Algebra 2 Quiz: Creating Solving One Variable Equations Inequalities

20 questions · exam conditions

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Creating Solving One Variable Equations InequalitiesQuestion 1 of 20

A medication dose starts at 200 mg. Each hour, 15% of the medication is eliminated (so 85% remains). Model the amount remaining after $t$ hours with an exponential equation and solve for when 50 mg remains.

Let $t$ = time in hours.

Model:

200(0.15)^t=50

. Then

t=\log_{0.15}(0.25)\approx 0.69

hours. It takes about 0.69 hours.

Model:

200(1.15)^t=50

. Then

t=\log_{1.15}(0.25)\approx -10.00

hours. It takes about -10 hours.

Model:

200(0.85)^t=50

. Then

t=\log_{0.85}(0.25)\approx 8.54

hours. It takes about 8.54 hours for 50 mg to remain.

Model:

200-0.15t=50

. Then

t=1000

hours. It takes 1000 hours.

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Algebra 2 Quiz

Algebra 2 Quiz: Creating Solving One Variable Equations Inequalities

Practice Creating Solving One Variable Equations Inequalities in Algebra 2 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 Creating Solving One Variable Equations Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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

Let $t$ = time in hours.

Model: $200(0.15)^t=50$ . Then $t=\log_{0.15}(0.25)\approx 0.69$ hours. It takes about 0.69 hours.
Model: $200(1.15)^t=50$ . Then $t=\log_{1.15}(0.25)\approx -10.00$ hours. It takes about -10 hours.
Model: $200(0.85)^t=50$ . Then $t=\log_{0.85}(0.25)\approx 8.54$ hours. It takes about 8.54 hours for 50 mg to remain. (correct answer)
Model: $200-0.15t=50$ . Then $t=1000$ hours. It takes 1000 hours.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential growth/decay problems arise when something grows or shrinks by a constant percent: 'eliminates 15% hourly' means 85% remains, so multiply by 0.85 each hour, giving formula amount = initial × (0.85)^t. To find when it reaches a specific value, set up equation like 200(0.85)^t = 50 and solve using logarithms: (0.85)^t = 50/200 = 0.25, so t = ln(0.25)/ln(0.85) ≈ 8.54 hours. The logarithm unlocks the exponent! Since 15% is eliminated each hour, 85% remains, so the medication amount after t hours is 200(0.85)^t mg, and we solve 200(0.85)^t = 50 by dividing both sides by 200 to get (0.85)^t = 0.25, then taking logarithms: t = log₀.₈₅(0.25) ≈ 8.54 hours. Choice C correctly models exponential decay as 200(0.85)^t (since 85% remains each hour) and solves using logarithms to find t ≈ 8.54 hours. Choice A incorrectly uses 0.15 as the base (which would mean only 15% remains each hour), B uses 1.15 which would represent 15% growth not decay, and D treats it as linear decay subtracting 0.15t instead of multiplying by (0.85)^t. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 2

A tank is being filled by two hoses. Hose A can fill the tank in 6 hours and Hose B can fill the same tank in 9 hours.

Write and solve an equation to find how long it takes to fill the tank if both hoses run together.

Let $t$ = the number of hours to fill the tank together.

Set up $6+9=t$ . Then $t=15$ hours. The tank fills in 15 hours.
Set up $\frac{1}{6}+\frac{1}{9}=t$ . Then $t=\frac{5}{18}$ hours. The tank fills in about $0.28$ hours.
Set up $\frac{1}{6}+\frac{1}{9}=\frac{1}{t}$ . Then $t=\frac{18}{5}$ hours. The tank fills in $3.6$ hours. (correct answer)
Set up $\frac{1}{6}-\frac{1}{9}=\frac{1}{t}$ . Then $t=18$ hours. The tank fills in 18 hours.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Work-rate problems lead to rational equations: if one worker completes a job in time t₁ and another in t₂, their combined rate is 1/t₁ + 1/t₂ (adding rates), which equals 1/t_combined. The reciprocals represent 'fraction of job per hour,' and adding these fractions gives the combined rate. Solving these rational equations requires finding LCD and often produces fractional time answers that make sense: 3.4 hours = 3 hours 24 minutes. For this tank-filling scenario, set up the equation as 1/6 + 1/9 = 1/t; finding a common denominator of 18 gives (3 + 2)/18 = 5/18, so 1/t = 5/18 and t = 18/5 = 3.6 hours, meaning together they fill the tank in 3 hours and 36 minutes. Choice A correctly sets up the equation by adding the rates and solves to find t = 18/5 hours, which is the accurate combined time. A common distractor like choice B forgets to take the reciprocal after adding the rates, leading to an unrealistically small time that doesn't make sense for combined effort. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 3

A community pool has two pumps that can fill it. Pump A can fill the pool in 6 hours, and Pump B can fill the pool in 8 hours.

Write and solve a rational equation to find how long it takes to fill the pool if both pumps run at the same time.

Let $t$ = the time (in hours) to fill the pool together.

Set up $6+8=t$ . Then $t=14$ hours, so it takes 14 hours together.
Set up $\frac{1}{6}+\frac{1}{8}=\frac{1}{t}$ . Then $t=\frac{14}{3}$ hours, so it takes about 4.67 hours together.
Set up $\frac{1}{6}+\frac{1}{8}=\frac{1}{t}$ . Then $t=\frac{24}{7}$ hours, so it takes about 3.43 hours together. (correct answer)
Set up $\frac{6}{t}+\frac{8}{t}=1$ . Then $t=14$ hours, so it takes 14 hours together.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Work-rate problems lead to rational equations: if one worker completes a job in time t₁ and another in t₂, their combined rate is 1/t₁ + 1/t₂ (adding rates), which equals 1/t_combined. The reciprocals represent 'fraction of job per hour,' and adding these fractions gives the combined rate. Solving these rational equations requires finding LCD and often produces fractional time answers that make sense: 3.4 hours = 3 hours 24 minutes. For this pool-filling scenario, set up the equation as 1/6 + 1/8 = 1/t; find a common denominator of 24 to get 4/24 + 3/24 = 7/24 = 1/t, so t = 24/7 ≈ 3.43 hours, meaning both pumps together fill the pool in about 3 hours 26 minutes. Choice B correctly sets up the equation by adding the rates and solves to find t = 24/7 hours, accurately determining the combined time. A common mistake, as in choice C, is adding the individual times instead of the rates, which overestimates the combined time since they work together faster. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 4

A ball is thrown upward from a platform. Its height (in feet) after $t$ seconds is modeled by $h(t)=-16t^2+48t+64$ . Write and solve an equation to find when the ball hits the ground.

Let $t$ = time in seconds (use the solution that makes sense in context).

Set $-16t^2+48t+64=0$ . Solutions: $t=4$ and $t=-1$ . The ball hits the ground at $t=4$ seconds. (correct answer)
Set $-16t^2+48t+64=0$ . Solutions: $t=1$ and $t=-4$ . The ball hits the ground at $t=1$ second.
Set $-16t^2+48t+64=0$ . Solutions: $t=4$ and $t=1$ . The ball hits the ground at $t=1$ second.
Set $-16t^2+48t+64=0$ . Solution: $t=2$ . The ball hits the ground at $t=2$ seconds.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Projectile motion problems lead to quadratic equations: the height function h(t) = -16t² + v₀t + h₀ represents vertical motion under gravity, where -16 is half the acceleration due to gravity (in ft/s²), v₀ is initial velocity, and h₀ is initial height. The ball hits the ground when h(t) = 0. Setting -16t² + 48t + 64 = 0 and dividing by -16: t² - 3t - 4 = 0. Factoring: (t-4)(t+1) = 0, giving t = 4 or t = -1. Since time cannot be negative in this context, t = 4 seconds. We can verify: h(4) = -16(16) + 48(4) + 64 = -256 + 192 + 64 = 0 ✓. Choice A correctly sets the height equation equal to zero (-16t² + 48t + 64 = 0) and solves to get t = 4 and t = -1, properly rejecting the negative time to conclude the ball hits the ground at t = 4 seconds. Choice B incorrectly states the solutions are t = 1 and t = -4, but substituting t = 1 gives h(1) = -16 + 48 + 64 = 96 feet, not 0. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 5

A number and its reciprocal have a sum of $\frac{10}{3}$ .

What equation represents this context? Solve it and interpret the solutions.

Let $x$ = the number (assume $x\ne 0$ ).

Set up $x+\frac{1}{x}=\frac{10}{3}$ . Then $3x^2-10x+3=0$ , so $x=3$ or $x=\frac{1}{3}$ . (correct answer)
Set up $x+\frac{1}{x}=\frac{10}{3}$ . Then $3x^2-10x+3=0$ , so $x=\frac{3}{10}$ only.
Set up $x-\frac{1}{x}=\frac{10}{3}$ . Then $3x^2-10x-3=0$ , so $x=3$ or $x=\frac{1}{3}$ .
Set up $\frac{1}{x}=\frac{10}{3}x$ . Then $x=\pm\sqrt{\frac{3}{10}}$ , so the number is $\pm\sqrt{\frac{3}{10}}$ .

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Reciprocal relationships often lead to rational equations that become quadratic: for x + 1/x = k, multiply by x to get x² - kx + 1 = 0, solved via quadratic formula, with solutions being reciprocals if valid. For this number, set up x + 1/x = 10/3, multiply by 3x to clear fractions: 3x² + 3 = 10x, rearrange to 3x² - 10x + 3 = 0, factor as (3x - 1)(x - 3) = 0, so x = 3 or x = 1/3, meaning the numbers are 3 and its reciprocal 1/3. Choice A correctly sets up the sum equation and solves to find both solutions x = 3 and x = 1/3. A mistake like in choice C changes addition to subtraction, altering the relationship and solutions. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 6

A community pool is being filled by two hoses. Hose A can fill the pool in 6 hours, and Hose B can fill the pool in 8 hours. If both hoses run at the same time, write and solve a rational equation to find how long it takes to fill the pool.

Let $t$ = the time (in hours) to fill the pool using both hoses together.

Set up $\frac{1}{6}+\frac{1}{8}=t$ . Then $t=\frac{7}{24}$ hour. The pool fills in $\frac{7}{24}$ hour.
Set up $\frac{1}{6}+\frac{1}{8}=\frac{1}{t}$ . Then $t=\frac{24}{7}$ hours $\approx 3.43$ hours. The pool fills in about 3.43 hours. (correct answer)
Set up $\frac{1}{6}-\frac{1}{8}=\frac{1}{t}$ . Then $t=24$ hours. The pool fills in 24 hours.
Set up $\frac{6+8}{2}=t$ . Then $t=7$ hours. The pool fills in 7 hours.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Work-rate problems lead to rational equations: if one worker completes a job in time t₁ and another in t₂, their combined rate is 1/t₁ + 1/t₂ (adding rates), which equals 1/t_combined. The reciprocals represent 'fraction of job per hour,' and adding these fractions gives the combined rate. Solving these rational equations requires finding LCD and often produces fractional time answers that make sense: 3.4 hours = 3 hours 24 minutes. Here, Hose A's rate is 1/6 pool per hour and Hose B's is 1/8, so combined 1/6 + 1/8 = (4+3)/24 = 7/24 = 1/t, thus t = 24/7 ≈ 3.43 hours, meaning the pool fills faster together as expected. Choice B correctly sets up the equation by adding rates and solves for t ≈ 3.43 hours, providing the accurate time to fill the pool. A common mistake, like in Choice A, is setting the sum of rates equal to t instead of 1/t, which gives an incorrect small fraction. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 7

A car rental company charges $45 per day plus a one-time fee of $30. You have at most $300 to spend. Create an inequality and find the maximum whole number of days you can rent the car.

Let $d$ = number of days.

Inequality: $45d+30\le 300$ . Maximum $d=6$ days. (correct answer)
Inequality: $45d+30\le 300$ . Maximum $d=7$ days.
Inequality: $45d-30\le 300$ . Maximum $d=7$ days.
Inequality: $45d+30\ge 300$ . Maximum $d=6$ days.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Budget constraint problems lead to linear inequalities: the total cost is one-time fee plus (daily rate × days), which must stay within budget, giving 30 + 45d ≤ 300. Solving: 45d ≤ 270, so d ≤ 6, meaning you can rent for at most 6 days. Since we need whole days, the maximum is exactly 6 days. Choice A correctly sets up the inequality as total cost ≤ budget (45d + 30 ≤ 300) and solves to get d ≤ 6, so maximum d = 6 days. Choice B has the correct inequality but incorrectly solves to get d = 7, C subtracts the one-time fee instead of adding it, and D uses ≥ instead of ≤ which would mean we want to spend at least $300. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 8

A rectangular patio has an area of $96\text{ m}^2$ . Its length is 4 meters more than its width.

Set up a quadratic equation and solve to find the patio's width and length.

Let $w$ = the width in meters.

Set up $w(w+4)=96$ . Then $w^2+4w-96=0$ , so $w=8$ (reject $-12$ ). The patio is $8\text{ m}$ by $12\text{ m}$ . (correct answer)
Set up $w(w-4)=96$ . Then $w^2-4w-96=0$ , so $w=12$ (reject $-8$ ). The patio is $12\text{ m}$ by $8\text{ m}$ .
Set up $w+(w+4)=96$ . Then $2w+4=96$ , so $w=46$ . The patio is $46\text{ m}$ by $50\text{ m}$ .
Set up $w(w+4)=96$ . Then $w^2+4w-96=0$ , so $w=12$ (reject $-8$ ). The patio is $12\text{ m}$ by $16\text{ m}$ .

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Area problems with related dimensions often lead to quadratic equations: if length is width plus a constant, set up area = width × (width + constant), resulting in w² + cw - area = 0, solved via factoring or quadratic formula, rejecting negative roots since dimensions are positive. For this patio, set up w(w + 4) = 96, expand to w² + 4w - 96 = 0, use quadratic formula w = [-4 ± √(16 + 384)]/2 = [-4 ± √400]/2 = [-4 ± 20]/2, so w = 8 or w = -12 (reject negative), meaning width 8 m and length 12 m. Choice A correctly sets up the quadratic with length as w + 4 and solves to find the dimensions 8 m by 12 m. A distractor like choice C mistakenly uses a linear equation by adding sides instead of multiplying for area, resulting in impossible dimensions. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 9

A streaming service has 5000 subscribers and grows by 12% each month.

Set up an exponential equation and solve for the number of months $t$ until it reaches 10,000 subscribers.

Let $t$ = time in months.

Set up $5000(1.12)^t=10000$ . Then $t=\log_{1.12}(2)\approx 6.1$ months, so it takes about 6 months. (correct answer)
Set up $5000(0.88)^t=10000$ . Then $t=\log_{0.88}(2)\approx 6.1$ months, so it takes about 6 months.
Set up $5000+0.12t=10000$ . Then $t=41666.7$ months, so it takes about 41,667 months.
Set up $5000(1.12)^t=10000$ . Then $t=\log_{12}(2)\approx 0.28$ months, so it takes about 0.28 months.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential growth/decay problems arise when something grows or shrinks by a constant percent: 'grows 8% yearly' means multiply by 1.08 each year, giving formula amount = initial × (1.08)^t. To find when it reaches a specific value, set up equation like 500(1.08)^t = 1000 and solve using logarithms: (1.08)^t = 2, so t = ln(2)/ln(1.08) ≈ 9 years. The logarithm unlocks the exponent! For this subscriber growth, set up 5000(1.12)^t = 10000, simplify to (1.12)^t = 2, then t = ln(2)/ln(1.12) ≈ 6.1 months, meaning it takes about 6 months to double subscribers. Choice A correctly sets up the exponential growth equation and solves using logarithms to find approximately 6.1 months. A common error, as in choice C, models it linearly, leading to an absurdly long time frame. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 10

A pump can fill a tank in 6 hours, and a second pump can fill the same tank in 8 hours. Write and solve an equation to find how long it takes to fill the tank if both pumps run together.

Let $t$ = time (in hours) to fill the tank together.

Set up $\frac{1}{6}+\frac{1}{8}=t$ . Then $t=\frac{7}{24}$ hour. The tank fills in $\frac{7}{24}$ hour.
Set up $\frac{1}{6}+\frac{1}{8}=\frac{1}{t}$ . Then $\frac{7}{24}=\frac{1}{t}$ so $t=\frac{24}{7}\approx 3.43$ hours. The tank fills in about $3.43$ hours. (correct answer)
Set up $\frac{1}{6}-\frac{1}{8}=\frac{1}{t}$ . Then $t=24$ hours. The tank fills in 24 hours.
Set up $\frac{6+8}{2}=t$ . Then $t=7$ hours. The tank fills in 7 hours.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Work-rate problems lead to rational equations: if one worker completes a job in time t₁ and another in t₂, their combined rate is 1/t₁ + 1/t₂ (adding rates), which equals 1/t_combined. The reciprocals represent 'fraction of job per hour,' and adding these fractions gives the combined rate. Solving these rational equations requires finding LCD and often produces fractional time answers that make sense: 3.4 hours = 3 hours 24 minutes. Here, pump 1 fills 1/6 of the tank per hour and pump 2 fills 1/8 per hour, so together they fill 1/6 + 1/8 = 4/24 + 3/24 = 7/24 of the tank per hour, meaning the equation is 1/6 + 1/8 = 1/t, which gives 7/24 = 1/t, so t = 24/7 ≈ 3.43 hours. Choice B correctly sets up the equation as rate₁ + rate₂ = combined rate (1/6 + 1/8 = 1/t) and solves to get t = 24/7 hours. Choice A incorrectly sets the sum of rates equal to time instead of 1/time, while C subtracts rates (which would apply if one pump emptied), and D averages the times rather than adding rates. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 11

A rectangular garden has an area of 150 m $^2$ . Its length is 5 meters more than its width. Write and solve an equation to find the garden's dimensions.

Let $w$ = width (in meters).

Set up $w(w-5)=150$ . Then $w=15$ so the garden is $15\text{ m by }10\text{ m}$ .
Set up $w(w+5)=150$ . Then $w=10$ (reject $w=-15$ ). Dimensions are $10$ m by $15$ m. (correct answer)
Set up $2w+5=150$ . Then $w=72.5$ . Dimensions are $72.5$ m by $77.5$ m.
Set up $w^2+5=150$ . Then $w=\sqrt{145}$ . Dimensions are $\sqrt{145}$ m by $\sqrt{145}+5$ m.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Area problems with rectangular dimensions lead to quadratic equations: if width is w and length is w+5, then area = width × length gives w(w+5) = 150, which expands to w² + 5w = 150 or w² + 5w - 150 = 0. Factoring gives (w+15)(w-10) = 0, so w = -15 or w = 10, but negative width is impossible, leaving w = 10 meters and length = 15 meters. Choice B correctly sets up the equation as width × length = area (w(w+5) = 150) and solves to get w = 10, rejecting the negative solution, giving dimensions 10m by 15m. Choice A incorrectly uses w-5 for length instead of w+5, C treats it as a perimeter problem using 2w+5 = 150, and D sets up w² + 5 = 150 which doesn't represent the area formula. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 12

A streaming service has 5000 subscribers and grows by 12% each month. Model the number of subscribers after $t$ months and solve for when it will reach 10,000 subscribers.

Let $t$ = number of months.

Model: $5000(1.12)^t=10000$ . Then $t=\log_{1.12}(2)\approx 6.12$ months. It will reach 10,000 subscribers in about 6.12 months. (correct answer)
Model: $5000(0.12)^t=10000$ . Then $t=\log_{0.12}(2)\approx -0.33$ months. It will reach 10,000 in about -0.33 months.
Model: $5000+1.12t=10000$ . Then $t\approx 4464.29$ months. It will reach 10,000 in about 4464 months.
Model: $5000(1.12)^t=10000$ . Then $t=\log_{1.12}(0.5)\approx 6.12$ months. It will reach 10,000 in about 6.12 months.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential growth/decay problems arise when something grows or shrinks by a constant percent: 'grows 12% monthly' means multiply by 1.12 each month, giving formula subscribers = initial × (1.12)^t. To find when it reaches a specific value, set up equation like 5000(1.12)^t = 10000 and solve using logarithms: (1.12)^t = 10000/5000 = 2, so t = ln(2)/ln(1.12) ≈ 6.12 months. The logarithm unlocks the exponent! The service starts with 5000 subscribers and grows by factor 1.12 each month (12% growth means multiply by 1 + 0.12 = 1.12), so after t months there are 5000(1.12)^t subscribers, and we solve 5000(1.12)^t = 10000 by dividing both sides by 5000 to get (1.12)^t = 2, then taking logarithms: t = log₁.₁₂(2) ≈ 6.12 months. Choice A correctly models exponential growth as 5000(1.12)^t and solves using logarithms to find t ≈ 6.12 months. Choice B incorrectly uses 0.12 as the base (which would mean the subscribers shrink to 12% each month), C treats it as linear growth adding 1.12t instead of multiplying by (1.12)^t, and D has the correct model but incorrectly calculates log₁.₁₂(0.5) instead of log₁.₁₂(2). Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 13

A print shop charges a one-time setup fee of $80 plus $12 per T-shirt. Create an inequality for the total cost to stay within a $500 budget, and find the maximum number of T-shirts that can be ordered.

Let $n$ = number of T-shirts.

Inequality: $12n+80\ge 500$ . Maximum $n=35$ shirts.
Inequality: $12n+80\le 500$ . Maximum $n=35$ shirts. (correct answer)
Inequality: $12n-80\le 500$ . Maximum $n=48$ shirts.
Inequality: $80n+12\le 500$ . Maximum $n=6$ shirts.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Budget constraint problems lead to linear inequalities: the total cost is setup fee plus (cost per item × quantity), which must stay within budget, giving 80 + 12n ≤ 500. Solving: 12n ≤ 420, so n ≤ 35, meaning you can order at most 35 T-shirts. Since we need whole shirts, the maximum is exactly 35. Choice B correctly sets up the inequality as total cost ≤ budget (12n + 80 ≤ 500) and solves to get n ≤ 35, so maximum n = 35 shirts. Choice A uses ≥ instead of ≤ which would mean we want to spend at least $500, C subtracts the setup fee instead of adding it, and D reverses the coefficients putting 80n + 12 which doesn't match the problem structure. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 14

A medication dose starts at 240 mg. Each hour, 20% of the medication is eliminated (so 80% remains). Model the amount remaining with an exponential equation and solve for how many hours it takes until 60 mg remains.

Let $t$ = time in hours.

Set up $240(0.8)^t=60$ . Then $t=\log_{0.8}(0.25)\approx 6.21$ . It takes about 6.21 hours. (correct answer)
Set up $240(1.2)^t=60$ . Then $t=\log_{1.2}(0.25)\approx 6.21$ . It takes about 6.21 hours.
Set up $240(0.2)^t=60$ . Then $t=\log_{0.2}(0.25)\approx 0.86$ . It takes about 0.86 hours.
Set up $240-0.2t=60$ . Then $t=900$ . It takes 900 hours.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential decay problems arise when something shrinks by a constant percent: 'loses 20% each hour' means 80% remains, so multiply by 0.8 each hour, giving formula amount = initial × (0.8)^t. To find when it reaches a specific value, set up equation like 240(0.8)^t = 60 and solve using logarithms: (0.8)^t = 0.25, so t = ln(0.25)/ln(0.8) ≈ 6.21 hours. The logarithm unlocks the exponent! Starting with 240 mg, if 20% is eliminated each hour, then 80% remains: after t hours we have 240(0.8)^t mg. Setting this equal to 60 mg: 240(0.8)^t = 60, so (0.8)^t = 60/240 = 0.25. Taking logarithms: t = log₀.₈(0.25) = ln(0.25)/ln(0.8) ≈ 6.21 hours. Choice A correctly models the exponential decay as 240(0.8)^t = 60 and solves using logarithms to get approximately 6.21 hours. Choice C incorrectly uses 0.2 as the base (240(0.2)^t = 60), which would mean only 20% remains each hour instead of 80% - this gives an unrealistic 0.86 hours for the medication to drop to 60 mg! Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 15

A medication dose starts at 180 mg and 20% is eliminated each hour. Model the amount remaining by an exponential equation and solve for how many hours until 60 mg remain.

Let $t$ = time in hours.

Set up $180(0.80)^t=60$ . Then $t=\log_{0.8}(1/3)\approx 4.92$ hours. About $4.92$ hours are needed. (correct answer)
Set up $180(1.20)^t=60$ . Then $t=\log_{1.2}(1/3)\approx 4.92$ hours. About $4.92$ hours are needed.
Set up $180(0.20)^t=60$ . Then $t=\log_{0.2}(1/3)\approx 0.68$ hours. About $0.68$ hours are needed.
Set up $180-0.20t=60$ . Then $t=600$ hours. About $600$ hours are needed.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential growth/decay problems arise when something grows or shrinks by a constant percent: 'eliminates 20% hourly' means 80% remains, so multiply by 0.80 each hour, giving formula amount = initial × (0.80)^t. To find when it reaches a specific value, set up equation like 180(0.80)^t = 60 and solve using logarithms: (0.80)^t = 1/3, so t = ln(1/3)/ln(0.80) ≈ 4.92 hours. The logarithm unlocks the exponent! Starting with 180 mg and losing 20% each hour means 80% remains, so the equation is 180(0.80)^t = 60. Dividing by 180: (0.80)^t = 60/180 = 1/3. Taking logarithms: t = log₀.₈₀(1/3) = ln(1/3)/ln(0.80) ≈ -1.099/-0.223 ≈ 4.92 hours. Choice A correctly models exponential decay with 180(0.80)^t = 60 and solves to get approximately 4.92 hours, which makes sense as the medication needs to drop to one-third of its initial dose. Choice C incorrectly uses 0.20 as the base (as if 20% remains instead of 80%), which would show much faster decay. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 16

A company charges a one-time setup fee of $75 plus $18 per shirt. Create an inequality for a budget of at most $500 and find the maximum number of shirts that can be ordered.

Let $n$ = the number of shirts (a whole number).

Inequality: $18n+75\ge 500$ . Solution: $n\ge 23.6$ , so maximum is 24 shirts.
Inequality: $18n+75\le 500$ . Solution: $n\le 23.6$ , so maximum is 23 shirts. (correct answer)
Inequality: $18n-75\le 500$ . Solution: $n\le 31.9$ , so maximum is 31 shirts.
Inequality: $18n+75\le 500$ . Solution: $n\le 25.0$ , so maximum is 25 shirts.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Budget constraint problems lead to linear inequalities: when costs have a fixed component plus a variable component, the total cost is fixed + (rate × quantity). Here, with a $75 setup fee and $18 per shirt, staying within a $500 budget means 75 + 18n ≤ 500. The inequality captures the constraint 'at most $500.' Setting up the inequality: total cost = setup fee + (cost per shirt × number of shirts) = 75 + 18n. The constraint 'at most $500' translates to 75 + 18n ≤ 500. Solving: 18n ≤ 425, so n ≤ 425/18 ≈ 23.61. Since n must be a whole number of shirts, the maximum is n = 23 shirts. Choice B correctly sets up the inequality 18n + 75 ≤ 500 and solves to get n ≤ 23.6, concluding that the maximum whole number of shirts is 23. Choice A incorrectly uses ≥ instead of ≤, which would mean spending at least $500 rather than at most $500. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 17

A contractor can paint a house in 12 hours. A second contractor can paint the same house in 18 hours. Write and solve an equation to find how long it takes them to paint the house working together.

Let $t$ = the time (in hours) for both contractors working together.

Set up $\frac{1}{12}+\frac{1}{18}=t$ . Then $t=\frac{5}{36}$ hours. It takes about $0.14$ hours.
Set up $\frac{1}{12}+\frac{1}{18}=\frac{1}{t}$ . Then $t=\frac{36}{5}=7.2$ hours. Working together, they finish in $7.2$ hours. (correct answer)
Set up $\frac{1}{12}+\frac{1}{18}=\frac{1}{t}$ . Then $t=\frac{30}{1}=30$ hours. Working together, they finish in $30$ hours.
Set up $\frac{1}{12}+\frac{1}{18}=\frac{1}{t}$ . Then $t=\frac{36}{7}\approx 5.14$ hours. Working together, they finish in about $5.14$ hours.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Work-rate problems lead to rational equations: if one worker completes a job in time t₁ and another in t₂, their combined rate is 1/t₁ + 1/t₂ (adding rates), which equals 1/t_combined. The reciprocals represent 'fraction of job per hour,' and adding these fractions gives the combined rate. Solving these rational equations requires finding LCD and often produces fractional time answers that make sense: 3.4 hours = 3 hours 24 minutes. The first contractor's rate is 1/12 (of the house per hour) and the second's is 1/18, so together: 1/12 + 1/18 = 1/t. Finding LCD of 36: 3/36 + 2/36 = 5/36 = 1/t, so t = 36/5 = 7.2 hours. Choice B correctly sets up the equation as rates adding to combined rate and solves to get t = 7.2 hours, which means they paint the house in 7 hours and 12 minutes working together. Choice A incorrectly sets up the equation as 1/12 + 1/18 = t (missing the reciprocal relationship), getting a nonsensical 0.14 hours. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 18

A streaming service charges a one-time setup fee of $25 plus $8 per month. Create an inequality for a budget of at most $105 and find the maximum number of months you can pay for.

Let $m$ = the number of months.

Set up $8m+25\le 105$ . Then $8m\le 80$ , so $m\le 10$ . Maximum is 10 months. (correct answer)
Set up $8m+25\ge 105$ . Then $m\ge 10$ . Maximum is 10 months.
Set up $8m-25\le 105$ . Then $8m\le 130$ , so $m\le 16.25$ . Maximum is 16 months.
Set up $\frac{8}{m}+25\le 105$ . Then $m\le 0.1$ . Maximum is 0 months.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Budget constraint problems typically lead to linear inequalities: total cost = fixed fee + (rate × quantity). Here, the total cost is $25 + $8m, and we need this to be at most $105, giving the inequality 25 + 8m ≤ 105. Solving: 8m ≤ 80, so m ≤ 10. Since m represents months (a discrete quantity), the maximum is 10 months. The total cost for m months is: setup fee + monthly charges = $25 + $8m. For a budget of at most $105: 25 + 8m ≤ 105. Subtracting 25: 8m ≤ 80. Dividing by 8: m ≤ 10. Since m must be a whole number of months, the maximum is 10 months. We can verify: 25 + 8(10) = 105 ✓. Choice A correctly sets up the inequality 8m + 25 ≤ 105 and solves to find m ≤ 10, giving a maximum of 10 months. Choice B incorrectly uses ≥ instead of ≤, which would mean we want to spend at least $105 rather than at most $105 - this reverses the problem's meaning! Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 19

A rectangular garden has a length that is 8 feet more than twice its width. If the area of the garden is 96 square feet, which equation could be used to find the width $w$ of the garden?

$2w^2 + 8w = 96$
$w(2w + 8) = 96$ (correct answer)
$2w + 8w = 96$
$w^2 + (2w + 8) = 96$

Explanation: The correct answer is B. If the width is

w

, then the length is

2w + 8

. The area formula gives us

w(2w + 8) = 96

. Choice A results from incorrectly distributing:

w(2w + 8) = 2w^2 + 8w

, but this doesn't equal the area. Choice C treats area as perimeter (adding length and width). Choice D incorrectly adds length and width instead of multiplying them for area.

Question 20

A rectangular parking lot is being designed where the length is 20 feet less than three times the width. If the perimeter must be at least 400 feet but no more than 600 feet, which inequality represents the constraint on the width $w$ ?

$400 \leq 4w - 20 \leq 600$
$400 \leq 6w - 40 \leq 600$
$400 \leq 8w - 40 \leq 600$ (correct answer)
$400 \leq 8w - 20 \leq 600$

Explanation: When you encounter word problems involving geometric constraints, start by translating the given relationships into algebraic expressions, then apply the constraint conditions. Here, you need to express the perimeter in terms of width

w

. The length is "20 feet less than three times the width," so

l = 3w - 20

. The perimeter of a rectangle is

P = 2l + 2w

. Substituting the expression for length:

P = 2(3w - 20) + 2w = 6w - 40 + 2w = 8w - 40

. Since the perimeter must be "at least 400 feet but no more than 600 feet," you get

400 \leq 8w - 40 \leq 600

, which is answer choice C. Let's examine why the other options are incorrect. Choice A gives

400 \leq 4w - 20 \leq 600

, which incorrectly uses

4w - 20

instead of

8w - 40

. This suggests the student forgot to double both the length and width when calculating perimeter. Choice B gives

400 \leq 6w - 40 \leq 600

, which only accounts for

6w - 40

– this comes from adding the length expression

3w - 20

2w

but forgetting to double the length. Choice D gives

400 \leq 8w - 20 \leq 600

, which correctly doubles the width (

4w

) and length coefficient (

6w

) but fails to double the constant term, giving

-20

instead of

-40

. Remember: when setting up perimeter problems, carefully track each step of substitution and ensure you're doubling both variable terms and constants when applying the perimeter formula

P = 2l + 2w