An investment account and a savings account are opened at the same time. The value of the … — SAT

An investment account and a savings account are opened at the same time. The value of the investment account, in dollars, is modeled by the function $I(t) = 5,000(1.08)^t$ , where $t$ is the number of years since the account was opened. The value of the savings account increases linearly by $\$400$ each year, starting with an initial deposit of $\$5,000$ . What is the difference, in dollars, between the value of the investment account and the value of the savings account $2$ years after they are opened?

32Answer
B
0
C
432
D
5,032

Answer

The difference between the values of the two accounts after 2 years is 32 dollars.

The correct answer of 32 is found by evaluating both account models at 2 years. The investment account value is

5,000(1.08)^2 = 5,832

, and the savings account value is

5,000 + 400(2) = 5,800

. Subtracting the two values yields a difference of 32.

Step-by-Step Solution

Calculate the value of the investment account after 2 years.

I(2) = 5,000(1.08)^2 = 5,000(1.1664) = 5,832

The investment account grows exponentially, so we substitute

t = 2

into the given exponential model.

Determine the linear equation for the savings account and calculate its value after 2 years.

S(t) = 5,000 + 400t

, so

S(2) = 5,000 + 400(2) = 5,800

The savings account grows linearly by a constant rate of

400 per year from an initial deposit of

5,000.

Find the difference between the two account values.

5,832 - 5,800 = 32

To find how much more the investment account is worth, subtract the savings account value from the investment account value.

Key Concept

Distinguishing between and calculating values for linear growth (constant addition per time unit) and exponential growth (constant percentage multiplication per time unit).

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