NBT Mathematics: Finance, Growth and Decay Flashcards

What is the total amount after investing $2,000 at 5% simple interest per year for 3 years?

Answer: $2,300

Simple interest = 2000 × 0.05 × 3 = $300, so the total is $2,300.

$1,000 is invested at 10% per year compounded annually. What is the value after 2 years?

Answer: $1,210

A = 1000(1.10)^2 = $1,210.

A car worth $20,000 depreciates at 15% per year on the reducing balance. What is it worth after 2 years?

Answer: $14,450

A = 20000(0.85)^2 = $14,450.

A town of 5,000 people grows by 4% each year. Which expression gives the population after n years?

Answer: 5000(1.04)^n

Growth of 4% per year means multiplying by 1.04 each year.

A radioactive sample of 80 g has a half-life of 8 days. How much remains after 24 days?

Answer: 10 g

24 days is 3 half-lives, so 80 × (1/2)^3 = 10 g.

The price of an item rises from $40 to $50. What is the percentage increase?

Answer: 25%

The increase of $10 divided by the original $40 is 25%.

A $12,000 machine depreciates by 10% of its original value each year (straight-line). What is its value after 4 years?

Answer: $7,200

Straight-line loss is $1,200 per year, so 12000 − 4800 = $7,200.