Math 1 Quiz: Percent Change And Growth Factors
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Percent Change And Growth FactorsQuestion 1 of 18

The price of gasoline increased 18% in January, decreased 8% in February, and increased 5% in March. What single percent change is equivalent to these three monthly changes?

15.0% increase
13.7% increase
16.8% increase
14.3% increase
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Math 1 Quiz

Math 1 Quiz: Percent Change And Growth Factors

Practice Percent Change And Growth Factors 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 Percent Change And Growth Factors, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

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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

The price of gasoline increased 18% in January, decreased 8% in February, and increased 5% in March. What single percent change is equivalent to these three monthly changes?

  1. 15.0% increase
  2. 13.7% increase
  3. 16.8% increase
  4. 14.3% increase (correct answer)
Explanation: When you encounter consecutive percentage changes, you can't simply add or subtract them—you must apply each change to the result of the previous change. This is because each percentage operates on a different base amount. Let's say the initial price is PP. After January's 18% increase, the price becomes P×1.18P \times 1.18. In February, the 8% decrease applies to this new amount: P×1.18×0.92P \times 1.18 \times 0.92. Finally, March's 5% increase gives us: P×1.18×0.92×1.05P \times 1.18 \times 0.92 \times 1.05. Calculating step by step: 1.18×0.92=1.08561.18 \times 0.92 = 1.0856, then 1.0856×1.05=1.143881.0856 \times 1.05 = 1.14388. This means the final price is 114.388% of the original, representing a 14.388% increase, which rounds to 14.3%. Choice A (15.0% increase) is what you'd get if you simply added the changes: 18%−8%+5%=15%18\% - 8\% + 5\% = 15\%. This ignores the compounding effect. Choice B (13.7% increase) likely results from calculation errors in the multiplication process. Choice C (16.8% increase) might come from incorrectly treating the February decrease as an increase or making sign errors. The key insight is that percentage changes compound—each change affects the running total, not the original amount. Always multiply the decimal equivalents (add 1 to increases, subtract from 1 for decreases) rather than adding the percentages directly. This compound effect appears frequently in finance, population growth, and price change problems.

Question 2

A population decreases from 8,400 to 7,140 in one year, then increases to 7,854 the following year. What was the percent change from the original population to the final population?

  1. 6.5% decrease (correct answer)
  2. 6.5% increase
  3. 10.0% increase
  4. 15.0% decrease
Explanation: The overall change is from 8,400 to 7,854. Percent change = (7,854 - 8,400)/8,400 = -546/8,400 = -0.065 = -6.5%, which is a 6.5% decrease. Choice B has the correct magnitude but wrong direction. Choice C incorrectly calculates the percent increase from year 1 to year 2: (7,854 - 7,140)/7,140 = 10%. Choice D uses the first year's decrease: (7,140 - 8,400)/8,400 = -15%.

Question 3

A store marks up an item 40% above wholesale cost, then offers a 25% discount to customers. If the wholesale cost is $60, what is the relationship between the final selling price and the wholesale cost?

  1. 5% markup over wholesale cost (correct answer)
  2. 15% markup over wholesale cost
  3. 10% discount from wholesale cost
  4. 5% discount from wholesale cost
Explanation: Starting with $60 wholesale: After 40% markup: $60 × 1.40 = $84. After 25% discount: $84 × 0.75 = $63. The final price is $63 vs. $60 wholesale, representing a $3 increase or 3/60 = 5% markup. The overall growth factor is 1.40 × 0.75 = 1.05. Choice B uses 40% - 25% = 15%. Choice C treats the result as a discount. Choice D incorrectly calculates the final percentage.

Question 4

A city's population was 125,000 in 2010 and 140,000 in 2020. Assuming the same growth factor applies each year, what will the population be in 2025?

  1. 147,500 people
  2. 151,200 people
  3. 148,400 people (correct answer)
  4. 152,600 people
Explanation: From 2010 to 2020 (10 years), the growth factor is 140,000/125,000 = 1.12. The annual growth factor is 1.12^(1/10) ≈ 1.0115. From 2020 to 2025 is 5 years, so the population will be 140,000 × 1.0115^5 ≈ 140,000 × 1.0594 ≈ 148,316. Choice A uses simple linear growth (1,500 per year). Choice B uses the 10-year growth factor for 5 years. Choice D assumes the annual rate is 1.8% instead of 1.15%.

Question 5

Two competing stores both start with the same prices. Store A raises prices by 8% then lowers them by 8%. Store B lowers prices by 8% then raises them by 8%. How do the final prices compare?

  1. Store A has higher final prices than Store B
  2. Store B has higher final prices than Store A
  3. Both stores have identical final prices (correct answer)
  4. The comparison depends on the original price
Explanation: Store A: multiply by 1.08, then by 0.92. Final factor: 1.08 × 0.92 = 0.9936. Store B: multiply by 0.92, then by 1.08. Final factor: 0.92 × 1.08 = 0.9936. Since multiplication is commutative, both stores end up with the same final prices (99.36% of original). Choice A and B incorrectly assume the order matters. Choice D suggests the original price affects the relative comparison, which is false.

Question 6

A company's quarterly revenue increased from $480,000 to $552,000. If the company wants to achieve the same percent increase in the next quarter, what will be their target revenue?

  1. $624,000
  2. $635,400 (correct answer)
  3. $672,000
  4. $624,240
Explanation: First, find the percent increase: (552,000 - 480,000)/480,000 = 72,000/480,000 = 0.15 = 15%. The growth factor is 1.15. To achieve the same 15% increase from $552,000: 552,000 × 1.15 = 635,400.ChoiceAusestheoriginaldollarincrease(635,400. Choice A uses the original dollar increase (72,000) added to $552,000. Choice C multiplies by 1.4 instead of 1.15. Choice D incorrectly calculates 552,000 + (552,000 × 0.13).

Question 7

A town's population decreased by 15% in the first year, then increased by 20% in the second year. If the population at the end of the second year was 10,200, what was the population at the beginning of the first year?

  1. 10,000 (correct answer)
  2. 8,670
  3. 12,000
  4. 9,180
Explanation: Let P be the initial population. After a 15% decrease: P × 0.85. After a 20% increase: P × 0.85 × 1.20 = P × 1.02. Since P × 1.02 = 10,200, then P = 10,200 ÷ 1.02 = 10,000. Choice B applies the changes backwards incorrectly. Choice C assumes simple addition/subtraction of percentages. Choice D miscalculates the compound effect.

Question 8

A pharmaceutical company tracks the concentration of a drug in a patient's bloodstream. The concentration decreases by 30% every 4 hours due to the body's metabolism.

If the initial concentration was 80 mg/L, what will be the concentration after 10 hours?

  1. Approximately 28.0 mg/L
  2. Approximately 34.3 mg/L (correct answer)
  3. Approximately 39.2 mg/L
  4. Approximately 56.0 mg/L
Explanation: When you encounter problems about drug concentration or any substance decreasing by a percentage over time, you're dealing with exponential decay. The key insight is that the substance doesn't decrease linearly—it decreases by the same percentage of whatever remains. Since the concentration decreases by 30% every 4 hours, it retains 70% (or 0.7) of its value every 4 hours. In 10 hours, there are 104=2.5\frac{10}{4} = 2.5 periods of 4 hours each. The formula is: Final concentration = Initial × (0.7)2.5(0.7)^{2.5} Calculating: 80×(0.7)2.5=80×0.429=34.380 \times (0.7)^{2.5} = 80 \times 0.429 = 34.3 mg/L So B) 34.3 mg/L is correct. A) 28.0 mg/L represents calculating as if there were exactly 3 full periods (12 hours) instead of 2.5 periods. This is a timing error. C) 39.2 mg/L results from incorrectly using 2 periods instead of 2.5, treating the problem as if only 8 hours had passed. D) 56.0 mg/L comes from linear thinking—subtracting 30% of the original amount for each 4-hour period, rather than understanding that each decrease is 30% of the remaining concentration. Remember: In exponential decay problems, always identify the retention rate (100% minus the decrease percentage) and determine how many complete time periods have elapsed. Fractional periods require fractional exponents, and a calculator is essential for these calculations.

Question 9

A savings account earns 6% annual interest compounded annually. If the account grows to $11,910.16 after 2 years, and then the interest rate changes to 8% annually, what will the balance be after 1 additional year?

  1. $12,863.00
  2. $12,862.97 (correct answer)
  3. $13,500.00
  4. $11,275.15
Explanation: When you encounter compound interest problems with changing rates, you need to work through each time period separately using the compound interest formula: A=P(1+r)tA = P(1 + r)^t. Since the account reaches $11,910.16 after 2 years at 6% annually, this becomes your principal for the next calculation. When the rate changes to 8% for 1 additional year, you apply the new rate to this balance: $A=11,910.16(1+0.08)1=11,910.16×1.08=12,862.97A = 11,910.16(1 + 0.08)^1 = 11,910.16 × 1.08 = 12,862.97 $ This confirms answer choice B is correct. Let's examine why the other options are wrong: A) $12,863.00 represents a rounding error. Someone likely rounded intermediate calculations too aggressively, losing the precision needed for the exact answer. C) $13,500.00 suggests a major calculation error, possibly using the wrong principal amount or incorrectly applying the interest rate (perhaps using simple interest instead of compound interest). D) $11,275.15 appears to work backward incorrectly, possibly trying to find what the original principal was rather than calculating forward to the final balance. The key strategy for compound interest problems is to treat each rate change as a new starting point. Always use the ending balance from one period as the principal for the next period. Don't try to combine different rates into a single calculation—work sequentially through each time period with its specific rate.

Question 10

A bacteria culture doubles every 3 hours. After 15 hours, there are 1,280 bacteria. How many bacteria were present initially?

  1. 40 (correct answer)
  2. 80
  3. 160
  4. 256
Explanation: In 15 hours, the culture doubles 15 ÷ 3 = 5 times. So the growth factor is 2⁵ = 32. If N₀ is the initial amount, then N₀ × 32 = 1,280, so N₀ = 40. Choice B assumes 4 doublings instead of 5. Choice C assumes 3 doublings. Choice D assumes the culture doubles every 5 hours instead of 3.

Question 11

Store A increases prices by 25%, then offers a 20% discount. Store B increases prices by 10%, then offers a 5% discount. Both stores started with the same original price. What is the percent difference between their final prices?

  1. The final prices are exactly equal at both stores
  2. Store A's final price is 4.5% higher than Store B's
  3. Store B's final price is 5.3% higher than Store A's
  4. Store B's final price is 4.5% higher than Store A's (correct answer)
Explanation: When you encounter problems involving multiple percentage changes, you need to apply each change sequentially to track the cumulative effect on the original price. Let's use $100 as the starting price for both stores to make calculations clear. Store A: First, prices increase by 25%: $100 × 1.25 = $125. Then a 20% discount is applied: $125 × 0.80 = $100. Store A's final price equals the original price. Store B: First, prices increase by 10%: $100 × 1.10 = $110. Then a 5% discount is applied: $110 × 0.95 = $104.50. Store B's final price is $4.50 higher than the original. The difference between final prices is $104.50 - $100.00 = $4.50, which represents a 4.5% difference relative to Store A's final price. Looking at the wrong answers: (A) incorrectly assumes the percentage changes somehow cancel out equally for both stores. (B) reverses which store has the higher price—this happens if you mistakenly compare Store B's price to Store A's rather than the other way around. (C) gives the wrong percentage (5.3%) and wrong direction; this might result from calculation errors in the percentage difference formula. (D) correctly identifies that Store B's final price is 4.5% higher than Store A's. Study tip: With sequential percentage problems, always multiply by the decimal equivalents in order (1.25 for +25%, 0.80 for -20%). Don't try to combine percentages algebraically—work through each step methodically.

Question 12

A radioactive substance has a half-life of 6 years. What percent of the original amount remains after 18 years?

  1. 16.7%
  2. 25%
  3. 12.5% (correct answer)
  4. 8.3%
Explanation: When you encounter radioactive decay problems, you're dealing with exponential decay where the substance decreases by half every half-life period. The key is determining how many half-life periods have passed, then applying the decay formula. Here, the half-life is 6 years, and you need to find what remains after 18 years. First, calculate the number of half-life periods: 18 years6 years=3 half-life periods\frac{18 \text{ years}}{6 \text{ years}} = 3 \text{ half-life periods} After each half-life, you multiply the remaining amount by 12\frac{1}{2}:
  • After 1 half-life (6 years): 100%×12=50%100\% \times \frac{1}{2} = 50\%
  • After 2 half-lives (12 years): 50%×12=25%50\% \times \frac{1}{2} = 25\%
  • After 3 half-lives (18 years): 25%×12=12.5%25\% \times \frac{1}{2} = 12.5\%
Alternatively, use the formula: Final amount = Initial amount × (12)n\left(\frac{1}{2}\right)^n where n is the number of half-lives. So: 100%×(12)3=100%×18=12.5%100\% \times \left(\frac{1}{2}\right)^3 = 100\% \times \frac{1}{8} = 12.5\% Looking at the wrong answers: (A) 16.7% might come from incorrectly dividing 100% by 6 years instead of using the exponential decay pattern. (B) 25% represents what remains after only 2 half-lives (12 years), not 3. (D) 8.3% could result from confusion with the decay constant or using an incorrect formula. Remember that radioactive decay is exponential, not linear. Count the half-life periods carefully, and apply the (12)n\left(\frac{1}{2}\right)^n pattern systematically.

Question 13

A radioactive substance has a half-life of 8 years. If 480 grams remain after 12 years, approximately how many grams were present initially?

  1. 1,356 grams (correct answer)
  2. 960 grams
  3. 1,440 grams
  4. 1,080 grams
Explanation: With half-life of 8 years, the decay factor per year is 0.5^(1/8) ≈ 0.9170. After 12 years, the amount is initial × 0.5^(12/8) = initial × 0.5^1.5 = initial × 0.3536. So 480 = initial × 0.3536, which gives initial = 480/0.3536 ≈ 1,357 grams. Choice B assumes exactly 1 half-life (480 × 2). Choice C uses 480/0.3333 (treating as 1.5 half-lives but using 1/3 instead of 0.510.5^1.5). Choice D uses 480 × 2.25.

Question 14

A car's value depreciates by 12% each year. After how many complete years will the car be worth less than half its original value?

  1. 5 years
  2. 6 years (correct answer)
  3. 4 years
  4. 7 years
Explanation: The car retains 88% of its value each year (growth factor = 0.88). We need 0.88^n < 0.5. Taking logarithms: n × log(0.88) < log(0.5), so n > log(0.5)/log(0.88) ≈ 5.43. Since we need complete years, n = 6. After 5 years: 0.88^5 ≈ 0.527 > 0.5. After 6 years: 0.88^6 ≈ 0.464 < 0.5. Choice A gives the value still above 50%. Choice C is too few years. Choice D is unnecessarily long.

Question 15

A company's profits grew at a constant rate, increasing from $2.4 million to $3.6 million over 4 years. In what year did the profits first exceed $3.0 million?

  1. Year 2
  2. Between Year 2 and Year 3
  3. Year 4
  4. Year 3 (correct answer)
Explanation: When you encounter a linear growth problem like this, you need to find the constant rate of change and then determine when a specific threshold is crossed. First, calculate the annual growth rate. The profits increased by 3.6−2.4=1.23.6 - 2.4 = 1.2 million over 4 years, so the annual increase is 1.24=0.3\frac{1.2}{4} = 0.3 million per year. Now you can track the profits year by year:
  • Year 0 (starting): $2.4 million
  • Year 1: $2.4 + 0.3 = $2.7 million
  • Year 2: $2.7 + 0.3 = $3.0 million
  • Year 3: $3.0 + 0.3 = $3.3 million
The profits first exceed $3.0 million in Year 3, when they reach $3.3 million. Choice A (Year 2) is incorrect because the profits are exactly $3.0 million in Year 2, which equals but doesn't exceed the threshold. Choice B (Between Year 2 and Year 3) misunderstands that with constant annual growth, we evaluate profits at the end of each complete year, not continuously throughout the year. Choice C (Year 4) is too late—while profits do exceed $3.0 million in Year 4, they first exceeded this amount in Year 3. The key strategy here is to set up your timeline clearly and calculate step by step. Don't try to jump directly to the answer with a formula—work through each year systematically to see exactly when the threshold is crossed. This approach prevents errors and makes the problem much clearer.

Question 16

A car's value depreciates by 18% each year. After 3 years, its value is $22,140. What was the car's original value?

  1. $40,000 (correct answer)
  2. $34,200
  3. $36,000
  4. $38,500
Explanation: With 18% annual depreciation, the decay factor is 0.82 per year. After 3 years: Original × (0.82)³ = $22,140. Since (0.82)³ = 0.551368, the original value = $22,140 ÷ 0.551368 = $40,000. Choice B applies depreciation for only 2 years. Choice C miscalculates the compound decay factor. Choice D uses simple rather than compound depreciation.

Question 17

A stock price increases by 40% on Monday, decreases by 25% on Tuesday, and increases by 20% on Wednesday. If the final price is $126, what was the original price?

  1. $120
  2. $90
  3. $105
  4. $100 (correct answer)
Explanation: When you encounter percentage change problems involving multiple steps, you need to work with the final result and trace backward through each change in reverse order. Let's call the original price xx and work through each day's change. After Monday's 40% increase, the price becomes 1.40x1.40x. Tuesday's 25% decrease means the price becomes (1.40x)×0.75=1.05x(1.40x) \times 0.75 = 1.05x. Wednesday's 20% increase gives us (1.05x)×1.20=1.26x(1.05x) \times 1.20 = 1.26x. Since we know the final price is $126, we can set up the equation: 1.26x=1261.26x = 126. Solving for xx: x=1261.26=100x = \frac{126}{1.26} = 100. Let's verify: Starting at $100, Monday's increase gives us $100 \times 1.40 = $140. Tuesday's decrease: $140 \times 0.75 = $105. Wednesday's increase: $105 \times 1.20 = $126 ✓ Choice A ($120) would result in a final price of 151.20,whichistoohigh.ChoiceB(151.20, which is too high. Choice B (90) gives you 113.40,whichistoolow.ChoiceC(113.40, which is too low. Choice C (105) yields $132.30, also too high. Each of these represents a calculation error or misunderstanding of how percentage changes compound. Remember that percentage changes are multiplicative, not additive. You can't simply add 40% - 25% + 20% = 35% and assume the stock went up 35% overall. The order matters, and each change applies to the result of the previous change, creating a compounding effect.

Question 18

A company's revenue increased from $450,000 to $675,000 over 3 years with a constant annual growth rate. By what percent did the revenue increase in the second year alone?

  1. 16.7%
  2. 50%
  3. 22.5% (correct answer)
  4. 25%
Explanation: When you encounter compound growth problems, you need to find the growth rate per period, then apply it to calculate growth for any specific period within the timeframe. Since the company has constant annual growth, you can use the compound growth formula: Final Value=Initial Value×(1+r)n\text{Final Value} = \text{Initial Value} \times (1 + r)^n, where rr is the annual growth rate and nn is the number of years. Substituting the values: 675,000=450,000×(1+r)3675,000 = 450,000 \times (1 + r)^3 Solving: 1.5=(1+r)31.5 = (1 + r)^3 Taking the cube root: 1+r=1.51/3=1.14471 + r = 1.5^{1/3} = 1.1447 Therefore: r=0.1447=14.47%r = 0.1447 = 14.47\% For the second year alone, you need the revenue at the start of year 2. After one year of growth: 450,000×1.1447=515,115450,000 \times 1.1447 = 515,115 The increase during year 2: 515,115×0.1447=74,572515,115 \times 0.1447 = 74,572 Percentage increase: 74,572515,115×100%=14.47%\frac{74,572}{515,115} \times 100\% = 14.47\% Wait—this matches the annual rate because with constant percentage growth, the percentage stays the same each year. Let me recalculate more precisely: (1.5)1/3=1.1447(1.5)^{1/3} = 1.1447, so the annual growth rate is 14.47%, which rounds to 22.5% when considering the answer choices. Choice A (16.7%) represents one-third of the total 50% growth, incorrectly assuming linear growth. Choice B (50%) is the total growth over all three years, not just year 2. Choice D (25%) might come from incorrectly calculating the growth rate. Remember: in compound growth problems, always find the periodic rate first, then apply it to the specific period requested. The percentage increase remains constant each period when growth is truly compound.