AP Precalculus: Exponential Models (Drill 1)

Question 1. A car purchased for $24,000 depreciates at a rate of 15% per year. Which of the following functions models the value \( V \), in dollars, of the car \( t \) years after purchase?

• A) \( V(t) = 24{,}000 \cdot (0.85)^t \) ✓
• B) \( V(t) = 24{,}000 \cdot (0.15)^t \)
• C) \( V(t) = 24{,}000 \cdot (1.15)^t \)
• D) \( V(t) = 24{,}000 \cdot (1.85)^t \)

Explanation: Choice A is correct. A 15% depreciation rate means the car retains \( 1 - 0.15 = 0.85 \) of its value each year. The model is \( V(t) = 24{,}000 \cdot (0.85)^t \).

Choice B is incorrect because \( b = 0.15 \) is the decay rate, not the decay factor. This would reduce the car to 15% of its value after one year, a loss of 85%.

Choice C is incorrect because \( b = 1.15 \) represents 15% growth, not depreciation.

Choice D is incorrect because \( b = 1.85 \) would represent 85% growth per year.

Question 2. A bacterial culture starts with 200 cells and doubles every 3 hours. Which of the following correctly represents the number of cells \( N \) after \( t \) hours?

• A) \( N(t) = 200 \cdot 2^{3t} \)
• B) \( N(t) = 200 \cdot 3^t \)
• C) \( N(t) = 200 \cdot 6^t \)
• D) \( N(t) = 200 \cdot 2^{t/3} \) ✓

Explanation: Choice D is correct. The culture completes \( t/3 \) doubling periods in \( t \) hours, so \( N(t) = 200 \cdot 2^{t/3} \). Check: \( N(3) = 200 \cdot 2^1 = 400 \) ✓. \( N(6) = 200 \cdot 2^2 = 800 \) ✓.

Choice A is incorrect because \( 2^{3t} \) implies 3t doubling periods in t hours, as if the culture doubles 3 times per hour.

Choice B is incorrect because \( b = 3 \) confuses the doubling period with the growth factor.

Choice C is incorrect because \( b = 6 \) multiplies the doubling factor (2) and period (3), which has no mathematical basis.

Question 3.

t (years)	P(t)
0	800
1	680
2	578
3	491.3

The table shows values of an exponential function \( P \). Which of the following is closest to the annual percent decrease represented in the table?

• A) 12%
• B) 17%
• C) 20%
• D) 15% ✓

Explanation: Choice D is correct. The ratio of consecutive outputs gives the base: \( 680/800 = 0.85 \), \( 578/680 = 0.85 \), \( 491.3/578 \approx 0.85 \). The base is \( b = 0.85 \), meaning a decrease of \( 1 - 0.85 = 0.15 \), or 15% per year.

Choice A is incorrect because 12% gives \( b = 0.88 \), and \( 800 \cdot 0.88 = 704 \neq 680 \).

Choice B is incorrect because 17% gives \( b = 0.83 \), and \( 800 \cdot 0.83 = 664 \neq 680 \).

Choice C is incorrect because 20% gives \( b = 0.80 \), and \( 800 \cdot 0.80 = 640 \neq 680 \).

Question 4. A radioactive substance has a half-life of 12 years. A sample initially contains 500 grams. Which of the following correctly represents the amount remaining, in grams, after \( t \) years?

• A) \( A(t) = 500 \cdot \left(\dfrac{1}{2}\right)^{12t} \)
• B) \( A(t) = 500 \cdot \left(\dfrac{1}{2}\right)^{t-12} \)
• C) \( A(t) = 500 \cdot \left(\dfrac{1}{2}\right)^{t/12} \) ✓
• D) \( A(t) = 500 \cdot 12^{t/2} \)

Explanation: Choice C is correct. In \( t \) years, \( t/12 \) half-life periods elapse. \( A(t) = 500 \cdot \left(\tfrac{1}{2}\right)^{t/12} \). Check: \( A(12) = 500 \cdot \tfrac{1}{2} = 250 \) ✓. \( A(24) = 125 \) ✓.

Choice A is incorrect because \( 12t \) implies 12t half-life periods per year, the substance would halve 12 times per year.

Choice B is incorrect because \( t - 12 \) shifts when decay begins; at \( t = 0 \) it gives a value larger than 500.

Choice D is incorrect because base 12 and exponent \( t/2 \) have no grounding in the half-life model, this grows rather than decays.

Question 5. A population of 1,000 animals grows according to the model \( P(t) = 1{,}000 \cdot (1.06)^t \), where \( t \) is measured in years. A student makes the following claim:

"Since the growth factor is 1.06, the population increases by exactly 60 animals every year."

Which of the following best evaluates the student's reasoning?

• A) The student is correct. A growth factor of 1.06 means 60 animals are added each year regardless of the current population size.
• B) The student is correct. The population increases by 6% of 1,000 = 60 animals each year, so the annual increase is always 60.
• C) The student is incorrect. A growth factor of 1.06 means the population increases by 60% each year, not 6%, so the annual increase is far larger than 60 animals according to the exponential model.
• D) The student is incorrect. A growth factor of 1.06 means the population increases by 6% of the current population each year, so the number of animals added grows over time. ✓

Explanation: Choice D is correct. A growth factor of \( b = 1.06 \) means the population is multiplied by 1.06 each year; it increases by 6% of the current population. In year 1: 6% of 1,000 = 60 animals. In year 2: 6% of 1,060 = 63.6 animals. The increase grows each year. The student confused exponential growth (percent of current value) with linear growth (fixed addition).

Choice A is incorrect because the annual increase is not fixed; it depends on the current population.

Choice B is incorrect for the same reason. The 6% applies to an ever-growing base, not always 1,000.

Choice C is incorrect because 1.06 represents 6% growth, not 60%.