be discarded, but because it exceeds the ball's total flight time
The reasoning is incorrect; $t = 0.5$ should be discarded since the ball starts above 64 feet from the building

Explanation: Both solutions are valid and meaningful. The ball reaches 64 feet at t = 0.5 seconds (on its way up) and again at t = 1.5 seconds (on its way down). The student's reasoning incorrectly assumes that a given height can only be reached once, but in projectile motion, objects typically pass through intermediate heights twice - once ascending and once descending.

Question 17

A rectangular field has an area of 300 square meters. The length is 10 meters more than the width. Setting up the equation $w(w + 10) = 300$ and solving gives $w \approx 14.14$ and $w \approx -24.14$ . A student keeps only the positive solution but rounds it to $w = 14$ meters, stating "fractional measurements aren't practical for field construction." Which assessment of this approach is most appropriate?

The approach is correct; practical construction requires integer measurements, and discarding the negative solution is appropriate
The approach is correct; rounding to practical measurements is more important than mathematical precision in applied problems
The approach is incorrect; both solutions should be kept since they represent different possible field orientations
The approach is partially correct; the negative solution should be discarded, but rounding creates an error since 14 × 24 ≠ 300 (correct answer)

Explanation: When solving real-world quadratic equations, you need to balance mathematical accuracy with practical interpretation. This problem tests whether you understand that mathematical solutions must still satisfy the original constraints. The student correctly discarded the negative solution (

w \approx -24.14

) since width cannot be negative. However, rounding

w \approx 14.14

w = 14

creates a significant error. Let's check: if width = 14 meters, then length = 24 meters, giving an area of

14 \times 24 = 336

square meters, not the required 300 square meters. This 36 square meter difference (12% error) is substantial and unacceptable. Choice A is wrong because while discarding the negative solution is correct, the rounding creates an unacceptable error. Choice B is incorrect because "practical measurements" don't justify solutions that fail to meet the problem's constraints—the field must have exactly 300 square meters. Choice C is wrong because negative dimensions have no physical meaning; there's only one valid orientation for a rectangular field with positive dimensions. Choice D correctly identifies that the student made the right decision about the negative solution but created an error through rounding. In applied problems, your final answer must still satisfy the original equation within reasonable tolerance. Study tip: Always verify your rounded answers by substituting back into the original equation. If rounding creates too large an error, keep more decimal places or express your answer as a fraction to maintain accuracy.

Question 18

A company's profit model is $P(x) = -2x^2 + 100x - 800$ where $x$ is the number of units sold (in hundreds) and $P$ is profit in thousands of dollars. When solving $P(x) = 450$ to find production levels yielding $450,000 profit, a student gets $ $x = 15$ $and$ $x = 35$ $. However, market research indicates that consumer demand drops significantly beyond 2000 units ($ $x = 20$ $). How should these solutions be interpreted?

Both solutions should be kept; market constraints don't affect the mathematical validity of profit calculations at different production levels (correct answer)
Only $x = 15$ should be kept since $x = 35$ represents an unrealistic production level given market demand constraints
Both solutions should be discarded since they don't account for the demand limitations built into the profit model
Only $x = 35$ should be kept since higher production levels indicate more efficient operations despite market challenges

Explanation: Both x = 15 and x = 35 are mathematically valid solutions to the profit equation. The market research about demand dropping beyond x = 20 represents additional business intelligence, but doesn't make either mathematical solution extraneous. The profit model P(x) already incorporates relevant economic factors; external market constraints are separate considerations that don't invalidate the mathematical solutions.

Question 19

A water tank drains according to the model $V(t) = 1000 - 50t + 2t^2$ where $V$ is volume in gallons and $t$ is time in minutes. When solving $V(t) = 0$ to find when the tank is empty, a student obtains $t = 30.41$ and $t = -5.41$ . The student keeps both solutions, reasoning that "the negative time represents how long ago the tank was last empty." Which evaluation of this reasoning is most accurate?

The reasoning is correct; negative time solutions often represent meaningful past events in mathematical models
The reasoning is incorrect; negative time has no physical meaning, but the mathematical interpretation shows the model's domain limitations
The reasoning is partially correct; while negative time can represent past events, this model only describes the current draining process (correct answer)
The reasoning is correct; both solutions should be kept since the quadratic model inherently describes the tank's behavior over all time

Explanation: While negative time can sometimes represent meaningful past events in mathematical contexts, this specific model V(t) = 1000 - 50t + 2t² describes the current draining process starting from an initial volume of 1000 gallons at t = 0. The negative solution doesn't represent when the tank was 'last empty' but rather extends the mathematical model beyond its intended domain. Only t = 30.41 minutes is contextually meaningful for this draining scenario.

Question 20

The area of a rectangular garden is 150 square feet. If the length is 5 feet more than twice the width, setting up the equation $(2w + 5) \cdot w = 150$ and solving yields $w = 7.5$ and $w = -10$ . A student claims that both solutions should be discarded because "negative width is impossible and 7.5 feet is too small for a practical garden." Which response most accurately addresses this reasoning?

The student is correct; both solutions should be discarded due to practical constraints, requiring a recalculation of the problem setup
The student is partially correct; $w = -10$ should be discarded due to physical impossibility, while $w = 7.5$ is mathematically and physically valid (correct answer)
The student is incorrect; both solutions are mathematically valid and the practicality of garden size is not relevant to mathematical interpretation
The student is incorrect; $w = 7.5$ should be kept while $w = -10$ represents the width measured in the opposite direction

Explanation: The negative solution w = -10 must be discarded because width cannot be negative in this physical context. However, w = 7.5 feet is both mathematically correct and physically meaningful, regardless of whether someone considers it 'too small' for practical purposes. Mathematical validity requires physical possibility, not subjective judgments about practicality.