Algebra Quiz: Recognize Percent Growth Or Decay
20 questions · exam conditions
0:00
Recognize Percent Growth Or DecayQuestion 1 of 20

A town's population is recorded each year:

Year tt: 0, 1, 2, 3 Population P(t)P(t): 20,000; 21,600; 23,328; 25,194.24

From the data, determine if there is constant percent change. If so, what is the percent rate per year?

Yes; exponential growth at 0.08% per year
Yes; exponential decay at 8% per year
No; it is linear because the population increases by 1,600 each year
Yes; exponential growth at 8% per year
← Back to quizzes

Algebra Quiz

Algebra Quiz: Recognize Percent Growth Or Decay

Practice Recognize Percent Growth Or Decay in Algebra 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 Recognize Percent Growth Or Decay, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.

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

A town's population is recorded each year:

Year tt: 0, 1, 2, 3 Population P(t)P(t): 20,000; 21,600; 23,328; 25,194.24

From the data, determine if there is constant percent change. If so, what is the percent rate per year?

  1. Yes; exponential growth at 0.08% per year
  2. Yes; exponential decay at 8% per year
  3. No; it is linear because the population increases by 1,600 each year
  4. Yes; exponential growth at 8% per year (correct answer)
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. From a table, to identify exponential with constant percent rate: divide consecutive y-values to find ratios. If y₂/y₁ = y₃/y₂ = y₄/y₃ = same number, that's your growth/decay factor b. If b > 1 (like 1.08), it's growth at (b-1)×100% = 8%. If 0 < b < 1 (like 0.92), it's decay at (1-b)×100% = 8% decay. Let's check if this is exponential by finding ratios: From year 0 to 1: 21,600/20,000 = 1.08. From year 1 to 2: 23,328/21,600 = 1.08. From year 2 to 3: 25,194.24/23,328 = 1.08. All ratios equal 1.08, confirming exponential! Since 1.08 is greater than 1, this is growth. The percent rate is 1.08 - 1 = 0.08 = 8%. Choice A correctly identifies this as exponential growth at 8% per year because all consecutive ratios equal 1.08 and 1.08 - 1 = 0.08 = 8% growth. Choice C confuses exponential with linear: it sees the pattern of increasing values and calculates the difference 21,600 - 20,000 = 1,600, but we need to check how they're changing. The differences are NOT constant (21,600 - 20,000 = 1,600; 23,328 - 21,600 = 1,728; 25,194.24 - 23,328 = 1,866.24), but the ratios ARE constant (all 1.08). Constant addition = linear, constant multiplication = exponential! The ratio test for exponential from a table: (1) Divide consecutive y-values: y₂/y₁, y₃/y₂, y₄/y₃, etc., (2) If all ratios are equal, it's exponential and that ratio is your growth/decay factor b, (3) If b > 1, it's growth; if 0 < b < 1, it's decay, (4) Calculate percent rate: r = b - 1, convert to percent. Example: ratios all equal 1.06 → exponential growth, 6% per period. Easy!

Question 2

A bacteria culture starts with 1000 cells and increases by 20% each hour. Which statement correctly identifies the growth factor bb and the percent rate rr per hour in the exponential form N(t)=a⋅btN(t)=a\cdot b^t?

  1. b=1.02b=1.02 and r=0.02r=0.02 (2% growth)
  2. b=0.80b=0.80 and r=−0.20r=-0.20 (20% decay)
  3. b=1.20b=1.20 and r=0.20r=0.20 (20% growth) (correct answer)
  4. b=20b=20 and r=20r=20 (20% growth)
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. Constant percent growth means the quantity is multiplied by the same factor greater than 1 each time period: if a population grows by 5% per year, it's multiplied by 1.05 each year (since 105% = 1 + 0.05 = 1.05 means 'keep all of what you had plus gain 5% more'). This creates exponential growth where the amount added each period gets larger because you're taking a percent of an increasing base! The context describes 'a bacteria culture starts with 1000 cells and increases by 20% each hour.' Key phrase: 'increases by 20% each hour' directly tells us this is exponential growth with a constant percent rate. Each hour, the quantity is multiplied by 1 + 0.20 = 1.20, making this exponential rather than linear. In one hour, you have 120% of what you started with (original 100% plus 20%). Choice B correctly identifies this as b=1.20 and r=0.20 (20% growth) because the growth factor includes the original 100% plus the 20% increase. Choice A says growth when it's actually decay (or vice versa): looking at the context, since it describes increase, this is growth, not decay. When the base is greater than 1, or when the context says 'increases,' that's exponential growth! To find percent rate from a growth/decay factor: (1) Identify b (the base or factor), (2) Subtract 1: r = b - 1, (3) Convert to percent: multiply by 100. Example: b = 1.12 → r = 0.12 → 12% growth. For decay: b = 0.95 → r = -0.05 → 5% decay (we usually state as positive '5% decay' rather than 'negative 5%'). The subtraction of 1 is the crucial step! Context language decoder for exponential: 'grows by X% per year,' 'decreases by X% per month,' 'X% interest compounded,' 'doubles every,' 'halves every,' 'increases X-fold' → all signal constant percent change (exponential). But 'adds $X per period' or 'increases by X units' → constant additive change (linear). The 'percent per period' pattern is the key giveaway!

Question 3

Which situation involves a constant percent rate of change (exponential), not a constant additive change (linear)?

(a) A gym membership costs $40 per month plus a one-time $20 sign-up fee.

(b) A car's value decreases by 10% each year.

(c) A water tank is filled at 3 gallons per minute.

(d) A plant grows 2 cm each week.​​

  1. Only (b) (correct answer)
  2. Only (a) and (c)
  3. Only (b) and (d)
  4. All of them
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. The key difference: linear growth adds the same amount each time (constant rate: +50, +50, +50), while exponential growth multiplies by the same percent each time (constant ratio: ×1.1, ×1.1, ×1.1). To check which: if differences are constant, it's linear; if ratios are constant, it's exponential. Example: 100, 150, 200, 250 has constant differences (+50) = linear. But 100, 110, 121, 133.1 has constant ratios (×1.1) = exponential! Let's contrast: (a) gym membership costs $40 per month involves adding the same amount each time—constant additive change. (b) car's value decreases by 10% each year involves multiplying by the same percent each time—constant multiplicative change. (c) water tank filled at 3 gallons per minute is adding 3 gallons each time. (d) plant grows 2 cm each week is adding 2 cm each time. Only (b) has constant ratios, while the others have constant differences. This is exponential decay! Choice A correctly identifies only (b) as involving constant percent rate because 'decreases by 10% each year' means the car retains 90% of its value each year—multiply by 0.90, which is exponential decay. Choice C sees 'percent' in option (b) and correctly identifies it as exponential, but mistakenly includes (d) which says 'grows 2 cm each week'—that's adding the same length each time, not multiplying by the same percent. Growing by a fixed amount is linear, not exponential! Context language decoder for exponential: 'grows by X% per year,' 'decreases by X% per month,' 'X% interest compounded,' 'doubles every,' 'halves every,' 'increases X-fold' → all signal constant percent change (exponential). But 'adds $X per period' or 'increases by X units' → constant additive change (linear). The 'percent per period' pattern is the key giveaway!

Question 4

A medication amount in the bloodstream is modeled by A(t)=80(0.90)tA(t)=80(0.90)^t, where tt is measured in hours. What is the percent rate of change per hour?

  1. 0.90% decay per hour
  2. 10% decay per hour (correct answer)
  3. 10% growth per hour
  4. 90% decay per hour
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. Constant percent decay means the quantity is multiplied by the same factor between 0 and 1 each time period: if a car depreciates by 15% per year, it's multiplied by 0.85 each year (since keeping 85% = 1 - 0.15 = 0.85 means 'you lose 15%'). The quantity shrinks exponentially, approaching but never quite reaching zero. Looking at the function A(t)=80(0.90)^t: the base 0.90 is less than 1, indicating exponential decay. The percent rate is calculated from r = 0.90 - 1 = -0.10 = 10% decay. This means each time t increases by 1, A is multiplied by 0.90, which is a 10% decrease. Choice C correctly identifies this as 10% decay per hour because the base < 1 and |r| = 0.10 confirms the rate. Choice B has the percent rate wrong: the base 0.90 doesn't mean 90% decay. When b = 0.90, we subtract 1 to get the rate: 0.90 - 1 = -0.10 = 10% decay. The base includes the remaining 90% (the '0.90'), so the decay is 10%, not 90%! To find percent rate from a growth/decay factor: (1) Identify b (the base or factor), (2) Subtract 1: r = b - 1, (3) Convert to percent: multiply by 100. Example: b = 1.12 → r = 0.12 → 12% growth. For decay: b = 0.95 → r = -0.05 → 5% decay (we usually state as positive '5% decay' rather than 'negative 5%'). The subtraction of 1 is the crucial step!

Question 5

Consider the function y=300(0.90)ty = 300(0.90)^t, where tt is measured in months. What is the constant percent change per month, and is it growth or decay?

  1. 0.90% decay per month
  2. 90% decay per month
  3. 10% growth per month
  4. 10% decay per month (correct answer)
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. Constant percent decay means the quantity is multiplied by the same factor between 0 and 1 each time period: if a car depreciates by 15% per year, it's multiplied by 0.85 each year (since keeping 85% = 1 - 0.15 = 0.85 means 'you lose 15%'). The quantity shrinks exponentially, approaching but never quite reaching zero. Looking at the function y = 300(0.90)^t: the base 0.90 is less than 1, indicating exponential decay. The percent rate is calculated from r = 0.90 - 1 = -0.10, but we state it as 10% decay (since decay rate is 1 - 0.90 = 0.10 = 10%). This means each time t increases by 1, y is multiplied by 0.90, which is a 10% decrease. Choice C correctly identifies this as 10% decay per month because the base 0.90 implies keeping 90%, so losing 10%. Choice B has the percent rate wrong: the base 0.90 doesn't mean 90% decay. When b = 0.90, we subtract from 1 to get the decay rate: 1 - 0.90 = 0.10 = 10% decay. The base represents the portion kept (90%), so the decay is the remaining 10%! To find percent rate from a growth/decay factor: (1) Identify b (the base or factor), (2) Subtract 1: r = b - 1, (3) Convert to percent: multiply by 100. Example: b = 1.12 → r = 0.12 → 12% growth. For decay: b = 0.95 → r = -0.05 → 5% decay (we usually state as positive '5% decay' rather than 'negative 5%'). The subtraction of 1 is the crucial step! Context language decoder for exponential: 'grows by X% per year,' 'decreases by X% per month,' 'X% interest compounded,' 'doubles every,' 'halves every,' 'increases X-fold' → all signal constant percent change (exponential). But 'adds $X per period' or 'increases by X units' → constant additive change (linear). The 'percent per period' pattern is the key giveaway!

Question 6

A medication amount in the bloodstream decreases by 20% every hour. Which function represents the amount A(t)A(t) after tt hours if A(0)=80A(0)=80 mg?

  1. A(t)=80(1.20)tA(t)=80(1.20)^t
  2. A(t)=80(0.80)tA(t)=80(0.80)^t (correct answer)
  3. A(t)=80−20tA(t)=80-20t
  4. A(t)=80−0.20tA(t)=80-0.20t
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. Constant percent decay means the quantity is multiplied by the same factor between 0 and 1 each time period: if a medication decreases by 20% per hour, it's multiplied by 0.80 each hour (since keeping 80% = 1 - 0.20 = 0.80 means 'you lose 20%'). The quantity shrinks exponentially, approaching but never quite reaching zero. The context describes 'decreases by 20% every hour.' Key phrase: 'decreases by 20%' directly tells us this is exponential decay with a constant percent rate. Each hour, the quantity is multiplied by 1 - 0.20 = 0.80, making this exponential rather than linear. In one hour, you have 80% of what you started with (original 100% minus 20%). Choice B correctly identifies this as A(t) = 80(0.80)^t because decreasing by 20% means multiplying by 0.80 each hour, and the initial amount is 80 mg. Choice A says growth when it's actually decay: looking at the context, since it describes 'decreases,' this is decay, not growth. When the context says 'decreases,' 'depreciates,' or 'decays,' that's exponential decay with a base between 0 and 1! Context language decoder for exponential: 'grows by X% per year,' 'decreases by X% per month,' 'X% interest compounded,' 'doubles every,' 'halves every,' 'increases X-fold' → all signal constant percent change (exponential). But 'adds $X per period' or 'increases by X units' → constant additive change (linear). The 'percent per period' pattern is the key giveaway!

Question 7

A population is modeled by P(t)=12,000⋅(0.98)tP(t)=12{,}000\cdot(0.98)^t, where tt is in years. Which statement is correct?

  1. Exponential growth at 2% per year
  2. Neither; the base 0.980.98 means 98% decay per year
  3. Linear decay at 0.98 people per year
  4. Exponential decay at 2% per year (correct answer)
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. Constant percent decay means the quantity is multiplied by the same factor between 0 and 1 each time period: if a car depreciates by 15% per year, it's multiplied by 0.85 each year (since keeping 85% = 1 - 0.15 = 0.85 means 'you lose 15%'). The quantity shrinks exponentially, approaching but never quite reaching zero. Looking at the function P(t)=12,000·(0.98)^t: the base 0.98 is less than 1, indicating exponential decay. The percent rate is calculated from r = 0.98 - 1 = -0.02 = 2% decay. This means each time t increases by 1, P is multiplied by 0.98, which is a 2% decrease. Choice B correctly identifies this as exponential decay at 2% per year because the base <1 and |r|=0.02 confirms the rate. Choice D has the percent rate wrong: the base 0.98 doesn't mean 98% decay. When b = 0.98, we subtract 1 to get the rate: 0.98 - 1 = -0.02 = 2% decay. The base includes the remaining 98% (the '0.98'), so the decay is 2%, not 98%! To find percent rate from a growth/decay factor: (1) Identify b (the base or factor), (2) Subtract 1: r = b - 1, (3) Convert to percent: multiply by 100. Example: b = 1.12 → r = 0.12 → 12% growth. For decay: b = 0.95 → r = -0.05 → 5% decay (we usually state as positive '5% decay' rather than 'negative 5%'). The subtraction of 1 is the crucial step!

Question 8

A company's number of users increases by 15% each month. Which function represents this situation if the company starts with 2,000 users at month t=0t=0?

  1. U(t)=2000(0.85)tU(t)=2000(0.85)^t
  2. U(t)=2000+0.15tU(t)=2000+0.15t
  3. U(t)=2000(1.15)tU(t)=2000(1.15)^t (correct answer)
  4. U(t)=2000(15)tU(t)=2000(15)^t
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. Constant percent growth means the quantity is multiplied by the same factor greater than 1 each time period: if a population grows by 5% per year, it's multiplied by 1.05 each year (since 105% = 1 + 0.05 = 1.05 means 'keep all of what you had plus gain 5% more'). This creates exponential growth where the amount added each period gets larger because you're taking a percent of an increasing base! The context describes 'increases by 15% each month.' Key phrase: 'increases by 15%' directly tells us this is exponential growth with a constant percent rate. Each month, the quantity is multiplied by 1 + 0.15 = 1.15, making this exponential rather than linear. In one month, you have 115% of what you started with (original 100% plus 15%). Choice C correctly identifies this as U(t) = 2000(1.15)^t because increasing by 15% means multiplying by 1.15 each month, and we start with 2000 users at t = 0. Choice A has the wrong base: 0.85 would mean decay by 15% (keeping only 85%), not growth by 15%. When something increases by 15%, we multiply by 1.15, not 0.85! Context language decoder for exponential: 'grows by X% per year,' 'decreases by X% per month,' 'X% interest compounded,' 'doubles every,' 'halves every,' 'increases X-fold' → all signal constant percent change (exponential). But 'adds $X per period' or 'increases by X units' → constant additive change (linear). The 'percent per period' pattern is the key giveaway!

Question 9

A car's value depreciates by 12% each year. After how many complete years will the car's value first drop below 50% of its original value?

  1. 4 years
  2. 5 years
  3. 6 years (correct answer)
  4. 7 years
Explanation: The car retains 88% = 0.88 of its value each year. We need (0.88)n<0.5(0.88)^n < 0.5. Testing values: (0.88)4≈0.599(0.88)^4 ≈ 0.599, (0.88)5≈0.527(0.88)^5 ≈ 0.527, (0.88)6≈0.464(0.88)^6 ≈ 0.464. After 6 years, the value first drops below 50%. Choice A uses 4 years where value is still 59.9%. Choice B uses 5 years where value is still 52.7%. Choice D overshoots the requirement.

Question 10

A company's quarterly revenue data shows the following pattern: Q1: $200,000, Q2: $240,000, Q3: $288,000, Q4: $345,600.

Based on the revenue pattern shown, what type of growth model best describes this company's performance, and what would be the projected revenue for Q5?

  1. Linear growth model with projected Q5 revenue of approximately $403,200
  2. Exponential growth model with projected Q5 revenue of approximately $414,720 (correct answer)
  3. Quadratic growth model with projected Q5 revenue of approximately $425,000
  4. Exponential decay model with projected Q5 revenue of approximately $380,000
Explanation: Checking the ratios: 240,000200,000=1.2\frac{240,000}{200,000} = 1.2, 288,000240,000=1.2\frac{288,000}{240,000} = 1.2, 345,600288,000=1.2\frac{345,600}{288,000} = 1.2. The revenue grows by 20% each quarter, indicating exponential growth. Q5 projection: 345,600×1.2=414,720345,600 \times 1.2 = 414,720. Choice A assumes linear growth (constant dollar increases). Choice C suggests quadratic growth (increasing rate of change). Choice D incorrectly identifies this as decay despite clear growth.

Question 11

A city's population grows by 8% every 2 years. What is the equivalent annual growth rate that would produce the same population after 2 years?

  1. Approximately 3.85% per year (correct answer)
  2. Exactly 4.00% per year
  3. Approximately 4.12% per year
  4. Approximately 4.25% per year
Explanation: If the population grows by 8% over 2 years, the growth factor is 1.08. For equivalent annual growth rate rr, we need (1+r)2=1.08(1 + r)^2 = 1.08. Solving: 1+r=1.08≈1.03851 + r = \sqrt{1.08} ≈ 1.0385, so r≈0.0385=3.85%r ≈ 0.0385 = 3.85\%. Choice B incorrectly halves the 8% rate. Choice C uses the approximation (1.04)2=1.0816(1.04)^2 = 1.0816 which is too high. Choice D further overestimates the required rate.

Question 12

A population of bacteria doubles every 3 hours. If the initial population is 500 bacteria, what is the percent increase in the population after 9 hours compared to the initial population?

  1. 200%
  2. 600%
  3. 700% (correct answer)
  4. 800%
Explanation: After 9 hours (3 doubling periods), the population grows by a factor of 23=82^3 = 8. The final population is 500×8=4000500 \times 8 = 4000. The percent increase is 4000−500500×100%=3500500×100%=700%\frac{4000 - 500}{500} \times 100\% = \frac{3500}{500} \times 100\% = 700\%. Choice A incorrectly uses only one doubling period. Choice B finds the final population as a percent of the initial (800%) but subtracts 100% incorrectly. Choice D gives the final population as a percent of initial without subtracting the original 100%.

Question 13

A savings account earns 4.5% interest compounded annually. A certificate of deposit earns 4.4% compounded quarterly. Which investment performs better over 3 years, and by approximately how much per $1000 invested?

  1. Savings account performs better by approximately $3.50 per $1000 invested
  2. Certificate of deposit performs better by approximately $2.80 per $1000 invested
  3. Savings account performs better by approximately $1.00 per $1000 invested
  4. Certificate of deposit performs better by approximately $1.40 per $1000 invested (correct answer)
Explanation: When comparing investments with different compounding frequencies, you need to calculate the final amount for each option using the compound interest formula: A=P(1+rn)ntA = P(1 + \frac{r}{n})^{nt}, where P is principal, r is annual rate, n is compounding frequency per year, and t is time in years. For the savings account: A=1000(1+0.0451)1×3=1000(1.045)3=$1,141.17A = 1000(1 + \frac{0.045}{1})^{1 \times 3} = 1000(1.045)^3 = \$1,141.17 For the certificate of deposit: A=1000(1+0.0444)4×3=1000(1.011)12=$1,142.55A = 1000(1 + \frac{0.044}{4})^{4 \times 3} = 1000(1.011)^{12} = \$1,142.55 The CD earns $1,142.55 - $1,141.17 = $1.38 more per $1000, which rounds to approximately $1.40. Choice A incorrectly states the savings account performs better and uses an inflated difference that might result from calculation errors. Choice B has the right winner (CD) but claims a $2.80 difference, likely from confusing annual with quarterly compounding or other computational mistakes. Choice C incorrectly identifies the savings account as the winner with a $1.00 difference, possibly from rounding errors or using simple rather than compound interest. Choice D correctly identifies that the certificate of deposit performs better by approximately $1.40. Remember that more frequent compounding can sometimes overcome a slightly lower interest rate. Always calculate both scenarios completely rather than assuming the higher annual rate automatically wins.

Question 14

A video game has 500 active players at the start of the month. The number of active players increases by 10% each month. Which statement best describes this change?

  1. Exponential growth at 10% per month (multiply by 1.101.10 each month) (correct answer)
  2. Neither, because percent change cannot be constant over time
  3. Linear growth because the number of players increases by a constant amount each month
  4. Exponential decay at 10% per month (multiply by 0.900.90 each month)
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. Constant percent growth means the quantity is multiplied by the same factor greater than 1 each time period: if a population grows by 5% per year, it's multiplied by 1.05 each year (since 105% = 1 + 0.05 = 1.05 means 'keep all of what you had plus gain 5% more'). This creates exponential growth where the amount added each period gets larger because you're taking a percent of an increasing base! The context describes 'the number of active players increases by 10% each month.' Key phrase: 'increases by 10%' directly tells us this is exponential growth with a constant percent rate. Each month, the quantity is multiplied by 1 + 0.10 = 1.10, making this exponential rather than linear. In one year, you have 1.10 × 100% of what you started with (original 100% plus 10%). Choice A correctly identifies this as exponential growth at 10% per month because the context percent language shows constant multiplicative change by 1.10. Choice C confuses exponential with linear: it sees the pattern of increasing values and assumes linear, but we need to check how they're changing. Showing that differences are not constant (e.g., first month +50 from 500, next +55 from 550) while ratios are constant (×1.10) means exponential! Constant addition = linear, constant multiplication = exponential! Context language decoder for exponential: 'grows by X% per year,' 'decreases by X% per month,' 'X% interest compounded,' 'doubles every,' 'halves every,' 'increases X-fold' → all signal constant percent change (exponential). But 'adds $X per period' or 'increases by X units' → constant additive change (linear). The 'percent per period' pattern is the key giveaway! Real-world clue: exponential growth/decay contexts involve processes where the amount of change depends on how much you currently have: 'Population grows by 5% per year' means a population of 1000 gains 50, but a population of 10,000 gains 500—the change is bigger when the base is bigger. That's exponential! Linear is when you add the same amount regardless of current size.

Question 15

The function y=200(1.05)ty = 200(1.05)^t models the number of bacteria after tt hours. What is the constant percent rate of change per hour?

  1. 5% decay per hour
  2. 0.05% growth per hour
  3. 5% growth per hour (correct answer)
  4. 105% growth per hour
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. Constant percent growth means the quantity is multiplied by the same factor greater than 1 each time period: if a population grows by 5% per year, it's multiplied by 1.05 each year (since 105% = 1 + 0.05 = 1.05 means 'keep all of what you had plus gain 5% more'). This creates exponential growth where the amount added each period gets larger because you're taking a percent of an increasing base! Looking at the function y = 200(1.05)^t: the base 1.05 is greater than 1, indicating exponential growth. The percent rate is calculated from r = 1.05 - 1 = 0.05 = 5% growth. This means each time t increases by 1, y is multiplied by 1.05, which is a 5% increase. Choice A correctly identifies this as 5% growth per hour because the base > 1 and r = 0.05 confirms the rate. Choice C has the percent rate wrong: the base 1.05 doesn't mean 105% growth. When b = 1.05, we subtract 1 to get the rate: 1.05 - 1 = 0.05 = 5% growth. The base includes the original 100% (the '1') plus the growth rate (the '0.05'), so it's 5%, not 105%! To find percent rate from a growth/decay factor: (1) Identify b (the base or factor), (2) Subtract 1: r = b - 1, (3) Convert to percent: multiply by 100. Example: b = 1.12 → r = 0.12 → 12% growth. For decay: b = 0.95 → r = -0.05 → 5% decay (we usually state as positive '5% decay' rather than 'negative 5%'). The subtraction of 1 is the crucial step!

Question 16

A savings account balance is shown below. From the table, determine whether the balance changes by a constant percent rate per month, and if so, find the percent rate.

tt (months): 0, 1, 2, 3, 4 Balance ($): 500, 525, 551.25, 578.8125, 607.753125

  1. Yes; exponential growth at 25% per month
  2. Yes; exponential growth at 5% per month (correct answer)
  3. No; the ratios are not constant, so it is neither exponential nor linear
  4. Yes; linear growth adding $25 each month
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. From a table, to identify exponential with constant percent rate: divide consecutive y-values to find ratios. If y₂/y₁ = y₃/y₂ = y₄/y₃ = same number, that's your growth/decay factor b. If b > 1 (like 1.05), it's growth at (b-1)×100% = 5%. If 0 < b < 1 (like 0.95), it's decay at (1-b)×100% = 5% decay. Let's check if this is exponential by finding ratios: From month 0 to 1: 525/500 = 1.05. From month 1 to 2: 551.25/525 = 1.05. From month 2 to 3: 578.8125/551.25 = 1.05. From month 3 to 4: 607.753125/578.8125 = 1.05. All ratios equal 1.05, confirming exponential! Since 1.05 is greater than 1, this is growth. The percent rate is 1.05 - 1 = 0.05 = 5%. Choice A correctly identifies this as exponential growth at 5% per month because all consecutive ratios equal 1.05, showing constant multiplicative change of 5% growth. Choice C has the percent rate wrong: the growth factor 1.05 means 5% growth, not 25%. When b = 1.05, we subtract 1 to get the rate: 1.05 - 1 = 0.05 = 5% growth. The base includes the original 100% (the '1') plus the growth rate (the '0.05'), so it's 5%, not 25%! The ratio test for exponential from a table: (1) Divide consecutive y-values: y₂/y₁, y₃/y₂, y₄/y₃, etc., (2) If all ratios are equal, it's exponential and that ratio is your growth/decay factor b, (3) If b > 1, it's growth; if 0 < b < 1, it's decay, (4) Calculate percent rate: r = b - 1, convert to percent. Example: ratios all equal 1.05 → exponential growth, 5% per period. Easy!

Question 17

A gym membership fee is tracked over months:

Month mm: 0, 1, 2, 3 Fee F(m)F(m): $40, $45, $50, $55

Does this represent constant percent change (exponential) or constant additive change (linear)?

  1. Constant percent growth (exponential)
  2. Constant additive growth (linear) (correct answer)
  3. Exponential decay at 5% per month
  4. Neither; it cannot be classified
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. The key difference: linear growth adds the same amount each time (constant rate: +50, +50, +50), while exponential growth multiplies by the same percent each time (constant ratio: ×1.1, ×1.1, ×1.1). To check which: if differences are constant, it's linear; if ratios are constant, it's exponential. Example: 100, 150, 200, 250 has constant differences (+50) = linear. But 100, 110, 121, 133.1 has constant ratios (×1.1) = exponential! Let's check if this is exponential by finding ratios: From month 0 to 1: 45/40 = 1.125. From month 1 to 2: 50/45 = 1.111... From month 2 to 3: 55/50 = 1.1. The ratios differ (1.125, 1.111, 1.1), so this is NOT exponential. Now check differences: 45 - 40 = 5; 50 - 45 = 5; 55 - 50 = 5. All differences equal $5, confirming linear growth! Choice B correctly identifies this as constant additive growth (linear) because the fee increases by exactly 5eachmonth,notbyaconstantpercent.ChoiceAconfuseslinearwithexponential:itseesthepatternofincreasingvaluesandassumesexponential,butweneedtocheckhowthey′rechanging.Thedifferencesareconstant(+5 each month, not by a constant percent. Choice A confuses linear with exponential: it sees the pattern of increasing values and assumes exponential, but we need to check how they're changing. The differences are constant (+5 each month), while the ratios are NOT constant. Constant addition = linear, constant multiplication = exponential! Exponential vs linear quick-check: calculate both differences AND ratios. If differences are constant (like +5, +5, +5), it's linear. If ratios are constant (like ×1.1, ×1.1, ×1.1), it's exponential. Can't be both! This two-part check prevents confusion between the types.

Question 18

The value of a collectible is recorded each month:

Month tt: 0, 1, 2, 3 Value ($): 200, 230, 264.5, 304.175

From the data, determine if there is constant percent change. If so, what is the percent rate per month?​​

  1. Exponential growth at 15% per month (correct answer)
  2. Exponential decay at 15% per month
  3. Linear growth: increases by $30 each month
  4. Exponential growth at 30% per month
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. From a table, to identify exponential with constant percent rate: divide consecutive y-values to find ratios. If y₂/y₁ = y₃/y₂ = y₄/y₃ = same number, that's your growth/decay factor b. If b > 1 (like 1.08), it's growth at (b-1)×100% = 8%. If 0 < b < 1 (like 0.92), it's decay at (1-b)×100% = 8% decay. Let's check if this is exponential by finding ratios: From month 0 to 1: 230/200 = 1.15. From month 1 to 2: 264.5/230 = 1.15. From month 2 to 3: 304.175/264.5 = 1.15. All ratios equal 1.15, confirming exponential! Since 1.15 is greater than 1, this is growth. The percent rate is 1.15 - 1 = 0.15 = 15%. Choice A correctly identifies this as exponential growth at 15% per month because all consecutive ratios equal 1.15, which represents multiplying by 1.15 each month—a 15% increase. Choice C confuses exponential with linear: it sees the first increase of $30 (from 200 to 230) and assumes linear growth, but we need to check how they're changing. The next increase is $34.50, then $39.675—the increases themselves are growing! Constant ratios (×1.15) mean exponential, not linear. The ratio test for exponential from a table: (1) Divide consecutive y-values: y₂/y₁, y₃/y₂, y₄/y₃, etc., (2) If all ratios are equal, it's exponential and that ratio is your growth/decay factor b, (3) If b > 1, it's growth; if 0 < b < 1, it's decay, (4) Calculate percent rate: r = b - 1, convert to percent. Example: ratios all equal 1.06 → exponential growth, 6% per period. Easy!

Question 19

A company's revenue over quarters is listed below:

Quarter tt: 0, 1, 2, 3, 4 Revenue ($): 1,000; 1,100; 1,210; 1,331; 1,464.1

From the table, determine if there is constant percent change and identify the percent rate per quarter.​​

  1. Exponential growth at 10% per quarter (correct answer)
  2. Exponential decay at 10% per quarter
  3. Linear growth: increases by $110 per quarter
  4. Exponential growth at 1.10% per quarter
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. From a table, to identify exponential with constant percent rate: divide consecutive y-values to find ratios. If y₂/y₁ = y₃/y₂ = y₄/y₃ = same number, that's your growth/decay factor b. If b > 1 (like 1.08), it's growth at (b-1)×100% = 8%. If 0 < b < 1 (like 0.92), it's decay at (1-b)×100% = 8% decay. Let's check if this is exponential by finding ratios: From quarter 0 to 1: 1,100/1,000 = 1.10. From quarter 1 to 2: 1,210/1,100 = 1.10. From quarter 2 to 3: 1,331/1,210 = 1.10. From quarter 3 to 4: 1,464.1/1,331 = 1.10. All ratios equal 1.10, confirming exponential! Since 1.10 is greater than 1, this is growth. The percent rate is 1.10 - 1 = 0.10 = 10%. Choice A correctly identifies this as exponential growth at 10% per quarter because all consecutive ratios equal 1.10, which means revenue is multiplied by 1.10 each quarter—a 10% increase. Choice D identifies the growth factor correctly as 1.10 but then mistakes the percent: the percent rate is NOT 1.10%, it's 10%! If b = 1.10, the percent growth is 10% (not 1.10%). Factor - 1 = rate! The 1.10 represents 110% of the original (100% + 10% growth). To find percent rate from a growth/decay factor: (1) Identify b (the base or factor), (2) Subtract 1: r = b - 1, (3) Convert to percent: multiply by 100. Example: b = 1.12 → r = 0.12 → 12% growth. For decay: b = 0.95 → r = -0.05 → 5% decay (we usually state as positive '5% decay' rather than 'negative 5%'). The subtraction of 1 is the crucial step!

Question 20

Which situation involves a constant percent rate of change (exponential) rather than a constant additive change (linear)?

  1. A bank account earns 2% interest each month (correct answer)
  2. A tank is filled by adding 5 liters every minute
  3. A runner increases distance by 1 mile each week
  4. A movie ticket price increases by $1 each year
Explanation: This question tests your ability to recognize exponential relationships—situations where a quantity grows or decays by a constant percent rate per time period, which is very different from linear growth where you add the same amount each time. The key difference: linear growth adds the same amount each time (constant rate: +50, +50, +50), while exponential growth multiplies by the same percent each time (constant ratio: ×1.1, ×1.1, ×1.1). To check which: if differences are constant, it's linear; if ratios are constant, it's exponential. Example: 100, 150, 200, 250 has constant differences (+50) = linear. But 100, 110, 121, 133.1 has constant ratios (×1.1) = exponential! Let's contrast: Options B, C, and D all describe adding the same amount each time—constant additive change. But option A describes earning 2% interest each month, which involves multiplying by the same percent each time—constant multiplicative change. The exponential case has the account balance multiplied by 1.02 each month, while the linear cases would have constant differences like +5 liters, +1 mile, or +$1. This is exponential growth! Choice A correctly identifies the bank account earning 2% interest as exponential because the balance is multiplied by 1.02 each month (constant percent change), not increased by a fixed dollar amount. Choices B, C, and D all describe linear situations: they involve adding a constant amount (5 liters, 1 mile, $1) rather than multiplying by a constant factor. Adding the same amount = linear, multiplying by the same percent = exponential! Context language decoder for exponential: 'grows by X% per year,' 'decreases by X% per month,' 'X% interest compounded,' 'doubles every,' 'halves every,' 'increases X-fold' → all signal constant percent change (exponential). But 'adds $X per period' or 'increases by X units' → constant additive change (linear). The 'percent per period' pattern is the key giveaway!