3.2 Financial & Distance Math: Interest, Profit/Loss, Speed & Time

3.2 Financial & Distance Math: Interest, Profit/Loss, Speed & Time

Financial mathematics and distance-rate-time motion problems represent core question types on the AFCT Arithmetic Reasoning subtest. Mastering these topics requires a strong command of specific formulas, unit consistency, and algebraic modeling.

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1. Simple Interest & Compound Interest Basics

Simple interest is calculated solely on the initial principal amount. The formula is:

$I = P \cdot r \cdot t$

• $I =$ Total interest earned or owed (in dollars)
• $P =$ Principal starting amount (in dollars)
• $r =$ Annual interest rate (expressed as a decimal, e.g., $6% = 0.06$)
• $t =$ Time duration (strictly in years)

Total Accumulated Balance

The total balance ($A$) accumulated after time $t$ is the principal plus interest:

$A = P + I = P(1 + r \cdot t)$

Time Conversions for $t$

If time is given in months or days, it must be converted to years before substituting into the formula:

• Months to Years: $t = \frac{\text{Number of Months}}{12}$
• Days to Years (Standard Bank Year): $t = \frac{\text{Number of Days}}{360}$ or $\frac{\text{Number of Days}}{365}$

Step-by-Step Worked Example 1: A service member deposits $6,000 into an account earning 5% simple annual interest. What is the total balance after 18 months?

• Step 1: Convert time into years: $t = \frac{18}{12} = 1.5\text{ years}$.
• Step 2: Convert rate to decimal: $r = 0.05$.
• Step 3: Calculate interest: $I = $6,000 \times 0.05 \times 1.5 = $450$.
• Step 4: Calculate total balance: $A = $6,000 + $450 = $6,450$.

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2. Cost Price, Selling Price, Profit & Loss

In retail and procurement math, transactions revolve around Cost Price ($CP$) and Selling Price ($SP$).

Finding Cost Price from Selling Price and Profit %

To determine the original Cost Price when Selling Price and Profit percentage are given:

$SP = CP \cdot (1 + \text{Profit \% in decimal}) \implies CP = \frac{SP}{1 + \text{Profit \%}}$

Step-by-Step Worked Example 2: A surplus vendor sells a piece of reconditioned equipment for $1,320, realizing a 10% profit over the original reconditioning cost. What was the cost price?

• Step 1: Express relationship: $SP = CP \times (1 + 0.10) = 1.10 \times CP$.
• Step 2: Substitute known value: $$1,320 = 1.10 \times CP$.
• Step 3: Solve for $CP$: $CP = \frac{1,320}{1.10} = $1,200$.

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3. Distance, Rate & Time Core Principles

The fundamental uniform motion formula connecting Distance ($d$), Rate/Speed ($r$), and Time ($t$) is:

$d = r \cdot t \quad \Longleftrightarrow \quad r = \frac{d}{t} \quad \Longleftrightarrow \quad t = \frac{d}{r}$

Unit Consistency Rule

Before plugging values into $d = r \cdot t$, ensure all units align. If rate is in miles per hour (mph), distance must be in miles, and time must be in hours. If time is given in minutes, convert to hours by dividing by 60.

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4. Relative Speed and Pursuit Word Problems

When two objects move simultaneously, solve using relative speed concepts.

Case 1: Moving in Opposite Directions (Towards or Away)

When two objects move towards each other (or away from each other in opposite directions), their rates add together because the distance between them changes at a faster combined rate:

$r_{\text{relative}} = r_1 + r_2 \implies \text{Time until Meeting} = \frac{\text{Initial Distance}}{r_1 + r_2}$

Case 2: Moving in the Same Direction (Pursuit / Catch-Up)

When a faster object pursues a slower object that started earlier or ahead, their rates subtract:

$r_{\text{relative}} = r_{\text{fast}} - r_{\text{slow}} \implies \text{Time to Catch Up} = \frac{\text{Head-Start Distance}}{r_{\text{fast}} - r_{\text{slow}}}$

Step-by-Step Worked Example 3 (Catch-Up Pursuit): Truck A leaves a logistics base traveling north at 40 mph. 2 hours later, Truck B leaves the same base traveling north along the same highway at 60 mph. How many hours after Truck B departs will it catch up to Truck A?

• Step 1: Calculate Truck A's head-start distance: $d_{\text{head-start}} = 40\text{ mph} \times 2\text{ hrs} = 80\text{ miles}$.
• Step 2: Calculate relative catch-up rate: $r_{\text{relative}} = 60\text{ mph} - 40\text{ mph} = 20\text{ mph}$.
• Step 3: Calculate time for Truck B to close the gap: $t = \frac{80\text{ miles}}{20\text{ mph}} = 4\text{ hours}$.

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5. Round-Trip Average Speed & The Harmonic Mean Trap

AFCT Critical Warning: If a vehicle travels to a destination at speed $v_1$ and returns along the exact same route at speed $v_2$, the average speed for the round trip is NOT the simple average $\frac{v_1 + v_2}{2}$.

Because the vehicle spends more time traveling at the slower speed, average speed is weighted by time. When distance is equal in both directions, average speed is calculated using the harmonic mean formula:

$\text{Average Speed} = \frac{2 \cdot v_1 \cdot v_2}{v_1 + v_2}$

Step-by-Step Worked Example 4: An aircraft flies from Base X to Base Y (distance 300 miles) at 150 mph. On the return flight, headwinds reduce its speed to 100 mph. What is the aircraft's average speed for the entire round trip?

• Step 1: Apply harmonic mean formula: $\text{Average Speed} = \frac{2 \times 150 \times 100}{150 + 100}$.
• Step 2: Simplify numerator and denominator: $\frac{30,000}{250} = 120\text{ mph}$.
• Verification via Total Distance / Total Time:
  • Time out: $\frac{300}{150} = 2\text{ hours}$. Time back: $\frac{300}{100} = 3\text{ hours}$. Total time $= 5\text{ hours}$.
  • Total distance $= 300 + 300 = 600\text{ miles}$.
  • $\text{Average Speed} = \frac{600\text{ miles}}{5\text{ hours}} = 120\text{ mph}$.