Math 3 Quiz: Solving And Explaining Steps
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Solving And Explaining StepsQuestion 1 of 20

A student is solving the system: 2x + 3y = 12 and x² + y² = 13 using substitution. They solve the first equation for x = (12 - 3y)/2 and substitute into the second. What should be their next steps to avoid common algebraic errors?

Substitute and expand carefully: ((12 - 3y)/2)² + y² = 13, so (144 - 72y + 9y²)/4 + y² = 13, then multiply through by 4
After substituting, multiply the entire equation by 4 first to clear fractions, then expand (12 - 3y)² systematically
Substitute and simplify: (12 - 3y)²/4 + y² = 13, so (144 - 72y + 9y²)/4 + 4y²/4 = 13, giving (144 - 72y + 13y²)/4 = 13
Substitute to get ((12 - 3y)/2)² + y² = 13, expand the numerator: (12 - 3y)² = 144 - 72y + 9y², then divide by 4
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Math 3 Quiz

Math 3 Quiz: Solving And Explaining Steps

Practice Solving And Explaining Steps in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Solving And Explaining Steps, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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Question 1

A student is solving the system: 2x + 3y = 12 and x² + y² = 13 using substitution. They solve the first equation for x = (12 - 3y)/2 and substitute into the second. What should be their next steps to avoid common algebraic errors?

  1. Substitute and expand carefully: ((12 - 3y)/2)² + y² = 13, so (144 - 72y + 9y²)/4 + y² = 13, then multiply through by 4
  2. After substituting, multiply the entire equation by 4 first to clear fractions, then expand (12 - 3y)² systematically
  3. Substitute and simplify: (12 - 3y)²/4 + y² = 13, so (144 - 72y + 9y²)/4 + 4y²/4 = 13, giving (144 - 72y + 13y²)/4 = 13 (correct answer)
  4. Substitute to get ((12 - 3y)/2)² + y² = 13, expand the numerator: (12 - 3y)² = 144 - 72y + 9y², then divide by 4
Explanation: When solving systems involving linear and quadratic equations, careful fraction handling is crucial to avoid algebraic mistakes that can derail your entire solution. After substituting x=12−3y2x = \frac{12 - 3y}{2} into x2+y2=13x^2 + y^2 = 13, you get (12−3y2)2+y2=13\left(\frac{12 - 3y}{2}\right)^2 + y^2 = 13. The key insight is recognizing that you need a common denominator before combining terms. Since the first term has denominator 4 when expanded, you should express y2y^2 as 4y24\frac{4y^2}{4} to match. This gives you 144−72y+9y24+4y24=13\frac{144 - 72y + 9y^2}{4} + \frac{4y^2}{4} = 13, which combines to 144−72y+13y24=13\frac{144 - 72y + 13y^2}{4} = 13. Option C follows this correct approach. Option A makes the error of not converting y2y^2 to the common denominator before adding, leading to the incorrect expression 144−72y+9y24+y2\frac{144 - 72y + 9y^2}{4} + y^2. Option B suggests clearing fractions first, but this approach is more prone to errors when dealing with squared expressions in the denominator. Option D incorrectly describes dividing the expanded numerator by 4, which isn't the proper way to handle the fraction. Study tip: When substituting expressions involving fractions into equations, always establish a common denominator before combining terms. This systematic approach prevents the mixing of fractional and whole number terms that often leads to algebraic errors in system problems.

Question 2

The concentration C of a drug in the bloodstream is modeled by C = 5te^(-0.4t), where t is time in hours and C is in mg/L.

To find when the concentration equals 3 mg/L, Ben sets up 5te^(-0.4t) = 3. This equation cannot be solved algebraically using elementary functions. What is the most appropriate method to explain the solution process?

  1. Use iterative approximation or graphing technology to find that t ≈ 0.73 hours and t ≈ 6.59 hours are the two solutions (correct answer)
  2. Apply numerical methods like Newton's method, starting with an initial guess based on the behavior of the function
  3. Rewrite as te^(-0.4t) = 0.6 and use the Lambert W function: t = -W(-0.24)/0.4 for the principal solution
  4. Use graphical analysis: plot y = 5te^(-0.4t) and y = 3, then identify intersection points as approximate solutions
Explanation: When you encounter transcendental equations like 5te−0.4t=35te^{-0.4t} = 3 that mix polynomial and exponential terms, they typically cannot be solved using standard algebraic methods. The question tests your understanding of appropriate solution strategies for such equations. The most appropriate method combines accessibility with completeness. Using iterative approximation or graphing technology provides a practical approach that reveals both solutions: t≈0.73t ≈ 0.73 hours and t≈6.59t ≈ 6.59 hours. This method is accessible to most students and clearly shows that the concentration reaches 3 mg/L at two different times—once as the drug concentration increases initially, and again as it decreases later. This makes biological sense for drug concentration models. Looking at the other options: Option B suggests Newton's method, which is mathematically sophisticated but requires calculus knowledge and doesn't emphasize finding all solutions. Option C mentions the Lambert W function, which is the exact analytical solution but involves advanced mathematics beyond typical coursework and only gives the principal solution initially. Option D proposes graphical analysis, which is helpful for visualization but less precise than iterative methods for determining actual numerical values. The key insight is that this concentration function has a maximum somewhere between the two solutions, creating the two intersection points with the horizontal line y=3y = 3. Remember: when facing transcendental equations in applied contexts, combining graphical visualization with numerical approximation methods gives you both conceptual understanding and practical solutions. Always consider whether multiple solutions might exist by thinking about the function's behavior.

Question 3

The temperature T (in °F) of a cooling object follows the equation T = 75 + 45e^(-0.2t), where t is time in minutes.

To find when the temperature reaches 90°F, Sarah sets up 75 + 45e^(-0.2t) = 90. Which solution method correctly handles the isolation and logarithmic steps?

  1. Subtract 75: 45e^(-0.2t) = 15, divide by 45: e^(-0.2t) = 1/3, take ln: -0.2t = ln(1/3), so t = -ln(1/3)/0.2 = ln(3)/0.2 ≈ 5.49 minutes
  2. Subtract 75: 45e^(-0.2t) = 15, take natural log: ln(45) - 0.2t = ln(15), so t = (ln(45) - ln(15))/0.2 ≈ 5.49 minutes
  3. Divide by 45 first: e^(-0.2t) = (90-75)/45 = 1/3, so -0.2t = ln(1/3) = -ln(3), giving t = ln(3)/0.2 ≈ 5.49 minutes
  4. Subtract 75: 45e^(-0.2t) = 15, divide by 45: e^(-0.2t) = 1/3, so -0.2t = -ln(3), and t = ln(3)/0.2 ≈ 5.49 minutes (correct answer)
Explanation: Choice D correctly isolates the exponential term first (45e^(-0.2t) = 15), then divides by 45 to get e^(-0.2t) = 1/3. Taking the natural logarithm: -0.2t = ln(1/3) = -ln(3), so t = ln(3)/0.2. Choice A shows correct steps but has an unnecessary complication. Choice B incorrectly takes the logarithm before isolating the exponential. Choice C has the right approach but shows ln(1/3) = -ln(3) without clear intermediate steps.

Question 4

The population P of bacteria in a culture is modeled by P = 1200/(1 + 5e^(-0.3t)), where t is time in hours.

To find when the population reaches 800, Jamie sets up 1200/(1 + 5e^(-0.3t)) = 800. Which sequence of algebraic steps correctly isolates the exponential term?

  1. Multiply both sides by (1 + 5e^(-0.3t)): 1200 = 800(1 + 5e^(-0.3t)), then 1200 = 800 + 4000e^(-0.3t), so 400 = 4000e^(-0.3t)
  2. Cross multiply: 1200 = 800(1 + 5e^(-0.3t)) = 800 + 4000e^(-0.3t), so 400 = 4000e^(-0.3t), giving e^(-0.3t) = 0.1 (correct answer)
  3. Divide both sides by 800: 1.5 = 1/(1 + 5e^(-0.3t)), so 1 + 5e^(-0.3t) = 1/1.5, giving 5e^(-0.3t) = 1/1.5 - 1
  4. Invert both sides: (1 + 5e^(-0.3t))/1200 = 1/800, so 1 + 5e^(-0.3t) = 1200/800 = 1.5, thus 5e^(-0.3t) = 0.5
Explanation: Choice B correctly shows the complete isolation process: cross multiply to get 1200 = 800 + 4000e^(-0.3t), subtract 800 to get 400 = 4000e^(-0.3t), then divide by 4000 to get e^(-0.3t) = 0.1. From here, taking ln gives -0.3t = ln(0.1), so t = -ln(0.1)/0.3. Choice A stops before the final division. Choice C uses an unnecessarily complex approach with reciprocals. Choice D incorrectly inverts the equation setup.

Question 5

The voltage across a capacitor is given by V = 12(1 - e^(-t/RC)), where R = 2000 Ω, C = 0.001 F, and t is time in seconds.

To find when the voltage reaches 8 volts, Lisa sets up 12(1 - e^(-t/2)) = 8. Which solution process correctly handles the exponential isolation and logarithmic conversion?

  1. After getting e^(-t/2) = 1/3, apply ln: -t/2 = ln(1/3), so t = -2ln(1/3) = 2ln(3), since ln(1/3) = -ln(3), giving t ≈ 2.20 seconds (correct answer)
  2. Divide by 12: 1 - e^(-t/2) = 2/3, so e^(-t/2) = 1 - 2/3 = 1/3, then -t/2 = ln(1/3), giving t = -2ln(1/3) ≈ 2.20 seconds
  3. Simplify: 1 - e^(-t/2) = 2/3, so e^(-t/2) = 1/3, take natural log: -t/2 = ln(1/3), multiply by -2: t = -2ln(1/3) = 2ln(3) ≈ 2.20 seconds
  4. Divide by 12: 1 - e^(-t/2) = 2/3, so e^(-t/2) = 1/3, take ln: -t/2 = ln(1/3) = -ln(3), so t = 2ln(3) ≈ 2.20 seconds
Explanation: When solving exponential equations like this capacitor voltage problem, you need to systematically isolate the exponential term, then apply logarithms to "undo" the exponential function. Starting with 12(1−e−t/2)=812(1 - e^{-t/2}) = 8, you first divide both sides by 12 to get 1−e−t/2=231 - e^{-t/2} = \frac{2}{3}. Then subtract 1 from both sides (or equivalently, subtract 23\frac{2}{3} from 1) to isolate the exponential: e−t/2=13e^{-t/2} = \frac{1}{3}. Now apply the natural logarithm to both sides: −t2=ln⁡(13)-\frac{t}{2} = \ln(\frac{1}{3}). Using the logarithm property that ln⁡(13)=−ln⁡(3)\ln(\frac{1}{3}) = -\ln(3), you get −t2=−ln⁡(3)-\frac{t}{2} = -\ln(3). Multiplying both sides by -2 gives t=2ln⁡(3)≈2.20t = 2\ln(3) \approx 2.20 seconds. Choice A correctly shows this complete process and explicitly demonstrates the key logarithm property that converts ln⁡(13)\ln(\frac{1}{3}) to −ln⁡(3)-\ln(3). Choice B stops at t=−2ln⁡(13)t = -2\ln(\frac{1}{3}) without showing the simplification to the cleaner form 2ln⁡(3)2\ln(3). Choice C has the right steps but doesn't show the logarithm property conversion. Choice D makes an error by writing −t2=ln⁡(13)=−ln⁡(3)-\frac{t}{2} = \ln(\frac{1}{3}) = -\ln(3) and then incorrectly jumping to t=2ln⁡(3)t = 2\ln(3) without showing the multiplication by -2. Remember: when solving exponential equations, always work methodically—isolate the exponential term first, then apply logarithms, and finally use logarithm properties to simplify your answer to its cleanest form.

Question 6

A water tank is being filled according to the equation h = 2√(t + 4) - 3, where h is the height of water in feet and t is time in minutes since filling began.

To find when the water height reaches 5 feet, Alex solves 2√(t + 4) - 3 = 5. Which step-by-step solution correctly identifies and addresses the key challenge in solving this radical equation?

  1. Add 3: 2√(t + 4) = 8, divide by 2: √(t + 4) = 4, square both sides: t + 4 = 16, so t = 12 minutes
  2. Add 3: 2√(t + 4) = 8, square both sides: 4(t + 4) = 64, so 4t + 16 = 64, giving t = 12 minutes
  3. Square both sides first: 4(t + 4) - 9 = 25, so 4t + 16 = 34, giving t = 4.5 minutes
  4. Isolate the radical term first: 2√(t + 4) = 8, then √(t + 4) = 4, square: t + 4 = 16, so t = 12, then verify solution (correct answer)
Explanation: Choice D correctly identifies that the key challenge is properly isolating the radical before squaring, and includes the crucial verification step. The solution gives t = 12, and checking: h = 2√(12 + 4) - 3 = 2√16 - 3 = 8 - 3 = 5 ✓. Choice A follows correct steps but omits verification. Choice B squares incorrectly without first dividing by 2. Choice C squares both sides prematurely, leading to a more complex equation.

Question 7

When solving the equation x⁴ - 13x² + 36 = 0 using substitution, a student lets u = x². After solving 4u² - 13u + 36 = 0, they get u = 4 and u = 9. What error did the student make, and what is the correct approach?

  1. The substitution equation should be u² - 13u + 36 = 0, giving u = 4 and u = 9, so x = ±2 and x = ±3 (correct answer)
  2. The student incorrectly wrote 4u² instead of u²; the correct equation is u² - 13u + 36 = 0, yielding the same u-values
  3. After finding u = 4 and u = 9, they forgot to substitute back: x² = 4 gives x = ±2, and x² = 9 gives x = ±3
  4. The error was in the quadratic formula application; solving u² - 13u + 36 = 0 correctly gives u = 4 and u = 9
Explanation: Choice A correctly identifies that if u = x², then x⁴ = u², so the equation becomes u² - 13u + 36 = 0, not 4u² - 13u + 36 = 0. Solving u² - 13u + 36 = 0 using factoring or the quadratic formula gives u = 4 and u = 9. Then x² = 4 yields x = ±2, and x² = 9 yields x = ±3. Choice B identifies the error but doesn't complete the solution. Choice C assumes the wrong quadratic was solved correctly. Choice D doesn't address the coefficient error.

Question 8

A projectile's height h (in meters) is given by h = -4.9t² + 24.5t + 2, where t is time in seconds.

To find when the projectile hits the ground (h = 0), Marcus uses the quadratic formula on -4.9t² + 24.5t + 2 = 0. Which calculation correctly applies the formula and interprets the results?

  1. t = (-24.5 ± √(24.5² - 4(-4.9)(2)))/2(-4.9) = (-24.5 ± √639.25)/(-9.8); only the positive solution t ≈ 5.08 seconds is physically meaningful
  2. t = (-24.5 ± √(24.5² + 4(4.9)(2)))/2(-4.9) = (-24.5 ± √639.25)/(-9.8); both solutions are valid since time can be negative in physics
  3. t = (-24.5 ± √(24.5² - 4(-4.9)(2)))/2(-4.9); this gives t ≈ -0.08 and t ≈ 5.08; both represent when height equals zero (correct answer)
  4. t = (24.5 ± √(24.5² - 4(-4.9)(2)))/2(-4.9) = (24.5 ± √639.25)/(-9.8); the positive discriminant confirms two real solutions exist
Explanation: When you encounter projectile motion problems, you're working with quadratic equations where the solutions represent specific moments in time. The key is correctly applying the quadratic formula and understanding what each solution means physically. For the equation −4.9t2+24.5t+2=0-4.9t^2 + 24.5t + 2 = 0, the quadratic formula gives us t=−b±b2−4ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=−4.9a = -4.9, b=24.5b = 24.5, and c=2c = 2. Substituting these values: t=−24.5±(24.5)2−4(−4.9)(2)2(−4.9)=−24.5±639.25−9.8t = \frac{-24.5 \pm \sqrt{(24.5)^2 - 4(-4.9)(2)}}{2(-4.9)} = \frac{-24.5 \pm \sqrt{639.25}}{-9.8} This calculation yields two solutions: t≈−0.08t \approx -0.08 seconds and t≈5.08t \approx 5.08 seconds. Both are mathematically valid solutions where the height equals zero, making choice C correct. Choice A incorrectly dismisses the negative solution as meaningless, but negative time can represent a valid mathematical solution (perhaps when the projectile would have been at ground level if we extended the motion backward). Choice B contains a sign error in the discriminant calculation, writing +4(4.9)(2)+4(4.9)(2) instead of −4(−4.9)(2)-4(-4.9)(2). Choice D has the wrong sign on the bb term in the numerator, writing +24.5+24.5 instead of −24.5-24.5. Remember: when applying the quadratic formula, double-check each term's sign carefully. The discriminant should be b2−4acb^2 - 4ac, and both solutions represent valid mathematical answers to when h=0h = 0, even if only one may be physically meaningful in context.

Question 9

A student is solving the rational equation (x + 2)/(x - 3) - 1/(x + 1) = 2 for a circuit analysis problem. After finding a common denominator and solving, they get x = 1 and x = -5. What should be the next step and why?

  1. Accept both solutions since they satisfy the algebraic manipulations and represent valid circuit configurations
  2. Check both solutions in the original equation to verify they don't make any denominator zero (correct answer)
  3. Reject x = -5 because it's negative, and circuit problems typically require positive values
  4. Use the quadratic formula to verify the solutions were found correctly before proceeding
Explanation: Choice B is correct because rational equations can introduce extraneous solutions during the solving process, especially when multiplying by expressions containing variables. Both x = 1 and x = -5 must be checked: neither makes x - 3 = 0 or x + 1 = 0, so both are mathematically valid. Choice A skips the essential verification step. Choice C incorrectly assumes negative values are automatically invalid. Choice D is unnecessary since the question states the solutions were already found algebraically.

Question 10

A company's profit P (in thousands of dollars) is modeled by the equation P = -2t² + 16t - 24, where t represents the number of years since the company started operations.

To find when the company first breaks even (profit equals zero), which sequence of steps correctly solves the equation -2t² + 16t - 24 = 0?

  1. Factor out -2: -2(t² - 8t + 12) = 0, then factor the quadratic: -2(t - 2)(t - 6) = 0, giving t = 2 or t = 6 (correct answer)
  2. Use the quadratic formula: t = (-16 ± √(256 + 192))/(-4), then t = (-16 ± √448)/(-4), giving t = 2 or t = 6
  3. Complete the square: -2(t² - 8t) = 24, then -2(t - 4)² + 32 = 24, so (t - 4)² = 4, giving t = 2 or t = 6
  4. Factor out -2: -2(t² - 8t + 12) = 0, then use quadratic formula on t² - 8t + 12 = 0, giving t = 4 ± 2
Explanation: Choice A correctly factors out the GCF of -2, then factors the resulting quadratic t² - 8t + 12 as (t - 2)(t - 6). Setting each factor equal to zero gives t = 2 and t = 6. Choice B has an error in the discriminant calculation (should be 256 - 192, not 256 + 192). Choice C makes an error in completing the square by not properly handling the constant term. Choice D correctly factors out -2 but then incorrectly applies the quadratic formula.

Question 11

A ball is thrown upward from a building. Its height above ground is modeled by h = -5t² + 20t + 60, where h is height in meters and t is time in seconds. To find when the ball hits the ground, which step-by-step solution is correct?

  1. Set h = 0: -5t² + 20t + 60 = 0, divide by -5: t² - 4t - 12 = 0, factor: (t - 6)(t + 2) = 0, so t = 6 seconds (reject t = -2) (correct answer)
  2. Set h = 60: -5t² + 20t + 60 = 60, simplify: -5t² + 20t = 0, factor: -5t(t - 4) = 0, so t = 0 or t = 4 seconds
  3. Set h = 0: -5t² + 20t + 60 = 0, use quadratic formula: t = (-20 ± √(400 + 1200))/(-10), so t = (20 ± 40)/10, giving t = 6 or t = -2, so t = 6
  4. Set h = 0: -5t² + 20t + 60 = 0, factor out -5: -5(t² - 4t - 12) = 0, so t² - 4t - 12 = 0, giving t = 2 ± √16 from completing the square
Explanation: Choice A correctly sets height equal to 0 for ground level, divides the entire equation by -5 to simplify, factors the resulting quadratic correctly as (t - 6)(t + 2) = 0, and properly rejects the negative time solution. Choice B incorrectly sets height equal to 60 (the initial height) instead of 0. Choice C uses the quadratic formula correctly but makes a sign error in the denominator. Choice D starts correctly but doesn't complete the solution process clearly.

Question 12

The number of bacteria in a culture is modeled by N = 500e^(0.3t), where t is time in hours. Scientists want to find when the population reaches 2000 bacteria.

Which sequence of steps correctly solves for the time when N = 2000?

  1. Set 500e^(0.3t) = 2000, divide by 500 to get e^(0.3t) = 4, take natural log: ln(e^(0.3t)) = ln(4), so 0.3t·ln(e) = ln(4), giving t = ln(4)/0.3
  2. Set 500e^(0.3t) = 2000, divide by 500 to get e^(0.3t) = 4, take log base 10: 0.3t = log(4), so t = log(4)/0.3 ≈ 2.01 hours
  3. Set 500e^(0.3t) = 2000, take natural log of both sides: ln(500) + 0.3t = ln(2000), so 0.3t = ln(2000) - ln(500), giving t = ln(4)/0.3
  4. Set 500e^(0.3t) = 2000, divide by 500 to get e^(0.3t) = 4, take natural log: 0.3t = ln(4), so t = ln(4)/0.3 ≈ 4.62 hours (correct answer)
Explanation: When you encounter exponential growth problems, you're solving equations where the variable appears in an exponent. The key is systematically isolating that variable using logarithms. To find when the bacteria population reaches 2000, you start by setting up the equation: 500e0.3t=2000500e^{0.3t} = 2000. The most efficient approach is to first isolate the exponential term by dividing both sides by 500, giving you e0.3t=4e^{0.3t} = 4. Then apply the natural logarithm to both sides. Since ln⁡(e0.3t)=0.3t\ln(e^{0.3t}) = 0.3t (the natural log and exponential function are inverses), you get 0.3t=ln⁡(4)0.3t = \ln(4). Finally, solve for t: t=ln⁡(4)0.3≈4.62t = \frac{\ln(4)}{0.3} \approx 4.62 hours. Option A makes a critical error by writing 0.3t⋅ln⁡(e)=ln⁡(4)0.3t \cdot \ln(e) = \ln(4). While ln⁡(e)=1\ln(e) = 1, this step is unnecessary and shows a misunderstanding of how logarithms work with exponents. Option B incorrectly uses log base 10 instead of the natural logarithm. When dealing with equations involving ee, you must use ln⁡\ln, not log⁡10\log_{10}. This leads to a wrong numerical answer. Option C takes the natural log of both sides before isolating the exponential term. While this eventually works out to the same answer, it's unnecessarily complicated and more prone to errors. Remember: when solving exponential equations with base ee, isolate the exponential term first, then apply the natural logarithm. This direct approach minimizes calculation errors and is faster on timed exams.

Question 13

The concentration C of a drug in the bloodstream (in mg/L) follows C = 12e^(-0.2t), where t is hours after injection.

To determine when the concentration drops to 3 mg/L, which solution method correctly explains each step?

  1. Set 12e^(-0.2t) = 3, take natural log: ln(12) - 0.2t = ln(3), solve: -0.2t = ln(3) - ln(12), so t = (ln(12) - ln(3))/0.2 = ln(4)/0.2
  2. Set 12e^(-0.2t) = 3, divide by 12: e^(-0.2t) = 0.25, take natural log: -0.2t = ln(0.25), so t = -ln(0.25)/0.2 ≈ 6.93 hours
  3. Set 12e^(-0.2t) = 3, divide by 12: e^(-0.2t) = 1/4, take natural log: -0.2t = ln(1/4) = -ln(4), so t = ln(4)/0.2 ≈ 6.93 hours (correct answer)
  4. Set 12e^(-0.2t) = 3, divide by 12: e^(-0.2t) = 1/4, take natural log: -0.2t = ln(1/4), so t = ln(1/4)/(-0.2) = ln(4)/0.2 ≈ 6.93 hours
Explanation: When you encounter exponential decay problems like drug concentration, you're solving equations where the variable appears in an exponent. The key is systematically isolating the exponential term before applying logarithms. To find when the concentration drops to 3 mg/L, you start with 12e−0.2t=312e^{-0.2t} = 3. First, isolate the exponential by dividing both sides by 12: e−0.2t=14e^{-0.2t} = \frac{1}{4}. Next, take the natural logarithm of both sides: −0.2t=ln⁡(14)-0.2t = \ln(\frac{1}{4}). Using the logarithm property ln⁡(1a)=−ln⁡(a)\ln(\frac{1}{a}) = -\ln(a), this becomes −0.2t=−ln⁡(4)-0.2t = -\ln(4). Finally, solve for t: t=−ln⁡(4)−0.2=ln⁡(4)0.2≈6.93t = \frac{-\ln(4)}{-0.2} = \frac{\ln(4)}{0.2} \approx 6.93 hours. This matches answer choice C exactly. Answer A makes an error by taking the natural logarithm before isolating the exponential term, incorrectly applying ln⁡(12e−0.2t)=ln⁡(12)−0.2t\ln(12e^{-0.2t}) = \ln(12) - 0.2t. Answer B correctly reaches e−0.2t=0.25e^{-0.2t} = 0.25 but writes this as a decimal instead of the fraction 14\frac{1}{4}, making the subsequent logarithm work less clear. Answer D makes a sign error when dividing by -0.2, incorrectly treating the operation as ln⁡(14)−0.2\frac{\ln(\frac{1}{4})}{-0.2} instead of recognizing that ln⁡(14)=−ln⁡(4)\ln(\frac{1}{4}) = -\ln(4). Always isolate the exponential term first in these problems, then apply natural logarithms. This systematic approach prevents the algebraic errors that create most wrong answer choices.

Question 14

A company's revenue R (in thousands) and cost C (in thousands) are given by R = 50x - x² and C = 20x + 100, where x is units produced (in hundreds). To find the break-even points where revenue equals cost, which solution process is correct?

  1. Set 50x - x² = 20x + 100, rearrange: -x² + 30x - 100 = 0, multiply by -1: x² - 30x + 100 = 0, factor: (x - 10)(x - 20) = 0, so x = 10 or x = 20 (correct answer)
  2. Set 50x - x² = 20x + 100, rearrange: -x² + 30x - 100 = 0, use quadratic formula: x = (30 ± √(900 - 400))/2, so x = (30 ± √500)/2 = 15 ± 5√5
  3. Set 50x - x² = 20x + 100, rearrange: x² - 30x + 100 = 0, factor: (x - 10)(x - 20) = 0, giving break-even at 1000 and 2000 units produced
  4. Set R = C: 50x - x² = 20x + 100, simplify: 30x - x² = 100, rearrange: x² - 30x + 100 = 0, factor as (x - 10)(x - 20) = 0, so x = 10 or x = 20
Explanation: Choice A correctly sets revenue equal to cost, rearranges to standard form by moving all terms to one side, multiplies by -1 to make the leading coefficient positive, and factors correctly to find x = 10 or x = 20 (representing 1000 or 2000 units). Choice B makes an error in the discriminant calculation (should be 900 - 400 = 500, but then incorrectly calculates the roots). Choice C has correct algebra but fails to show the important step of multiplying by -1. Choice D shows the work clearly but doesn't explicitly show the multiplication by -1 step that makes factoring easier.

Question 15

The temperature T (in °F) of a cooling object follows Newton's law of cooling: T = 75 + 125e^(-0.1t), where t is time in minutes. To find when the temperature reaches 100°F, which solution approach correctly shows all steps?

  1. Set T = 100: 75 + 125e^(-0.1t) = 100, subtract 75: 125e^(-0.1t) = 25, divide by 125: e^(-0.1t) = 1/5, take ln: -0.1t = ln(0.2), so t = -10ln(0.2)
  2. Set T = 100: 75 + 125e^(-0.1t) = 100, subtract 75: 125e^(-0.1t) = 25, divide by 125: e^(-0.1t) = 0.2, take ln: -0.1t = ln(0.2), so t = ln(5)/0.1
  3. Set 75 + 125e^(-0.1t) = 100, isolate exponential: 125e^(-0.1t) = 25, so e^(-0.1t) = 1/5, take natural log: -0.1t = ln(1/5) = -ln(5), therefore t = 10ln(5)
  4. Set T = 100: 75 + 125e^(-0.1t) = 100, rearrange: 125e^(-0.1t) = 25, divide: e^(-0.1t) = 0.2, take ln: -0.1t = ln(0.2), solve: t = -ln(0.2)/0.1 = ln(5)/0.1 (correct answer)
Explanation: Choice D correctly isolates the exponential term by subtracting 75 and dividing by 125, takes the natural logarithm properly, and correctly handles the negative signs to get t = ln(5)/0.1 ≈ 16.1 minutes. Choice A makes an error in the final step by not properly handling the negative sign. Choice B has the wrong final expression, showing ln(5)/0.1 but deriving it from ln(0.2). Choice C correctly identifies that ln(1/5) = -ln(5) but then makes an error in solving for t, getting 10ln(5) instead of ln(5)/0.1.

Question 16

The height h (in feet) of a projectile launched upward is given by h = -16t² + 64t + 80, where t is time in seconds after launch.

To find when the projectile returns to its initial height of 80 feet, which solution process correctly explains each step?

  1. Set -16t² + 64t + 80 = 80, simplify to -16t² + 64t = 0, factor out -16t to get -16t(t - 4) = 0, so t = 0 or t = 4 seconds (correct answer)
  2. Set -16t² + 64t + 80 = 0, factor out -16 to get -16(t² - 4t - 5) = 0, factor the quadratic as (t - 5)(t + 1) = 0, so t = 5 seconds
  3. Set -16t² + 64t + 80 = 80, simplify to -16t² + 64t = 0, divide by -16 to get t² - 4t = 0, so t(t - 4) = 0, giving t = 0 or t = 4
  4. Set -16t² + 64t = -80, add 80 to both sides to get -16t² + 64t + 80 = 0, use quadratic formula to get t = 0 or t = 4 seconds
Explanation: Choice A correctly sets the height equal to 80 feet, simplifies by subtracting 80 from both sides, factors out the GCF of -16t, and identifies both solutions t = 0 (launch time) and t = 4 (return time). Choice B incorrectly sets the height equal to 0 instead of 80. Choice C has the correct setup and algebra but doesn't show the factoring as clearly. Choice D starts with an incorrect equation setup by moving 80 to the wrong side initially.

Question 17

Water is draining from a cylindrical tank. The volume V (in liters) remaining after t minutes is modeled by V = 500 - 25t - 0.5t².

To find when the tank is completely empty, which step-by-step solution correctly shows the process?

  1. Set V = 0: 500 - 25t - 0.5t² = 0, multiply by -2: t² + 50t - 1000 = 0, use quadratic formula: t = (-50 ± √(2500 + 4000))/2, so t = (-50 ± √6500)/2
  2. Set V = 0: 500 - 25t - 0.5t² = 0, multiply by -2: t² + 50t - 1000 = 0, factor: (t + 70)(t - 20) = 0, so t = 20 minutes (reject t = -70) (correct answer)
  3. Set V = 0: -0.5t² - 25t + 500 = 0, divide by -0.5: t² + 50t - 1000 = 0, complete the square: (t + 25)² = 1625, so t = -25 ± √1625
  4. Set V = 0: 500 - 25t - 0.5t² = 0, rearrange: 0.5t² + 25t - 500 = 0, multiply by 2: t² + 50t - 1000 = 0, giving t = 20 or t = -50 minutes
Explanation: Choice B correctly sets V = 0, multiplies by -2 to eliminate decimals and make the leading coefficient positive, then factors the quadratic correctly as (t + 70)(t - 20) = 0, giving t = 20 minutes as the physical solution. Choice A sets up correctly but doesn't complete the calculation or factoring. Choice C rearranges unnecessarily and uses completing the square when factoring is simpler. Choice D makes an error in the factorization, getting (t - 20)(t + 50) instead of (t - 20)(t + 70).

Question 18

When solving the absolute value equation |2x - 5| = |x + 3| for a distance optimization problem, what is the most systematic approach to ensure all solutions are found?

  1. Graph both expressions y = |2x - 5| and y = |x + 3| and find intersection points visually
  2. Square both sides to eliminate absolute values: (2x - 5)² = (x + 3)², then solve the resulting quadratic
  3. Consider four cases based on the signs of (2x - 5) and (x + 3): both positive, both negative, and mixed signs
  4. Set up two equations: 2x - 5 = x + 3 and 2x - 5 = -(x + 3), then solve each separately (correct answer)
Explanation: Choice D is the most systematic algebraic approach. When |A| = |B|, either A = B or A = -B. This gives: 2x - 5 = x + 3 (so x = 8) and 2x - 5 = -(x + 3) (so 3x = 2, x = 2/3). Both solutions check in the original equation. Choice B works but is more complex. Choice C is theoretically complete but unnecessarily complicated. Choice A is graphical rather than algebraic and may lack precision.

Question 19

A student solving log₃(x + 6) - log₃(x - 2) = 2 makes the following steps: log₃((x + 6)/(x - 2)) = 2, then (x + 6)/(x - 2) = 9. What should be the next steps to complete the solution correctly?

  1. Cross multiply: x + 6 = 9(x - 2) = 9x - 18, so 8x = 24, giving x = 3; check: log₃(9) - log₃(1) = 2 ✓
  2. Cross multiply: x + 6 = 9x - 18, so 24 = 8x, giving x = 3; verify x > 2 to ensure log₃(x - 2) is defined
  3. Solve: x + 6 = 9x - 18, so x = 3; check both x + 6 > 0 and x - 2 > 0, then verify in original equation (correct answer)
  4. Cross multiply to get x + 6 = 9x - 18, solve for x = 3, then substitute back to confirm both logarithms exist and equal 2
Explanation: Choice C provides the most complete solution process. After getting x = 3, we must verify the domain restrictions: x + 6 > 0 (satisfied since 9 > 0) and x - 2 > 0 (satisfied since 1 > 0), then check the original equation: log₃(9) - log₃(1) = 2 - 0 = 2 ✓. Choice A checks the equation but not domain restrictions. Choice B checks only one domain condition. Choice D is incomplete about what specifically to verify.

Question 20

When solving the exponential equation 3^(2x-1) = 27^(x-2) for a population growth model, what is the most appropriate first step and why?

  1. Take the natural logarithm of both sides because it directly linearizes exponential equations with any base
  2. Rewrite 27 as 3³ to create a common base, since 27^(x-2) = (3³)^(x-2) = 3^(3x-6) (correct answer)
  3. Convert to logarithmic form as (2x-1)log(3) = (x-2)log(27) to isolate the variable terms
  4. Substitute u = 3^x to transform the equation into polynomial form for easier algebraic manipulation
Explanation: Choice B is correct because recognizing that 27 = 3³ allows us to rewrite the equation as 3^(2x-1) = 3^(3x-6), making it possible to equate exponents directly: 2x-1 = 3x-6, which gives x = 5. Choice A would work but is unnecessarily complex. Choice C is algebraically valid but more complicated than needed. Choice D is inappropriate since this equation doesn't fit the substitution pattern for polynomial transformation.