A laboratory uses two types of equipment, Instrument A and Instrument B, which both have a… — SAT

A laboratory uses two types of equipment, Instrument A and Instrument B, which both have an initial value of $V$ dollars when purchased at the same time ( $t = 0$ years). The value of Instrument A decreases linearly at a constant rate over time. At $t = 3$ years, the value of Instrument A is $\$12,000$ , and at $t = 6$ years, its value is $\$8,000$ . The value of Instrument B decreases exponentially at a constant annual rate. At $t = 2$ years, the value of Instrument B is $\$4,000$ . What is the value, in dollars, of Instrument B at $t = 4$ years?

Answer: 1000 dollars

Answer

1000

The correct answer is 1000. By defining the value of Instrument A linearly, we obtain the system

V - 3d = 12000

and

V - 6d = 8000

. Solving this system yields an initial value of

V = 16000

. Instrument B decreases exponentially with the model

V_B(t) = 16000 \cdot b^t

. Given

V_B(2) = 4000

, we have

16000 \cdot b^2 = 4000

, which gives

b^2 = 0.25

. The value at

t = 4

is then calculated as

V_B(4) = V_B(2) \cdot b^2 = 4000 \cdot 0.25 = 1000

Step-by-Step Solution

Set up a system of linear equations for Instrument A's value.

V - 3d = 12000

and

V - 6d = 8000

Instrument A decreases linearly, meaning its value is modeled by

V_A(t) = V - dt

Solve for the constant annual rate of decrease,

d

, and the initial purchase value,

V

d = \frac{4000}{3}

and

V = 16000

Subtracting the two linear equations eliminates

V

, allowing us to solve for

d

and subsequently substitute it back to find

V

Apply the initial value

V = 16000

to Instrument B's exponential decay model at

t = 2

16000 \cdot b^2 = 4000

Instrument B decreases exponentially starting from the same initial value

V

, so its model is

V_B(t) = V \cdot b^t

Solve for the 2-year decay factor,

b^2

b^2 = 0.25

Dividing both sides of the equation by

16000

isolates the term representing the growth or decay factor over a two-year interval.

Calculate the value of Instrument B at

t = 4

years.

1000

Multiply the value at

t = 2

by the 2-year decay factor

b^2

, which yields

4000 \cdot 0.25 = 1000

Key Concept

Distinguishing between linear decay (constant absolute rate of change) and exponential decay (constant percentage rate of change) to set up and solve coupled system models.

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic