DAT Quantitative Reasoning: Compound Interest and Depreciation

DAT High-Yield Concept

Compound Interest Formula: $A = P\left(1 + \frac{r}{n}\right)^{nt}$ , where $A$ is total accumulated amount, $P$ is principal, $r$ is annual interest rate (decimal), $n$ is compounding periods per year, and $t$ is time in years.
Present Value Equation: Solved by dividing future amount by the growth factor: $P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}$
Declining-Balance Depreciation: $BV = P(1 - r)^t$ where $BV$ is book value after $t$ years with annual depreciation rate $r$ .
Exponential Base Criterion: Growth occurs when the repeated base factor is greater than 1 ( $b > 1$ ), whereas decay occurs when the base lies strictly between 0 and 1 ( $0 < b < 1$ ), as in half-life models $a\left(\frac{1}{2}\right)^t$ .

Compound Interest and Present Value

Compound interest generates exponential growth because interest earned during each period is added to the principal, earning additional interest in subsequent periods. In contrast to simple interest, where interest is earned solely on the initial principal via $A = P(1 + rt)$ (covered in depth in its own lesson), compounding recalculates interest on the updated balance at every compounding interval $n$ .

Key Variable Definitions:

$P$ : Principal or initial present value.
$r$ : Nominal annual interest rate expressed as a decimal (e.g., $6\% = 0.06$ ).
$n$ : Compounding periods per year (annual $n=1$ , semiannual $n=2$ , quarterly $n=4$ , monthly $n=12$ ).
$t$ : Time elapsed in years.

As the compounding frequency $n$ increases, interest accrues more frequently, slightly increasing the total future yield $A$ for any given period.

Solving for Present Value

To determine the initial deposit $P$ needed to reach a targeted future financial goal $A$ , rearrange the compound interest formula by dividing the total future amount by the compound growth factor $\left(1 + \frac{r}{n}\right)^{nt}$ .

Worked Example: Present Value Calculation

Problem: A dental student wants to accumulate $12,155.06 in a savings account after 2 years to purchase clinical equipment. The account pays an annual interest rate of 10% compounded semiannually. What principal $P$ must be deposited today?

Step-by-Step Solution:

Identify given values: $A = \$12,155.06$ , $r = 0.10$ , $n = 2$ , and $t = 2$ .
Calculate periodic interest rate: $\frac{r}{n} = \frac{0.10}{2} = 0.05$ .
Calculate total compounding periods: $nt = 2 \times 2 = 4$ .
Substitute into the present value formula: $P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}} = \frac{12155.06}{(1 + 0.05)^4} = \frac{12155.06}{(1.05)^4}$
Evaluate the growth factor: $(1.05)^4 = 1.21550625$ .
Compute principal: $P = \frac{12155.06}{1.21550625} = 10000$ .

Answer: The present value required is $10,000.

Common DAT Exam Trap: Forgetting to divide $r$ by $n$ or failing to multiply $t$ by $n$ . Always ensure the exponent represents the total number of periods $nt$ , not just years $t$ .

Depreciation and Exponential Decay

Depreciation models the loss of monetary value of equipment or assets over time. In declining-balance depreciation, an asset decreases in value by a fixed annual percentage $r$ . The book value $BV$ after $t$ years is calculated using the exponential decay formula:

BV = P(1 - r)^t

Growth vs. Decay Base Criterion: An exponential model

y = a \cdot b^t

exhibits exponential growth when the repeated base factor

b > 1

. It exhibits exponential decay when

0 < b < 1

. For example, half-life decay models take the form

a\left(\frac{1}{2}\right)^t

, where the quantity reduces by half with each time unit.

Exponential decay (Source: Svjo · CC BY-SA 4.0 · Wikimedia Commons)

Worked Example: Depreciation Book Value

Problem: A dental clinic buys a 3D intraoral scanner for $40,000. The device depreciates at an annual rate of 20%. What is the book value $BV$ of the scanner after 3 years?

Step-by-Step Solution:

Identify parameters: Initial principal $P = \$40,000$ , annual rate $r = 0.20$ , and time $t = 3$ .
Calculate the decay factor: $1 - r = 1 - 0.20 = 0.80$ .
Apply the depreciation equation: $BV = P(1 - r)^t = 40000 \times (0.80)^3$
Compute $(0.80)^3 = 0.512$ .
Multiply by initial value: $BV = 40000 \times 0.512 = 20480$ .

Answer: The book value after 3 years is $20,480.

Common DAT Exam Trap: Confusing constant percentage depreciation with linear (straight-line) depreciation. Linear decay subtracts a fixed dollar amount each year, whereas exponential decay applies the percentage reduction to the remaining balance.

Worked Example: Exponential Half-Life Decay

Problem: A radioactive tracer used in oral diagnostic imaging has an initial mass of 80 mg and a half-life of 1 hour. What mass remains after 3 hours?

Step-by-Step Solution:

Identify initial amount $a = 80\text{ mg}$ and base factor $b = \frac{1}{2}$ .
Set up the decay function: $M(t) = a\left(\frac{1}{2}\right)^t = 80\left(\frac{1}{2}\right)^3$
Evaluate exponent: $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$ .
Calculate remaining mass: $M(3) = 80 \times \frac{1}{8} = 10\text{ mg}$ .

Answer: 10 mg remains after 3 hours.

Comparing Growth and Decay Models

Mnemonic - "PLUS grows, MINUS shrinks": In financial exponential models, $1 + \text{rate}$ increases total value (Compound Interest), while $1 - \text{rate}$ reduces asset value (Depreciation).

Model Type	Formula	Base Factor ( $b$ )	Behavior Over Time	DAT Application
Compound Interest	$A = P\left(1 + \frac{r}{n}\right)^{nt}$	$b = 1 + \frac{r}{n} > 1$	Accelerating growth (curving upward)	Investments, savings, accrued interest
Declining Depreciation	$BV = P(1 - r)^t$	$b = 1 - r < 1$	Decelerating drop toward zero	Equipment valuation, vehicle resale
Half-Life Decay	$N(t) = a\left(\frac{1}{2}\right)^t$	$b = \frac{1}{2}$	Halves every period $t$	Radioactive isotopes, drug elimination

Critical Thinking Questions

If interest rate $r$ and time $t$ remain constant, how does compounding monthly ( $n = 12$ ) alter the future balance $A$ compared to compounding semiannually ( $n = 2$ )?
An equipment item depreciates by 50% every 4 years. How can you write its book value equation using base $\frac{1}{2}$ as a function of time $t$ in years?
Why is present value $P$ always strictly less than future value $A$ whenever $r > 0$ and $t > 0$ ?

References

Sources returned by Google Search grounding during lesson verification.

DAT Quantitative Reasoning Formulas - Effortless Math — effortlessmath.com
QR Formulas Cheat Sheet | DAT Bootcamp — bootcamp.com

Compound Interest and Depreciation

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