AP Environmental Science · Calculation Toolkit · 2026 Exam

Math & Calculations

A focused calculation toolkit for AP APES — covering every high-frequency quantitative skill with formulas, step-by-step worked examples, common traps, and practice problems. Calculators are permitted on both sections. Reference materials are available in Bluebook for FRQ 3.

7 Calculation Categories Toolkit · Not the Full Exam ⚠ Calculator Permitted ⚡ FRQ 3 Reference Sheet Provided Unit Conversions Included

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CALC

🌿 Ecology: NPP, Energy Flow & Biomagnification

CALC — NPP/GPP/R; 10% rule food chain; biomagnification concentration factor FRQ — Show your work; include units; state formula before substituting 🔥 R = respiration consumed BY producers, not decomposers

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1.1 Net Primary Productivity (NPP)

NPP = GPP − R

NPP = Net Primary Productivity (energy/biomass available to consumers)
GPP = Gross Primary Productivity (total energy fixed by photosynthesis)
R = Respiration by producers (energy used by plants themselves for metabolism)

GPP = NPP + R | R = GPP − NPP

Rearranged versions — know all three forms. Units: kcal/m²/yr, g C/m²/yr, or kJ/m²/yr

▶ Worked Example 1 — Finding NPP

A forest ecosystem has a GPP of 8,500 kcal/m²/yr. Producers use 3,200 kcal/m²/yr for their own respiration. Calculate the NPP and explain its ecological significance.

Write the formula: NPP = GPP − R

Substitute values: NPP = 8,500 − 3,200

Calculate: NPP = 5,300 kcal/m²/yr

Significance: NPP represents the energy available to all consumers (herbivores, carnivores, decomposers) in the ecosystem. GPP energy is locked up in producer metabolism; only NPP flows to higher trophic levels.

▶ Worked Example 2 — Finding GPP from NPP and Respiration %

A grassland ecosystem has an NPP of 600 g C/m²/yr. If producers use 40% of GPP for respiration, what is the GPP?

If R = 40% of GPP, then NPP = 60% of GPP. So: NPP = 0.60 × GPP

Rearrange: GPP = NPP ÷ 0.60 = 600 ÷ 0.60

Answer: GPP = 1,000 g C/m²/yr. Check: R = 1,000 − 600 = 400 = 40% of 1,000. ✓

1.2 The 10% Rule — Trophic Energy Transfer

Energy at Trophic Level_n+1 = Energy at Trophic Level_n × 0.10

Only ~10% of energy at one trophic level is transferred to the next level. The other ~90% is lost as heat (metabolism, movement, waste). This is why food chains rarely exceed 4–5 levels.

Key insightTo sustain 1 kg of top predator requires roughly 10 kg of prey, 100 kg of primary consumers, and 1,000 kg of producers. Explains why eating lower on the food chain is more efficient.

Ecological efficiencyThe actual transfer efficiency varies from ~5–20% in real ecosystems. AP exam uses exactly 10% unless told otherwise.

Why energy is "lost"Lost to: cellular respiration (heat); material not consumed (bone, cellulose, roots); material consumed but not assimilated (feces); material assimilated but used for respiration not growth.

▶ Worked Example 3 — Food Chain Energy Calculation

A meadow ecosystem has 500,000 kcal of energy available at the producer level. Calculate: (a) energy available to primary consumers; (b) energy available to secondary consumers; (c) energy available to tertiary consumers. (d) How many kcal of producer energy are needed to support 50 kcal of tertiary consumer biomass?

Primary consumers (herbivores): 500,000 × 0.10 = 50,000 kcal

Secondary consumers: 50,000 × 0.10 = 5,000 kcal

Tertiary consumers: 5,000 × 0.10 = 500 kcal

Work backwards: 50 kcal tertiary → 500 kcal secondary → 5,000 kcal primary → 50,000 kcal producers needed. Or: 50 × 10 × 10 × 10 = 50,000 kcal.

1.3 Biomagnification Concentration Factor

Concentration Factor = Concentration in organism ÷ Concentration in environment (water)

Top predators tend to accumulate much higher concentrations because they consume large amounts of prey over their lifetimes, and virtually all the fat-soluble toxin in each prey item transfers upward. As a rough illustration, this can parallel the ~10% energy transfer rule, but biomagnification factors vary and should not be treated as a fixed calculation.

▶ Worked Example 4 — Biomagnification

DDT concentration in water is 0.0002 ppm. Small fish have 2 ppm. Large fish have 18 ppm. (a) What is the concentration factor from water to small fish? (b) What concentration would you expect in an osprey that eats large fish?

Concentration factor (water → small fish): 2 ppm ÷ 0.0002 ppm = 10,000×

Large fish to osprey: multiply by ~10: 18 × 10 = ~180 ppm (actual values vary; use 10× unless given real data).

Overall from water to osprey: ~180 ÷ 0.0002 = 900,000× concentration factor across all trophic levels.

Common Calculation Mistakes

❌ Confusing GPP and NPP. GPP includes respiration; NPP is what's left after producers use energy for themselves. When a question says "energy available to consumers," it wants NPP, not GPP.

❌ Forgetting to include units in your answer. "5,300" earns no credit without "kcal/m²/yr."

❌ Applying the 10% rule in the wrong direction. Going UP the food chain: multiply by 0.1 each level. Working backwards from a higher level: divide by 0.1 (= multiply by 10) for each level below.

Practice Calculation · Ecology

A tropical rainforest has a GPP of 9,200 kcal/m²/yr and an NPP of 4,600 kcal/m²/yr. (a) How much energy do producers use for respiration? (b) If herbivores (primary consumers) are 10% efficient and carnivores (secondary consumers) are also 10% efficient, how much energy is available at the secondary consumer level?

(a) Respiration: R = GPP − NPP = 9,200 − 4,600 = 4,600 kcal/m²/yr. (In this case, exactly 50% of GPP is used for respiration.)

(b) Secondary consumers:
• Primary consumers = NPP × 0.10 = 4,600 × 0.10 = 460 kcal/m²/yr
• Secondary consumers = 460 × 0.10 = 46 kcal/m²/yr

CALC

👥 Population & Growth Calculations

CALC — Rule of 70; population growth rate; per capita birth/death rate 🔥 Rule of 70: use the PERCENT rate (e.g. 2%), not decimal (0.02)

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2.1 Population Growth Rate

Growth Rate (%) = [(Births − Deaths + Immigration − Emigration) ÷ Initial Population] × 100

Also written as: r = (b − d) + (i − e) per 1,000 or as a decimal/percent.
For closed populations (no migration): r = b − d

▶ Worked Example 1 — Population Growth Rate

A country has a population of 50 million. In one year: 1,200,000 births, 400,000 deaths, 300,000 immigrants, 100,000 emigrants. Calculate the annual growth rate (%).

Net change = Births − Deaths + Immigration − Emigration
= 1,200,000 − 400,000 + 300,000 − 100,000 = 1,000,000

Growth rate = (Net change ÷ Initial population) × 100
= (1,000,000 ÷ 50,000,000) × 100 = 2.0%

2.2 Rule of 70 — Doubling Time

Doubling Time (years) = 70 ÷ Growth Rate (%)

Growth Rate must be in percent (e.g., 2, not 0.02). The “70” comes from ln(2) × 100 ≈ 69.3, rounded to 70 for simplicity.
Also applies to any exponentially growing quantity: money, energy use, CO₂ concentrations, etc.

Growth Rate (%) = 70 ÷ Doubling Time (years)

Rearranged: if you know the doubling time, solve for the rate.

▶ Worked Example 2 — Rule of 70

(a) A population is growing at 3.5% per year. How long until it doubles? (b) A country's population doubled in 35 years. What was its approximate annual growth rate?

Doubling time = 70 ÷ 3.5 = 20 years

Growth rate = 70 ÷ 35 = 2.0% per year

▶ Worked Example 3 — Population Size After Doubling

A city has 500,000 people and is growing at 2% per year. (a) What is its doubling time? (b) What will its population be after two doubling periods?

Doubling time = 70 ÷ 2 = 35 years

After 1 doubling (35 years): 500,000 × 2 = 1,000,000
After 2 doublings (70 years): 1,000,000 × 2 = 2,000,000

2.3 Population Density

Population Density = Number of individuals ÷ Area (km² or hectares)

Units: individuals per km², individuals per hectare, or people per km². Used to compare population pressure on resources and habitat.

2.4 Per Capita Birth Rate and Death Rate

Per Capita Birth Rate = (Births in a year ÷ Total population) × 1,000

Expressed as births per 1,000 population per year. Similarly for death rate and infant mortality rate.
Natural Increase Rate = Birth Rate − Death Rate (per 1,000 per year)

▶ Worked Example 4 — Per Capita Rates

A country of 10 million has 250,000 births and 150,000 deaths per year. Calculate: (a) birth rate per 1,000; (b) death rate per 1,000; (c) natural increase rate per 1,000; (d) natural increase as a percentage.

Birth rate = (250,000 ÷ 10,000,000) × 1,000 = 25 per 1,000

Death rate = (150,000 ÷ 10,000,000) × 1,000 = 15 per 1,000

Natural increase = 25 − 15 = 10 per 1,000 per year

As %: 10 per 1,000 = 10/1,000 × 100 = 1.0% per year

2.5 Total Fertility Rate (TFR) and Replacement Level

Replacement TFR2.1 children per woman (in developed nations). Slightly above 2.0 because some children die before reaching reproductive age. In developing nations with higher child mortality, replacement TFR may be ~2.3–2.5.

TFR < 2.1Population declining (if no immigration). Examples: Japan (~1.2), South Korea (~0.8), most of Europe (~1.5). Population momentum means decline is delayed by large youth cohorts.

TFR > 2.1Population growing. Sub-Saharan Africa: TFR ~4–6. Niger: ~7.0 (world's highest). Fast-growing populations with high youth dependency ratios.

Calculation AP trapTFR is the average number of children per woman over her lifetime, NOT per 1,000 population. Do not confuse with birth rate per 1,000.

Common Calculation Mistakes

❌ Rule of 70: use percent, not decimal. Growth rate of 2% → doubling time = 70 ÷ 2 = 35 years. If you use 0.02: 70 ÷ 0.02 = 3,500 years — completely wrong.

❌ Forgetting immigration and emigration in the full growth rate formula. In closed-population scenarios you can ignore them, but if the question gives immigration/emigration data, include them.

❌ Applying the Rule of 70 to a rate already expressed per 1,000. Convert to percent first: 25 per 1,000 = 2.5%. Then doubling time = 70 ÷ 2.5 = 28 years.

Practice Calculation · Populations

Country A has a birth rate of 40 per 1,000 and a death rate of 10 per 1,000. Country B has a birth rate of 12 per 1,000 and a death rate of 9 per 1,000. Ignoring migration, calculate (a) the annual growth rate (%) for each country, and (b) the doubling time for each country.

Country A:
Natural increase = 40 − 10 = 30 per 1,000 = 3.0% per year
Doubling time = 70 ÷ 3.0 = ~23 years

Country B:
Natural increase = 12 − 9 = 3 per 1,000 = 0.3% per year
Doubling time = 70 ÷ 0.3 = ~233 years

Country A doubles its population in ~23 years; Country B in ~233 years — a 10× difference in doubling time from a 10× difference in growth rate.

CALC

⚡ Energy Calculations: EROI, Efficiency & Waste Heat

CALC — EROI ratio; waste heat from thermal efficiency; energy unit conversions; per capita consumption 🔥 Fuel input = Electricity output ÷ Efficiency (NOT the other way!)

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3.1 Energy Return on Investment (EROI)

EROI = Energy Output ÷ Energy Input

Higher EROI = more efficient energy source. EROI <1 = energy sink (more energy spent extracting than obtained).
No units — EROI is a dimensionless ratio, expressed as X:1 (e.g., 20:1 means 20 units out per 1 unit invested).

▶ Worked Example 1 — EROI Calculation and Comparison

A wind farm produces 500,000 MJ of electricity over its lifetime. It required 25,000 MJ of energy to manufacture, install, and maintain. (a) Calculate the EROI. (b) How does this compare to corn ethanol with an EROI of 1.3:1?

EROI = Energy output ÷ Energy input = 500,000 ÷ 25,000 = 20:1

Wind EROI (20:1) is ~15× higher than corn ethanol (1.3:1). Corn ethanol barely produces net energy; wind delivers 20 units for every 1 invested. From a pure energy standpoint, wind is vastly more efficient than corn ethanol.

Energy Source	Typical EROI	Net Energy Delivered per Unit Invested
Hydropower	~40–50:1	40–50 units out per 1 in
Early conventional oil (1930s)	~100:1	Historically highest EROI
Modern conventional oil	~20:1	Declining as easy deposits exhausted
Wind (onshore)	~20–25:1	High and improving
Coal	~18–30:1	Declining with deeper mining
Solar PV	~8–20:1	Rapidly improving with technology
Nuclear	~5–15:1	High energy density; significant infrastructure cost
Oil sands / Tar sands	~3–5:1	Energy-intensive extraction
Corn ethanol	~1.3:1	Near energy break-even; ~23% net energy gain
EROI <1	<1	Energy sink — NOT a viable energy source

3.2 Thermal Efficiency & Waste Heat

Efficiency (%) = (Useful Energy Output ÷ Total Energy Input) × 100

Fuel Input = Electricity Output ÷ Efficiency (decimal)

Rearranged forms:
Waste Heat = Fuel Input − Electricity Output
Waste Heat = Electricity Output × (1 − Efficiency) ÷ Efficiency

▶ Worked Example 2 — Coal Plant Waste Heat

A coal-fired power plant generates 800 MW of electricity at 33% thermal efficiency. (a) How much fuel energy input is required? (b) How much energy is released as waste heat?

Fuel input = Electricity output ÷ Efficiency
= 800 MW ÷ 0.33 = 2,424 MW fuel input

Waste heat = Fuel input − Electricity output
= 2,424 − 800 = 1,624 MW waste heat
That's 67% of all fuel energy wasted as heat — fundamental thermodynamic limit.

▶ Worked Example 3 — Natural Gas Combined Cycle (CCGT)

A natural gas combined-cycle plant produces 1,000 MW at 55% efficiency. How much waste heat is produced? How does this compare to the coal plant above (same electricity output at 33%)?

Fuel input = 1,000 ÷ 0.55 = 1,818 MW

Waste heat = 1,818 − 1,000 = 818 MW waste heat (45% of input)

Comparison: Coal at 33% efficiency produces ~1,624 MW of waste heat for 800 MW of electricity. CCGT at 55% produces 818 MW waste for 1,000 MW electricity. CCGT wastes ~50% less energy per unit electricity produced.

▶ Worked Example 4 — LED vs. Incandescent Efficiency

An incandescent bulb converts 5% of electricity to visible light; an LED converts 45%. Both bulbs use 60W and 10W respectively to produce equivalent light output. (a) How much energy does each waste as heat per hour? (b) What percentage less electricity does the LED use?

Incandescent: 60W × 0.95 = 57W wasted as heat
LED: 10W × 0.55 = 5.5W wasted as heat

Energy reduction = (60 − 10) ÷ 60 × 100 = 83% less electricity

3.3 Per Capita Energy Consumption

Per Capita Consumption = Total Consumption ÷ Population

Common energy units: GJ/person/year (gigajoules), kWh/person/year, BTU/person/year.
US: ~290 GJ/person/yr. Germany: ~150 GJ/person/yr. Sub-Saharan Africa: ~15–30 GJ/person/yr.

Practice Calculation · Energy

A nuclear power plant produces 500 MW of electricity at 33% thermal efficiency. (a) What is the total fuel energy (thermal power) input required? (b) How much energy is released as waste heat? (c) If a cogeneration system captures 60% of the waste heat for district heating, how much useful total energy does the plant now deliver (electricity + captured heat)?

(a) Fuel input: 500 MW ÷ 0.33 = 1,515 MW thermal input

(b) Waste heat: 1,515 − 500 = 1,015 MW waste heat

(c) Cogeneration:
• Captured heat = 1,015 × 0.60 = 609 MW
• Total useful energy = 500 MW (electricity) + 609 MW (heat) = 1,109 MW
• New overall efficiency = 1,109 ÷ 1,515 = 73% (up from 33%)
This is why cogeneration is so valuable — it roughly doubles the useful energy extracted from the same fuel.

Common Calculation Mistakes

❌ The most common energy calculation error: Dividing electricity output BY efficiency (correct) vs. multiplying by efficiency (wrong). If a plant is 33% efficient and produces 1,000 MW electricity, fuel input = 1,000 ÷ 0.33 = 3,030 MW. NOT 1,000 × 0.33 = 330 MW (that would be the WASTE portion, which makes no sense as "input").

❌ EROI is unitless. Do not attach units to an EROI ratio. "20:1" or just "20" is the answer — not "20 MJ:1 MJ."

❌ Corn ethanol EROI ~1.3:1 means it produces net energy, but barely — only 30% more energy than went in. Students sometimes say EROI <1 when they mean corn ethanol; corn ethanol is above 1 (barely), so it is technically a net energy source, just extremely inefficient.

CALC

⚙ Concentration, Dilution & Pollution Calculations

CALC — ppm/ppb conversions; dilution factor; LD₅₀ comparisons; BOD 🔥 ppm = mg/L in water; ppb = μg/L in water — these conversions are tested

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4.1 ppm, ppb, and Concentration Conversions

ppm = mg/L ppb = μg/L ppt = ng/L

In water (assuming density ≈ 1 g/mL = 1 kg/L):
1 ppm = 1 mg per liter = 1 mg/kg = 1 g per million grams
1 ppb = 1 μg per liter = 1 μg/kg = 1 g per billion grams
1 ppm = 1,000 ppb | 1 ppb = 0.001 ppm

Analogy for Scale

1 ppm = 1 inch in 16 miles
1 ppb = 1 inch in 16,000 miles (roughly the circumference of Earth at the equator)
1 ppt = 1 second in 31,700 years

Illustrative current example: PFAS drinking water MCL set at 4 ppt by EPA (2024). This illustrates how biologically significant some contaminants are at vanishingly small concentrations — use this for scale intuition, not as a required exam fact (MCLs can change).

In Air vs. Water

In air: ppm and ppb usually refer to volume/volume (mL gas per million mL air). CO₂ at 425 ppm = 425 mL per million mL of air.
In water: ppm usually refers to mass/volume = mg/L.
The AP exam will usually specify the context; use mg/L for water problems unless told otherwise.

▶ Worked Example 1 — ppm/ppb Conversion

A lake water sample contains 0.005 mg/L of mercury. (a) Express this in ppb. (b) A fish in the lake has 2.0 ppm mercury in its tissue. How many times more concentrated is the mercury in the fish than in the water?

0.005 mg/L = 0.005 ppm. Convert to ppb: 0.005 ppm × 1,000 = 5 ppb

Fish: 2.0 ppm = 2,000 ppb. Concentration factor = 2,000 ppb ÷ 5 ppb = 400× more concentrated in the fish. This is bioaccumulation.

4.2 Dilution Calculations

C₁V₁ = C₂V₂

C₁ = initial concentration V₁ = initial volume
C₂ = final concentration V₂ = final volume
Used when a pollutant is diluted into a larger volume (river mixing with polluted discharge, pesticide dilution).

▶ Worked Example 2 — River Dilution

A factory discharges 1,000 L/day of wastewater containing 500 ppm of a pesticide into a river flowing at 99,000 L/day (clean water). What is the downstream pesticide concentration?

Total volume = 1,000 + 99,000 = 100,000 L/day

C₁V₁ = C₂V₂
500 ppm × 1,000 L = C₂ × 100,000 L

C₂ = (500 × 1,000) ÷ 100,000 = 5 ppm

Dilution factor = 100,000 ÷ 1,000 = 100. Concentration reduced by a factor of 100 (from 500 ppm to 5 ppm).

4.3 LD₅₀ (Lethal Dose 50%) — Comparing Toxicity

LD₅₀ (mg/kg) = Dose that kills 50% of test population

Lower LD₅₀ = MORE toxic (less substance needed to kill). Higher LD₅₀ = less toxic (more substance needed).
Units: mg of substance per kg of body weight. To find lethal dose for an animal of known weight:
Lethal Dose (mg) = LD₅₀ (mg/kg) × Body Weight (kg)

▶ Worked Example 3 — Comparing Toxicity with LD₅₀

Substance A has an LD₅₀ of 5 mg/kg; Substance B has an LD₅₀ of 500 mg/kg. (a) Which is more toxic? (b) What dose of each would be lethal to 50% of a group of 20 kg animals?

Substance A (LD₅₀ = 5 mg/kg) is 100× more toxic than Substance B (LD₅₀ = 500 mg/kg). It takes only 5 mg/kg to kill half the population versus 500 mg/kg for Substance B.

Substance A: 5 mg/kg × 20 kg = 100 mg
Substance B: 500 mg/kg × 20 kg = 10,000 mg (10 g)

4.4 Per Capita Waste & Resource Consumption

Per Capita = Total ÷ Population | Total = Per Capita × Population

Used for: waste generation (kg/person/day), water use (L/person/day), energy consumption (GJ/person/yr), CO₂ emissions (tonnes/person/yr), food calories (kcal/person/day).

US MSW generation~2.1 kg/person/day. Total US: ~2.1 × 330 million × 365 = ~253 million tonnes/yr.

US water use~380 L/person/day domestic use. Agricultural sector: ~70% of total US water withdrawals.

US CO₂ per capita~14–16 tonnes CO₂/person/yr. Global average: ~4.8 tonnes. Sub-Saharan Africa: ~0.8 tonnes.

Common Calculation Mistakes

❌ Lower LD₅₀ = MORE toxic. Students get this backwards. LD₅₀ is the dose needed to kill half the population — if a substance is very toxic, you need very little of it (low dose = low LD₅₀ number). Botulinum toxin LD₅₀ ~0.001 μg/kg (extremely toxic). Table salt LD₅∐ ~3,000 mg/kg (relatively non-toxic).

❌ In ppm/ppb conversions: 1 ppm = 1 mg/L in water. Not 1 g/L (that's 1,000 ppm). Not 1 μg/L (that's 1 ppb). The milli- prefix (mg = 10³ grams) connects to the "per million" in ppm: 1 mg in 1 L water = 1 mg in 1,000 g = 1 g in 1,000,000 g = 1 part per million.

CALC

📈 Percent Change, Percent, & Data Interpretation

CALC — Percent change vs. percentage point change; reading graphs; interpreting data tables 🔥 Percent change ≠ percentage point change — a very common AP trap

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5.1 Percent Change

Percent Change = [(New Value − Old Value) ÷ Old Value] × 100

Positive result = increase. Negative result = decrease.
Always divide by the OLD (original) value, not the new value.

▶ Worked Example 1 — Percent Change

(a) Atmospheric CO₂ was 280 ppm pre-industrial; it is 425 ppm today. What is the percent change? (b) A forest lost 400 of its 2,000 species of birds after deforestation. What percent of bird species were lost?

% Change = (425 − 280) ÷ 280 × 100 = 145 ÷ 280 × 100 = ~52% increase

% Lost = (400 ÷ 2,000) × 100 = 20% of bird species lost

5.2 Percentage Point Change vs. Percent Change — Critical Distinction

This Is One of the Most Common AP Calculation Errors

A percentage point change is the arithmetic difference between two percentages.
A percent change is the relative change compared to the original.

Example: Renewable energy goes from 10% to 15% of the global energy mix.
• Percentage point change = 15% − 10% = 5 percentage points
• Percent change = (15 − 10) ÷ 10 × 100 = 50% increase (relative to the original 10%)

The same shift is "5 percentage points" or "50% relative increase." They are BOTH correct but describe different things. FRQs will specify which to calculate. If it says "percent change," use the formula. If it says "percentage point change," just subtract.

5.3 Efficiency as a Percentage

Efficiency (%) = (Useful Output ÷ Total Input) × 100

Applies to energy systems, agricultural yield, water treatment removal rates, recycling rates, etc.

▶ Worked Example 2 — Multiple Efficiency Applications

(a) A scrubber removes 9,000 kg of SO₂ from 10,000 kg entering the flue. What is its removal efficiency? (b) A wastewater treatment plant removes 94,500 mg/L of BOD from influent containing 100,000 mg/L. What is BOD removal efficiency?

Scrubber efficiency = (9,000 ÷ 10,000) × 100 = 90% SO₂ removal

BOD removal = (94,500 ÷ 100,000) × 100 = 94.5% BOD removal

Practice Calculation · Percent Change

In 1970, there were 417 breeding pairs of bald eagles in the contiguous US. By 2006, there were 9,789 breeding pairs. (a) What is the percent change in bald eagle breeding pairs? (b) If there are currently ~350,000 individual bald eagles (all ages), and the population was 417 breeding pairs × 2 = 834 adults in 1970, what is the percent change in adult individuals from 1970 to present?

(a) Breeding pairs:
% Change = (9,789 − 417) ÷ 417 × 100 = 9,372 ÷ 417 × 100 = ~2,247% increase
(The bald eagle population increased by approximately 23× in breeding pairs.)

(b) Adult individuals (approximate):
1970: 834 adults; 2024: ~350,000 total (using the given figure)
% Change = (350,000 − 834) ÷ 834 × 100 = 349,166 ÷ 834 × 100 = ~41,866% increase
This dramatic recovery followed DDT ban in 1972 and ESA protections — one of the greatest wildlife conservation success stories.

CALC

📊 Logarithmic Scales: pH and Decibels

CALC — pH scale: 0.1 unit drop = 26% more acidic; dB scale: +10 dB = 10× more intense 🔥 Never add, subtract, or directly compare logarithmic values with arithmetic

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6.1 pH Scale — Ocean Acidification and Acid Rain

pH = −log₁₀[H⁺] [H⁺] = 10^−pH

The pH scale is logarithmic and inverse: each 1 unit decrease in pH = 10× increase in H⁺ concentration.
A drop of 0.1 pH units = 10^0.1 = 1.26 → ~26% increase in H⁺ concentration.
Lower pH = more acidic = more H⁺. Higher pH = more alkaline = less H⁺.

▶ Worked Example 1 — Ocean Acidification

Ocean pH dropped from 8.2 (pre-industrial) to 8.1 today. (a) How many times more acidic is the current ocean vs. pre-industrial? (b) If CO₂ emissions continue and pH drops further to 7.9 by 2100, how much more acidic will the ocean be compared to today (pH 8.1)?

ΔpH = 8.2 − 8.1 = 0.1 units
Factor = 10^0.1 = ~1.26 (26% more acidic)

ΔpH = 8.1 − 7.9 = 0.2 units
Factor = 10^0.2 = ~1.58 (58% more acidic than today)

Compared to pre-industrial: ΔpH = 8.2 − 7.9 = 0.3 units. Factor = 10^0.3 = ~2.0 (100% more acidic = twice as acidic)

▶ Worked Example 2 — Acid Rain pH Comparison

Normal rain has pH 5.6. Acid rain in the Adirondacks has been measured at pH 4.2. How many times more acidic is the acid rain than normal rain?

ΔpH = 5.6 − 4.2 = 1.4 pH units

Factor = 10^1.4 = ~25× more acidic

Acid rain at pH 4.2 has approximately 25 times more H⁺ than normal rain at pH 5.6.

6.2 Decibel Scale — Noise Pollution

+10 dB = 10× sound intensity +20 dB = 100× +30 dB = 1,000×

The decibel scale is logarithmic: every +10 dB multiplies sound intensity by 10.
Sound Intensity Factor = 10^ΔdB/10 where ΔdB is the difference in decibels.
Note: perceived loudness roughly doubles every +10 dB (psychoacoustic) while intensity increases 10×.

▶ Worked Example 3 — Decibel Comparisons

(a) How many times more intense is a 90 dB lawnmower than a 60 dB normal conversation? (b) A highway creates 80 dB noise. A new sound barrier reduces noise to 65 dB at nearby homes. How much has sound intensity been reduced (as a factor)?

ΔdB = 90 − 60 = 30 dB
Factor = 10^30/10 = 10³ = 1,000× more intense

ΔdB = 80 − 65 = 15 dB
Reduction factor = 10^15/10 = 10^1.5 = ~32× reduction in sound intensity

ΔdB	Intensity Factor	Perceived Loudness Change	AP Example
+10 dB	10×	~2× louder	60 dB conversation → 70 dB vacuum cleaner = 10× more intense
+20 dB	100×	~4× louder	60 dB → 80 dB traffic = 100× more intense
+30 dB	1,000×	~8× louder	60 dB → 90 dB factory = 1,000× more intense
+40 dB	10,000×	~16× louder	40 dB quiet room → 80 dB traffic = 10,000× more intense

Common Logarithmic Scale Mistakes

❌ NEVER add, subtract, or compare logarithmic values arithmetically. pH 6 is NOT twice as acidic as pH 3. pH 6 = 10⁻⁶ M H⁺; pH 3 = 10⁻³ M H⁺. pH 3 is 10³ = 1,000 times more acidic than pH 6. Similarly, 80 dB is NOT twice as loud/intense as 40 dB — it is 10,000 times more intense.

❌ For pH: remember the ocean dropping from 8.2 to 8.1 is only 0.1 pH units, but that's 10^0.1 = 1.26 = 26% more acidic. Students call this "a tiny change" — but 26% more H⁺ is biologically enormous for organisms that evolved in a stable alkaline ocean.

❌ For acid rain: normal rain pH is ~5.6 (not 7.0). CO₂ naturally dissolves in rain to form weak carbonic acid. If students use pH 7.0 as the "normal" baseline, they will incorrectly report that rain with pH 6.0 is "acidic rain" when it's actually less acidic than normal rain.

CALC

🔨 Unit Conversions — Quick Reference

CALC — Energy unit conversions; area/land conversions; mass and volume conversions 🔥 Always carry units through your calculation to catch errors

Mastery:

○ Not Started

◑ Reviewing

✓ Mastered

⚡ Energy Units

1 kilowatt-hour (kWh)= 3,600,000 J = 3.6 MJ

1 BTU (British Thermal Unit)= 1,055 J ≈ 1 kJ

1 kilocalorie (kcal, "food calorie")= 4,184 J ≈ 4.2 kJ

1 quad (quadrillion BTU)= 10¹⁵ BTU = 1.055 × 10¹⁸ J

1 GJ (gigajoule)= 10⁹ J = 278 kWh

1 barrel of oil≈ 6.1 GJ ≈ 1,700 kWh

1 tonne of coal≈ 29 GJ

1 MW (megawatt)= 1,000 kW = 10⁶ W

1 MW for 1 year= 8,760 MWh = 8.76 GWh

🌝 Area & Land Units

1 hectare (ha)= 10,000 m² = 2.47 acres

1 acre= 0.405 ha = 4,047 m²

1 km²= 100 ha = 247 acres = 0.386 mi²

1 square mile (mi²)= 2.59 km² = 640 acres = 259 ha

Total Earth surface= 510 million km²

Total Earth land area= 149 million km²

1 billion hectares= 10 million km²

⚖ Mass & Weight Units

1 tonne (metric ton)= 1,000 kg = 2,205 lbs

1 kilogram (kg)= 2.205 lbs = 1,000 g

1 pound (lb)= 0.454 kg = 454 g

1 short ton (US)= 2,000 lbs = 907 kg

1 milligram (mg)= 10⁻³ g = 10⁻⁶ kg

1 microgram (μg)= 10⁻⁶ g = 10⁻³ mg

1 nanogram (ng)= 10⁻⁹ g

💧 Volume & Water Units

1 liter (L)= 1,000 mL = 1,000 cm³ = 1 dm³

1 m³= 1,000 L = 264 gallons

1 US gallon= 3.785 L

1 acre-foot= 1,234 m³ = 325,853 gallons

1 km³ (cubic km)= 10⁹ m³ = 10¹² L = 1 trillion liters

1 person uses ~380 L/day= ~100 gallons/day domestic

🏤 Concentration Quick Reference

1 ppm (in water)= 1 mg/L = 1 mg/kg

1 ppb (in water)= 1 μg/L = 0.001 ppm

1 ppt (in water)= 1 ng/L = 0.001 ppb

1 ppm = 1,000 ppb= 1,000,000 ppt

PFAS EPA MCL (illustrative, 2024)= 4 ppt = 0.004 ppb = 0.000004 ppm

CO₂ in atmosphere~425 ppm = 425,000 ppb

🌍 Temperature & CO₂ Emissions

°F to °C°C = (°F − 32) ÷ 1.8

°C to °F°F = (°C × 1.8) + 32

1 gallon gasoline burned≈ 8.9 kg CO₂ emitted

1 tonne coal burned≈ 2.5 tonnes CO₂

1 kWh US avg grid electricity≈ 0.39 kg CO₂ (2023)

Global CO₂ emissions≈ 37 billion tonnes CO₂/yr

▶ Worked Example — Multi-Step Unit Conversion

A city converts 10,000 streetlights from 250-watt HPS (high-pressure sodium) to 80-watt LED equivalents. The streetlights run 4,000 hours per year. (a) How many kWh per year does the city save? (b) If electricity costs $0.12/kWh, how much money is saved annually? (c) At 0.39 kg CO₂ per kWh, how many kg of CO₂ does this avoid?

Power saved per light = 250 − 80 = 170 watts = 0.17 kW
Total savings = 0.17 kW × 10,000 lights × 4,000 hrs = 6,800,000 kWh/yr = 6.8 million kWh/yr

Annual cost savings = 6,800,000 kWh × $0.12/kWh = $816,000/yr

CO₂ avoided = 6,800,000 kWh × 0.39 kg CO₂/kWh = 2,652,000 kg CO₂ = 2,652 tonnes CO₂/yr

Exam Prep

⚠ Top Calculation Mistakes — All Categories

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Rule of 70: use the PERCENT rate (e.g., 2), NOT the decimal (0.02)Using the decimal rate gives a nonsensical answer. Growth rate of 2% per year: doubling time = 70 ÷ 2 = 35 years (correct). If you use 0.02: 70 ÷ 0.02 = 3,500 years (completely wrong). The Rule of 70 is specifically designed for the percent form of the growth rate.
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Waste heat calculation: Fuel Input = Electricity Output ÷ Efficiency (NOT × Efficiency)This is the most common energy calculation error. If a plant is 33% efficient and produces 1,000 MW, the fuel input needed is 1,000 ÷ 0.33 = 3,030 MW. Students often multiply instead: 1,000 × 0.33 = 330 MW (this is the fraction that IS electricity, not the fuel input). Then waste heat = 3,030 − 1,000 = 2,030 MW, NOT 670 MW.
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pH is inverse and logarithmic: lower pH = MORE acidic = MORE H⁺; 0.1 pH drop = 26% more acidicTwo common errors: (1) thinking lower pH = less acidic; (2) thinking pH changes are linear. A drop of 0.1 pH units = 10^0.1 = 1.26 times more H⁺ = 26% more acidic. A drop of 1.0 pH unit = 10 times more acidic. A drop of 2.0 units = 100 times. Never compare pH values arithmetically.
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In the 10% rule, apply 0.10 (not 10%) when multiplying; and divide by 0.10 (= multiply by 10) when working backwardsGoing up the food chain: Energy₂ = Energy₁ × 0.10. Working backwards from top: Energy at level below = Energy at level above × 10. Example: 50 kcal at secondary consumer → divide 0.10 twice to find producers: 50 ÷ 0.10 = 500 (primary consumers) ÷ 0.10 = 5,000 kcal (producers).
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Percent change ÷ OLD value; percentage point change = simple subtractionIf renewable energy share rises from 10% to 15%: percent change = (15−10)÷10 ×100 = 50% increase (relative). Percentage point change = 15−10 = 5 percentage points (absolute). Both are correct but describe different things. The AP exam will specify which to calculate.
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Lower LD₅∐ = MORE toxic (not less). LD₅∐ is the dose that KILLS 50%The number in LD₅∐ is the amount needed to kill half the test population. If a substance is very toxic, only a tiny amount is needed → low LD₅∐ number. Botulinum toxin: LD₅∐ ~0.001 μg/kg (extremely toxic). Table salt: LD₅∐ ~3,000 mg/kg (low toxicity). Don't confuse "low number" with "low toxicity."
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Decibels: +10 dB = 10× more intense, NOT twice as intenseThe dB scale is logarithmic. A 90 dB lawnmower vs. a 60 dB conversation: 30 dB difference = 10^(30/10) = 10³ = 1,000 times more intense sound energy. Perceived loudness is approximately 8 times louder (not 10 times — intensity and perceived loudness scale differently). Never do arithmetic with dB values directly; always convert to intensity factors using powers of 10.
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ppm and ppb confusion: 1 ppm = 1 mg/L in water; 1 ppb = 1 μg/L (1,000 times smaller than ppm)1 ppm = 1 mg per liter of water. 1 ppb = 1 microgram (μg) per liter = 1,000 times smaller than 1 mg. As a scale illustration: the PFAS EPA MCL of 4 ppt = 0.004 ppb = 0.000004 ppm (use this for scale intuition only — specific MCLs are not AP exam facts to memorize). Getting unit prefixes wrong by one step produces an answer off by a factor of 1,000.

Exam Strategy

★ Calculation Exam Strategy — AP APES Math Tips

Scope of Calculations on the AP APES Exam

APES is not a math-heavy exam. Official College Board weighting places Mathematical Routines at only 6–9% of MCQ — roughly 5–7 questions out of 80. In the FRQ section, calculations typically appear in FRQ 3 only, as 1–2 parts of a multi-part question. Use this toolkit to master the handful of formulas that do appear — do not let calculation anxiety distort how much time you invest in math vs. content knowledge.

FRQ 3 Reference Materials: Beginning with the 2026 exam, reference materials are available for APES in Bluebook during FRQ 3. College Board describes the reference sheet as explaining how to enter symbolic notation for calculations. Access it through the Bluebook test preview before exam day — do not encounter it for the first time under exam conditions. Having the sheet does not replace understanding: you still need to set up the problem correctly, define variables, and carry units to the final answer.

6–9%

Calc Qs in MCQ

1–2

Calc Parts in FRQ 3

✅

Calculator Permitted

✅

FRQ 3 Ref Sheet

Steps for Full Credit

The 5-Step FRQ Calculation Method

Step	What to Write	Why It Matters
1. State the formula	Write out the formula you will use: "NPP = GPP − R"	Partial credit may be awarded for the correct formula even if arithmetic is wrong
2. Identify variables	Label what each number represents: "GPP = 8,500 kcal/m²/yr; R = 3,200 kcal/m²/yr"	Shows the grader you understand what you're plugging in; catches substitution errors
3. Substitute values	Show the substitution: "NPP = 8,500 − 3,200"	Earns points even if you make an arithmetic error in the final step
4. Calculate	Show your arithmetic: "NPP = 5,300"	Use your calculator; show intermediate steps for complex calculations
5. State units and answer	"NPP = 5,300 kcal/m²/yr"	Missing units may cost a point even if the number is correct; restate the answer clearly

Most Likely Calculations to Appear on the AP Exam

Calculation Type	Frequency	Key Formula	Unit Connection
NPP / GPP / R	Very High	NPP = GPP − R	Unit 1
10% Rule (food chain energy)	Very High	Energy₂ = Energy₁ × 0.10	Units 1, 5
Rule of 70 (doubling time)	Very High	T₂ = 70 ÷ r(%)	Unit 3
Population growth rate	High	r = (B−D+I−E) ÷ N × 100	Unit 3
Waste heat / thermal efficiency	High	Input = Output ÷ Efficiency; Waste = Input − Output	Unit 6
EROI	High	EROI = E_out ÷ E_in	Unit 6
Percent change	High	% Change = (New−Old)÷Old × 100	All Units
pH / ocean acidification	Medium-High	ΔpH of 0.1 = 26% more acidic	Units 7, 9
ppm/ppb conversions	Medium	1 ppm = 1 mg/L; 1 ppb = 1 μg/L	Units 7, 8
Biomagnification concentration	Medium	Concentration factor = Conc_organism ÷ Conc_water	Units 1, 8
LD₅∐ comparisons	Medium	Lower LD₅∐ = more toxic	Unit 8
Decibel comparisons	Low-Medium	+10 dB = 10× intensity	Unit 7
Per capita consumption	Low-Medium	Per capita = Total ÷ Population	Units 5, 6

Final Calculation Strategy Tips

1. Always carry units through every step. Units cancel like fractions. If your final answer has the wrong units, you made an error somewhere. This alone will catch ~30% of calculation mistakes.

2. Sanity-check your answer. Does 3,000 years for a doubling time make sense for a country with 2% growth? No — you probably used the decimal instead of the percent. Does 0.00001 kcal for an apex predator make sense in a 10% rule problem? No — you probably multiplied instead of divided when working backwards.

3. Show ALL work on FRQs, even if the answer seems obvious. You can receive partial credit for correct setup, correct formula, and correct substitution even if you make an arithmetic error. A blank answer earns 0 even if you got the right number mentally.

4. Read the question carefully for what is being asked. "How much energy is LOST as waste heat" is different from "How much FUEL INPUT is needed." "Percent change" is different from "percentage points." The difference between right and wrong is often in the last five words of the question.