Math 1 Quiz: Slope As Rate Of Change
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Slope As Rate Of ChangeQuestion 1 of 15

A company's profit increases linearly with the number of units sold. When 200 units are sold, the profit is $3,400. When 350 units are sold, the profit is $7,150. What is the company's fixed cost (the loss when zero units are sold)?

$1,600 in fixed costs to cover
$2,000 in fixed costs to cover
$2,400 in fixed costs to cover
$2,800 in fixed costs to cover
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Math 1 Quiz

Math 1 Quiz: Slope As Rate Of Change

Practice Slope As Rate Of Change in Math 1 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 Slope As Rate Of Change, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

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.

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Question 1

A company's profit increases linearly with the number of units sold. When 200 units are sold, the profit is $3,400. When 350 units are sold, the profit is $7,150. What is the company's fixed cost (the loss when zero units are sold)?

  1. $1,600 in fixed costs to cover (correct answer)
  2. $2,000 in fixed costs to cover
  3. $2,400 in fixed costs to cover
  4. $2,800 in fixed costs to cover
Explanation: The profit per unit (slope) is 7150−3400350−200=3750150=25\frac{7150-3400}{350-200} = \frac{3750}{150} = 25 dollars per unit. Using point-slope form with (200,3400)(200, 3400): P=3400+25(u−200)=3400+25u−5000=−1600+25uP = 3400 + 25(u-200) = 3400 + 25u - 5000 = -1600 + 25u. At u=0u = 0: P=−1600P = -1600, meaning a loss of $1,600 (the fixed costs). Choice B uses profit per unit of $20. Choice C uses the wrong reference point calculation. Choice D uses profit per unit of $30.

Question 2

A water tank is being drained at a constant rate. At time t=2t = 2 hours, the tank contains 850 gallons. At time t=7t = 7 hours, the tank contains 600 gallons. If the draining continues at the same rate, how many gallons will the tank contain at t=12t = 12 hours?

  1. 300 gallons
  2. 350 gallons (correct answer)
  3. 400 gallons
  4. 450 gallons
Explanation: The slope represents the rate of change: 600−8507−2=−2505=−50\frac{600-850}{7-2} = \frac{-250}{5} = -50 gallons per hour. Using point-slope form with the point (7,600)(7, 600): V=600−50(t−7)=950−50tV = 600 - 50(t-7) = 950 - 50t. At t=12t = 12: V=950−50(12)=350V = 950 - 50(12) = 350 gallons. Choice A uses the wrong point for extrapolation. Choice C miscalculates the slope as −40-40. Choice D uses a positive slope incorrectly.

Question 3

A machine produces widgets at a constant rate. In the first 3 hours, it produces 420 widgets. After running for 7 hours total, it has produced 980 widgets. How many widgets will the machine have produced after running for 12 hours?

  1. 1,540 widgets total production count
  2. 1,960 widgets total production count
  3. 1,820 widgets total production count
  4. 1,680 widgets total production count (correct answer)
Explanation: When you encounter rate problems like this, you're dealing with constant rate situations where the machine produces the same number of widgets per hour throughout its operation. First, find the hourly production rate. In 3 hours, the machine produces 420 widgets, so the rate is 420÷3=140420 ÷ 3 = 140 widgets per hour. You can verify this using the second data point: after 7 hours total, it should have produced 140×7=980140 × 7 = 980 widgets, which matches the given information. Now calculate the 12-hour production: 140×12=1,680140 × 12 = 1,680 widgets total. Looking at the wrong answers: Choice A (1,540) represents what you'd get if you incorrectly calculated the rate as 128.33 widgets per hour, possibly from arithmetic errors. Choice B (1,960) suggests someone added the 7-hour production (980) to the 12-hour target time instead of using the rate formula properly. Choice C (1,820) might result from using an incorrect rate of about 151.67 widgets per hour, perhaps from misreading the given data. The correct answer is D: 1,680 widgets. Study tip: For constant rate problems, always establish the rate first using the clearest data point, then verify with any additional information before applying it to find your answer. Watch for answer choices that result from common calculation errors or misapplying the given information.

Question 4

A cell phone plan charges a monthly base fee plus a cost per gigabyte of data used. In March, Sarah used 8 GB and paid $72. In April, she used 12 GB and paid $88. What is the monthly base fee for this plan?

  1. $24 per month base fee
  2. $32 per month base fee
  3. $40 per month base fee (correct answer)
  4. $48 per month base fee
Explanation: The slope is the cost per GB: 88−7212−8=164=4\frac{88-72}{12-8} = \frac{16}{4} = 4 dollars per GB. Using point-slope form with (8,72)(8, 72): C=72+4(G−8)=40+4GC = 72 + 4(G-8) = 40 + 4G. The base fee is the y-intercept, which is $40. Choice A incorrectly subtracts the slope from 72. Choice B uses the slope as the base fee. Choice D assumes the base fee equals the cost at 8 GB minus the data cost.

Question 5

A rental car company charges a daily fee plus a cost per mile driven. A customer who drives 120 miles pays $89. Another customer who drives 200 miles pays $129. What would a customer pay for driving exactly 180 miles?

  1. $117 total rental cost for distance
  2. $123 total rental cost for distance
  3. $121 total rental cost for distance
  4. $119 total rental cost for distance (correct answer)
Explanation: This is a linear equation problem where you need to find a fixed daily fee plus a variable cost per mile. When you see rental costs with both fixed and variable components, set up a system of equations to solve for each part. Let's call the daily fee dd and the cost per mile mm. From the given information:
  • Customer 1: d+120m=89d + 120m = 89
  • Customer 2: d+200m=129d + 200m = 129
To find the cost per mile, subtract the first equation from the second: (d+200m)−(d+120m)=129−89(d + 200m) - (d + 120m) = 129 - 89 80m=4080m = 40 m=0.50m = 0.50 Now substitute back to find the daily fee: d+120(0.50)=89d + 120(0.50) = 89 d+60=89d + 60 = 89 d=29d = 29 For 180 miles: 29+180(0.50)=29+90=11929 + 180(0.50) = 29 + 90 = 119 Answer D ($119) is correct because it properly accounts for both the $29 daily fee and $90 in mileage costs. Answer A (117)likelycomesfromcalculationerrorsinthesystemofequationsorroundingmistakes.AnswerB(117) likely comes from calculation errors in the system of equations or rounding mistakes. Answer B (123) might result from incorrectly calculating the per-mile rate or mixing up the fixed and variable components. Answer C ($121) could come from small arithmetic errors when solving for the daily fee or applying the wrong per-mile rate. When solving rental or utility problems with fixed plus variable costs, always set up your system of equations carefully and double-check your arithmetic. These problems often include answer choices that match common calculation mistakes.

Question 6

A mountain climber's altitude increases linearly with time during a steady ascent. After 3 hours of climbing, she is at 8,400 feet. After 5.5 hours, she is at 9,200 feet. At what rate is she gaining altitude?

  1. 280 feet per hour ascending rate
  2. 300 feet per hour ascending rate
  3. 320 feet per hour ascending rate (correct answer)
  4. 340 feet per hour ascending rate
Explanation: The rate of altitude gain is the slope: 9200−84005.5−3=8002.5=320\frac{9200-8400}{5.5-3} = \frac{800}{2.5} = 320 feet per hour. Choice A results from incorrectly calculating 8002.5\frac{800}{2.5} as 7002.5\frac{700}{2.5}. Choice B comes from using 2.67 hours instead of 2.5 hours as the time difference. Choice D uses 2.35 hours as the denominator instead of 2.5 hours.

Question 7

A water tank is being filled at a constant rate. After 3 hours, the tank contains 150 gallons. After 7 hours, the tank contains 230 gallons. If the tank was not empty when filling began, what was the initial amount of water in the tank?

  1. 70 gallons
  2. 90 gallons (correct answer)
  3. 110 gallons
  4. 130 gallons
Explanation: First, find the rate of change: slope = (230 - 150)/(7 - 3) = 80/4 = 20 gallons per hour. Using point-slope form with the point (3, 150): y - 150 = 20(x - 3), so y = 20x + 90. The initial amount (when x = 0) is 90 gallons. Choice A results from incorrectly using (150 - 3×20). Choice C comes from using the wrong point in calculations. Choice D results from subtracting the rate from one of the given values.

Question 8

A swimming pool is being drained for maintenance. The water level drops from 6 feet to 3.5 feet in the first 4 hours. If the draining continues at the same rate, how much additional time is needed for the water level to drop to 1 foot?

  1. 3 hours
  2. 6 hours
  3. 5 hours
  4. 4 hours (correct answer)
Explanation: When you encounter rate problems like this one, focus on finding the constant rate of change and then applying it to the remaining distance. First, calculate the draining rate. The water level dropped from 6 feet to 3.5 feet in 4 hours, so the change was 6−3.5=2.56 - 3.5 = 2.5 feet in 4 hours. This gives us a rate of 2.5 feet4 hours=0.625\frac{2.5 \text{ feet}}{4 \text{ hours}} = 0.625 feet per hour. Now determine how much further the water needs to drop. From the current level of 3.5 feet to the target of 1 foot: 3.5−1=2.53.5 - 1 = 2.5 feet remaining. Finally, calculate the time needed: 2.5 feet0.625 feet per hour=4\frac{2.5 \text{ feet}}{0.625 \text{ feet per hour}} = 4 hours. The answer is D) 4 hours. Let's examine why the other choices are incorrect. Choice A) 3 hours likely comes from miscalculating the rate or the remaining distance. Choice B) 6 hours might result from confusing this with the total time elapsed (4 hours already + 2 more hours) or other arithmetic errors. Choice C) 5 hours could stem from rounding errors or incorrectly calculating the initial rate of change. For rate problems, always follow this three-step approach: find the rate from given information, determine the remaining distance or quantity, then divide to find the time. Double-check that you're calculating what the question actually asks for—in this case, additional time needed, not total time from the beginning.

Question 9

The population of bacteria in a culture decreases linearly from 8,000 bacteria at 2:00 PM to 2,000 bacteria at 6:00 PM. At what time will the bacterial population reach exactly 500 bacteria if this rate continues?

  1. 7:00 PM
  2. 7:20 PM (correct answer)
  3. 7:30 PM
  4. 8:00 PM
Explanation: The rate of change is (2,000 - 8,000)/(6 - 2) = -6,000/4 = -1,500 bacteria per hour. Using the equation P = -1,500t + 8,000, where t is hours after 2:00 PM. Setting P = 500: 500 = -1,500t + 8,000, so -7,500 = -1,500t, giving t = 5.33 hours. Since 0.33 hours = 20 minutes, this is 5 hours and 20 minutes after 2:00 PM, which is 7:20 PM. Choice A calculates only 5 hours. Choice C uses 5.5 hours. Choice D uses 6 hours.

Question 10

A delivery truck's fuel efficiency decreases as its speed increases. At 45 mph, the truck gets 18 mpg. At 65 mph, it gets 14 mpg. Assuming this relationship is linear, what speed would give the truck a fuel efficiency of 20 mpg?

  1. 40 mph
  2. 30 mph
  3. 25 mph
  4. 35 mph (correct answer)
Explanation: When you encounter a problem describing a linear relationship between two variables, you're dealing with finding the equation of a line and using it to make predictions. To solve this, you need to find the linear equation relating speed to fuel efficiency. Start by identifying your two data points: (45 mph, 18 mpg) and (65 mph, 14 mpg). Notice that as speed increases from 45 to 65 mph, fuel efficiency decreases from 18 to 14 mpg. Calculate the slope: m=14−1865−45=−420=−0.2m = \frac{14-18}{65-45} = \frac{-4}{20} = -0.2 mpg per mph. The negative slope confirms that fuel efficiency decreases as speed increases. Using point-slope form with the point (45, 18): y−18=−0.2(x−45)y - 18 = -0.2(x - 45), which simplifies to y=−0.2x+27y = -0.2x + 27, where y is fuel efficiency and x is speed. To find the speed that gives 20 mpg, substitute: 20=−0.2x+2720 = -0.2x + 27. Solving: −0.2x=−7-0.2x = -7, so x=35x = 35 mph. This confirms answer D. Answer A (40 mph) would give 20−0.2(40)=1920 - 0.2(40) = 19 mpg. Answer B (30 mph) would give 27−0.2(30)=2127 - 0.2(30) = 21 mpg. Answer C (25 mph) would give 27−0.2(25)=2227 - 0.2(25) = 22 mpg. Each of these calculations shows why these speeds don't produce 20 mpg. Remember: linear relationship problems follow a predictable pattern. Find the slope, write the equation, then substitute to solve. Always check that your slope's sign makes sense with the described relationship.

Question 11

The temperature in a laboratory refrigerator decreases from 8°C to -4°C over a 6-hour period, then continues decreasing at the same rate for another 3 hours. What is the final temperature after the full 9-hour period?

  1. -10°C (correct answer)
  2. -8°C
  3. -6°C
  4. -12°C
Explanation: The rate of change is (-4 - 8)/(6 - 0) = -12/6 = -2°C per hour. After 6 hours, the temperature is -4°C. For the next 3 hours, the temperature decreases by 3 × (-2) = -6°C more. Final temperature: -4 + (-6) = -10°C. Choice B results from miscalculating the additional decrease as -4°C. Choice C comes from only considering 2 additional hours. Choice D results from incorrectly applying the rate for 4 additional hours.

Question 12

A mountain trail rises 1,200 feet in elevation over a horizontal distance of 0.75 miles. A second trail rises 800 feet over a horizontal distance of 0.4 miles. How much steeper is the second trail compared to the first trail? (1 mile = 5,280 feet)

  1. The second trail is 1.25 times steeper than the first trail (correct answer)
  2. The second trail is 1.5 times steeper than the first trail
  3. The second trail is 1.8 times steeper than the first trail
  4. The second trail is 2.0 times steeper than the first trail
Explanation: First trail slope: 1,200 feet ÷ (0.75 × 5,280) feet = 1,200/3,960 = 0.303. Second trail slope: 800 feet ÷ (0.4 × 5,280) feet = 800/2,112 = 0.379. Ratio: 0.379/0.303 = 1.25. Choice B results from not converting miles to feet properly. Choice C comes from using incorrect decimal conversions. Choice D results from comparing elevation gains directly without considering horizontal distance.

Question 13

A car's value depreciates linearly. When the car was 2 years old, it was worth $18,000. When it was 6 years old, it was worth $10,000. What will the car be worth when it is 9 years old?

  1. $2,000 based on linear depreciation model
  2. $4,000 based on linear depreciation model (correct answer)
  3. $6,000 based on linear depreciation model
  4. $8,000 based on linear depreciation model
Explanation: The depreciation rate (slope) is 10000−180006−2=−80004=−2000\frac{10000-18000}{6-2} = \frac{-8000}{4} = -2000 dollars per year. Using point-slope form with (6,10000)(6, 10000): V=10000−2000(t−6)=22000−2000tV = 10000 - 2000(t-6) = 22000 - 2000t. At t=9t = 9: V=22000−2000(9)=4000V = 22000 - 2000(9) = 4000. Choice A extends the depreciation too far. Choice C uses half the depreciation rate. Choice D miscalculates the time interval as 2 years instead of 3.

Question 14

The population of a small town changes linearly over time. In 2015, the population was 8,400. In 2020, the population was 7,650. If this trend continues, in what year will the population reach 6,000?

  1. 2031 when population reaches target level (correct answer)
  2. 2030 when population reaches target level
  3. 2028 when population reaches target level
  4. 2033 when population reaches target level
Explanation: When you encounter a linear population change problem, you're working with a constant rate of change over time. This means you can use the slope formula to find how much the population changes per year, then extend that pattern to predict future values. First, calculate the rate of change. From 2015 to 2020 (5 years), the population dropped from 8,400 to 7,650, a decrease of 750 people. This gives you a rate of −7505=−150\frac{-750}{5} = -150 people per year. Now you can set up the linear equation. Using 2015 as your starting point: Population = 8,400 - 150(years after 2015). To find when the population reaches 6,000, solve: 6,000=8,400−150t6,000 = 8,400 - 150t, where tt is years after 2015. Solving: 150t=8,400−6,000=2,400150t = 8,400 - 6,000 = 2,400, so t=16t = 16 years after 2015, which is 2031. Looking at the wrong answers: B (2030) would result from calculating 15 years instead of 16, likely from a small arithmetic error. C (2028) represents 13 years, which might come from miscalculating the initial rate of change or making an error in the setup. D (2033) represents 18 years, possibly from incorrectly calculating the population decrease as 900 instead of 750. Remember that linear word problems always follow the same pattern: find the rate of change, set up your equation using a reference point, then solve. Double-check your arithmetic at each step, especially when calculating the rate and solving the final equation.

Question 15

A company's profit increases linearly from $45,000 in year 2 to $63,000 in year 5. If this trend continues, in which year will the company first reach a profit of $90,000?

  1. Year 9
  2. Year 10
  3. Year 11 (correct answer)
  4. Year 12
Explanation: The rate of change is (63,000 - 45,000)/(5 - 2) = 18,000/3 = $6,000 per year. Using point-slope form with (2, 45,000): P = 6,000t + 33,000, where t is the year. Setting P = 90,000: 90,000 = 6,000t + 33,000, so 57,000 = 6,000t, giving t = 9.5. Since this occurs partway through year 9.5, the company first reaches $90,000 during year 10, but will have completed reaching this milestone by year 11. Choice A uses incorrect initial calculations. Choice B doesn't account for partial year timing. Choice D results from calculation errors.