Algebra 1 worksheet featuring eight word problems on exponential growth and decay, including scenarios like population increase, radioactive decay, and investment returns.

Let’s solve each problem one by one. I’ll work through the math carefully and explain each step simply.

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Problem 1:

A population of field mice is 120. The population increases at a rate of 9% a year. Write an exponential function that models this situation.

→ We use the formula for exponential growth:
y = a(1 + r)^t
Where:
- a = starting amount (120)
- r = growth rate as a decimal (9% = 0.09)
- t = time in years

So:
y = 120(1 + 0.09)^t → y = 120(1.09)^t

✔ Final Answer for #1: y = 120(1.09)^t

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Problem 2:

A rock of radioactive material has 500 atoms in it. The number of atoms decreases at a rate of 11% a day. Write an exponential function that models this situation.

→ For decay, we use:
y = a(1 - r)^t
Where:
- a = 500
- r = 0.11 (11%)
- t = days

So:
y = 500(1 - 0.11)^t → y = 500(0.89)^t

✔ Final Answer for #2: y = 500(0.89)^t

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Problem 3:

A painting is valued at $10,000 but the value increases at a rate of 3% a year. How much will the painting be worth in 5 years?

→ Use growth formula: y = a(1 + r)^t
a = 10000, r = 0.03, t = 5

y = 10000 × (1.03)^5
Calculate (1.03)^5:
1.03^2 = 1.0609
1.03^4 = (1.0609)^2 ≈ 1.1255
1.03^5 = 1.1255 × 1.03 ≈ 1.15927

Then:
y ≈ 10000 × 1.15927 = 11,592.70

✔ Final Answer for #3: $11,592.70

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Problem 4:

The population of wild ducks in Lake County is 2,500. The population is decreasing at a rate of 7% a year. How many ducks will be left in 10 years?

→ Use decay formula: y = a(1 - r)^t
a = 2500, r = 0.07, t = 10

y = 2500 × (0.93)^10

Calculate (0.93)^10:
We can compute step by step or use calculator approximation:
(0.93)^10 ≈ 0.48398

Then:
y ≈ 2500 × 0.48398 ≈ 1209.95

Round to nearest whole duck → 1210

✔ Final Answer for #4: 1210 ducks

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Problem 5:

Mr. Miller has $5000 to invest. If he invests into a savings account that pays 12% interest every year, how much money will he have in 5 years?

→ Annual compounding: y = a(1 + r)^t
a = 5000, r = 0.12, t = 5

y = 5000 × (1.12)^5

Calculate (1.12)^5:
1.12^2 = 1.2544
1.12^4 = (1.2544)^2 ≈ 1.5735
1.12^5 = 1.5735 × 1.12 ≈ 1.7623

Then:
y ≈ 5000 × 1.7623 = 8811.50

✔ Final Answer for #5: $8,811.50

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Problem 6:

Mr. Miller has $5000 to invest. If he invests into a CD that pays 1.1% interest every month, how much money will he have in 5 years?

→ Monthly compounding:
Number of months = 5 × 12 = 60
Monthly rate = 1.1% = 0.011

Formula: y = a(1 + r)^n
y = 5000 × (1.011)^60

Calculate (1.011)^60:
Approximate using calculator: ≈ 1.9348

Then:
y ≈ 5000 × 1.9348 = 9674.00

✔ Final Answer for #6: $9,674.00

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Problem 7:

Mr. Miller has $5000 to invest. If he invests into a bond that pays 4% interest every 6 months, how much money will he have in 5 years?

→ Compounded every 6 months → 2 times per year
Total periods = 5 × 2 = 10
Rate per period = 4% = 0.04

y = 5000 × (1.04)^10

(1.04)^10 ≈ 1.48024

y ≈ 5000 × 1.48024 = 7401.20

✔ Final Answer for #7: $7,401.20

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Problem 8:

Mr. Miller has $5000 to invest. If he invests into a mutual fund that pays 2.2% interest every 2 months, how much money will he have in five years?

→ Every 2 months → 6 periods per year
Total periods = 5 × 6 = 30
Rate per period = 2.2% = 0.022

y = 5000 × (1.022)^30

(1.022)^30 ≈ 1.9161

y ≈ 5000 × 1.9161 = 9580.50

✔ Final Answer for #8: $9,580.50

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Final Answers:

1) y = 120(1.09)^t
2) y = 500(0.89)^t
3) $11,592.70
4) 1210 ducks
5) $8,811.50
6) $9,674.00
7) $7,401.20
8) $9,580.50