Population and Ecology Math Worksheet Generator
Exponential and logistic growth, mark-recapture, Simpson’s index, energy transfer, and Hardy-Weinberg. No AI, no credits.
Grades 8 – 12No account needed
Problem sets for exponential and logistic growth, population density, mark and recapture (Lincoln index), Simpson’s diversity index, the 10 percent energy transfer rule, and Hardy-Weinberg equilibrium. Every answer is computed by two independent methods (closed form versus iterative, recessive frequency versus allele counts). No AI, so it never uses a generation credit.
What students practice
- Biology
- Ecology
- Population dynamics
Live sample sheet
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Name: ____________________Date: ____________
Population and Ecology Math
1. A study counts 252 trees in a 18 hectare forest. What is the population density?
2. In a mark-recapture study, 34 fish are marked and released. Later, 93 fish are captured, of which 12 are marked. What is the estimated population size?
3. A study counts 440 trees in a 10 hectare forest. What is the population density?
4. A population of 90 bacteria grows exponentially at a rate of 0.077 per hour. What is the population size after 9 hours?
5. A population of 55 bacteria grows exponentially at a rate of 0.072 per hour. What is the population size after 14 hours?
6. A population of 52 bacteria grows exponentially at a rate of 0.085 per hour. What is the population size after 5 hours?
7. A study counts 64 trees in a 6 hectare forest. What is the population density?
8. A population of 26 bacteria grows exponentially at a rate of 0.081 per hour. What is the population size after 13 hours?
9. A population of 25 bacteria grows exponentially at a rate of 0.086 per hour. What is the population size after 12 hours?
10. In a mark-recapture study, 45 fish are marked and released. Later, 84 fish are captured, of which 5 are marked. What is the estimated population size?
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Answer Key — Population and Ecology Math
1. A study counts 252 trees in a 18 hectare forest. What is the population density?
Solution:
Knowns: N = 252, A = 18 ha
Density = N / A = 252 / 18 = 14.00 trees/ha
Check (exact fraction): 14 = 14 — agrees
Answer: 14.00; trees/hectare
2. In a mark-recapture study, 34 fish are marked and released. Later, 93 fish are captured, of which 12 are marked. What is the estimated population size?
Solution:
Knowns: M = 34, C = 93, R = 12
Lincoln index: N = (M × C) / R = (34 × 93) / 12
N = 3162 / 12 = 263.5
Check (exact fraction): 527/2 = 263.5 — agrees
Answer: 263.5; fish
3. A study counts 440 trees in a 10 hectare forest. What is the population density?
Solution:
Knowns: N = 440, A = 10 ha
Density = N / A = 440 / 10 = 44.00 trees/ha
Check (exact fraction): 44 = 44 — agrees
Answer: 44.00; trees/hectare
4. A population of 90 bacteria grows exponentially at a rate of 0.077 per hour. What is the population size after 9 hours?
Solution:
Knowns: N₀ = 90, r = 0.077/hr, t = 9 hr
Exponential growth: N(t) = N₀ × e^(rt)
N(9) = 90 × e^(0.077 × 9) = 90 × e^0.693
N(9) = 90 × 1.9997 = 180.0
Check (iterative): step-by-step calculation = 180.0 — agrees
Answer: 180.0; organisms
5. A population of 55 bacteria grows exponentially at a rate of 0.072 per hour. What is the population size after 14 hours?
Solution:
Knowns: N₀ = 55, r = 0.072/hr, t = 14 hr
Exponential growth: N(t) = N₀ × e^(rt)
N(14) = 55 × e^(0.072 × 14) = 55 × e^1.008
N(14) = 55 × 2.7401 = 150.7
Check (iterative): step-by-step calculation = 150.7 — agrees
Answer: 150.7; organisms
6. A population of 52 bacteria grows exponentially at a rate of 0.085 per hour. What is the population size after 5 hours?
Solution:
Knowns: N₀ = 52, r = 0.085/hr, t = 5 hr
Exponential growth: N(t) = N₀ × e^(rt)
N(5) = 52 × e^(0.085 × 5) = 52 × e^0.425
N(5) = 52 × 1.5296 = 79.5
Check (iterative): step-by-step calculation = 79.5 — agrees
Answer: 79.5; organisms
7. A study counts 64 trees in a 6 hectare forest. What is the population density?
Solution:
Knowns: N = 64, A = 6 ha
Density = N / A = 64 / 6 = 10.67 trees/ha
Check (exact fraction): 32/3 = 10.666666666666666 — agrees
Answer: 10.67; trees/hectare
8. A population of 26 bacteria grows exponentially at a rate of 0.081 per hour. What is the population size after 13 hours?
Solution:
Knowns: N₀ = 26, r = 0.081/hr, t = 13 hr
Exponential growth: N(t) = N₀ × e^(rt)
N(13) = 26 × e^(0.081 × 13) = 26 × e^1.053
N(13) = 26 × 2.8662 = 74.5
Check (iterative): step-by-step calculation = 74.5 — agrees
Answer: 74.5; organisms
9. A population of 25 bacteria grows exponentially at a rate of 0.086 per hour. What is the population size after 12 hours?
Solution:
Knowns: N₀ = 25, r = 0.086/hr, t = 12 hr
Exponential growth: N(t) = N₀ × e^(rt)
N(12) = 25 × e^(0.086 × 12) = 25 × e^1.032
N(12) = 25 × 2.8067 = 70.2
Check (iterative): step-by-step calculation = 70.2 — agrees
Answer: 70.2; organisms
10. In a mark-recapture study, 45 fish are marked and released. Later, 84 fish are captured, of which 5 are marked. What is the estimated population size?
Solution:
Knowns: M = 45, C = 84, R = 5
Lincoln index: N = (M × C) / R = (45 × 84) / 5
N = 3780 / 5 = 756.0
Check (exact fraction): 756 = 756 — agrees
Answer: 756.0; fish
Answers computed by code and double-checked by two independent methods.
Assumptions and Conventions
Assumptions:
- Exponential growth assumes unlimited resources and no predation.
- Logistic growth assumes a carrying capacity K and density-dependent growth.
- Hardy-Weinberg assumes no mutation, no migration, no selection, random mating, and large population.
- Lincoln index assumes marked individuals mix randomly and marks are not lost.
- Simpson's index: D = 1 - Σ(n(n-1)/(N(N-1))). Higher D = more diverse.
- 10% energy transfer rule: ~10% of energy passes to each trophic level.
Conventions:
- Course: biology
- Difficulty: standard
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Frequently asked questions
Do students need an account to use the Population and Ecology Math?
No. Students never need an account. They open a shared link or join with a code.
Can I print the Population and Ecology Math without signing up?
Yes. This tool computes answers by code, so you can generate, preview and print worksheet without an account. A free account adds saving and student links.
How are the answers computed?
Every answer is computed by code and verified by a second independent method. No AI is used.
What grades is the Population and Ecology Math for?
This tool fits Grades 8 – 12.
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Name: ____________________Date: ____________
Population and Ecology Math
1. In a mark-recapture study, 44 fish are marked and released. Later, 88 fish are captured, of which 26 are marked. What is the estimated population size?
2. A study counts 206 trees in a 2 hectare forest. What is the population density?
3. In a mark-recapture study, 100 fish are marked and released. Later, 129 fish are captured, of which 16 are marked. What is the estimated population size?
4. In a mark-recapture study, 92 fish are marked and released. Later, 147 fish are captured, of which 21 are marked. What is the estimated population size?
5. A population of 53 bacteria grows exponentially at a rate of 0.037 per hour. What is the population size after 17 hours?
6. In a mark-recapture study, 72 fish are marked and released. Later, 124 fish are captured, of which 25 are marked. What is the estimated population size?
7. A study counts 254 trees in a 2 hectare forest. What is the population density?
8. A population of 28 bacteria grows exponentially at a rate of 0.078 per hour. What is the population size after 6 hours?
9. A population of 15 bacteria grows exponentially at a rate of 0.067 per hour. What is the population size after 19 hours?
10. A study counts 145 trees in a 1 hectare forest. What is the population density?
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Answer Key — Population and Ecology Math
1. In a mark-recapture study, 44 fish are marked and released. Later, 88 fish are captured, of which 26 are marked. What is the estimated population size?
Solution:
Knowns: M = 44, C = 88, R = 26
Lincoln index: N = (M × C) / R = (44 × 88) / 26
N = 3872 / 26 = 148.9
Check (exact fraction): 1936/13 = 148.92307692307693 — agrees
Answer: 148.9; fish
2. A study counts 206 trees in a 2 hectare forest. What is the population density?
Solution:
Knowns: N = 206, A = 2 ha
Density = N / A = 206 / 2 = 103.00 trees/ha
Check (exact fraction): 103 = 103 — agrees
Answer: 103.00; trees/hectare
3. In a mark-recapture study, 100 fish are marked and released. Later, 129 fish are captured, of which 16 are marked. What is the estimated population size?
Solution:
Knowns: M = 100, C = 129, R = 16
Lincoln index: N = (M × C) / R = (100 × 129) / 16
N = 12900 / 16 = 806.3
Check (exact fraction): 3225/4 = 806.25 — agrees
Answer: 806.3; fish
4. In a mark-recapture study, 92 fish are marked and released. Later, 147 fish are captured, of which 21 are marked. What is the estimated population size?
Solution:
Knowns: M = 92, C = 147, R = 21
Lincoln index: N = (M × C) / R = (92 × 147) / 21
N = 13524 / 21 = 644.0
Check (exact fraction): 644 = 644 — agrees
Answer: 644.0; fish
5. A population of 53 bacteria grows exponentially at a rate of 0.037 per hour. What is the population size after 17 hours?
Solution:
Knowns: N₀ = 53, r = 0.037/hr, t = 17 hr
Exponential growth: N(t) = N₀ × e^(rt)
N(17) = 53 × e^(0.037 × 17) = 53 × e^0.629
N(17) = 53 × 1.8757 = 99.4
Check (iterative): step-by-step calculation = 99.4 — agrees
Answer: 99.4; organisms
6. In a mark-recapture study, 72 fish are marked and released. Later, 124 fish are captured, of which 25 are marked. What is the estimated population size?
Solution:
Knowns: M = 72, C = 124, R = 25
Lincoln index: N = (M × C) / R = (72 × 124) / 25
N = 8928 / 25 = 357.1
Check (exact fraction): 8928/25 = 357.12 — agrees
Answer: 357.1; fish
7. A study counts 254 trees in a 2 hectare forest. What is the population density?
Solution:
Knowns: N = 254, A = 2 ha
Density = N / A = 254 / 2 = 127.00 trees/ha
Check (exact fraction): 127 = 127 — agrees
Answer: 127.00; trees/hectare
8. A population of 28 bacteria grows exponentially at a rate of 0.078 per hour. What is the population size after 6 hours?
Solution:
Knowns: N₀ = 28, r = 0.078/hr, t = 6 hr
Exponential growth: N(t) = N₀ × e^(rt)
N(6) = 28 × e^(0.078 × 6) = 28 × e^0.468
N(6) = 28 × 1.5968 = 44.7
Check (iterative): step-by-step calculation = 44.7 — agrees
Answer: 44.7; organisms
9. A population of 15 bacteria grows exponentially at a rate of 0.067 per hour. What is the population size after 19 hours?
Solution:
Knowns: N₀ = 15, r = 0.067/hr, t = 19 hr
Exponential growth: N(t) = N₀ × e^(rt)
N(19) = 15 × e^(0.067 × 19) = 15 × e^1.273
N(19) = 15 × 3.5716 = 53.6
Check (iterative): step-by-step calculation = 53.6 — agrees
Answer: 53.6; organisms
10. A study counts 145 trees in a 1 hectare forest. What is the population density?
Solution:
Knowns: N = 145, A = 1 ha
Density = N / A = 145 / 1 = 145.00 trees/ha
Check (exact fraction): 145 = 145 — agrees
Answer: 145.00; trees/hectare
Answers computed by code and double-checked by two independent methods.
Assumptions and Conventions
Assumptions:
- Exponential growth assumes unlimited resources and no predation.
- Logistic growth assumes a carrying capacity K and density-dependent growth.
- Hardy-Weinberg assumes no mutation, no migration, no selection, random mating, and large population.
- Lincoln index assumes marked individuals mix randomly and marks are not lost.
- Simpson's index: D = 1 - Σ(n(n-1)/(N(N-1))). Higher D = more diverse.
- 10% energy transfer rule: ~10% of energy passes to each trophic level.
Conventions:
- Course: biology
- Difficulty: standard
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Population and Ecology Math - Biology - Population growth
Growth models, mark-recapture, diversity, energy transfer, and Hardy-Weinberg. No AI, no credits.