AP® Environmental Science Beta

Formulas you must memorize

The exam gives you no formula sheet. These are the equations its calculations use, in our own words: what each means, when to use it, and the trap that costs points.

Three habits that save points

Write the setup, then the answer. The calculation question gives one point for the setup and one for the answer with its units, so the setup must be on the page.
Brackets and units in the setup. $\frac{(150 - 120)}{120} \times 100$ , not 150 - 120 / 120 × 100; and units that visibly cancel.
Read the question to the end. "In what year" means add the doubling time to the start year; "in millions" means give the scale.

Percents and change

Formula and a worked line	What it is and when to use it	The trap	Learn it in
$\text{percent change}$ $= \frac{\text{new} - \text{old}}{\text{old}} \times 100$ Worked: $\frac{(150 - 120)\,\text{fish}}{120\,\text{fish}} \times 100$ $= 25\,\%$	How much a quantity grew or shrank, as a percent of where it started. Any "by what percent did it change" question.	Brackets around the difference, divide by the OLD value, then × 100. A fall comes out negative.	Math you need Percents and Percent ChangePractice it: Calculation studio
$\text{part}$ $= \frac{\text{percent}}{100} \times \text{whole}$ Worked: $0.40 \times 6{,}000\,\text{m}^3$ $= 2{,}400\,\text{m}^3$	A percent of a quantity. Energy passed on, water lost, the share of a budget.	Turn the percent into a decimal first: 10% is 0.10. "40% less" means 60% is left.	Math you need Percents and Percent ChangePractice it: Calculation studio

Populations

Formula and a worked line	What it is and when to use it	The trap	Learn it in
$\text{rate of natural increase }(\%)$ $= \frac{\text{CBR} - \text{CDR}}{10}$ Worked: $\frac{30 - 10}{10}$ $= 2.0\,\%\text{ per year}$	Growth from births and deaths, when both rates are per 1,000 people. Turning crude birth and death rates into a growth rate in percent.	Divide by 10, not 100: the rates are per 1,000.	Unit 3 Human Population DynamicsPractice it: Population growth, Age structure
$\text{growth }(\%)$ $= \frac{(\text{B} + \text{I}) - (\text{D} + \text{E})}{\text{population}} \times 100$ Worked: $\frac{(1{,}100{,}000 + 100{,}000) - (400{,}000 + 50{,}000)}{50{,}000{,}000} \times 100$ $= 1.5\,\%$	Growth from counts, with migration: B births, I immigrants, D deaths, E emigrants. When the scenario gives numbers of people rather than rates.	Immigration adds, emigration subtracts; divide by the population at the start.	Unit 3 Human Population DynamicsPractice it: Population growth
$\text{doubling time}$ $\approx \frac{70}{\text{growth rate }(\%)}$ Worked: $\frac{70}{1.4} = 50\,\text{years}$	The rule of 70: how long a population (or anything growing at a steady rate) takes to double. A growth rate in percent per year, and a question about doubling.	Use the rate in percent (2, not 0.02).	Math you need Growth Rates, Doubling Time and the Rule of 70Practice it: Population growth, Calculation studio
$\text{year it doubles}$ $= \text{start year}$ $+ \frac{70}{\text{growth rate }(\%)}$ Worked: $2025 + \frac{70}{1.4} = 2075$	The calendar year a population doubles. Any question that asks "in what year".	Add the doubling time to the start year; 50 years is not the answer to "in what year".	Math you need Growth Rates, Doubling Time and the Rule of 70Practice it: Population growth, Calculation studio
$N = N_0 \times 2^{t / T_d}$ Worked: $1{,}000\,\text{bacteria} \times 2^{30 / 10}$ $= 8{,}000\,\text{bacteria}$	A population after a number of doubling times. How big something gets after a stated time at a steady growth rate.	Count the doublings first: t divided by the doubling time.	Unit 3 Population Growth and Resource AvailabilityPractice it: Population growth
$N = N_0 \times \left(1 + \frac{r}{100}\right)^{t}$ Worked: $5{,}000\,\text{people} \times (1.02)^{10}$ $= 6{,}095\,\text{people}$	Exponential growth at r percent a year, year by year. A projection a set number of years ahead.	r is a percent, so divide by 100; t is the exponent, not a multiplier.	Unit 3 Population Growth and Resource AvailabilityPractice it: Population growth
$\text{TFR} \approx 2.1$ Worked: $3.4 - 2.1$ $= 1.3\,\text{children per woman above 2.1}$	Replacement-level fertility: the average number of children per woman that keeps a population the same size over time. Judging whether a population will grow or shrink in the long run.	Not 2.0: a little extra makes up for children who do not survive to have children.	Unit 3 Total Fertility RatePractice it: Age structure
$\text{TFR}$ $= \frac{\text{sum of age-specific birth rates per 1,000 women} \times 5}{1{,}000}$ Worked: $\frac{(20 + 110 + 130 + 90 + 40 + 10 + 0) \times 5}{1{,}000}$ $= 2.0\,\text{children per woman}$	Total fertility rate from birth rates in 5-year age bands. A table of births per 1,000 women by age group.	Multiply by the 5 years in each band, then divide by 1,000.	Unit 3 Total Fertility RatePractice it: Age structure

Energy flow and productivity

Formula and a worked line	What it is and when to use it	The trap	Learn it in
$\text{NPP} = \text{GPP} - \text{respiration}$ Worked: $24{,}000 - 9{,}000\,\text{kcal/m}^2\text{/yr}$ $= 15{,}000\,\text{kcal/m}^2\text{/yr}$	Net primary productivity: the energy producers store, after their own respiration. What is left for consumers, from GPP and respiration.	NPP is never larger than GPP; GPP = NPP + respiration.	Unit 1 Primary ProductivityPractice it: Energy flow
$E_{\text{next level}} = E \times 0.10$ Worked: $15{,}000\,\text{kcal} \times 0.10 \times 0.10$ $= 150\,\text{kcal}$	The 10% rule: about a tenth of the energy at one trophic level reaches the next. Energy at the top of a food chain from the energy at the bottom.	Multiply by 0.10 once per arrow: producers to secondary consumers is two steps, × 0.01.	Unit 1 Energy Flow and the 10% RulePractice it: Energy flow, Calculation studio

Energy, efficiency and emissions

Formula and a worked line	What it is and when to use it	The trap	Learn it in
$\text{efficiency}$ $= \frac{\text{useful output}}{\text{input}} \times 100$ Worked: $\frac{350\,\text{MWh}}{1{,}000\,\text{MWh}} \times 100$ $= 35\,\%$	The share of input energy that comes out as useful work. Power plants, engines, light bulbs.	Output over input, never the other way round.	Unit 6 Energy ConservationPractice it: Energy chain
$\text{overall efficiency}$ $= e_1 \times e_2 \times e_3 \times \ldots$ Worked: $0.35 \times 0.90 \times 0.90 \times 100$ $= 28.35\,\%$	Efficiencies in a chain multiply. Fuel to plant to wires to motor.	Multiply the decimals; do not add or average the percents.	Unit 6 Fossil FuelsPractice it: Energy chain
$\text{energy (kWh)}$ $= \text{power (kW)} \times \text{time (h)}$ Worked: $1{,}500\,\text{W} \times 4\,\text{h} \times \frac{1\,\text{kW}}{1{,}000\,\text{W}}$ $= 6\,\text{kWh}$	Energy used by a device running at a steady power. Electricity use from wattage and hours.	Watts to kilowatts: divide by 1,000.	Math you need Unit Conversions and Dimensional AnalysisPractice it: Energy chain, Calculation studio
$\text{cost}$ $= \text{quantity} \times \text{price per unit}$ Worked: $13{,}600\,\text{kWh} \times \frac{$0.12}{1\,\text{kWh}}$ $= $1{,}632$	What a quantity costs. Electricity, water or fuel bills.	Keep the units: kWh × $/kWh gives$ ; multiply, do not divide by the price.	Unit 6 Energy ConservationPractice it: Calculation studio
$\text{savings}$ $= (\text{use before} - \text{use after}) \times \text{price}$ Worked: $(25{,}600 - 12{,}000)\,\text{kWh} \times \frac{$0.12}{1\,\text{kWh}}$ $= $1{,}632$	Money (or energy) saved by a change. An upgrade, a new bulb, a new irrigation method.	Subtract first, then multiply; the cost of the new use alone is not the saving.	Unit 6 Energy ConservationPractice it: Calculation studio
$\text{emissions}$ $= \text{quantity} \times \text{emission factor}$ Worked: $13{,}600\,\text{kWh} \times \frac{0.4\,\text{kg CO}_2}{1\,\text{kWh}}$ $= 5{,}440\,\text{kg CO}_2$	Carbon dioxide (or another gas) released, from how much fuel or electricity is used. Emissions, or emissions saved by a change.	The factor's units must cancel what you multiply: kWh × kg CO₂/kWh.	Unit 6 Energy ConservationPractice it: Energy chain
$\text{value in new unit}$ $= \text{value} \times \frac{\text{new unit}}{\text{old unit}}$ Worked: $25\,\text{gal} \times \frac{3.785\,\text{L}}{1\,\text{gal}}$ $= 94.6\,\text{L}$	Unit conversion by dimensional analysis: each factor is a fraction equal to 1. Gallons to liters, hectares to square meters, BTU to kWh, and any chain of them.	Put the unit you want to get rid of in the denominator, so it cancels.	Math you need Unit Conversions and Dimensional AnalysisPractice it: Calculation studio

Toxicology and half-life

Formula and a worked line	What it is and when to use it	The trap	Learn it in
$\text{number of half-lives}$ $= \frac{\text{time}}{t_{1/2}}$ Worked: $\frac{24\,\text{days}}{8\,\text{days}}$ $= 3\,\text{half-lives}$	How many times a radioactive or breaking-down substance has halved. The first step of every half-life question.	Divide the time by the half-life, not the other way round.	Unit 6 Nuclear PowerPractice it: Dose and response, Calculation studio
$\text{amount left}$ $= \text{amount} \times \left(\frac{1}{2}\right)^{t / t_{1/2}}$ Worked: $80\,\text{mg} \times \left(\frac{1}{2}\right)^{24/8}$ $= 10\,\text{mg}$	What is left after time t. Radioactive decay, or a pesticide breaking down.	Three half-lives leave one eighth, not one third or one sixth.	Unit 6 Nuclear PowerPractice it: Dose and response, Calculation studio
$\text{LD}_{50}$ Worked: $\text{dose at }50\%\text{ deaths}$ $= 20\,\text{mg/kg}$	The dose that kills half of a test population, read off a dose-response curve. Comparing how toxic two substances are.	A lower LD50 means a MORE toxic substance; read across from 50% to the curve, then down to the dose.	Unit 8 Lethal Dose 50% (LD50)Practice it: Dose and response
$\text{dose}$ $= \text{dose per kg} \times \text{body mass (kg)}$ Worked: $25\,\text{mg/kg} \times 12\,\text{kg} = 300\,\text{mg}$	A whole-body dose from a dose in mg per kg. How much of a substance would reach a stated dose.	Convert body mass to kilograms first.	Unit 8 Lethal Dose 50% (LD50)Practice it: Dose and response
$C_{\text{top}} = C \times f^{\,n}$ Worked: $0.04\,\text{ppm} \times 10^{3} = 40\,\text{ppm}$	Biomagnification: concentration multiplies by a factor at each step up a food chain. The concentration in a top predator.	Bioaccumulation is one organism over time; biomagnification is up the chain.	Unit 8 Bioaccumulation and BiomagnificationPractice it: Dose and response
$\text{magnification}$ $= \frac{\text{higher concentration}}{\text{lower concentration}}$ Worked: $\frac{0.2\,\text{ppm}}{0.002\,\text{ppm}}$ $= 100\,\text{times}$	How many times more concentrated a substance is at one level than another. "How many times greater" questions, and comparing LD50s.	Divide, do not subtract; the answer has no unit ("times").	Unit 8 Bioaccumulation and BiomagnificationPractice it: Dose and response

Land, water and people

Formula and a worked line	What it is and when to use it	The trap	Learn it in
$\text{per capita}$ $= \frac{\text{total}}{\text{population}}$ Worked: $\frac{4.5 \times 10^{9}\,\text{L}}{3 \times 10^{6}\,\text{people}}$ $= 1{,}500\,\text{L per person}$	An amount per person. Water use, energy use, waste or footprint per person.	Scale words first: 4.5 billion is 4,500,000,000.	Math you need Rates and Per-Capita ValuesPractice it: Calculation studio
$\text{population density}$ $= \frac{\text{number}}{\text{area}}$ Worked: $\frac{1{,}200{,}000\,\text{people}}{400\,\text{km}^2}$ $= 3{,}000\,\text{people/km}^2$	How crowded a population is. People per square kilometer, trees per hectare.	Area units must match the question (km² or hectares).	Unit 3 Population Growth and Resource AvailabilityPractice it: Calculation studio
$\text{water volume}$ $= \text{depth} \times \text{area}$ Worked: $0.5\,\text{m} \times 1{,}200{,}000\,\text{m}^2$ $= 600{,}000\,\text{m}^3$	Irrigation water from the depth applied and the field area. Water applied to a field or stored in a wetland.	Put depth and area in matching units (m and m²) before you multiply.	Unit 5 Irrigation MethodsPractice it: Calculation studio
$\text{water per area}$ $= \frac{\text{volume}}{\text{area}}$ Worked: $\frac{9{,}000\,\text{m}^3}{1.5\,\text{ha}}$ $= 6{,}000\,\text{m}^3\text{/ha}$	Water use per hectare (or per acre). Comparing irrigation methods or farms of different sizes.	Volume over area, and keep both units in the answer (m³/ha).	Unit 5 Irrigation MethodsPractice it: Calculation studio

Numbers and conversions to know

Scale words. thousand = $10^3$ , million = $10^6$ , billion = $10^9$ , trillion = $10^{12}$ . Multiply by adding exponents and divide by subtracting them. Scientific Notation and Metric Prefixes
Metric prefixes. kilo- = 1,000; mega- = 1,000,000; giga- = 1,000,000,000; milli- = one thousandth. 1 kWh = 1,000 Wh. Scientific Notation and Metric Prefixes
Dimensional analysis. Write each conversion as a fraction so the unit you do not want cancels: $5,$ $ext{kWh} imes rac{0.9, ext{kg}}{1, ext{kWh}}$ $= 4.5,$ $ext{kg}$ . Unit Conversions and Dimensional Analysis
Show the setup. On the calculation question, one point is for the setup and one for the answer with units. Type the setup with brackets and units, then the answer. Percents and Percent Change

On exam day

No formula sheet. Bluebook shows only how to type math (fractions, powers, brackets) and a reminder to show work and units. Every formula above must be memorized.
Calculator. A four-function (with square root) or scientific calculator, or the Desmos scientific calculator in Bluebook. Graphing calculators are not allowed from 2027.

Practice with it

Free unit practice tests · Formula flashcards · Free notes for every topic · Full practice exams