Algebra Quiz: Recognize Percent Growth Or Decay

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.

Question 1

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$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!

Question 2

1. Exponential growth at 2% per year
2. Neither; the base $0.98$ 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 3

1. Exponential growth at 15% per year (multiply by $1.15$ each year)
2. Linear decrease by \3{,}000$ each year
3. Neither; depreciation is always linear
4. Exponential decay at 15% per year (multiply by $0.85$ each 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. The context describes 'a car worth $20,000 depreciates by 15$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 4

1. Exponential growth: increases 102% per year
2. Exponential growth: increases 2% per year (correct answer)
3. Linear growth: increases by 2 people per year
4. Exponential decay: decreases 2% per 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. Constant percent growth means the quantity is multiplied by the same factor greater than 1 each time period: if a population grows by 2% per year, it's multiplied by 1.02 each year (since 102% = 1 + 0.02 = 1.02 means 'keep all of what you had plus gain 2% 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 P(t) = 12000(1.02)^t: the base 1.02 is greater than 1, indicating exponential growth. The percent rate is calculated from r = 1.02 - 1 = 0.02 = 2% growth. This means each time t increases by 1, P is multiplied by 1.02, which is a 2% increase. Choice C correctly identifies this as exponential growth: increases 2% per year because the base 1.02 means the population grows by 2% annually. Choice D has the percent rate wrong: the base 1.02 doesn't mean 102% growth. When b = 1.02, we subtract 1 to get the rate: 1.02 - 1 = 0.02 = 2% growth. The base includes the original 100% (the '1') plus the growth rate (the '0.02'), so it's 2%, not 102%! 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 = 1.02 → r = 0.02 → 2% growth. The subtraction of 1 is the crucial step!

Question 5

1. $A(t)=80(1.20)^t$
2. $A(t)=80(0.80)^t$ (correct answer)
3. $A(t)=80-20t$
4. $A(t)=80-0.20t$

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 6

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 7

1. $b=1.02$ and $r=0.02$ (2% growth)
2. $b=0.80$ and $r=-0.20$ (20% decay)
3. $b=1.20$ and $r=0.20$ (20% growth) (correct answer)
4. $b=20$ and $r=20$ (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 8

1. Exponential growth at 10% per year
2. Linear decay: decreases by 1,000 per year
3. Exponential decay at 10% per year (correct answer)
4. Exponential decay at 90% per 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. 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: 9,000/10,000 = 0.90. From year 1 to 2: 8,100/9,000 = 0.90. From year 2 to 3: 7,290/8,100 = 0.90. All ratios equal 0.90, confirming exponential! Since 0.90 is less than 1, this is decay. The percent rate is 0.90 - 1 = -0.10 = 10% decay. Choice C correctly identifies this as exponential decay at 10% per year because all consecutive ratios equal 0.90, meaning the population is multiplied by 0.90 each year—it retains 90% and loses 10%. Choice D says decay when it's actually decay (or vice versa): looking at the base 0.90, since 0.90 < 1, this is decay, not growth. When the base is between 0 and 1, or when the context shows decreasing values, that's exponential decay! But choice D also mistakes 0.90 for meaning 90% decay—the decay rate is 10%, not 90%! 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 9

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 $40 per month involves adding the same amount each time—constant additive change. (b) car's value decreases by 10$X per period' or 'increases by X units' → constant additive change (linear). The 'percent per period' pattern is the key giveaway!

Question 10

1. Plan 1 is exponential; Plan 2 is linear
2. Both plans are linear because they increase each month
3. Plan 1 is linear (constant additive change); Plan 2 is exponential growth (constant percent change) (correct answer)
4. Both plans are exponential because they increase 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. 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: Plan 1 involves adding the same amount each time—constant additive change. But Plan 2 involves multiplying by the same percent each time—constant multiplicative change. The exponential case has constant ratios (like ×1.05 each month), while a linear case would have constant differences (like +50 each month). This is exponential growth for Plan 2 and linear for Plan 1! Choice C correctly identifies this as Plan 1 linear (constant additive change) and Plan 2 exponential growth (constant percent change) because adding a fixed amount is linear, while percent increase is exponential. Choice D says both are exponential because they increase each month, but that's confusing the outcome (increase) with the mechanism: 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. 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 11

1. 1.20% growth per week
2. 20% growth per week (correct answer)
3. 20% decay per week
4. 120% growth per week

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 description: a quantity is multiplied by 1.20 each week, so the base 1.20 is greater than 1, indicating exponential growth. The percent rate is calculated from r = 1.20 - 1 = 0.20 = 20% growth. This means each week, the quantity is multiplied by 1.20, which is a 20% increase. Choice A correctly identifies this as 20% growth per week because the base >1 and r=0.20 confirms the rate. Choice D has the percent rate wrong: the base 1.20 doesn't mean 120% growth. When b = 1.20, we subtract 1 to get the rate: 1.20 - 1 = 0.20 = 20% growth. The base includes the original 100% (the '1') plus the growth rate (the '0.20'), so it's 20%, not 120%! 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 12

1. Exponential growth: it multiplies by $1.10$ each hour. (correct answer)
2. Exponential decay: it decreases 10% each hour.
3. Neither: constant percent change does not match an exponential function.
4. Linear growth: it adds 10 mg each 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 10% per hour, it's multiplied by 1.10 each hour (since 110% = 1 + 0.10 = 1.10 means 'keep all of what you had plus gain 10% 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 'grows by a constant percent rate of 10% each hour.' Key phrase: 'grows by 10%' directly tells us this is exponential growth with a constant percent rate. Each hour, the quantity is multiplied by 1 + 0.10 = 1.10, making this exponential rather than linear. In one hour, you have 110% of what you started with (original 100% plus 10%). Choice C correctly identifies this as exponential growth because it states the bacteria multiplies by 1.10 each hour, which is exactly what 10% growth means. Choice A confuses exponential with linear: it sees the pattern of increasing values and assumes linear addition, but 10% growth means multiplying by 1.10, not adding 10 mg. Constant addition = linear, constant multiplication = exponential! Real-world clue: exponential growth/decay contexts involve processes where the amount of change depends on how much you currently have: 'Bacteria grows by 10% per hour' means 200 mg gains 20 mg in the first hour, but 220 mg gains 22 mg in the second hour—the change is bigger when the base is bigger. That's exponential!

Question 13

1. Yes; exponential growth at 5% per time unit
2. Yes; exponential decay at 10% per time unit
3. Yes; exponential growth at 10% per time unit
4. No; it is linear (constant additive change), not constant percent change (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 t = 0 to 1: 55/50 = 1.1. From t = 1 to 2: 60/55 = 1.091... From t = 2 to 3: 65/60 = 1.083... The ratios differ (1.1, 1.091, 1.083), so this is NOT exponential. Now check differences: 55 - 50 = 5; 60 - 55 = 5; 65 - 60 = 5. All differences equal 5, confirming linear growth! Choice C correctly identifies this as linear (constant additive change), not constant percent change, because the quantity increases by exactly 5 each time period, not by a constant percent. Choice A says exponential growth at 10% when it's actually linear: it calculates only the first ratio 55/50 = 1.1 and concludes it's exponential without checking others. To confirm constant percent change, you must verify that ALL consecutive ratios are equal! If even one ratio differs, it's not exponential growth/decay. Always check at least 3 intervals. 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 14

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 $(0.88)^n < 0.5$. Testing values: $(0.88)^4 ≈ 0.599$, $(0.88)^5 ≈ 0.527$, $(0.88)^6 ≈ 0.464$. 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 15

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: $\frac{240,000}{200,000} = 1.2$, $\frac{288,000}{240,000} = 1.2$, $\frac{345,600}{288,000} = 1.2$. The revenue grows by 20% each quarter, indicating exponential growth. Q5 projection: $345,600 \times 1.2 = 414,720$. 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 16

1. Fund A grows faster because 6% annually exceeds 6% annually from Fund B
2. Fund B grows faster because it compounds more frequently at the same effective rate
3. Both funds have identical growth because they both average 6% per year
4. Fund B grows faster because its annual rate is approximately 6.09% versus Fund A's 6% (correct answer)

Explanation: When comparing investment funds with different compounding frequencies, you need to calculate the effective annual rate to make a fair comparison. The key insight is that more frequent compounding at the same nominal rate produces higher returns. Fund A is straightforward: 6% annually means exactly 6% growth per year. Fund B compounds every 6 months at 3% per period, so it compounds twice yearly. To find Fund B's effective annual rate, use the compound interest formula: $(1 + 0.03)^2 - 1 = 1.0609 - 1 = 0.0609$ or 6.09%. Since 6.09% > 6%, Fund B grows faster. Looking at the wrong answers: Choice A incorrectly assumes Fund B also has a 6% annual rate, missing that 3% every 6 months compounds to more than 6% annually. Choice B correctly identifies that Fund B grows faster due to more frequent compounding, but wrongly claims both funds have the same effective rate—they don't. Choice C falls into the trap of thinking that 3% twice per year simply averages to 6% annually, ignoring the compounding effect entirely. Choice D correctly identifies that Fund B's effective annual rate (6.09%) exceeds Fund A's rate (6%), making it the faster-growing investment. Study tip: Whenever you see different compounding frequencies, always convert to the same time period for comparison. More frequent compounding at a given rate always beats less frequent compounding, even if the rates look similar at first glance.

Question 17

1. 80 grams, which is 25.0% of the original amount
2. 40 grams, which is 12.5% of the original amount (correct answer)
3. 53.33 grams, which is 16.67% of the original amount
4. 160 grams, which is 50.0% of the original amount

Explanation: When you encounter radioactive decay problems, you're dealing with exponential decay where the amount halves at regular intervals. The key is understanding what "half-life" means and systematically working through each time period. Half-life means that every 6 hours, exactly half of the substance remains. Starting with 320 grams, let's trace what happens:

• After 6 hours (1 half-life): $320 ÷ 2 = 160$ grams
• After 12 hours (2 half-lives): $160 ÷ 2 = 80$ grams
• After 18 hours (3 half-lives): $80 ÷ 2 = 40$ grams

Question 18

1. 33.75 mg/L (correct answer)
2. 45.0 mg/L
3. 40.0 mg/L
4. 20.0 mg/L

Explanation: The concentration retains 75% = 0.75 every 4 hours. After 12 hours (3 periods of 4 hours): $80 \times (0.75)^3 = 80 \times 0.421875 = 33.75$ mg/L. Choice B incorrectly uses only 2 decay periods. Choice C subtracts 25% only once. Choice D subtracts 25% three times linearly: $80 - 3(0.25 \times 80) = 80 - 60 = 20$.

Question 19

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 $r$, we need $(1 + r)^2 = 1.08$. Solving: $1 + r = \sqrt{1.08} ≈ 1.0385$, so $r ≈ 0.0385 = 3.85\%$. Choice B incorrectly halves the 8% rate. Choice C uses the approximation $(1.04)^2 = 1.0816$ which is too high. Choice D further overestimates the required rate.

Question 20

1. 200%
2. 600%
3. 700% (correct answer)
4. 800%

Explanation: After 9 hours (3 doubling periods), the population grows by a factor of $2^3 = 8$. The final population is $500 \times 8 = 4000$. The percent increase is $\frac{4000 - 500}{500} \times 100\% = \frac{3500}{500} \times 100\% = 700\%$. 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%.