At an agricultural research station, the nitrogen level of two soil plots, Plot A and Plot… — SAT

At an agricultural research station, the nitrogen level of two soil plots, Plot A and Plot B, is monitored over several weeks. The nitrogen level of Plot A increases linearly at a constant rate, and the nitrogen level of Plot B increases exponentially at a constant percentage rate. At week $t = 0$ , Plot A contains $150\text{ grams}$ of nitrogen and Plot B contains $80\text{ grams}$ of nitrogen. At week $t = 2$ , both plots contain the same amount of nitrogen, which is $180\text{ grams}$ . At week $t = 4$ , Plot B contains $N\text{ grams}$ of nitrogen and Plot A contains $A\text{ grams}$ of nitrogen. What is the value of $N - A$ ?

Answer: 195 grams

Answer

195

The correct answer is

195

. At

t = 0

, Plot A contains

150\text{ grams}

and Plot B contains

80\text{ grams}

. Since Plot A increases linearly and reaches

180\text{ grams}

t = 2

, its weekly rate of change is

\frac{180 - 150}{2} = 15\text{ grams per week}

. Thus, at week

t = 4

, Plot A contains

150 + 15(4) = 210\text{ grams}

. Since Plot B increases exponentially and reaches

180\text{ grams}

t = 2

, its weekly growth factor

b

satisfies

80b^2 = 180

, which gives

b^2 = 2.25

and

b = 1.5

. Thus, at week

t = 4

, Plot B contains

80(1.5)^4 = 405\text{ grams}

. The difference

N - A

405 - 210 = 195

Step-by-Step Solution

Find the nitrogen level of Plot A at week

t = 4

using a linear model.

A = 210

Plot A grows linearly from an initial

150\text{ grams}

t = 0

180\text{ grams}

t = 2

. The rate of increase is

\frac{180 - 150}{2} = 15\text{ grams per week}

. Thus, at

t = 4

, the nitrogen level is

150 + 15(4) = 210\text{ grams}

Find the nitrogen level of Plot B at week

t = 4

using an exponential model.

N = 405

Plot B grows exponentially from an initial

80\text{ grams}

t = 0

180\text{ grams}

t = 2

. The weekly growth factor

b

satisfies

80b^2 = 180

, so

b^2 = 2.25

and

b = 1.5

. Thus, at

t = 4

, the nitrogen level is

80(1.5)^4 = 80(5.0625) = 405\text{ grams}

Calculate the difference between the two nitrogen levels at week

t = 4

195

Subtracting

A

from

N

yields

N - A = 405 - 210 = 195

Key Concept

Modeling linear growth (constant rate of change) and exponential growth (constant percentage rate of change or growth factor) over time.

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