Math 1 Quiz: Making Predictions From Rate
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Making Predictions From RateQuestion 1 of 20

A spring extends 0.75 inches for every 2 pounds of weight added to it. If the spring is currently 8.5 inches long with 6 pounds attached, what is the best prediction for the spring's length when 20 pounds total are attached?

The spring will extend to approximately 13.75 inches, assuming it remains within its elastic limit
The spring will extend to approximately 15.25 inches, assuming it remains within its elastic limit
The spring will extend to approximately 16.00 inches, assuming it remains within its elastic limit
The spring will extend to approximately 18.50 inches, assuming it remains within its elastic limit
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Math 1 Quiz

Math 1 Quiz: Making Predictions From Rate

Practice Making Predictions From Rate 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 Making Predictions From Rate, 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.

All questions

Question 1

A spring extends 0.75 inches for every 2 pounds of weight added to it. If the spring is currently 8.5 inches long with 6 pounds attached, what is the best prediction for the spring's length when 20 pounds total are attached?

  1. The spring will extend to approximately 13.75 inches, assuming it remains within its elastic limit (correct answer)
  2. The spring will extend to approximately 15.25 inches, assuming it remains within its elastic limit
  3. The spring will extend to approximately 16.00 inches, assuming it remains within its elastic limit
  4. The spring will extend to approximately 18.50 inches, assuming it remains within its elastic limit
Explanation: Additional weight added: 20 - 6 = 14 pounds. Since 0.75 inches per 2 pounds, the rate is 0.375 inches per pound. Additional extension: 14 × 0.375 = 5.25 inches. Final length: 8.5 + 5.25 = 13.75 inches. Choice B incorrectly uses 0.75 inches per pound rate. Choice C adds the total weight in inches to current length. Choice D uses 20 pounds × 0.75 inches per 2 pounds = 7.5 inches, then adds incorrectly.

Question 2

The number of subscribers to a streaming service increases by 15% each month. If there are currently 24,000 subscribers, what is the most reliable prediction for the number of subscribers after 6 months?

  1. There will likely be approximately 55,596 subscribers, assuming growth conditions remain favorable throughout the period (correct answer)
  2. There will likely be approximately 45,600 subscribers, assuming growth conditions remain favorable throughout the period
  3. There will likely be approximately 39,600 subscribers, assuming growth conditions remain favorable throughout the period
  4. There will likely be approximately 35,400 subscribers, assuming growth conditions remain favorable throughout the period
Explanation: With 15% monthly growth, the subscriber count multiplies by 1.15 each month. After 6 months: 24,000 × (1.15)^6 = 24,000 × 2.3182 ≈ 55,596 subscribers. Choice B uses simple interest (24,000 + 6 × 3,600 = 45,600). Choice C uses 10% monthly growth rate. Choice D uses 8% monthly growth rate.

Question 3

A car's fuel efficiency decreases by 2.5 miles per gallon for every 10,000 miles driven. If a car gets 32 mpg when new, what is the most reasonable prediction for its fuel efficiency after 75,000 miles?

  1. The car will likely achieve approximately 13.25 mpg, assuming normal wear patterns and maintenance (correct answer)
  2. The car will likely achieve approximately 15.75 mpg, assuming normal wear patterns and maintenance
  3. The car will likely achieve approximately 18.25 mpg, assuming normal wear patterns and maintenance
  4. The car will likely achieve approximately 20.75 mpg, assuming normal wear patterns and maintenance
Explanation: After 75,000 miles, there are 75,000 ÷ 10,000 = 7.5 intervals of 10,000 miles. The total decrease is 7.5 × 2.5 = 18.75 mpg. Final efficiency: 32 - 18.75 = 13.25 mpg. Choice B uses 6.5 intervals instead of 7.5. Choice C uses 5.5 intervals. Choice D uses 4.5 intervals (common error of excluding the final partial interval).

Question 4

A machine produces widgets at a rate that decreases linearly from 120 widgets per hour at the start of a shift to 90 widgets per hour at the end of an 8-hour shift. What is the best prediction for the production rate exactly 5 hours into the shift?

  1. The machine will produce approximately 101.25 widgets per hour, assuming mechanical wear follows the linear pattern (correct answer)
  2. The machine will produce approximately 105.00 widgets per hour, assuming mechanical wear follows the linear pattern
  3. The machine will produce approximately 108.75 widgets per hour, assuming mechanical wear follows the linear pattern
  4. The machine will produce approximately 112.50 widgets per hour, assuming mechanical wear follows the linear pattern
Explanation: The rate decreases from 120 to 90 widgets/hour over 8 hours, a total decrease of 30 widgets/hour. The hourly rate of decrease is 30 ÷ 8 = 3.75 widgets/hour per hour. After 5 hours: 120 - 5 × 3.75 = 120 - 18.75 = 101.25 widgets/hour. Choice B uses 3 widgets/hour decrease per hour. Choice C uses 2.5 widgets/hour decrease per hour. Choice D uses 1.5 widgets/hour decrease per hour.

Question 5

A car's speed decreases at a constant rate of 8 mph every 30 seconds while braking. If the car is traveling at 65 mph when braking begins, what is the best prediction for its speed after 2 minutes and 15 seconds of braking?

  1. The car will be traveling at approximately 29 mph, assuming consistent braking force throughout the period (correct answer)
  2. The car will be traveling at approximately 37 mph, assuming consistent braking force throughout the period
  3. The car will be traveling at approximately 41 mph, assuming consistent braking force throughout the period
  4. The car will be traveling at approximately 45 mph, assuming consistent braking force throughout the period
Explanation: 2 minutes 15 seconds = 135 seconds. This represents 135 ÷ 30 = 4.5 intervals of 30 seconds. Total speed decrease: 4.5 × 8 = 36 mph. Final speed: 65 - 36 = 29 mph. Choice B uses 3.5 intervals (105 seconds). Choice C uses 3 intervals (90 seconds). Choice D uses 2.5 intervals (75 seconds).

Question 6

A biology experiment tracks the growth of algae in a controlled environment. The algae culture doubles in mass every 36 hours under optimal conditions.

If the algae culture has a mass of 12 grams at the start of day 4, what is the most reasonable prediction for its mass at the end of day 7?

  1. The culture will have a mass of approximately 48.00 grams, assuming optimal growth conditions are maintained
  2. The culture will have a mass of approximately 32.72 grams, assuming optimal growth conditions are maintained (correct answer)
  3. The culture will have a mass of approximately 96.00 grams, assuming optimal growth conditions are maintained
  4. The culture will have a mass of approximately 192.0 grams, assuming optimal growth conditions are maintained
Explanation: When you encounter exponential growth problems, you need to carefully track time intervals and apply the growth formula systematically. This algae doubles every 36 hours, so you must calculate how many doubling periods occur between the start of day 4 and the end of day 7. From the start of day 4 to the end of day 7 represents 3.5 days, or 84 hours. Since the culture doubles every 36 hours, the number of doubling periods is 8436=2.33\frac{84}{36} = 2.33 periods. Using the exponential growth formula: Final mass = Initial mass × 2n2^{n}, where n is the number of doubling periods. Therefore: Final mass = 12×22.33=12×5.04≈32.7212 × 2^{2.33} = 12 × 5.04 ≈ 32.72 grams. This confirms answer B is correct. Answer A (48.00 grams) assumes exactly 2 doubling periods, ignoring the additional 0.33 period. Answer C (96.00 grams) incorrectly calculates 3 complete doubling periods (12×23=9612 × 2^3 = 96), likely from miscounting the days as exactly 3 full periods. Answer D (192.0 grams) represents 4 complete doubling periods (12×24=19212 × 2^4 = 192), suggesting a significant error in time calculation. The key trap here is precision in time calculation. Always convert your time period to the same units as the doubling time, then divide carefully. Many students round to whole doubling periods, but fractional periods require using exponents with decimals, not just repeated multiplication.

Question 7

A research study tracks the height of a bamboo plant over several weeks. The data shows that the plant grows at an average rate of 2.8 inches per week during its rapid growth phase.

If the bamboo plant is 14.5 inches tall at the start of week 3, and the growth rate remains constant, what is the best prediction for when the plant will first exceed 35 inches in height?

  1. During week 10, assuming environmental factors continue to support consistent growth throughout this period (correct answer)
  2. During week 11, assuming environmental factors continue to support consistent growth throughout this period
  3. During week 12, assuming environmental factors continue to support consistent growth throughout this period
  4. During week 13, assuming environmental factors continue to support consistent growth throughout this period
Explanation: The plant needs to grow from 14.5 inches to just over 35 inches, requiring more than 35 - 14.5 = 20.5 inches of additional growth. At 2.8 inches per week, this takes 20.5 ÷ 2.8 ≈ 7.32 weeks. Since we start at the beginning of week 3, the plant will exceed 35 inches during week 3 + 7.32 ≈ week 10. Choice B rounds 7.32 up to 8 weeks. Choice C uses 8.5 weeks. Choice D uses 9 weeks (common error of significant over-estimation).

Question 8

The depth of water in a reservoir decreases at a constant rate of 0.8 feet per week during a drought. If the current depth is 45.6 feet, what is the best prediction for when the water depth will first drop below the critical level of 32 feet?

  1. The critical level will be reached in approximately 17 weeks, assuming drought conditions persist (correct answer)
  2. The critical level will be reached in approximately 18 weeks, assuming drought conditions persist
  3. The critical level will be reached in approximately 19 weeks, assuming drought conditions persist
  4. The critical level will be reached in approximately 20 weeks, assuming drought conditions persist
Explanation: The water needs to drop from 45.6 feet to just below 32 feet, requiring slightly more than 45.6 - 32 = 13.6 feet decrease. At 0.8 feet per week, this takes 13.6 ÷ 0.8 = 17 weeks exactly to reach 32 feet, so it drops below 32 feet during week 17. Choice B rounds up assuming it takes a full additional week. Choice C uses 15.2 feet of decrease needed. Choice D uses 16 feet of decrease needed.

Question 9

The temperature in a laboratory increases at an average rate of 3.2°F per hour between 8 AM and 2 PM. If the temperature at 10 AM is 68°F, what is the most reliable prediction for the temperature at 1 PM?

  1. The temperature will likely be around 77.6°F, provided heating conditions remain consistent during this period (correct answer)
  2. The temperature will likely be around 81.2°F, provided heating conditions remain consistent during this period
  3. The temperature will likely be around 84.8°F, provided heating conditions remain consistent during this period
  4. The temperature will likely be around 88.4°F, provided heating conditions remain consistent during this period
Explanation: From 10 AM to 1 PM is 3 hours. At a rate of 3.2°F per hour, the temperature increase is 3 × 3.2 = 9.6°F. Starting from 68°F at 10 AM: 68 + 9.6 = 77.6°F. Choice B adds 4 hours of increase instead of 3. Choice C uses the total 6-hour period from 8 AM to 2 PM. Choice D calculates from 8 AM to 1 PM (5 hours of increase from an assumed starting point).

Question 10

A company's profit increases linearly from $45,000 in January to $63,000 in April. Based on this rate of change, what is the best prediction for the company's profit in July, assuming the trend continues?

  1. The profit will be approximately $81,000, assuming market conditions and business operations remain stable (correct answer)
  2. The profit will be approximately $78,000, assuming market conditions and business operations remain stable
  3. The profit will be approximately $75,000, assuming market conditions and business operations remain stable
  4. The profit will be approximately $72,000, assuming market conditions and business operations remain stable
Explanation: From January to April is 3 months with an increase of $63,000 - $45,000 = $18,000. The monthly rate is $18,000 ÷ 3 = $6,000 per month. From April to July is 3 more months, so the profit increases by another 3 × $6,000 = $18,000. July profit: $63,000 + $18,000 = $81,000. Choice B uses 2.5 months from April to July. Choice C uses 2 months from April to July. Choice D uses the total increase from January to April as the monthly rate.

Question 11

A population of bacteria grows at a constant rate of 1,200 bacteria per hour. If there are 8,500 bacteria at time t=5t = 5 hours, how many bacteria were present initially (at t=0t = 0), and how many will be present at t=12t = 12 hours?

  1. Initially: 1,500; At 12 hours: 15,900
  2. Initially: 3,500; At 12 hours: 17,900
  3. Initially: 2,500; At 12 hours: 16,400
  4. Initially: 2,500; At 12 hours: 16,900 (correct answer)
Explanation: When you encounter a problem about constant growth rates, you're dealing with linear relationships. The key insight is that "constant rate" means the same amount is added each time period, creating a straight-line pattern. Since the bacteria grow at 1,200 per hour, you can write the population as: P(t)=P0+1200tP(t) = P_0 + 1200t, where P0P_0 is the initial population. Given that P(5)=8500P(5) = 8500, you can substitute: 8500=P0+1200(5)=P0+60008500 = P_0 + 1200(5) = P_0 + 6000. Solving gives P0=2500P_0 = 2500 bacteria initially. For the population at t=12t = 12: P(12)=2500+1200(12)=2500+14400=16900P(12) = 2500 + 1200(12) = 2500 + 14400 = 16900 bacteria. Choice A incorrectly calculates the initial population. If you started with 1,500, then at t=5t = 5 you'd have 1500+6000=75001500 + 6000 = 7500, not 8,500. The final calculation of 15,900 follows from this wrong starting point. Choice B makes an error in finding the initial population, getting 3,500 instead of 2,500. This might come from a calculation mistake when solving 8500=P0+60008500 = P_0 + 6000. The final answer of 17,900 correctly applies the growth rate to this incorrect initial value. Choice C gets the initial population right (2,500) but makes an arithmetic error in the final calculation, arriving at 16,400 instead of 16,900. Study tip: In linear growth problems, always set up the equation Final=Initial+rate×time\text{Final} = \text{Initial} + \text{rate} \times \text{time}, then work backwards from known information to find unknowns. Double-check your arithmetic, especially in multi-step calculations.

Question 12

A scientist observes that the pH level of a solution changes at a constant rate. The data shows the pH was 6.8 at the 15-minute mark and 5.2 at the 40-minute mark. If the experiment continues at this rate, when will the pH level reach 4.0?

  1. 52 minutes
  2. 59 minutes (correct answer)
  3. 55 minutes
  4. 62 minutes
Explanation: When you see a problem about something changing "at a constant rate," you're dealing with a linear relationship. This means you can use the slope formula and linear equations to predict future values. First, find the rate of change (slope) using the two given points: (15 minutes, 6.8 pH) and (40 minutes, 5.2 pH). The slope is 5.2−6.840−15=−1.625=−0.064\frac{5.2 - 6.8}{40 - 15} = \frac{-1.6}{25} = -0.064 pH units per minute. Now use the point-slope form to find when pH = 4.0. Using the first point: 4.0=6.8+(−0.064)(t−15)4.0 = 6.8 + (-0.064)(t - 15). Solving: 4.0=6.8−0.064t+0.964.0 = 6.8 - 0.064t + 0.96, so 4.0=7.76−0.064t4.0 = 7.76 - 0.064t. Therefore 0.064t=3.760.064t = 3.76, giving us t=58.75t = 58.75 minutes, which rounds to 59 minutes. Choice A (52 minutes) likely comes from calculation errors or using an incorrect rate. Choice C (55 minutes) might result from rounding the rate incorrectly early in the process. Choice D (62 minutes) could come from sign errors when setting up the equation or from using the wrong reference point. The correct answer is B) 59 minutes. For linear rate problems, always identify your two points clearly, calculate the slope precisely, and double-check your algebra. Remember that "constant rate" means linear relationship, so these problems are really just applied slope calculations in disguise.

Question 13

The depth of water in a swimming pool decreases at a constant rate due to evaporation and small leaks. The depth was 4.2 feet on day 10 and 3.6 feet on day 25. If this rate continues and the pool is considered empty when it reaches 0.5 feet deep, on what day will the pool need to be refilled?

  1. Day 108
  2. Day 95
  3. Day 102 (correct answer)
  4. Day 87
Explanation: When you encounter a word problem about constant rates of change, you're dealing with linear relationships. The key is to find the rate of change (slope) and use it to create an equation that models the situation. First, let's find the rate at which the water depth decreases. Between day 10 and day 25 (15 days), the depth changed from 4.2 feet to 3.6 feet, a decrease of 0.6 feet. So the rate is −0.6 feet15 days=−0.04 feet per day\frac{-0.6 \text{ feet}}{15 \text{ days}} = -0.04 \text{ feet per day}. Now we can write a linear equation. Using the point (10, 4.2), the equation is: depth = 4.2 - 0.04(day - 10). Simplifying: depth = 4.6 - 0.04(day). To find when the pool reaches 0.5 feet deep, we solve: 0.5 = 4.6 - 0.04(day). Rearranging: 0.04(day) = 4.1, so day = 4.10.04=102.5\frac{4.1}{0.04} = 102.5. Since we're looking for the day it needs refilling, this occurs on day 102. Looking at the wrong answers: Choice A (108) likely comes from calculation errors in the division. Choice B (95) might result from using the wrong reference point or sign error. Choice D (87) could come from incorrectly calculating the rate or using 3.6 - 4.2 = 0.6 instead of the negative change. Remember that linear word problems follow a pattern: identify two points, calculate the rate of change, write the equation, then solve for your target value. Always double-check that your rate has the correct sign!

Question 14

A car's value depreciates linearly. After 2 years, the car is worth $18,400. After 5 years, it's worth $13,300. Based on this depreciation rate, what will the car be worth after 8 years, and what was its original purchase price?

  1. 8 years: $7,900; Original: $21,800
  2. 8 years: $8,200; Original: $22,100
  3. 8 years: $8,200; Original: $21,800 (correct answer)
  4. 8 years: $9,100; Original: $23,200
Explanation: When you encounter linear depreciation problems, you're working with a straight line where the car loses the same dollar amount each year. The key is finding the rate of change (slope) and using it to work backward and forward in time. First, calculate the annual depreciation rate. From year 2 to year 5 (3 years), the car's value dropped from $18,400 to $13,300, a decrease of $5,100. This gives us $5,1003=1,700\frac{5,100}{3} = 1,700 $ per year. To find the original price, work backward from year 2. Since the car depreciates $1,700 annually, after 2 years it has lost 2×1,700=3,4002 \times 1,700 = 3,400. Therefore, the original price was 18,400+3,400=21,80018,400 + 3,400 = 21,800. For the value after 8 years, the car will have depreciated for 8 years total: 21,800−(8×1,700)=21,800−13,600=8,20021,800 - (8 \times 1,700) = 21,800 - 13,600 = 8,200. Answer A gets the original price wrong ($21,800 vs $21,800) and miscalculates the 8-year value as $7,900 instead of $8,200. Answer B correctly finds the 8-year value but inflates the original price to $22,100, likely from calculation errors. Answer D significantly overestimates both values, possibly from using an incorrect depreciation rate. Study tip: In linear depreciation problems, always calculate the rate first using the two given points, then use that consistent rate to find other values. Double-check by ensuring your depreciation rate works for both given data points.

Question 15

A company's profit increases linearly over time. In month 4, the profit was $28,000. In month 9, the profit was $43,000. Based on this trend, what was the profit in month 1, and what will it be in month 15?

  1. Month 1: $19,000; Month 15: $61,000 (correct answer)
  2. Month 1: $22,000; Month 15: $58,000
  3. Month 1: $16,000; Month 15: $64,000
  4. Month 1: $25,000; Month 15: $55,000
Explanation: The rate of change is 43000−280009−4=150005=3000\frac{43000-28000}{9-4} = \frac{15000}{5} = 3000 dollars per month. Using point-slope form with month 4: P=3000m+bP = 3000m + b, so 28000=3000(4)+b28000 = 3000(4) + b, giving b=16000b = 16000. Therefore P=3000m+16000P = 3000m + 16000. For month 1: P=3000(1)+16000=19000P = 3000(1) + 16000 = 19000. For month 15: P=3000(15)+16000=61000P = 3000(15) + 16000 = 61000.

Question 16

An elevator moves between floors at a constant rate. It takes 18 seconds to travel from floor 3 to floor 8, and 30 seconds to travel from floor 3 to floor 13. How long will it take the elevator to travel from floor 7 to floor 15?

  1. 20.8 seconds
  2. 19.2 seconds (correct answer)
  3. 18.4 seconds
  4. 21.6 seconds
Explanation: This problem tests your understanding of constant rate motion and how to find unit rates from given information. When you see an elevator or any object moving at a constant speed, you need to establish the rate per unit distance first. Start by finding the elevator's speed in seconds per floor. From floor 3 to floor 8 is 5 floors in 18 seconds, giving us 185=3.6\frac{18}{5} = 3.6 seconds per floor. Let's verify this rate: from floor 3 to floor 13 is 10 floors, which should take 10×3.6=3610 \times 3.6 = 36 seconds. But we're told it takes 30 seconds, so let me recalculate. Actually, let's work backwards from both pieces of information. If 5 floors take 18 seconds and 10 floors take 30 seconds, we can find the rate: 30−1810−5=125=2.4\frac{30-18}{10-5} = \frac{12}{5} = 2.4 seconds per floor. Now for floor 7 to floor 15: that's 8 floors at 2.4 seconds per floor, so 8×2.4=19.28 \times 2.4 = 19.2 seconds. Looking at the wrong answers: A) 20.8 seconds likely comes from incorrectly calculating the rate or making an arithmetic error. C) 18.4 seconds might result from using the wrong number of floors or misapplying the rate. D) 21.6 seconds could come from using an incorrect rate calculation or adding unnecessary time. Strategy tip: In constant rate problems, always find the unit rate first, then verify it works for all given information before applying it to find your answer. Double-checking your rate prevents cascading errors.

Question 17

The value of a cryptocurrency decreases at a rate of 12% per day for several consecutive days. If the value starts at $850, what is the most reasonable prediction for its value after exactly 4 days?

  1. The value will likely be approximately $513.58, assuming market volatility follows the established pattern (correct answer)
  2. The value will likely be approximately $442.00, assuming market volatility follows the established pattern
  3. The value will likely be approximately $388.40, assuming market volatility follows the established pattern
  4. The value will likely be approximately $334.82, assuming market volatility follows the established pattern
Explanation: With a 12% decrease per day, 88% of the value remains each day. After 4 days: $850 × (0.88)^4 = $850 × 0.6043 ≈ 513.58.ChoiceBincorrectlyuseslineardecrease(513.58. Choice B incorrectly uses linear decrease (850 - 4 × $102 = $442). Choice C uses a 15% daily decrease rate. Choice D uses a 20% daily decrease rate.

Question 18

A population of bacteria decreases at a rate of 8% per hour. If the initial population is 2,400 bacteria, which prediction about the population after 5 hours is most reasonable?

  1. The population will be approximately 1,440 bacteria, assuming environmental conditions remain stable
  2. The population will be approximately 1,584 bacteria, assuming environmental conditions remain stable (correct answer)
  3. The population will be approximately 1,200 bacteria, assuming environmental conditions remain stable
  4. The population will be approximately 960 bacteria, assuming environmental conditions remain stable
Explanation: When you encounter exponential decay problems, you're dealing with a situation where a quantity decreases by a constant percentage over regular time intervals. The key formula is: Final Amount = Initial Amount × (1 - decay rate)^time. Here, the bacteria population decreases by 8% per hour, so 92% (or 0.92) remains each hour. Starting with 2,400 bacteria, after 5 hours you calculate: 2400×(0.92)5=2400×0.6591≈1,5822400 × (0.92)^5 = 2400 × 0.6591 ≈ 1,582 bacteria. This matches answer choice B most closely at 1,584 bacteria. Let's examine why the other answers miss the mark. Answer A (1,440 bacteria) represents what you'd get with a linear decrease of 8% of the original amount each hour (2,400 - 5 × 192 = 1,440), but exponential decay compounds, so each hour you lose 8% of the remaining population, not the original. Answer C (1,200 bacteria) would result from a 10% decay rate over 5 hours, showing a calculation error in the decay percentage. Answer D (960 bacteria) represents an even steeper decline that doesn't match the given 8% rate. The most common trap in exponential decay problems is treating them like linear decreases. Remember that with exponential decay, you're always taking the percentage from the remaining amount, not the original amount. Always use the formula (1 - rate)^time, and double-check that your decay rate is converted properly to decimal form.

Question 19

A savings account balance grows exponentially at a rate of 6% per year. If the account has $3,200 after 2 years, what is the best prediction for the balance after 8 years total?

  1. The balance will be approximately $5,091, assuming the interest rate and compounding method remain unchanged (correct answer)
  2. The balance will be approximately $4,544, assuming the interest rate and compounding method remain unchanged
  3. The balance will be approximately $4,032, assuming the interest rate and compounding method remain unchanged
  4. The balance will be approximately $3,776, assuming the interest rate and compounding method remain unchanged
Explanation: From year 2 to year 8 is 6 additional years. With 6% annual growth, the balance multiplies by (1.06)^6 = 1.4185 over 6 years. Final balance: $3,200 × 1.4185 ≈ $5,091. Choice B uses 4% annual growth rate. Choice C uses 5% annual growth rate. Choice D uses 3% annual growth rate.

Question 20

The temperature in a laboratory freezer decreases at a constant rate of 2.5°C2.5°C per hour. If the temperature is −8°C-8°C at 2:00 PM, at what time will the temperature first reach −23°C-23°C?

  1. 8:00 PM (correct answer)
  2. 7:30 PM
  3. 8:30 PM
  4. 9:00 PM
Explanation: The temperature needs to decrease by −23−(−8)=−15°C-23 - (-8) = -15°C. At a rate of −2.5°C-2.5°C per hour, this will take 152.5=6\frac{15}{2.5} = 6 hours. Starting from 2:00 PM, adding 6 hours gives 8:00 PM.