Interpreting Solutions in Context - Algebra 3

1. A bakery's weekly profit (in dollars) is $P(x)=-x^2+100x-1600$, where $x$ is the price of a cake in dollars. The owner can charge at most $50 per cake. A student solves $P(x)=0$ and obtains $x=20$ or $x=80$. Which interpretation is correct?
   Answer: The bakery breaks even at $20 only; $x=80$ is outside the allowable price range and must be rejected. 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 $P(x)=0$ gives $-x^2+100x-1600=0$, which factors as $-(x-20)(x-80)=0$, so the mathematical break-even prices are $x=20$ and $x=80$. However, the owner "can charge at most $50 per cake," so $x=80$ is not an allowable price. The only valid break-even price is $x=20$, 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 $50 price limit, so it cannot be correct even though the algebra is right. Saying it breaks even at $80 only because $20 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 $50 confuses the maximum allowed price with a break-even point; in fact, $P(50)=-50^2+100(50)-1600=900$, 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.
2. 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?
   Answer: No such solution can be made, because mixing 10% and 30% solutions can only produce a concentration between 10% and 30%, never 50%. This is a mixture problem, so the key is to remember two separate constraints: the total volume equation, $x+y=20$, and the pure-acid equation, $0.10x+0.30y=10$. Solving those equations does give $x=-20$ and $y=40$, 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.
3. Which statement correctly interprets the solutions?
   Answer: Only $x=18$ is valid; $x=-24$ satisfies the equation algebraically but cannot represent the number of students. The value $x=-24$ satisfies the equation algebraically, but a negative number of students is impossible. The valid original group size is 18 students. The other options either accept an impossible negative count or incorrectly reject the valid positive solution.
4. Which statement correctly interprets these solutions?
   Answer: Only $t=2/3$ is valid; $t=-4$ does not satisfy the original equation and is outside the domain $t \ge 0$. Substituting $t=-4$ gives $|-4-3|=7$ on the left but $2(-4)+1=-7$ on the right, so it is not a solution. It also violates the domain $t \ge 0$. The value $t=2/3$ satisfies the original equation and is the only valid time.
5. Which conclusion is correct?
   Answer: At least 47 items must be sold; 46 items produce a loss, and 47 items produce a profit. At the break-even value, 46 items gives a cost of $902 and revenue of $897, so it is a loss. At 47 items, cost is $914 and revenue is $916.50, so it is a profit. Since items must be whole, the minimum profitable sales level is 47.
6. 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?
   Answer: No valid whole-number solution exists; ticket counts must be whole numbers, and no whole-number combination gives exactly $64. Whenever a word problem involves counting tickets, people, or objects, the variables must be whole numbers. So even though the system $x+y=25,\quad 5x+2y=64$ is mathematically consistent, its solution $x=\frac{14}{3},\ y=\frac{61}{3}$ cannot describe actual ticket sales. Substitute into the revenue equation: if there were 25 tickets, then $y=25-x$, so revenue is $5x+2(25-x)=50+3x.$ Setting this equal to 64 gives $3x=14,\quad x=\frac{14}{3}.$ Since $x$ 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 $64$. The "round to 5 adult and 20 student tickets" choice is tempting, but rounding gives revenue $5(5)+2(20)=65,$ 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.
7. Which statement correctly describes these solutions?
   Answer: Neither value is valid; since $\sqrt{t+1} \ge 0$, $h(t) \ge 4$ for all $t \ge 0$, so $h(t)=2$ has no solution. The correct isolation is $\sqrt{t+1}=-2$, 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.
8. A student solves for $v$ and then finds the other root of the quadratic is $t=-1.25$. Which conclusion is correct?
   Answer: $v=60$ ft/s, and $t=-1.25$ is outside the domain $t \ge 0$, so the only impact time is 5 seconds. Using $h(5)=0$ gives $-16(25)+5v+100=0$, so $v=60$ ft/s. The resulting quadratic has roots $5$ and $-1.25$, but negative time is not in the model's domain. The negative root is not a second physical impact time.
9. A rectangular garden has perimeter 100 feet and area 600 square feet. Let $w$ be the width, defined as the shorter side. A student solves $w(50-w)=600$ and gets $w=20$ or $w=30$. Which statement correctly interprets the solutions?
   Answer: The width is 20 feet; $w=30$ would make the width longer than the length, contradicting the definition of width. 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 $w+l=50$, so $l=50-w$. Substituting into the area condition gives $w(50-w)=600$, which becomes $w^2-50w+600=0$, or $(w-20)(w-30)=0$. Algebraically, both $20$ and $30$ satisfy the area and perimeter equations, so it is tempting to say either value works. But the problem specifically defines $w$ as the shorter side. If $w=30$, then the length is $50-30=20$, making the width longer than the length—a direct contradiction of the definition. Therefore the only valid width is $20$ feet. The choice saying the width could be either value ignores this defining constraint. The choice choosing $30$ 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.
10. A bakery's profit (in dollars) is $P(x)=-x^2+50x-100$, where $x$ is the number of cakes sold per day. A student solves $P(x)=700$ and finds no real solution. Which interpretation is correct?
    Answer: The bakery can never make $700 in profit, because its maximum possible daily profit is $525. Whenever you see a quadratic used as a real-world model, ask whether the discriminant supports the conclusion about real values. Here, solving $P(x)=700$ means solving $-x^2+50x-100=700$, or $x^2-50x+800=0$. Its discriminant is $(-50)^2-4(1)(800)=2500-3200=-700$, so there are no real solutions for $x$. No number of cakes sold can produce exactly $700. What does the profit function actually reach? Since $P(x)$ opens downward, its maximum is at the vertex $x=-b/(2a)=25$. Then $P(25)=-625+1250-100=525$. 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 $x$-value with an answer to $P(x)=700$; 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 $P(x)=\text{target}$, 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.
11. An account balance (in dollars) after $t$ years is $B(t)=3000(1.05)^t$. A student solves $B(t)=10000$ and obtains $t\approx24.6$. Which statement correctly interprets this result?
    Answer: The account grows from $3,000 to $10,000 over approximately 24.6 years. When you see an exponential balance model like $B(t)=3000(1.05)^t$, 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 $B(t)$ is the balance in dollars after $t$ years.So solving $3000(1.05)^t=10000$ does not ask for a yearly dollar increase or an interest rate; it asks forthe time $t$ at which the balance reaches $10000$.Here$t\approx24.6$ means the account grows from $3000$ to $10000$ over approximately $24.6$ 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 $t$ for the rate; the interest rate is5%, already visible inside $1.05$. 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.
12. Which statement correctly interprets the prices?
    Answer: Both prices are valid; either price produces $2800 in weekly revenue. Substituting either price into $R(p)$ gives $2800, so both are valid in context. A parabola can take the same value at two different prices because the graph decreases after the vertex; being above the maximizing price does not make the price invalid.
13. Which statement best evaluates the student's report?
    Answer: The report is incorrect: at $t=8$ the car is still worth more than $10,000, and the threshold is crossed at about $t=8.06$, so the first whole-year mark below $10,000 is $t=9$. Solving $V(t)=10000$ gives the continuous time at which the value equals $10,000, about $t=8.06$. Since the base $0.88$ is less than 1, a smaller exponent gives a larger value, so at $t=8$ the car is still worth more than $10,000; it first falls below the threshold just after $t=8.06$. Therefore the first whole-year mark at which the value is below the threshold is $t=9$ years. Rounding 8.06 down treats the crossing time as if it were at $t=8$, while the sign-error answer reverses the comparison of the powers.
14. A company's revenue from selling $n$ widgets is $R(n)=80n$, and its cost is $C(n)=50n+200$. A student solves $R=C$ and gets $n=6.6667$. Widgets can only be sold in whole numbers. What is the correct interpretation?
    Answer: The exact break-even point is 6.6667 widgets, so the company first reaches or exceeds break-even when it sells 7 whole widgets. 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: $80n = 50n + 200$ gives $n = 6.6667$. 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 $n$ where revenue is at least cost. Check the neighbors: at $n=6$, $R=480$ and $C=500$ — a loss. At $n=7$, $R=560$ and $C=550$ — 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.
15. A ball is launched straight upward from a height of 64 feet. Its height after $t$ seconds is $h(t)=-16t^2+48t+64$. A student solves $h(t)=0$ and obtains $t=4$ and $t=-1$. Which statement correctly interprets this result in the context of the ball's motion?
    Answer: The ball hits the ground at $t=4$ s; $t=-1$ is extraneous because negative time has no meaning here. When you solve $h(t)=0$, 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 $h(t)=-16(t-4)(t+1)$, so the roots are $t=4$ and $t=-1$. Since time cannot be negative, only $t=4$ 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 $t=1.5$ seconds, not 4, and launch height is at $t=0$, not -1. The choice claiming the ball hits the ground at 3 seconds because the $+64$ term is the vertex height misunderstands that $+64$ 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.
16. Which interpretation is correct?
    Answer: Sales reach $100,000 about 15.26 years after 2015, so around 2030; the equation correctly uses thousands of dollars. Since $S(t)$ is measured in thousands of dollars, $100,000 corresponds to $S=100$. The solution $t=15.26$ is already in years, so the year is about $2015+15.26\approx2030$. The other answers confuse the units of the output with the units of the input.
17. Which statement correctly interprets these results?
    Answer: Only $t=8$ satisfies the original equation; $t=0$ is extraneous, so the deficit reaches 0 in 2028. Squaring both sides produces valid candidates, but each candidate must be checked in the original equation. At $t=0$, the original equation gives $\sqrt{9}=3$ but $t-3=-3$, so $t=0$ is extraneous. At $t=8$, $\sqrt{25}=5$ and $t-3=5$, so $t=8$ is valid. Thus only the 2028 interpretation is correct. The other choices either accept both squared-equation roots without checking, reject the valid root, or incorrectly discard both because squaring is not reversible.
18. A student solves $x^2=(x-7)^2+(x-1)^2$ and obtains $x=8-\sqrt{14}\approx3.26$ and $x=8+\sqrt{14}\approx12.74$. Which interpretation is correct?
    Answer: Only $12.74$ feet is possible; $3.26$ feet would make $x-7$ negative, so that root is outside the context. For $x=3.26$, the leg $x-7$ is about $-3.74$ 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.
19. Let $x$ be the time, in hours, for pipe B alone. The equation $\frac{1}{x-5}+\frac{1}{x}=\frac{1}{6}$ has solutions $x=2$ and $x=15$. How long does pipe A alone take?
    Answer: $10$ hours, because $x=15$ is the only valid pipe-B time and pipe A is 5 hours faster If $x=2$, then pipe A's time is $x-5=-3$ hours, which is impossible. Thus the valid pipe-B time is $x=15$, and pipe A alone takes $15-5=10$ hours. The negative value comes from using the invalid root, and 15 hours is pipe B's time, not pipe A's.
20. A student solves this equation and reports $t=11.6$. What is the correct interpretation?
    Answer: It takes about 11.6 years, because $t$ was defined in years; the exponent $12t$ counts months. The variable $t$ was defined as years, so the solution $t=11.6$ is already in years. The exponent $12t$ is the number of monthly compounding periods, not the number of years. The initial investment cancels when solving for doubling time.

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