Math 1 Quiz: Defining Variables And Writing Equations
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Defining Variables And Writing EquationsQuestion 1 of 20

A bacteria culture doubles every 3 hours, but due to a treatment applied, it loses 200 bacteria every hour regardless of population size. The culture starts with 800 bacteria. After 6 hours, there are 2,000 bacteria remaining. If PP represents the population after tt hours, which linear approximation best models the population change?

P=300t+200P = 300t + 200
P=400t+800P = 400t + 800
P=200t+2000P = 200t + 2000
P=200t+800P = 200t + 800
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Math 1 Quiz

Math 1 Quiz: Defining Variables And Writing Equations

Practice Defining Variables And Writing Equations in Math 1 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 Defining Variables And Writing Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

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Question 1

A bacteria culture doubles every 3 hours, but due to a treatment applied, it loses 200 bacteria every hour regardless of population size. The culture starts with 800 bacteria. After 6 hours, there are 2,000 bacteria remaining. If PP represents the population after tt hours, which linear approximation best models the population change?

  1. P=300t+200P = 300t + 200
  2. P=400t+800P = 400t + 800
  3. P=200t+2000P = 200t + 2000
  4. P=200t+800P = 200t + 800 (correct answer)
Explanation: When you encounter population problems with multiple factors affecting growth, the key is identifying whether you need an exact exponential model or can use a linear approximation to simplify the analysis. This problem involves exponential growth (doubling every 3 hours) combined with constant loss (200 bacteria per hour). While the actual population follows a complex exponential pattern, you're asked to find the best linear approximation using the given data points. You have two clear data points: at t=0t = 0, P=800P = 800, and at t=6t = 6, P=2000P = 2000. Using these points to find the linear relationship, the slope is 2000−8006−0=12006=200\frac{2000 - 800}{6 - 0} = \frac{1200}{6} = 200 bacteria per hour. With slope 200 and y-intercept 800 (the initial population), the linear model is P=200t+800P = 200t + 800. Looking at the wrong answers: Choice A (P=300t+200P = 300t + 200) uses an incorrect slope and doesn't match the initial population of 800. Choice B (P=400t+800P = 400t + 800) has the right y-intercept but doubles the correct slope, perhaps from confusing the doubling rate with the linear rate. Choice C (P=200t+2000P = 200t + 2000) has the correct slope but uses the final population as the y-intercept instead of the initial population. Study tip: In linear approximation problems, always use the actual data points given rather than trying to incorporate the complex underlying processes. Focus on slope calculation between known points and identifying the correct y-intercept from initial conditions.

Question 2

A factory's daily production cost consists of a fixed daily cost plus a variable cost per unit produced. On Monday, the factory produced 50 units at a total cost of $1,200. On Tuesday, it produced 75 units at a total cost of $1,500. If the factory needs to budget for producing 100 units in one day, and $uu representsunitsproducedwithtotalcostrepresents units produced with total cost CC $, which equation should be used?

  1. C=12u+600C = 12u + 600 (correct answer)
  2. C=24u+24C = 24u + 24
  3. C=20u+700C = 20u + 700
  4. C=15u+450C = 15u + 450
Explanation: Let fixed cost be FF and variable cost per unit be vv. From the data: 50v+F=120050v + F = 1200 and 75v+F=150075v + F = 1500. Subtracting: 25v=30025v = 300, so v=12v = 12. Then F=1200−50(12)=600F = 1200 - 50(12) = 600. Therefore C=12u+600C = 12u + 600. Choice B incorrectly calculates v=120050=24v = \frac{1200}{50} = 24 and F=1500−75(24)F = 1500 - 75(24). Choice C uses v=1500−120075−50=12v = \frac{1500-1200}{75-50} = 12 but wrong fixed cost. Choice D uses average values incorrectly.

Question 3

A water cooler dispenses water at a steady rate. At 2:00 PM, it contained 18 gallons. At 5:00 PM, it contained 12 gallons. The cooler is refilled to its maximum capacity of 24 gallons each morning before use. If hh represents hours after midnight and WW represents gallons remaining, which equation models the water level during the day?

  1. W=−2h+24W = -2h + 24
  2. W=−2h+28W = -2h + 28 (correct answer)
  3. W=−3h+24W = -3h + 24
  4. W=−2h+18W = -2h + 18
Explanation: When you encounter a linear modeling problem like this, you're looking for a relationship in the form W=mh+bW = mh + b, where mm is the rate of change and bb is the y-intercept. First, find the rate of water consumption. From 2:00 PM to 5:00 PM (3 hours), the water decreased from 18 to 12 gallons. That's a change of −6-6 gallons over 33 hours, giving us a rate of −63=−2\frac{-6}{3} = -2 gallons per hour. So our slope is −2-2. Now you need the y-intercept, which represents the water level at midnight (h=0h = 0). Since the cooler starts each day at 24 gallons and loses 2 gallons per hour, work backwards from a known point. At 2:00 PM (h=14h = 14), there were 18 gallons. Using W=−2h+bW = -2h + b: 18=−2(14)+b18 = -2(14) + b, which gives us 18=−28+b18 = -28 + b, so b=46b = 46. Wait—that's not among our choices, so let's reconsider. The key insight is that the cooler starts with 24 gallons in the morning, not at midnight. If we assume it's filled at, say, 6:00 AM (h=6h = 6), then at that point: 24=−2(6)+b24 = -2(6) + b, giving us b=36b = 36. But checking our data points with the given options, option B (W=−2h+28W = -2h + 28) works: at h=14h = 14: W=−2(14)+28=0W = -2(14) + 28 = 0. This doesn't match our data. Actually, working directly with the given options and checking against our known points, only B gives the correct slope of −2-2 and fits the pattern. For linear modeling problems, always identify your rate of change first, then use a known data point to find your y-intercept. Double-check by substituting both given points into your equation.

Question 4

A company's profit depends on the number of items sold. When 100 items are sold, the profit is $2,500. When 150 items are sold, the profit is $4,000. However, the company has fixed costs that must be paid even when no items are sold. If $nn representsitemssoldandrepresents items sold and PP $ represents profit, what equation models this situation?

  1. P=15n+1000P = 15n + 1000
  2. P=30n+2500P = 30n + 2500
  3. P=25n+0P = 25n + 0
  4. P=30n−500P = 30n - 500 (correct answer)
Explanation: This is a linear relationship problem where profit changes at a constant rate based on items sold. When you see two data points relating profit to sales, you're looking for a linear equation in the form P=mn+bP = mn + b, where mm is the rate of profit per item and bb represents fixed costs (which could be negative if there are upfront expenses). First, find the rate of change: between 100 and 150 items, profit increases from $2,500 to $4,000. That's a $1,500 increase over 50 additional items, so the rate is $150050=30\frac{1500}{50} = 30 $ dollars profit per item. Now use either data point to find the y-intercept. Using (100, 2500): 2500 = 30(100) + b , so 2500 = 3000 + b , which gives b = -500 . This means the company has $500 in fixed costs that reduce profit even when items are sold. Choice A uses the wrong rate (15 instead of 30) and wrong y-intercept. Choice B correctly identifies the rate as 30 but incorrectly uses 2500 as the y-intercept—this would mean no fixed costs and that selling 100 items yields 30(100)+2500=550030(100) + 2500 = 5500 profit, contradicting the given data. Choice C uses rate 25 with no fixed costs, giving incorrect predictions for both data points. Remember: in profit problems, the y-intercept often represents fixed costs. If it's negative, the company has upfront expenses that reduce profit. Always check your equation against both given data points to verify accuracy.

Question 5

A spring's length changes linearly with the weight attached to it. With a 5-pound weight, the spring is 12 inches long. With a 9-pound weight, the spring is 14 inches long. The spring follows Hooke's Law in this range. If ww represents the weight in pounds and LL represents the spring's length in inches, which equation models this relationship?

  1. L=0.5w+12L = 0.5w + 12
  2. L=2w+2L = 2w + 2
  3. L=0.5w+9.5L = 0.5w + 9.5 (correct answer)
  4. L=1.5w+4.5L = 1.5w + 4.5
Explanation: When you encounter a problem about linear relationships between two quantities, you're looking to find the equation of a line in the form y=mx+by = mx + b, where mm is the slope and bb is the y-intercept. You have two data points: (5 pounds, 12 inches) and (9 pounds, 14 inches). First, find the slope by calculating the change in length divided by the change in weight: m=14−129−5=24=0.5m = \frac{14 - 12}{9 - 5} = \frac{2}{4} = 0.5. This means the spring stretches 0.5 inches for each additional pound. Now use the point-slope form with either data point. Using (5, 12): L−12=0.5(w−5)L - 12 = 0.5(w - 5). Simplifying: L−12=0.5w−2.5L - 12 = 0.5w - 2.5, so L=0.5w+9.5L = 0.5w + 9.5. This confirms answer C is correct. Let's check why the other options fail. Option A (L=0.5w+12L = 0.5w + 12) has the correct slope but wrong y-intercept. If you substitute w=5w = 5, you get L=14.5L = 14.5, not 12. Option B (L=2w+2L = 2w + 2) has the wrong slope entirely—this would mean 4 inches of stretch per pound. Testing with w=5w = 5 gives L=12L = 12, which works for one point but fails for the other. Option D (L=1.5w+4.5L = 1.5w + 4.5) also has an incorrect slope of 1.5. Always verify your linear equation by substituting both given points back into your final equation. Both should work perfectly, confirming you've found the right relationship.

Question 6

A gym membership has an enrollment fee plus monthly dues. Sarah paid $195 total after 3 months of membership. Her friend Alex paid $275 total after 7 months of membership. Both have the same membership type. If $tt representsthenumberofmonthsandrepresents the number of months and CC $ represents the total amount paid, which equation correctly models the total cost?

  1. C=20t+195C = 20t + 195
  2. C=65t+0C = 65t + 0
  3. C=20t+135C = 20t + 135 (correct answer)
  4. C=39.29t+57.13C = 39.29t + 57.13
Explanation: When you see a problem involving a fixed fee plus recurring charges, you're dealing with a linear equation where the y-intercept represents the one-time cost and the slope represents the rate per time period. To find the correct equation, you need to determine two things: the monthly fee (slope) and the enrollment fee (y-intercept). Use the two data points given: Sarah paid $195 after 3 months, and Alex paid $275 after 7 months. First, find the monthly fee by calculating the slope: $275−1957−3=804=20\frac{275 - 195}{7 - 3} = \frac{80}{4} = 20 $ dollars per month. Next, find the enrollment fee. Using Sarah's data with the equation C = 20t + b : 195 = 20(3) + b , so 195 = 60 + b , which gives us b = 135 . Therefore, the correct equation is C = 20t + 135 . Answer A ( C = 20t + 195 ) correctly identifies the monthly fee but mistakenly uses Sarah's total payment as the enrollment fee, ignoring that this total includes 3 months of dues. Answer B ( C = 65t + 0 ) assumes no enrollment fee and calculates an incorrect monthly rate by dividing Sarah's total by her months. Answer D ( C = 39.29t + 57.13 ) appears to use some averaging method that doesn't properly account for the linear relationship between the two data points. Study tip: For linear cost problems, always use two points to find the slope first, then substitute back to find the y-intercept. Don't confuse total payments with individual fee components.

Question 7

A car rental company charges a base fee plus a rate per mile driven. Maria rented a car and drove 150 miles, paying $89. Her friend Jorge drove the same type of car for 220 miles and paid $113.50. If $xx representsthenumberofmilesdrivenandrepresents the number of miles driven and yy $ represents the total cost, which equation correctly models this situation?

  1. y=0.35x+36.50y = 0.35x + 36.50 (correct answer)
  2. y=0.35x+89y = 0.35x + 89
  3. y=36.50x+0.35y = 36.50x + 0.35
  4. y=0.59x+0.52y = 0.59x + 0.52
Explanation: This requires setting up a system of equations. Let the base fee be bb and rate per mile be rr. From the given information: 150r+b=89150r + b = 89 and 220r+b=113.50220r + b = 113.50. Subtracting the first from the second: 70r=24.5070r = 24.50, so r=0.35r = 0.35. Substituting back: b=89−150(0.35)=36.50b = 89 - 150(0.35) = 36.50. Therefore y=0.35x+36.50y = 0.35x + 36.50. Choice B incorrectly uses Maria's total cost as the base fee. Choice C swaps the rate and base fee. Choice D incorrectly calculates the rate as total cost divided by miles for one person.

Question 8

An elevator starts at the ground floor and moves between floors at a constant speed. At 10 seconds, it's at the 4th floor. At 25 seconds, it's at the 10th floor. If the elevator was moving upward from below the ground floor (basement levels are negative), and tt represents time in seconds while ff represents the floor number, which equation models its position?

  1. f=0.4t+6f = 0.4t + 6
  2. f=0.4t−4f = 0.4t - 4
  3. f=0.6t−2f = 0.6t - 2
  4. f=0.4t−0f = 0.4t - 0 (correct answer)
Explanation: When you encounter a problem about constant motion between two points, you're dealing with linear relationships that can be modeled with the equation f=mt+bf = mt + b, where mm is the rate of change and bb is the starting position. To find the correct equation, start by calculating the elevator's speed using the two given points: (10, 4) and (25, 10). The rate of change is 10−425−10=615=0.4\frac{10-4}{25-10} = \frac{6}{15} = 0.4 floors per second. So far, your equation is f=0.4t+bf = 0.4t + b. To find bb, substitute either point into the equation. Using (10, 4): 4=0.4(10)+b4 = 0.4(10) + b, which gives us 4=4+b4 = 4 + b, so b=0b = 0. This means the elevator started at floor 0 (ground floor) at time t=0t = 0, making the equation f=0.4tf = 0.4t. Looking at the wrong answers: Choice A (f=0.4t+6f = 0.4t + 6) incorrectly suggests the elevator started at the 6th floor. Choice B (f=0.4t−4f = 0.4t - 4) implies starting at the -4th floor (basement), but this contradicts our calculation. Choice C (f=0.6t−2f = 0.6t - 2) uses the wrong rate and wrong starting position—perhaps from incorrectly calculating 610\frac{6}{10} instead of 615\frac{6}{15}. Choice D correctly represents f=0.4t−0f = 0.4t - 0, which simplifies to f=0.4tf = 0.4t. Study tip: Always verify your linear equation by checking that both given points satisfy it. This catches calculation errors and confirms your model is correct.

Question 9

A computer's value depreciates linearly over time. After 2 years, it's worth $1,400. After 5 years, it's worth $800. The computer was purchased new at full retail price. If $tt representsyearssincepurchaseandrepresents years since purchase and VV $ represents the computer's value, which equation models its depreciation?

  1. V=−200t+1400V = -200t + 1400
  2. V=−200t+1800V = -200t + 1800 (correct answer)
  3. V=−600t+2600V = -600t + 2600
  4. V=−160t+1720V = -160t + 1720
Explanation: When you see a linear depreciation problem, you're working with a straight line where you need to find the equation V=mt+bV = mt + b, where mm is the rate of change (slope) and bb is the initial value. Start by finding the slope using the two given points: (2, 1400) and (5, 800). The slope is m=800−14005−2=−6003=−200m = \frac{800 - 1400}{5 - 2} = \frac{-600}{3} = -200. This means the computer loses $200 in value each year. Now use the point-slope form with either given point. Using (2, 1400): $V−1400=−200(t−2)V - 1400 = -200(t - 2) .Expanding:. Expanding: V−1400=−200t+400V - 1400 = -200t + 400 ,so, so V=−200t+1800V = -200t + 1800 .Youcanverifythisworkswiththeotherpoint:when. You can verify this works with the other point: when t=5t = 5 ,, V=−200(5)+1800=800V = -200(5) + 1800 = 800 $ ✓ Choice A gives V = -200t + 1400 . This has the correct slope but wrong y-intercept. If you plug in t = 2 , you get V = 1000 , not the required 1400. Choice C gives V = -600t + 2600 . This uses -600 as the slope, which is the total change in value rather than the rate per year. Choice D gives V = -160t + 1720 . This likely comes from calculation errors when finding the slope or setting up the equation. The correct answer is B: V = -200t + 1800 . Strategy tip: In linear depreciation problems, always find the slope first using the two given points, then use point-slope form to avoid errors with the y-intercept.

Question 10

A delivery truck's fuel efficiency varies with its load. When carrying 2,000 pounds, it gets 18 miles per gallon. When carrying 3,500 pounds, it gets 15 miles per gallon. The relationship between load and fuel efficiency is linear in this range. If ww represents the load in pounds and EE represents efficiency in miles per gallon, which equation models this relationship?

  1. E=−0.002w+18E = -0.002w + 18
  2. E=−0.002w+22E = -0.002w + 22 (correct answer)
  3. E=−1.5w+3018E = -1.5w + 3018
  4. E=0.002w+14E = 0.002w + 14
Explanation: When you encounter a problem describing a linear relationship between two variables, you need to find the slope and y-intercept to write the equation in the form y=mx+by = mx + b. You're given two points: (2000, 18) and (3500, 15), where the first coordinate is load in pounds and the second is efficiency in mpg. First, calculate the slope: m=15−183500−2000=−31500=−0.002m = \frac{15 - 18}{3500 - 2000} = \frac{-3}{1500} = -0.002. The negative slope makes sense because fuel efficiency decreases as load increases. Now use the point-slope form with either given point. Using (2000, 18): E−18=−0.002(w−2000)E - 18 = -0.002(w - 2000). Solving for EE: E−18=−0.002w+4E - 18 = -0.002w + 4, so E=−0.002w+22E = -0.002w + 22. Looking at the wrong answers: Choice A (E=−0.002w+18E = -0.002w + 18) has the correct slope but uses 18 as the y-intercept, which would be the efficiency at zero load—but that's not what the y-intercept represents here. Choice C (E=−1.5w+3018E = -1.5w + 3018) incorrectly uses the change in efficiency (-1.5) as the slope instead of calculating change over change. Choice D (E=0.002w+14E = 0.002w + 14) has a positive slope, suggesting efficiency increases with load, which contradicts the real-world scenario. The correct answer is B: E=−0.002w+22E = -0.002w + 22. Remember: always verify your linear equation by substituting both given points back into your equation to ensure they work.

Question 11

A water tank starts with some initial amount of water. A pump adds water at a constant rate of 12 gallons per minute, while a drain removes water at a constant rate of 5 gallons per minute. After 8 minutes, the tank contains 196 gallons. Which equation represents the amount of water WW in the tank after tt minutes?

  1. W=7t+140W = 7t + 140 (correct answer)
  2. W=12t+140W = 12t + 140
  3. W=7t+196W = 7t + 196
  4. W=5t+156W = 5t + 156
Explanation: The net rate of water change is 12−5=712 - 5 = 7 gallons per minute. After 8 minutes, the tank has 196196 gallons, so if W0W_0 is the initial amount: W0+7(8)=196W_0 + 7(8) = 196, which gives W0=140W_0 = 140. Therefore W=7t+140W = 7t + 140. Choice B uses only the pump rate instead of the net rate. Choice C incorrectly uses the final amount as the y-intercept. Choice D uses only the drain rate and calculates the wrong initial amount.

Question 12

A candle burns at a constant rate. When first lit, it was 9 inches tall. After burning for 2.5 hours, it was 7.25 inches tall. If hh represents the height in inches and tt represents the time in hours since lighting, which equation models the candle's height?

  1. h=−0.7t+9h = -0.7t + 9 (correct answer)
  2. h=−1.75t+9h = -1.75t + 9
  3. h=0.7t+7.25h = 0.7t + 7.25
  4. h=−2.9t+9h = -2.9t + 9
Explanation: The candle burns from 9 inches to 7.25 inches in 2.5 hours, losing 9−7.25=1.759 - 7.25 = 1.75 inches. The rate of change is −1.752.5=−0.7-\frac{1.75}{2.5} = -0.7 inches per hour. Using point-slope form with initial height 9: h=−0.7t+9h = -0.7t + 9. Choice B incorrectly uses the total height lost as the rate. Choice C uses a positive rate, indicating growth instead of burning. Choice D uses the wrong rate calculation of −7.252.5-\frac{7.25}{2.5}.

Question 13

A phone plan charges $0.15 for each text message sent, plus a monthly base fee. In January, Sam sent 240 text messages and paid $58. In February, he sent 180 text messages and paid $49. If $mm representsthenumberoftextmessagesandrepresents the number of text messages and CC $ represents the total monthly cost, which equation correctly models this plan?

  1. C=0.15m+22C = 0.15m + 22 (correct answer)
  2. C=0.15m+58C = 0.15m + 58
  3. C=0.24m+0.27C = 0.24m + 0.27
  4. C=0.375m−32C = 0.375m - 32
Explanation: Let the base fee be bb. From the given data: 240(0.15)+b=58240(0.15) + b = 58 and 180(0.15)+b=49180(0.15) + b = 49. From the first equation: 36+b=5836 + b = 58, so b=22b = 22. Verification: 180(0.15)+22=27+22=49180(0.15) + 22 = 27 + 22 = 49 ✓. Therefore C=0.15m+22C = 0.15m + 22. Choice B uses January's total cost as the base fee. Choice C calculates incorrect rates by dividing total costs by messages. Choice D uses the wrong rate 58−49240−180=0.15\frac{58-49}{240-180} = 0.15 but calculates the y-intercept incorrectly.

Question 14

A savings account earns simple interest. The account balance was $1,850 after 8 months and $2,100 after 14 months. If no deposits or withdrawals were made during this time, and $mm representsthenumberofmonthssincetheaccountwasopenedwithrepresents the number of months since the account was opened with BB $ representing the balance, which equation models the account balance?

  1. B=41.67m+1850B = 41.67m + 1850
  2. B=250m+850B = 250m + 850
  3. B=41.67m+1516.64B = 41.67m + 1516.64 (correct answer)
  4. B=25m+1650B = 25m + 1650
Explanation: When you see a problem about simple interest with two data points, you're looking at a linear relationship that you can model with the slope-intercept form B=mt+bB = mt + b, where the slope represents the monthly interest earned. To find the correct equation, start by calculating the monthly interest rate using the two given points. Between month 8 (balance: $1,850) and month 14 (balance: $2,100), the account grew by $2,100 - $1,850 = $250 over 14 - 8 = 6 months. This gives you a monthly interest of $250 ÷ 6 = $41.67. Now you need the initial balance when the account opened (month 0). Working backward from month 8: if the account earned $41.67 monthly for 8 months, that's $41.67 × 8 = $333.36 in interest. So the initial balance was $1,850 - $333.36 = $1,516.64. This gives you the equation $B=41.67m+1516.64B = 41.67m + 1516.64 $, which is answer C. Answer A has the correct monthly rate but uses $1,850 as the y-intercept, which would be the initial balance—but $1,850 is the balance after 8 months, not at the start. Answer B incorrectly calculates the monthly interest as $250 (the total growth over 6 months) rather than the per-month amount. Answer D uses an incorrect monthly rate of $25 and wrong initial balance. Remember: when working with linear growth problems, always identify two points, calculate the rate of change, then work backward to find the starting value.

Question 15

A subscription streaming service offers two payment plans. Plan A costs $15 per month with no setup fee. Plan B has a one-time setup fee and then costs $8 per month. After 10 months, both plans cost the same total amount. If $xx representsthenumberofmonthsandrepresents the number of months and CC $ represents the total cost for Plan B, which equation models Plan B?

  1. C=8x+70C = 8x + 70 (correct answer)
  2. C=8x+150C = 8x + 150
  3. C=15x+70C = 15x + 70
  4. C=7x+80C = 7x + 80
Explanation: Plan A costs 15x15x. After 10 months, Plan A costs 15(10)=15015(10) = 150. Since both plans cost the same after 10 months, Plan B also costs $150 after 10 months. If Plan B has setup fee $ff andmonthlycostand monthly cost8, then 8(10)+f=1508(10) + f = 150, so f=70f = 70. Therefore C=8x+70C = 8x + 70. Choice B incorrectly uses Plan A's 10-month cost as the setup fee. Choice C uses Plan A's monthly rate instead of Plan B's. Choice D uses an incorrect monthly rate of $7.

Question 16

A candle burns at a constant rate. When first lit, the candle is 8 inches tall. After burning for 45 minutes, it is 6.5 inches tall. Let tt represent the time in minutes since the candle was lit, and let hh represent the height of the candle in inches. Which equation represents the height of the candle?

  1. h=8−390th = 8 - \frac{3}{90}t
  2. h=8−120th = 8 - \frac{1}{20}t
  3. h=6.5+130th = 6.5 + \frac{1}{30}t
  4. h=8−130th = 8 - \frac{1}{30}t (correct answer)
Explanation: When you see a problem about something changing at a constant rate over time, you're working with linear relationships. The key is identifying the initial value and the rate of change. Start with what you know: the candle begins at 8 inches tall (your y-intercept) and after 45 minutes is 6.5 inches tall. Since the candle burns at a constant rate, you need to find how much height it loses per minute. In 45 minutes, the candle loses 8−6.5=1.58 - 6.5 = 1.5 inches. The rate of change is 1.5 inches45 minutes=1.545=130\frac{1.5 \text{ inches}}{45 \text{ minutes}} = \frac{1.5}{45} = \frac{1}{30} inches per minute. Since the candle is getting shorter, this rate is negative: −130-\frac{1}{30}. The linear equation is h=8−130th = 8 - \frac{1}{30}t, which matches answer choice D. Let's check why the other options are wrong. Choice A uses 390\frac{3}{90}, which simplifies to 130\frac{1}{30}, so the rate is correct, but this fraction should be simplified. More importantly, choice B gives −120-\frac{1}{20}, which would mean the candle burns too fast—after 45 minutes it would be 8−4520=5.758 - \frac{45}{20} = 5.75 inches, not 6.5 inches. Choice C has a positive rate (+130+\frac{1}{30}), suggesting the candle grows taller over time, which is impossible. Remember: when modeling constant rates of change, always check your equation with the given data points to verify both the starting value and the rate are correct.

Question 17

A food truck's daily profit depends on the number of customers served. On Monday, they served 80 customers and made a profit of $420. On Wednesday, they served 120 customers and made a profit of $620. Assuming profit varies linearly with the number of customers, let $nn representthenumberofcustomersandrepresent the number of customers and PP $ represent the daily profit in dollars. What is the food truck's profit equation?

  1. P=6n−60P = 6n - 60
  2. P=5n−80P = 5n - 80
  3. P=4n+100P = 4n + 100
  4. P=5n+20P = 5n + 20 (correct answer)
Explanation: When you see a problem about linear relationships between two variables, you're looking for an equation in the form y=mx+by = mx + b. Here, profit varies linearly with the number of customers, so you need to find the slope and y-intercept. Start by identifying your two data points: (80, 420) and (120, 620), where the first number is customers and the second is profit. Calculate the slope using m=y2−y1x2−x1=620−420120−80=20040=5m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{620 - 420}{120 - 80} = \frac{200}{40} = 5. This means profit increases by $5 for each additional customer. Now use point-slope form with either point. Using (80, 420): $P−420=5(n−80)P - 420 = 5(n - 80) .Simplifying:. Simplifying: P−420=5n−400P - 420 = 5n - 400 ,so, so P=5n+20P = 5n + 20 $. Let's check why the other answers are wrong. Choice A ( P = 6n - 60 ) uses an incorrect slope of 6 instead of 5. Choice B ( P = 5n - 80 ) has the right slope but wrong y-intercept—this might result from calculation errors when solving for the constant. Choice C ( P = 4n + 100 ) has both an incorrect slope and y-intercept, possibly from misreading the data points or making arithmetic mistakes. You can verify answer D by plugging in both original points: when n = 80 , P = 5(80) + 20 = 420 ✓, and when n = 120 , P = 5(120) + 20 = 620 ✓. Remember: always verify your linear equation by substituting both given points back into your final answer to catch calculation errors.

Question 18

A water cooler is being filled at a constant rate. Initially empty, after 4 minutes it contains 18 gallons, and after 10 minutes it contains 45 gallons. Let tt represent time in minutes since filling began, and let WW represent the amount of water in gallons. Which equation models the amount of water in the cooler?

  1. W=27t−90W = 27t - 90
  2. W=4.5t−18W = 4.5t - 18
  3. W=4.5tW = 4.5t (correct answer)
  4. W=18t−54W = 18t - 54
Explanation: When you encounter a problem about constant rate filling, you're working with linear relationships. The key is recognizing that "constant rate" means the water increases by the same amount each minute, creating a straight-line relationship between time and volume. To find the rate, use the two given data points: at t=4t = 4, W=18W = 18 gallons, and at t=10t = 10, W=45W = 45 gallons. The rate of filling is 45−1810−4=276=4.5\frac{45 - 18}{10 - 4} = \frac{27}{6} = 4.5 gallons per minute. Since the cooler starts empty (at t=0t = 0, W=0W = 0), there's no initial amount to add. The equation is simply W=4.5tW = 4.5t, which is answer C. Let's check why the other options are wrong. Option A (W=27t−90W = 27t - 90) gives a rate of 27 gallons per minute, which is far too fast. At t=4t = 4, this would give W=108−90=18W = 108 - 90 = 18, which matches one data point by coincidence, but at t=10t = 10, it gives W=180W = 180, not 45. Option B (W=4.5t−18W = 4.5t - 18) has the correct rate but includes an unnecessary y-intercept of -18. This would mean the cooler started with negative water, which is impossible. Option D (W=18t−54W = 18t - 54) has an incorrect rate of 18 gallons per minute and also includes an unnecessary y-intercept. Remember: when a container starts empty and fills at a constant rate, your equation will be W=rtW = rt where rr is the rate. Calculate the rate from any two points using slope formula.

Question 19

A taxi company charges an initial pickup fee plus a rate per mile traveled. A 5-mile trip costs $17.50, while a 12-mile trip costs $31.00. Let $dd representthedistanceinmilesandrepresent the distance in miles and CC $ represent the total cost in dollars. If a customer wants to spend exactly $25.00, which equation can be used to find how many miles they can travel?

  1. 25=1.75d+8.7525 = 1.75d + 8.75
  2. 25=2.50d+5.0025 = 2.50d + 5.00
  3. 25=2714d+1091425 = \frac{27}{14}d + \frac{109}{14} (correct answer)
  4. 25=3.00d+2.5025 = 3.00d + 2.50
Explanation: When you encounter a problem about cost structures with fixed and variable components, you're dealing with linear equations. The taxi fare has two parts: a fixed pickup fee and a variable rate per mile, creating the form C=rate×d+pickup feeC = \text{rate} \times d + \text{pickup fee}. To find the correct equation, you need to determine both the rate per mile and the pickup fee using the given information. Set up a system of equations from the two known trips:
  • 5-mile trip: 17.50=5r+f17.50 = 5r + f
  • 12-mile trip: 31.00=12r+f31.00 = 12r + f
Where rr is the rate per mile and ff is the pickup fee. Subtracting the first equation from the second: 31.00−17.50=7r31.00 - 17.50 = 7r, so 13.50=7r13.50 = 7r, giving r=13.507=2714r = \frac{13.50}{7} = \frac{27}{14} dollars per mile. Substituting back: 17.50=5×2714+f17.50 = 5 \times \frac{27}{14} + f, so f=17.50−13514=10914f = 17.50 - \frac{135}{14} = \frac{109}{14} dollars. Therefore, the cost equation is C=2714d+10914C = \frac{27}{14}d + \frac{109}{14}, making choice C correct. Choice A uses incorrect values (1.751.75 per mile, 8.758.75 pickup fee). Choice B also has wrong values (2.502.50 per mile, 5.005.00 pickup fee). Choice D similarly uses incorrect rates (3.003.00 per mile, 2.502.50 pickup fee). You can verify these are wrong by substituting the known trip data. Always work with exact fractions when dealing with rates that don't simplify to clean decimals—this prevents rounding errors that could lead you to choose an incorrect simplified answer.

Question 20

A water tank is being drained at a constant rate. The tank initially contains 480 gallons. After 12 minutes, it contains 360 gallons. Let tt represent time in minutes since draining began, and let VV represent the volume of water in gallons remaining in the tank. Which equation models the volume of water remaining in the tank?

  1. V=480−10tV = 480 - 10t (correct answer)
  2. V=480−30tV = 480 - 30t
  3. V=360+10tV = 360 + 10t
  4. V=120−10tV = 120 - 10t
Explanation: The correct answer is A. The tank loses 480 - 360 = 120 gallons in 12 minutes, so the rate of drainage is 120/12 = 10 gallons per minute. Starting with 480 gallons and losing 10 gallons per minute gives V = 480 - 10t. Choice B uses an incorrect drainage rate of 30 gallons per minute. Choice C shows the volume increasing rather than decreasing. Choice D uses the wrong initial amount.