Algebra 3 Quiz: Interpreting Solutions In Context
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Interpreting Solutions In ContextQuestion 1 of 20

A botanist measures a plant's height each day for 10 days and fits the model h(t)=2+0.5th(t)=2+0.5t, where tt is days after planting and hh is inches. The model is valid only for 0≤t≤100\le t\le10. A student uses the model to compute h(20)=12h(20)=12 and concludes the plant will be 12 inches tall 20 days after planting. Which statement best evaluates this conclusion?

The conclusion is valid because the model gives exactly 12 inches at t=20t=20.
The conclusion is invalid because the model was built only through day 10, so day 20 is outside its valid domain.
The conclusion is invalid because the plant can never grow taller than 10 inches.
The conclusion is valid only if the plant's growth rate is exactly 0.5 inch per day for all 20 days, which the model guarantees.
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Algebra 3 Quiz

Algebra 3 Quiz: Interpreting Solutions In Context

Practice Interpreting Solutions In Context in Algebra 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 Solutions In Context, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 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

A botanist measures a plant's height each day for 10 days and fits the model h(t)=2+0.5th(t)=2+0.5t, where tt is days after planting and hh is inches. The model is valid only for 0≤t≤100\le t\le10. A student uses the model to compute h(20)=12h(20)=12 and concludes the plant will be 12 inches tall 20 days after planting. Which statement best evaluates this conclusion?

  1. The conclusion is valid because the model gives exactly 12 inches at t=20t=20.
  2. The conclusion is invalid because the model was built only through day 10, so day 20 is outside its valid domain. (correct answer)
  3. The conclusion is invalid because the plant can never grow taller than 10 inches.
  4. The conclusion is valid only if the plant's growth rate is exactly 0.5 inch per day for all 20 days, which the model guarantees.
Explanation: When you see a model with a stated domain, treat that domain as a hard boundary. The formula may be calculable anywhere, but the model's validity stops at the edge. Here, h(t)=2+0.5th(t)=2+0.5t was fit using data from days 0 to 10. The student plugs in t=20t=20 and gets 12, but that calculation is pure arithmetic—it doesn't represent a real prediction. The correct evaluation is that the conclusion is invalid because the model was built only through day 10, so day 20 is outside its valid domain. Now for the wrong choices. The claim that "the conclusion is valid because the model gives exactly 12 inches at t=20t=20" confuses a correct computation with a valid application—getting a number doesn't make it a trustworthy forecast. The statement "the conclusion is invalid because the plant can never grow taller than 10 inches" is factually wrong; the model itself predicts heights above 10 inches for t>16t>16, and there is no physical limit of 10 inches in the problem. Finally, the choice "the conclusion is valid only if the plant's growth rate is exactly 0.5 inch per day for all 20 days, which the model guarantees" is a trap: the model assumes a constant rate inside its domain, but it does not guarantee that the rate continues beyond day 10—that would be an unverified extrapolation. Your takeaway: before plugging any value into a model, always check the valid domain. If the input falls outside it, the output is a hypothetical number, not a conclusion. On the exam, spotting phrases like "valid only for" or "restricted domain" should immediately signal a potential extrapolation trap.

Question 2

A rectangular garden has perimeter 100 feet and area 600 square feet. Let ww be the width, defined as the shorter side. A student solves w(50−w)=600w(50-w)=600 and gets w=20w=20 or w=30w=30. Which statement correctly interprets the solutions?

  1. The width could be either 20 feet or 30 feet, because both values satisfy the area and perimeter equations.
  2. The width is 30 feet; choosing the larger root gives the required area while keeping the perimeter fixed.
  3. There is no valid width, because the equation produces two widths and a rectangle cannot have two different widths.
  4. The width is 20 feet; w=30w=30 would make the width longer than the length, contradicting the definition of width. (correct answer)
Explanation: When a geometry word problem defines a variable for you, the equation alone doesn't finish the job—you must also honor the verbal definition. Here, the perimeter condition gives w+l=50w+l=50, so l=50−wl=50-w. Substituting into the area condition gives w(50−w)=600w(50-w)=600, which becomes w2−50w+600=0w^2-50w+600=0, or (w−20)(w−30)=0(w-20)(w-30)=0. Algebraically, both 2020 and 3030 satisfy the area and perimeter equations, so it is tempting to say either value works. But the problem specifically defines ww as the shorter side. If w=30w=30, then the length is 50−30=2050-30=20, making the width longer than the length—a direct contradiction of the definition. Therefore the only valid width is 2020 feet. The choice saying the width could be either value ignores this defining constraint. The choice choosing 3030 as the larger root commits the same error, flipping the roles of width and length. And the choice claiming there is no valid width misunderstands the meaning of the two roots: they are two possible candidate values, not two widths a single rectangle must have at the same time. The key habit: after solving, check every solution against all stated conditions—especially words like "shorter side" or "longer side." On exam problems, the algebraic roots are only part of the story; the definition of the variable is the final filter.

Question 3

An account balance (in dollars) after tt years is B(t)=3000(1.05)tB(t)=3000(1.05)^t. A student solves B(t)=10000B(t)=10000 and obtains t≈24.6t\approx24.6. Which statement correctly interprets this result?

  1. The balance increases by about $24.60 each year until it reaches $10,000.
  2. The account grows from $3,000 to $10,000 over approximately 24.6 years. (correct answer)
  3. The balance was approximately $10,000 about 24.6 years ago.
  4. The annual interest rate is approximately $24.60 when the balance reaches $10,000.
Explanation: When you see an exponential balance model like B(t)=3000(1.05)tB(t)=3000(1.05)^t, ask yourself what the parts mean: the $3000 is the starting balance, the $1.05 means the balance grows by 5% each year, and the output $B(t)isthebalanceindollarsafteris the balance in dollars aftertyears.Sosolvingyears.So solving 3000(1.05)t=100003000(1.05)^t=10000 doesnotaskforayearlydollarincreaseoraninterestrate;itasksforthetimedoes not ask for a yearly dollar increase or an interest rate; it asks forthe timetatwhichthebalancereachesat which the balance reaches10000.Here.Heret\approx24.6meanstheaccountgrowsfrommeans the account grows from3000toto10000overapproximatelyover approximately24.6$ years, so that's the correct interpretation. The distractor saying "The balance increases by about $24.60 each year until it reaches $10,000" confuses the solved time value with a dollar amount; the growth is5% per year, not a flat $24.60 addition. The statement “The balance was approximately $10,000 about24.6 years ago" would involve a negative time, but this solution is for the future, not the past. And "The annual interest rate is approximately $24.60” mistakes $tfortherate;theinterestrateis5for the rate; the interest rate is5%, already visible inside1.05. Whenever you solve $$B(t)=\text{target}$$, identify whether your answer is a time, a rate, or a dollar amount.In exponential growth models, the exponent t$ is usually a length of time—so say "grows from initial value to target in that many years," not "by that many dollars." That unit-check alone will save you on interpretation questions.

Question 4

A bakery's weekly profit (in dollars) is P(x)=−x2+100x−1600P(x)=-x^2+100x-1600, where xx is the price of a cake in dollars. The owner can charge at most $50 per cake. A student solves $P(x)=0P(x)=0 andobtainsand obtains x=20x=20 oror x=80x=80 $. Which interpretation is correct?

  1. The bakery breaks even at $20 only; $x=80x=80 $ is outside the allowable price range and must be rejected. (correct answer)
  2. The bakery breaks even at $20 and $80, so it earns a profit for every price between those values.
  3. The bakery breaks even at $80 only, because $20 is too low to cover the cost of making a cake.
  4. The bakery breaks even at $50, because that is the highest price the owner is allowed to charge.
Explanation: When you see a quadratic profit function, the first thing to recognize is that its zeros are break-even prices—prices where profit equals zero. But you must always check those zeros against the real-world domain given in the problem. Here, solving P(x)=0P(x)=0 gives −x2+100x−1600=0-x^2+100x-1600=0, which factors as −(x−20)(x−80)=0-(x-20)(x-80)=0, so the mathematical break-even prices are x=20x=20 and x=80x=80. However, the owner "can charge at most $50 per cake,” so $x=80x=80 isnotanallowableprice.Theonlyvalidbreak−evenpriceisis not an allowable price. The only valid break-even price is x=20x=20 ,makingthecorrectinterpretation“Thebakerybreaksevenat, making the correct interpretation “The bakery breaks even at 20 only; x=80 is outside the allowable price range and must be rejected." Now look at the other choices. Saying the bakery breaks even at both 20 and 80 ignores the 50pricelimit,soitcannotbecorrecteventhoughthealgebraisright.Sayingitbreaksevenat50 price limit, so it cannot be correct even though the algebra is right. Saying it breaks even at 80 only because 20istoolowconfuses“pricetoolowtocovercost”withthefactthatprofitisexactlyzeroat20 is too low confuses “price too low to cover cost” with the fact that profit is exactly zero at 20—there, the bakery covers its costs perfectly. And saying it breaks even at 50confusesthemaximumallowedpricewithabreak−evenpoint;infact,50 confuses the maximum allowed price with a break-even point; in fact, P(50)=−502+100(50)−1600=900P(50)=-50^2+100(50)-1600=900 $, so the bakery earns $900 at that price. Your takeaway: always solve the equation, then check whether every solution fits the problem's domain. A valid algebraic answer can be an invalid real-world answer.

Question 5

A bakery's profit (in dollars) is P(x)=−x2+50x−100P(x)=-x^2+50x-100, where xx is the number of cakes sold per day. A student solves P(x)=700P(x)=700 and finds no real solution. Which interpretation is correct?

  1. The bakery can never make $700 in profit, because its maximum possible daily profit is $525. (correct answer)
  2. The bakery makes $700 in profit at two different sales levels, but neither level is real.
  3. The bakery makes $700 in profit when it sells 25 cakes per day.
  4. The solution is invalid, because every quadratic equation must have at least one real solution.
Explanation: Whenever you see a quadratic used as a real-world model, ask whether the discriminant supports the conclusion about real values. Here, solving P(x)=700P(x)=700 means solving −x2+50x−100=700-x^2+50x-100=700, or x2−50x+800=0x^2-50x+800=0. Its discriminant is (−50)2−4(1)(800)=2500−3200=−700(-50)^2-4(1)(800)=2500-3200=-700, so there are no real solutions for xx. No number of cakes sold can produce exactly $700. What does the profit function actually reach? Since P(x)P(x) opens downward, its maximum is at the vertex x=−b/(2a)=25x=-b/(2a)=25. Then P(25)=−625+1250−100=525P(25)=-625+1250-100=525. The bakery’s maximum possible daily profit is $525, so $700 is impossible. The “two sales levels” choice misreads the negative discriminant: it means zero real solutions, not two imaginary sales levels. The “25 cakes per day” choice confuses the vertex’s xx-value with an answer to P(x)=700P(x)=700; at 25 cakes the profit is only $525. The "every quadratic must have a real solution" choice is simply false—quadratics can have zero, one, or two real solutions depending on the discriminant. For exam day: when a quadratic word problem asks whether a certain target is possible, rewrite the equation as P(x)=targetP(x)=\text{target}, check the discriminant, and then compare the target to the vertex's maximum/minimum value. That combination tells you both whether and where the target can be reached.

Question 6

A manufacturer's daily profit (in dollars) is P(x)=−x2+30x−200P(x)=-x^2+30x-200, where xx is the number of units produced, in hundreds. A student solves P(x)=0P(x)=0 and gets x=10x=10 or x=20x=20. Which statement correctly interprets the break-even solutions and the profit behavior?

  1. The company breaks even at 10 and 20 units and has a positive profit for production levels between those amounts.
  2. The company breaks even at 1,000 and 2,000 units but has a positive profit for production levels below 1,000 or above 2,000 units.
  3. The company breaks even at 1,000 and 2,000 units but has a positive profit exactly at those production levels and loses money elsewhere.
  4. The company breaks even at 1,000 and 2,000 units and has a positive profit for production levels between those amounts. (correct answer)
Explanation: Whenever you see a profit function like P(x)=−x2+30x−200P(x)=-x^2+30x-200, remember that xx is measured in hundreds of units. So before interpreting solutions, you must convert them: x=10x=10 means 1,000 units, and x=20x=20 means 2,000 units. The break-even points are where profit equals zero, so the company breaks even at 1,000 and 2,000 units. Now examine the profit behavior. Since the coefficient of x2x^2 is negative, this parabola opens downward. That means profit is positive between the two roots and negative outside them. So the correct interpretation is: the company breaks even at 1,000 and 2,000 units and has a positive profit for production levels between those amounts. The choice saying "breaks even at 10 and 20 units" makes the common unit error — it treats xx as the actual unit count instead of hundreds. The choice that says "positive profit below 1,000 or above 2,000" gets the conversion right but flips the parabola's behavior, perhaps thinking the positive coefficient would open upward. The choice that says "positive profit exactly at those production levels and loses money elsewhere" confuses break-even (zero profit) with profit maximization — at the break-even points profit is exactly zero, not positive. Your study tip: always check the units in word problems, and quickly sketch the parabola's direction using the sign of the x2x^2 coefficient. That one check will prevent both the conversion trap and the sign-region trap.

Question 7

A chemist wants 20 liters of a 50% acid solution. She has a 10% acid solution and a 30% acid solution. Let x be the liters of 10% solution and y be the liters of 30% solution. Solving the system x + y = 20, 0.10x + 0.30y = 10 yields x = -20, y = 40. What is the correct interpretation?

  1. Mix -20 liters of 10% solution with 40 liters of 30% solution; the negative amount means 20 liters of 10% solution must be removed.
  2. Use 20 liters of the 30% solution, because 30% is the closest available concentration to 50%.
  3. No such solution can be made, because mixing 10% and 30% solutions can only produce a concentration between 10% and 30%, never 50%. (correct answer)
  4. Use 40 liters of 30% solution, and no 10% solution, because negative amounts mean none of that ingredient should be used.
Explanation: This is a mixture problem, so the key is to remember two separate constraints: the total volume equation, x+y=20x+y=20, and the pure-acid equation, 0.10x+0.30y=100.10x+0.30y=10. Solving those equations does give x=−20x=-20 and y=40y=40, but in a real-world chemistry context, amounts cannot be negative. A negative amount of solution has no physical meaning, so the algebraic result means the situation is impossible. More directly, mixing a 10% solution with a 30% solution can only produce concentrations between 10% and 30%, so obtaining a 50% solution is impossible regardless of the amounts. "Mix -20 liters … negative amount means 20 liters must be removed" treats a negative volume as a real action, but negative amounts are not allowed in mixture problems. "Use 20 liters of the 30% solution" overlooks that 20 liters of 30% acid contains only 6 liters of pure acid, not the 10 liters needed for 50%. "Use 40 liters of 30% solution and no 10% solution" ignores the total-volume requirement of 20 liters and also cannot produce 50% acid. All three wrong answers try to make the impossible arithmetic work in a physical setting when it simply cannot. On exam day, check whether the target concentration lies between the two concentrations you are mixing. If it does not, stop: no mixture of those solutions can ever reach it.

Question 8

A school sold 25 tickets to a fundraiser. Adult tickets cost $5 each and student tickets cost $2 each. Total revenue was $64. A student lets x be adult tickets and y be student tickets, solves x + y = 25, 5x + 2y = 64, and gets x = 14/3 ≈ 4.67 and y = 61/3 ≈ 20.33. What is the correct interpretation?

  1. About 5 adult tickets and 20 student tickets were sold, because x ≈ 4.67 rounds to 5 and y ≈ 20.33 rounds to 20.
  2. The school sold about 4.67 adult tickets and 20.33 student tickets, since fractions of tickets can occur in averaged data.
  3. The solution is valid, but the revenue must have been $65 because the extra $1 is just a rounding error.
  4. No valid whole-number solution exists; ticket counts must be whole numbers, and no whole-number combination gives exactly $64. (correct answer)
Explanation: Whenever a word problem involves counting tickets, people, or objects, the variables must be whole numbers. So even though the system x+y=25,5x+2y=64x+y=25,\quad 5x+2y=64 is mathematically consistent, its solution x=143, y=613x=\frac{14}{3},\ y=\frac{61}{3} cannot describe actual ticket sales. Substitute into the revenue equation: if there were 25 tickets, then y=25−xy=25-x, so revenue is 5x+2(25−x)=50+3x.5x+2(25-x)=50+3x. Setting this equal to 64 gives 3x=14,x=143.3x=14,\quad x=\frac{14}{3}. Since xx must be a whole number of adult tickets, no valid whole-number combination exists. That is why the correct interpretation is that no whole-number combination gives exactly 6464. The "round to 5 adult and 20 student tickets" choice is tempting, but rounding gives revenue 5(5)+2(20)=65,5(5)+2(20)=65, not 64, and rounding a fractional solution doesn't fix the fact that ticket counts must be integers. The "fractions of tickets can occur in averaged data" choice confuses this situation with an average; here we are told actual tickets were sold, not averaged. The "solution is valid, but revenue must have been $65” choice misreads the algebra: the solution is valid only as a fraction, not as ticket counts, and the $1 difference is not rounding error — the original equations were exact. On exam day, after solving a system, always check whether the variables represent countable items. If the answer is fractional, try integer values near it and verify both equations.

Question 9

A ball is launched straight upward from a height of 64 feet. Its height after tt seconds is h(t)=−16t2+48t+64h(t)=-16t^2+48t+64. A student solves h(t)=0h(t)=0 and obtains t=4t=4 and t=−1t=-1. Which statement correctly interprets this result in the context of the ball's motion?

  1. The ball reaches its maximum height at t=4t=4 s and returns to its launch height at t=−1t=-1 s.
  2. The ball hits the ground at t=3t=3 s, because the +64+64 term represents the vertex height and should be ignored.
  3. The ball hits the ground at t=4t=4 s; t=−1t=-1 is extraneous because negative time has no meaning here. (correct answer)
  4. The ball hits the ground at both t=−1t=-1 s and t=4t=4 s, so it was already on the ground 1 second before launch.
Explanation: When you solve h(t)=0h(t)=0, you are finding the times when the ball's height is zero—that is, when it hits the ground. But the domain of the situation is time after launch, so negative values have no physical meaning. Here, factoring gives h(t)=−16(t−4)(t+1)h(t)=-16(t-4)(t+1), so the roots are t=4t=4 and t=−1t=-1. Since time cannot be negative, only t=4t=4 seconds is valid; the ball hits the ground at 4 seconds. The negative root is extraneous, a mathematical solution that doesn't fit the real-world context. Now look at the wrong choices. The statement about reaching maximum height at 4 seconds and returning to launch height at -1 seconds confuses roots with the vertex—the vertex actually occurs at t=1.5t=1.5 seconds, not 4, and launch height is at t=0t=0, not -1. The choice claiming the ball hits the ground at 3 seconds because the +64+64 term is the vertex height misunderstands that +64+64 is the initial height, not the vertex; the ground time comes from the quadratic formula. Finally, saying the ball hits the ground at both -1 and 4 seconds ignores that negative time has no meaning in this context—the ball didn't exist before launch. Your takeaway: whenever a quadratic models a real-world quantity like time, always check that your solutions fall within the reasonable domain. If one root is negative, discard it as extraneous—the physical answer is the positive root.

Question 10

A company's revenue from selling nn widgets is R(n)=80nR(n)=80n, and its cost is C(n)=50n+200C(n)=50n+200. A student solves R=CR=C and gets n=6.6667n=6.6667. Widgets can only be sold in whole numbers. What is the correct interpretation?

  1. The company breaks even by selling 6.67 widgets, meaning 6 whole widgets plus two-thirds of another widget.
  2. The company breaks even after selling 6 widgets, because 6.6667 rounds down to 6.
  3. The company breaks even at exactly 6.6667 widgets, and selling 7 widgets also breaks even because 7 rounds to 6.6667.
  4. The exact break-even point is 6.6667 widgets, so the company first reaches or exceeds break-even when it sells 7 whole widgets. (correct answer)
Explanation: Whenever you solve an equation that models a real-world quantity restricted to whole numbers, the algebraic solution may give a decimal, but you must interpret it in context. Here, setting revenue equal to cost: 80n=50n+20080n = 50n + 200 gives n=6.6667n = 6.6667. That is the exact break-even point if fractional widgets were possible. However, widgets are sold only in whole numbers, so you need the smallest integer nn where revenue is at least cost. Check the neighbors: at n=6n=6, R=480R=480 and C=500C=500 — a loss. At n=7n=7, R=560R=560 and C=550C=550 — a profit. Thus the company first reaches or exceeds break-even at 7 widgets, so the correct interpretation is that the exact point is 6.6667, and 7 whole widgets are needed to actually break even or profit. The choice claiming "6 whole widgets plus two-thirds of another" is impossible because you cannot sell a fraction of a widget. The choice that says "rounds down to 6" ignores that at 6 the company still loses money, so it doesn't break even. The choice claiming "selling 7 also breaks even because 7 rounds to 6.6667" is wrong because 7 rounds to 7, not 6.6667, and at 7 revenue exceeds cost. Remember: for discrete quantities, after finding the fractional solution, test the integer just below and just above. Use the ceiling when you need the first integer that meets or exceeds the threshold.

Question 11

The time tt (in hours) after which a certain chemical reaction reaches completion satisfies t+2=t−4\sqrt{t+2}=t-4. A student squares both sides, solves, and obtains t=7t=7 and t=2t=2. Which interpretation is correct?

  1. The reaction reaches completion at t=7t=7 hours; t=2t=2 does not satisfy the original equation, so it is extraneous. (correct answer)
  2. The reaction reaches completion at t=2t=2 hours and t=7t=7 hours, so it completes at both times.
  3. The reaction reaches completion at t=2t=2 hours only, because t=7t=7 hours exceeds the scheduled duration of the experiment.
  4. The reaction never reaches completion, because squaring introduced an extraneous solution and made the equation unsolvable.
Explanation: When you solve an equation with a square root, your first move is often to square both sides — but squaring can create extra "solutions" that do not actually work in the original equation. That is exactly the situation here. Start with t+2=t−4\sqrt{t+2}=t-4. Squaring gives t+2=(t−4)2t+2=(t-4)^2, which simplifies to t2−9t+14=0t^2-9t+14=0, or (t−7)(t−2)=0(t-7)(t-2)=0. So the candidate solutions are t=7t=7 and t=2t=2. Now check each one in the original equation. For t=7t=7: 9=3\sqrt{9}=3 and 7−4=37-4=3, so it works. For t=2t=2: 4=2\sqrt{4}=2 but 2−4=−22-4=-2, so it fails. Therefore the reaction reaches completion at t=7t=7 hours, and t=2t=2 is extraneous. The choice saying the reaction completes at both times ignores this essential check. The choice saying t=2t=2 is correct because t=7t=7 exceeds the scheduled duration brings in an assumption that has no mathematical basis — the original equation alone determines valid times. And the choice saying the reaction never completes misunderstands extraneous solutions: squaring did not make the equation unsolvable; it just produced one extra root that you can identify and discard by substitution. Whenever you square both sides, remember to verify every candidate in the original equation. On this exam, extraneous roots from squaring are a common trap — checking your solutions is not optional.

Question 12

A biologist models a population (in hundreds) by P(t)=100+40t−5t2P(t)=100+40t-5t^2, where t≥0t\ge 0 is years from now. She asks when the population will return to its current level of 100. Solving P(t)=100P(t)=100 gives t=0t=0 and t=8t=8. Which interpretation is correct?

  1. The population returns to 100 both now and in 8 years, so both answers are correct.
  2. The population returns to 100 in 0 years, meaning it has not yet left that level.
  3. The population returns to 100 in 8 years; t=0t=0 is only the present time, not a future return. (correct answer)
  4. The population never returns to 100, because having two solutions shows that the model is inconsistent.
Explanation: When you see a quadratic model like P(t)=100+40t−5t2P(t)=100+40t-5t^2 and a question about when a value "returns" to its starting level, remember that time is part of the context: the model begins at t=0t=0, which is the present, not a future event. Solving P(t)=100P(t)=100 correctly gives t=0t=0 and t=8t=8 because P(0)=100P(0)=100 is the initial condition built into the equation. The question asks when the population will return to its current level, so it must first leave that level. The population is above 100 between those two times, and it comes back down to 100 at t=8t=8 years. Therefore, the meaningful answer is 8 years from now. The choice saying "both now and in 8 years" is tempting but overlooks the meaning of "return": now is not a return, it is the starting point. The choice saying "in 0 years, meaning it has not yet left that level" is also wrong, because the population does rise above 100 immediately after t=0t=0. Finally, the idea that two solutions show the model is inconsistent is a misunderstanding; quadratic equations often have two solutions, and here one is the initial time and the other is the future return time. Your takeaway: when a time-word problem gives t=0t=0 as a solution, ask whether the question is about the initial moment or about a change after that moment. "Return" always means a future event.

Question 13

A tank can be filled by pipe A alone in 5 fewer hours than pipe B alone. Working together, the two pipes fill the tank in 6 hours.

Let xx be the time, in hours, for pipe B alone. The equation 1x−5+1x=16\frac{1}{x-5}+\frac{1}{x}=\frac{1}{6} has solutions x=2x=2 and x=15x=15. How long does pipe A alone take?

  1. −3-3 hours, because finding pipe A's time from the invalid root gives 2−52-5
  2. 22 hours, because it is a positive solution of the equation
  3. 1010 hours, because x=15x=15 is the only valid pipe-B time and pipe A is 5 hours faster (correct answer)
  4. 1515 hours, because it is the only solution that keeps both denominators positive
Explanation: If x=2x=2, then pipe A's time is x−5=−3x-5=-3 hours, which is impossible. Thus the valid pipe-B time is x=15x=15, and pipe A alone takes 15−5=1015-5=10 hours. The negative value comes from using the invalid root, and 15 hours is pipe B's time, not pipe A's.

Question 14

A drone's height above its landing pad is h(t)=4+t+1h(t)=4+\sqrt{t+1} meters for t≥0t \ge 0. A student solves h(t)=2h(t)=2 by first squaring both sides and obtains t=3t=3 and t=35t=35.

Which statement correctly describes these solutions?

  1. Both t=3t=3 and t=35t=35 are valid because they make the squared equation true.
  2. Only t=3t=3 is valid, because t=35t=35 gives a height much greater than 2 meters.
  3. Only t=35t=35 is valid, because t=3t=3 is less than the drone's launch time.
  4. Neither value is valid; since t+1≥0\sqrt{t+1} \ge 0, h(t)≥4h(t) \ge 4 for all t≥0t \ge 0, so h(t)=2h(t)=2 has no solution. (correct answer)
Explanation: The correct isolation is t+1=−2\sqrt{t+1}=-2, which has no real solution because a square root cannot be negative. Squaring without isolating introduced two roots, but neither satisfies the original equation. The model's minimum height is 4 meters, so a height of 2 meters is impossible.

Question 15

A right-triangular garden has hypotenuse xx feet and legs x−7x-7 feet and x−1x-1 feet.

A student solves x2=(x−7)2+(x−1)2x^2=(x-7)^2+(x-1)^2 and obtains x=8−14≈3.26x=8-\sqrt{14}\approx3.26 and x=8+14≈12.74x=8+\sqrt{14}\approx12.74. Which interpretation is correct?

  1. Both root values satisfy the equation, so either length is possible.
  2. Only 3.263.26 feet is possible, because the smaller positive root is the only realistic size for a garden.
  3. Neither value is possible because the side lengths would have to be whole numbers.
  4. Only 12.7412.74 feet is possible; 3.263.26 feet would make x−7x-7 negative, so that root is outside the context. (correct answer)
Explanation: For x=3.26x=3.26, the leg x−7x-7 is about −3.74-3.74 feet, which is impossible. The larger root makes both legs positive and satisfies the Pythagorean relationship, so it is the only valid hypotenuse length. Satisfying the equation alone is not enough.

Question 16

A ball is launched upward from a 14.7-meter cliff. Its height after tt seconds is modeled by h(t)=−4.9t2+19.6t+14.7h(t)=-4.9t^2+19.6t+14.7.

A student solves h(t)=0h(t)=0 and obtains t=2+7t=2+\sqrt{7} and t=2−7t=2-\sqrt{7}. Which statement correctly interprets the solutions?

  1. Both times are valid because substituting either time into h(t)h(t) gives 0.
  2. The ball reaches the ground at t=2−7t=2-\sqrt{7} seconds and then is launched at t=0t=0.
  3. The value t=2−7t=2-\sqrt{7} is an extraneous solution introduced by the quadratic formula and must be rejected.
  4. Only t=2+7t=2+\sqrt{7} lies in the time domain t≥0t \ge 0, so it is the impact time; the negative value is algebraically valid but has no physical meaning. (correct answer)
Explanation: Both values satisfy the equation, but the model applies only after launch, so time must be nonnegative. Since 2−7≈−0.652-\sqrt{7} \approx -0.65, it is not a valid impact time. The quadratic formula did not introduce an extraneous root; the negative root is simply outside the context.

Question 17

A robot's distance from a wall, in meters, after tt seconds is d=∣t−3∣d=|t-3|, where t≥0t \ge 0. A student solves ∣t−3∣=2t+1|t-3|=2t+1 and obtains t=−4t=-4 and t=2/3t=2/3.

Which statement correctly interprets these solutions?

  1. Only t=2/3t=2/3 is valid; t=−4t=-4 does not satisfy the original equation and is outside the domain t≥0t \ge 0. (correct answer)
  2. Both values are valid because each came from one of the two absolute-value cases.
  3. Only t=−4t=-4 is valid, because it is the value that makes t−3t-3 negative.
  4. Neither value is valid because an absolute value cannot equal a linear expression.
Explanation: Substituting t=−4t=-4 gives ∣−4−3∣=7|-4-3|=7 on the left but 2(−4)+1=−72(-4)+1=-7 on the right, so it is not a solution. It also violates the domain t≥0t \ge 0. The value t=2/3t=2/3 satisfies the original equation and is the only valid time.

Question 18

A ball is thrown upward from a 100-foot cliff. Its height is modeled by h(t)=−16t2+vt+100h(t)=-16t^2+vt+100, where vv is the initial upward speed in ft/s. The ball hits the ground after 5 seconds.

A student solves for vv and then finds the other root of the quadratic is t=−1.25t=-1.25. Which conclusion is correct?

  1. v=60v=60 ft/s, and t=−1.25t=-1.25 is a second time the ball is at ground level before launch.
  2. v=60v=60 ft/s, and t=−1.25t=-1.25 is outside the domain t≥0t \ge 0, so the only impact time is 5 seconds. (correct answer)
  3. v=80v=80 ft/s, and t=−1.25t=-1.25 is the time before launch when the ball would have been on the ground.
  4. v=100v=100 ft/s, and the negative root is extraneous because the quadratic has only one positive root.
Explanation: Using h(5)=0h(5)=0 gives −16(25)+5v+100=0-16(25)+5v+100=0, so v=60v=60 ft/s. The resulting quadratic has roots 55 and −1.25-1.25, but negative time is not in the model's domain. The negative root is not a second physical impact time.

Question 19

A car purchased for $28,000 loses 12% of its value each year. Its value after tt years is V(t)=28000(0.88)tV(t)=28000(0.88)^t, where tt is measured from the date of purchase. A student solves V(t)=10000V(t)=10000, obtains t≈8.06t\approx 8.06, and reports: "Since 8.06 rounds to 8, the value first drops below $10,000 after 8 years."

Which statement best evaluates the student's report?

  1. The report is correct because 8.068.06 rounds to 8, and at the 8-year mark the car has just fallen below the $10,000 threshold.
  2. The report is correct because the 12% yearly decay is applied in one annual step, so the continuous solution t=8.06t=8.06 is not relevant to whole-year reporting.
  3. The report is incorrect: at t=8t=8 the car is still worth more than $10,000, and the threshold is crossed at about t=8.06t=8.06, so the first whole-year mark below $10,000 is t=9t=9. (correct answer)
  4. The report is incorrect: because 8<8.068<8.06, 0.888<0.888.060.88^8<0.88^{8.06}, so the car's value at the 8-year mark is already below the threshold.
Explanation: Solving V(t)=10000V(t)=10000 gives the continuous time at which the value equals $10,000, about t=8.06t=8.06. Since the base 0.880.88 is less than 1, a smaller exponent gives a larger value, so at t=8t=8 the car is still worth more than $10,000; it first falls below the threshold just after t=8.06t=8.06. Therefore the first whole-year mark at which the value is below the threshold is t=9t=9 years. Rounding 8.06 down treats the crossing time as if it were at t=8t=8, while the sign-error answer reverses the comparison of the powers.

Question 20

A store's annual sales in thousands of dollars are modeled by S(t)=42+3.8tS(t)=42+3.8t, where tt is years after 2015. A manager wants to know when sales reach $100,000 and solves $42+3.8t=10042+3.8t=100 ,getting, getting t≈15.26t\approx15.26 $.

Which interpretation is correct?

  1. Sales reach $100,000 about 15.26 years after 2015, so around 2030; the equation correctly uses thousands of dollars. (correct answer)
  2. Sales reach $100,000 about 15.26 thousand years after 2015, because $tt $ is in thousands.
  3. The equation should have used 100,000, so the time is about 26,304 years after 2015.
  4. The value 15.26 is in thousands of dollars, so the year cannot be determined from this calculation.
Explanation: Since S(t)S(t) is measured in thousands of dollars, $100,000 corresponds to $S=100S=100 .Thesolution. The solution t=15.26t=15.26 isalreadyinyears,sotheyearisaboutis already in years, so the year is about 2015+15.26≈20302015+15.26\approx2030 $. The other answers confuse the units of the output with the units of the input.