Q1: A population of bacteria doubles every 3 hours. If the initial population is 500 bacteria, what is the population after 9 hours? (a) 1,500 (b) 2,000 (c) 4,000 (d) 4,500
Solution:
Ans: (c) Explanation: Since the bacteria doubles every 3 hours, in 9 hours there are \(9 \div 3 = 3\) doubling periods. Using the exponential growth formula \(P(t) = P_0 \cdot 2^n\) where \(n\) is the number of doublings: \(P(9) = 500 \cdot 2^3 = 500 \cdot 8 = 4000\) bacteria. Option (a) assumes linear growth, (b) and (d) are incorrect calculations of the doubling pattern.
Ans: (b) Explanation:Exponential decay occurs when the base of the exponential function is between 0 and 1. In option (b), the base is 0.8, which satisfies \(0 < 0.8="">< 1\).="" option="" (a)="" and="" (d)="" have="" bases="" greater="" than="" 1,="" representing="">exponential growth. Option (c) is a linear function, not exponential.
Q3: A car depreciates in value by 15% each year. If the car is worth $20,000 today, what will it be worth in 2 years? (a) $13,000 (b) $14,450 (c) $17,000 (d) $16,150
Solution:
Ans: (b) Explanation: Using the exponential decay formula \(V(t) = V_0(1 - r)^t\) where \(r = 0.15\) and \(t = 2\): \(V(2) = 20000(1 - 0.15)^2 = 20000(0.85)^2 = 20000(0.7225) = 14450\) dollars. Option (a) incorrectly subtracts 15% twice linearly, (c) subtracts 15% once only, and (d) uses incorrect calculations.
Q4: What is the growth factor in the exponential function \(f(x) = 450(1.08)^x\)? (a) 450 (b) 1.08 (c) 0.08 (d) 8
Solution:
Ans: (b) Explanation: In the exponential function \(f(x) = a \cdot b^x\), the growth factor is represented by \(b\). In this function, \(b = 1.08\). Option (a) is the initial value, option (c) is the growth rate (as a decimal), and option (d) is the growth rate as a percentage without proper conversion.
Q5: A radioactive substance has a half-life of 5 years. Starting with 80 grams, how much remains after 15 years? (a) 5 grams (b) 10 grams (c) 20 grams (d) 40 grams
Solution:
Ans: (b) Explanation:Half-life means the substance reduces to half its amount in the given time period. In 15 years, there are \(15 \div 5 = 3\) half-life periods. Using \(A(t) = A_0 \cdot \left(\frac{1}{2}\right)^n\): \(A(15) = 80 \cdot \left(\frac{1}{2}\right)^3 = 80 \cdot \frac{1}{8} = 10\) grams. Option (d) represents 1 half-life, (c) represents 2 half-lives, and (a) is not a correct calculation.
Q6: Which equation represents an initial amount of 1200 growing at a rate of 5% per year for \(t\) years? (a) \(y = 1200(0.95)^t\) (b) \(y = 1200(1.5)^t\) (c) \(y = 1200(1.05)^t\) (d) \(y = 1200 + 0.05t\)
Solution:
Ans: (c) Explanation: For exponential growth at rate \(r\), the formula is \(y = a(1 + r)^t\). With \(a = 1200\) and \(r = 0.05\), we get \(y = 1200(1.05)^t\). Option (a) represents decay, (b) uses 50% instead of 5%, and (d) represents linear growth, not exponential.
Q7: An investment of $5,000 grows exponentially according to \(A(t) = 5000(1.06)^t\) where \(t\) is in years. What does the value 1.06 represent? (a) The investment grows by 6% each year (b) The investment loses 6% each year (c) The initial investment amount (d) The investment grows by 60% each year
Solution:
Ans: (a) Explanation: The base 1.06 can be written as \(1 + 0.06\), where 0.06 represents a 6% growth rate. This means the investment increases by 6% annually. Option (b) would require a base less than 1, (c) confuses the base with the coefficient 5000, and (d) misinterprets the decimal value.
Q8: Which situation best represents exponential decay? (a) A plant growing 2 inches each week (b) Water draining from a tank at a constant rate (c) The temperature of hot coffee cooling over time (d) A runner increasing speed by 0.5 mph each lap
Solution:
Ans: (c) Explanation:Exponential decay occurs when a quantity decreases by a percentage of its current value over equal time intervals. Hot coffee cooling follows Newton's Law of Cooling, which is exponential decay-the temperature difference decreases proportionally. Options (a) and (d) represent linear growth, and (b) represents linear decay at a constant rate.
Section B: Fill in the Blanks
Q9: The general form of an exponential growth function is \(y = a(1 + r)^t\), where \(a\) represents the __________ value.
Solution:
Ans: initial Explanation: In the exponential growth formula, \(a\) represents the initial value or starting amount before any growth occurs. The variable \(r\) is the growth rate and \(t\) is time.
Q10: If a quantity decreases by 20% each year, the decay factor is __________.
Solution:
Ans: 0.8 or 0.80 Explanation: The decay factor is calculated as \(1 - r\) where \(r\) is the decay rate. Since \(r = 0.20\), the decay factor is \(1 - 0.20 = 0.80\). This means the quantity retains 80% of its value each year.
Q11: The time it takes for a substance to reduce to half its original amount is called the __________.
Solution:
Ans: half-life Explanation:Half-life is a term commonly used in radioactive decay and other exponential decay contexts to describe the time period required for a quantity to decrease to 50% of its initial value.
Q12: In the function \(f(x) = 250(0.75)^x\), the value 250 represents the __________ amount.
Solution:
Ans: initial Explanation: In an exponential function of the form \(f(x) = a \cdot b^x\), the coefficient \(a\) is the initial amount when \(x = 0\). Here, when \(x = 0\), \(f(0) = 250(0.75)^0 = 250\).
Q13: If an exponential function has a base greater than 1, it represents exponential __________.
Solution:
Ans: growth Explanation: When the base of an exponential function \(b\) satisfies \(b > 1\), the function represents exponential growth because the output increases as the exponent increases. If \(0 < b="">< 1\),="" it="" represents="" exponential="" decay.="">
Q14: The exponential function \(y = 1000(1.12)^t\) represents a growth rate of __________ percent.
Solution:
Ans: 12 Explanation: The growth factor is 1.12, which can be written as \(1 + 0.12\). The growth rate \(r = 0.12\), which equals 12% when expressed as a percentage.
Section C: Word Problems
Q15: A city's population is 50,000 and is increasing at a rate of 3% per year. Write an exponential function to model the population after \(t\) years, and find the population after 5 years.
Solution:
Ans:
The exponential growth function is \(P(t) = P_0(1 + r)^t\)
Where \(P_0 = 50000\) and \(r = 0.03\)
So the function is: \(P(t) = 50000(1.03)^t\)
For \(t = 5\) years:
\(P(5) = 50000(1.03)^5\)
\(P(5) = 50000(1.159274)\)
\(P(5) = 57963.7\) Final Answer: The function is \(P(t) = 50000(1.03)^t\) and the population after 5 years is approximately 57,964 people.
Q16: A laptop costs $1,200 and depreciates at 18% per year. What will the laptop be worth after 3 years?
Solution:
Ans:
Using the exponential decay formula: \(V(t) = V_0(1 - r)^t\)
Where \(V_0 = 1200\), \(r = 0.18\), and \(t = 3\)
\(V(3) = 1200(1 - 0.18)^3\)
\(V(3) = 1200(0.82)^3\)
\(V(3) = 1200(0.551368)\)
\(V(3) = 661.64\) Final Answer: The laptop will be worth approximately $661.64 after 3 years.
Q17: A scientist starts with 300 grams of a radioactive element that has a half-life of 8 years. How much of the element will remain after 24 years?
Solution:
Ans:
Number of half-life periods: \(n = \frac{24}{8} = 3\) periods
Using the half-life formula: \(A(t) = A_0 \left(\frac{1}{2}\right)^n\)
Where \(A_0 = 300\) grams and \(n = 3\)
\(A(24) = 300 \left(\frac{1}{2}\right)^3\)
\(A(24) = 300 \cdot \frac{1}{8}\)
\(A(24) = 37.5\) Final Answer: 37.5 grams of the radioactive element will remain after 24 years.
Q18: An investment account starts with $8,000 and grows at 4.5% annually. How much money will be in the account after 6 years?
Solution:
Ans:
Using the exponential growth formula: \(A(t) = A_0(1 + r)^t\)
Where \(A_0 = 8000\), \(r = 0.045\), and \(t = 6\)
\(A(6) = 8000(1.045)^6\)
\(A(6) = 8000(1.302260)\)
\(A(6) = 10418.08\) Final Answer: The account will have approximately $10,418.08 after 6 years.
Q19: A social media post gets 200 views initially. The views increase by 25% each hour. How many views will the post have after 4 hours?
Solution:
Ans:
Using the exponential growth formula: \(V(t) = V_0(1 + r)^t\)
Where \(V_0 = 200\), \(r = 0.25\), and \(t = 4\)
\(V(4) = 200(1.25)^4\)
\(V(4) = 200(2.44140625)\)
\(V(4) = 488.28\) Final Answer: The post will have approximately 488 views after 4 hours.
Q20: A cup of hot chocolate is 180°F and cools exponentially. After 5 minutes, its temperature is 150°F. If the room temperature is 70°F, does the temperature difference from room temperature represent exponential decay? Calculate the decay factor.
Solution:
Ans:
Yes, this represents exponential decay (Newton's Law of Cooling).
Initial temperature difference: \(T_0 - T_{room} = 180 - 70 = 110\)°F
Temperature difference after 5 minutes: \(T_5 - T_{room} = 150 - 70 = 80\)°F
Using \(\Delta T = \Delta T_0 \cdot b^t\):
\(80 = 110 \cdot b^5\)
\(b^5 = \frac{80}{110} = 0.7273\)
\(b = (0.7273)^{\frac{1}{5}}\)
\(b \approx 0.9406\) Final Answer: Yes, the temperature difference represents exponential decay with a decay factor of approximately 0.94 per minute.
The document Worksheet (with Solutions): Exponential Growth & Decay is a part of the Grade 9 Course Mathematics: Algebra 1.
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