Exponential Growth And Decay Worksheet - Fill Online, Printable ... - Free Printable

You've uploaded a worksheet titled “Exponential Growth and Decay Word Problems” from “GPS Advanced Algebra – Unit 3.” Below, I’ll solve each of the 7 problems step-by-step with clear explanations.

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Problem 1

> Find a bank account balance if the account starts with $100, has an annual interest rate of 4%, and the money is left in the account for 12 years.

Solution:

This is exponential growth with compound interest. The formula is:

\[
A = P(1 + r)^t
\]

Where:
- \( A \) = final amount
- \( P \) = principal (initial amount) = $100
- \( r \) = annual interest rate = 4% = 0.04
- \( t \) = time in years = 12

Plug in:

\[
A = 100(1 + 0.04)^{12} = 100(1.04)^{12}
\]

Calculate \( (1.04)^{12} \):

\[
(1.04)^{12} \approx 1.601032
\]

So,

\[
A \approx 100 \times 1.601032 = 160.10
\]

✔ Answer: $160.10

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Problem 2

> In 1985, there were 285 cell phone subscribers in the small town of Centerville. The number of subscribers increased by 75% per year after 1985. How many cell phone subscribers were in Centerville in 1994?

Solution:

This is exponential growth. From 1985 to 1994 is 9 years.

Formula:

\[
A = P(1 + r)^t
\]

Where:
- \( P = 285 \)
- \( r = 75\% = 0.75 \)
- \( t = 9 \)

\[
A = 285(1 + 0.75)^9 = 285(1.75)^9
\]

Calculate \( (1.75)^9 \):

\[
(1.75)^9 \approx 155.78
\]

Then:

\[
A \approx 285 \times 155.78 \approx 44,397.3
\]

Since we’re talking about people, round to nearest whole number.

✔ Answer: 44,397 subscribers

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Problem 3

> Bacteria can multiply at an alarming rate when each bacteria splits into two new cells, thus doubling. If we start with only one bacteria which can double every hour, how many bacteria will we have by the end of one day?

Solution:

This is doubling every hour → exponential growth with base 2.

One day = 24 hours.

Starting with 1 bacterium:

\[
A = 1 \cdot 2^{24}
\]

\[
2^{24} = 16,777,216
\]

✔ Answer: 16,777,216 bacteria

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Problem 4

> Each year the local country club sponsors a tennis tournament. Play starts with 128 participants. During each round, half of the players are eliminated. How many players remain after 5 rounds?

Solution:

This is exponential decay — halving each round.

Formula:

\[
A = P \cdot \left(\frac{1}{2}\right)^t
\]

Where:
- \( P = 128 \)
- \( t = 5 \) rounds

\[
A = 128 \cdot \left(\frac{1}{2}\right)^5 = 128 \cdot \frac{1}{32} = 4
\]

✔ Answer: 4 players remain

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Problem 5

> The population of Winnemucca, Nevada, can be modeled by \( P = 6191(1.04)^t \) where \( t \) is the number of years since 1990. What was the population in 1990? By what percent did the population increase by each year?

Solution:

The model is:

\[
P = 6191(1.04)^t
\]

- When \( t = 0 \) (year 1990), \( P = 6191(1.04)^0 = 6191 \cdot 1 = 6191 \)

✔ Population in 1990: 6,191

- The growth factor is 1.04 → this means a 4% increase per year.

✔ Annual percent increase: 4%

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Problem 6

> You have inherited land that was purchased for $30,000 in 1960. The value of the land increased by approximately 5% per year. What is the approximate value of the land in the year 2011?

Solution:

Time from 1960 to 2011 = 51 years

Exponential growth:

\[
A = P(1 + r)^t
\]

Where:
- \( P = 30,000 \)
- \( r = 5\% = 0.05 \)
- \( t = 51 \)

\[
A = 30000(1.05)^{51}
\]

Calculate \( (1.05)^{51} \):

Using calculator: \( (1.05)^{51} \approx 12.040 \)

Then:

\[
A \approx 30000 \times 12.040 = 361,200
\]

✔ Answer: Approximately $361,200

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Problem 7

> During normal breathing, about 12% of the air in the lungs is expelled after one breath. Write an exponential decay model for the amount of the original air left in the lungs if the initial amount of air in the lungs is 500 mL. How much of the original air is present after 240 breaths?

Solution:

Each breath removes 12%, so 88% remains after each breath.

Exponential decay model:

\[
A = 500 \cdot (0.88)^t
\]

Where \( t \) = number of breaths.

After 240 breaths:

\[
A = 500 \cdot (0.88)^{240}
\]

Calculate \( (0.88)^{240} \):

This is a very small number. Using calculator:

\[
(0.88)^{240} \approx 1.05 \times 10^{-13}
\]

Then:

\[
A \approx 500 \times 1.05 \times 10^{-13} \approx 5.25 \times 10^{-11} \text{ mL}
\]

That’s essentially zero for practical purposes — but mathematically:

✔ Model: \( A = 500(0.88)^t \)
✔ After 240 breaths: ≈ 5.25 × 10⁻¹¹ mL (negligible)

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## ✔ Final Answers Summary:

1. $160.10
2. 44,397 subscribers
3. 16,777,216 bacteria
4. 4 players
5. 1990 population: 6,191; Annual increase: 4%
6. Approx. $361,200
7. Model: \( A = 500(0.88)^t \); After 240 breaths: ~5.25 × 10⁻¹¹ mL

Let me know if you’d like graphs or more detailed steps for any problem!