Digital SAT Math · Problem-Solving & Data Analysis

Percent growth/decay in context

A percent sentence becomes a multiplier, then a model. “Grows 8% each year” is not ×0.08 and not +8 each year — it is ×1.08, repeated once per year. This page is the hinge between one-shot percent arithmetic and full exponential algebra: turn the words into $a \cdot b^{t}$ , decide whether the story is linear or exponential at all, and keep the time unit inside the exponent.

On the test


Domain	Problem-Solving and Data Analysis (score report)
CB skill	Percentages (context items that compound over time), adjacent to Advanced Math nonlinear models
What it looks like	A short growth or decay story — population, price, membership, investment — asking for a model, a later value, or the meaning of a factor
Often asked	“Which expression models…?”, “What is the value after…?”, “Which best interprets…?”, “By what percent each year…?”
Format	Multiple choice and student-produced response
Calculator	Allowed; useful for $1.04^{2}$ and friends, useless if the base is already wrong

Recognition cues: each year / every month tied to a percent; increases by 6% of its previous value; decreases by 8% each year; doubles every…; grows exponentially; compounded annually.

What this page owns

Percentages owns one change done once (including reverse percent and successive factors on a single trip). Exponential functions owns pure $a \cdot b^{x}$ algebra, tables, and curves. This page owns the context translation: percent sentence → multiplier, linear vs exponential from the words, and time-unit placement when the period is not already “one step of $t$ .”

Pattern recognition

Almost every item is one of these:

Build — story gives a start and a percent per period; write $y = a(1 \pm r)^{t}$ .
Evaluate — model or story is known; find the amount after several periods.
Interpret — a printed base or coefficient; say what it means in the story’s units.
Classify — same story told with add a fixed amount vs multiply by a percent; pick the right family.
Period shift — the percent is annual but $t$ is months (or doubles every 4 weeks); the exponent absorbs the conversion.

Method

Fix the time unit. What does one step of the exponent mean — one year, one month, one 4-week doubling period? Write it down before you write $b$ .
Turn the percent into a multiplier. Growth of $r$ : $b = 1 + r$ . Decay of $r$ : $b = 1 - r$ . Up 8% → $1.08$ ; down 8% → $0.92$ . Never use $0.08$ as the base for an 8% change.
Write $a \cdot b^{t}$ . $a$ is the amount when $t = 0$ . Check one period: plug $t = 1$ and confirm you get exactly one multiplication by $b$ .
If the period changes, change the exponent — not the rate. Annual 6% with $m$ in months → $a(1.06)^{m/12}$ . Doubling every 4 weeks with $w$ in weeks → $a \cdot 2^{w/4}$ . Dividing the annual rate by 12 and calling that the monthly multiplier is wrong.
Answer the asked quantity — a model, a dollar amount, a count, or a percent — and sanity-check direction (growth should rise; decay should fall).

Words in the stem	Math move
increases by 6% each year	$b = 1.06$
decreases by 6% each year	$b = 0.94$
is 6% of its previous value	$b = 0.06$ — rare; usually a trap wording
increases by 40 each year	linear: $+40t$ , not exponential
doubles every $p$ periods	factor $2$ , exponent $t/p$
after $m$ months when the rate is annual	exponent $m/12$ , base still $1 + r_{\text{annual}}$

Worked example 1 — percent becomes a multiplier

Stem. A community garden starts with 320 seedlings. Each year the number of seedlings is 5% greater than the year before. (a) Write an expression for the number of seedlings after $t$ years. (b) How many seedlings are there after 2 years?

Step 1 — time unit. One step of $t$ is one year.

Step 2 — multiplier. Up 5% means keep 100% and add 5%: $b = 1.05$ . Not $0.05$ .

Step 3 — model.

S(t) = 320(1.05)^{t}

Step 4 — evaluate at $t = 2$ .

S(2) = 320(1.05)^{2} = 320(1.1025) = 352.8

so about 353 seedlings if the stem asks for a whole number; exact is 352.8 if it does not.

Check. Year 1: $320 \times 1.05 = 336$ . Year 2: $336 \times 1.05 = 352.8$ . Each jump is 5% of the current count, not of 320.

Trap watch. $320(0.05)^{t}$ collapses immediately. $320 + 16t$ adds a fixed 16 (5% of the original) each year — agrees at $t = 1$ only. Adding 5% + 5% = 10% to the start gives 352, close enough to look right and still wrong.

Worked example 2 — decay is $1 - r$ , not $r$

Stem. A laptop bought for $1,200 loses 8% of its resale value each year. Which function models the resale value $V(t)$ in dollars $t$ years after the purchase, and what is $V(2)$ ?

Step 1 — multiplier. Losing 8% means keeping 92%: $b = 1 - 0.08 = 0.92$ .

Step 2 — model.

V(t) = 1200(0.92)^{t}

Step 3 — two years.

V(2) = 1200(0.92)^{2} = 1200(0.8464) = 1015.68

Check. Year 1: $1200 \times 0.92 = 1104$ (lost $96, which is 8% of 1200). Year 2: $1104 \times 0.92 = 1015.68$ (lost $88.32, which is 8% of 1104 — not of 1200).

Trap watch. $1200(0.08)^{t}$ keeps only 8% a year and is nearly zero by year 2. $1200(1.08)^{t}$ grows the laptop. $1200 - 96t$ subtracts a fixed $96 forever and hits zero in 12.5 years, while a percent loss never quite does.

Worked example 3 — linear vs exponential, and a period in the exponent

Stem. (a) A park has 400 oak trees. Plan L plants 50 more trees each year. Plan E increases the oak count by 12% of the current count each year. Write a model for each plan. (b) Separately: a moth population doubles every 3 weeks. There are 90 moths now. Write a model for the population $w$ weeks from now.

(a) Classify first. Plan L adds a fixed 50 → linear:

L(t) = 400 + 50t

Plan E multiplies by a percent of the current value → exponential, $b = 1.12$ :

E(t) = 400(1.12)^{t}

Plan	Per year	Family	Model
L	+50 trees	linear	$400 + 50t$
E	×1.12	exponential	$400(1.12)^{t}$

At $t = 1$ both have grown, but not to the same place: $L(1) = 450$ , $E(1) = 448$ . At $t = 5$ , $L(5) = 650$ while $E(5) = 400(1.12)^{5} \approx 704.9$ — the percent plan has pulled ahead. Two points never decide the family; the words do.

(b) Period in the exponent. Factor $2$ , period $3$ weeks, variable $w$ in weeks:

M(w) = 90 \cdot 2^{w/3}

Check: $M(3) = 90 \cdot 2^{1} = 180$ , one doubling in 3 weeks, as stated.

Trap watch for (b). $90 \cdot 2^{w}$ ignores the period. $90 \cdot 2^{3w}$ triples the doubling rate. $90(1 + 2/3)^{w}$ invents a daily-style rate from the period — the same family of error as annual÷12.

Practice

Answer before you open the explanation. Two items are student-produced response (type the number, no choices), matching the real test. Every wrong choice below is a named trap from the method: wrong base, added percents, annual÷12, or linear instinct on a multiplicative story.

12 questions — 10 multiple choice, 2 student-produced response. Every wrong choice has its own explanation.

Question 1 Warm-up

A neighborhood newsletter starts with 800 subscribers. Each month the number of subscribers is 5% greater than the number the month before. Which expression gives the number of subscribers after t months?

Show the answer Choice C

Why it is right

An increase of 5% each month means the count is multiplied by 1.05 once per month: keep the original 100% and add 5%. Starting from 800 and repeating that multiplication for t months gives 800(1.05)^t. Check at t = 1: 800 × 1.05 = 840, which is exactly 40 more than 800 and therefore 5% of 800 — one growth step, as the sentence requires.

Why each other choice fails

Choice A: Uses the percent itself as the base. Multiplying by 0.05 each month keeps only 5% of the list and collapses the count immediately; an 5% rise needs the multiplier 1.05, not 0.05.
Choice B: Adds a fixed 40 subscribers every month (5% of the original 800) rather than 5% of the current count. That is a linear model; it matches the exponential only at t = 1 and then falls behind.
Choice D: A base of 0.95 is a 5% decrease each month, the opposite direction from the growth the stem describes.

Question 2 Standard

A used scooter is purchased for $2,400. Each year after the purchase, its resale value is 8% less than its resale value the year before. Which function models the resale value V(t), in dollars, t years after the purchase?

Show the answer Choice A

Why it is right

Losing 8% of the current value means keeping the other 92%, so each year's value is 0.92 times the previous year's value. Repeating that multiplication t times gives V(t) = 2400(0.92)^t. Check the first year: 2400 × 0.92 = 2208, a loss of 192, which is exactly 8% of 2400. The second year loses 8% of 2208, not of 2400, which is what the exponential form encodes.

Why each other choice fails

Choice B: Reads 'decays 8%' as multiply by 0.08. That keeps only 8% of the value each year — an 92% collapse — and V(1) = 192, not a mild 8% loss.
Choice C: A base greater than 1 makes the value grow. 1.08 grows the scooter by 8% a year; a decrease of 8% needs 1 − 0.08 = 0.92.
Choice D: Subtracts a fixed $192 every year (8% of the original price) rather than 8% of the current value. It agrees with the correct model at t = 1 and then drifts, and it reaches $0 in a finite time while a percent loss never quite does.

Question 3 Standard

A wildlife sanctuary has 1,000 sandhill cranes. The population is expected to increase by 10% each year for the next several years. According to this model, how many cranes should the sanctuary have after 3 years?

Show the answer Choice C

Why it is right

Each year multiplies the population by 1.10. After 3 years the count is 1000 × (1.10)^3. Because 1.10^2 = 1.21 and 1.10^3 = 1.331, the population is 1000 × 1.331 = 1331. Stepping year by year confirms it: 1000 → 1100 → 1210 → 1331.

Why each other choice fails

Choice A: Adds the percents across periods: 3 × 10% = 30%, then 1000 × 1.30 = 1300. Percent growth compounds, so the factors multiply (1.10^3 = 1.331), not the percents.
Choice B: Applies one year of growth only (1000 × 1.10 = 1100) and stops. The stem asks for 3 years, which is three multiplications by 1.10.
Choice D: Uses the percent 0.10 as the base: 1000 × (0.10)^3 = 1. An 10% rise needs the multiplier 1.10, not 0.10.

Question 4 Standard

A campus bike-share program has 200 bicycles in week 0. Each week the number of bicycles is 6% greater than the number the week before. Which equation models the number of bicycles, y, after x weeks?

Show the answer Choice D

Why it is right

The story multiplies the current count by a percent each week, so the model is exponential. A 6% increase means the multiplier is 1.06, and the starting amount is 200, giving y = 200(1.06)^x. At x = 1 the model returns 212, which is 12 more than 200 and therefore exactly 6% of 200 — one growth step matching the sentence.

Why each other choice fails

Choice A: Treats 0.06 as an additive rate of bicycles per week. Adding six-hundredths of a bicycle each week is neither the linear nor the exponential reading of '6% greater,' and after one week it is nowhere near a 6% rise.
Choice B: Uses the percent as the base. Multiplying by 0.06 each week keeps only 6% of the fleet; a 6% rise needs the base 1.06.
Choice C: Adds a fixed 12 bicycles every week (6% of the original 200). That is the linear trap: it agrees with the exponential at x = 1 and then understates growth, because later weeks should take 6% of a larger fleet.

Question 5 Standard

The population of a town, P(t), t years after 2018, is modeled by P(t) = 5000(0.92)^t. Which of the following is the best interpretation of the number 0.92 in this model?

Show the answer Choice B

Why it is right

In y = a·b^t the base b is the per-period multiplier. Here b = 0.92 means each year's population is 0.92 times — 92% of — the previous year's population. Equivalently, the town loses 8% of its current population each year, because 1 − 0.92 = 0.08. Checking one step: P(1) = 5000 × 0.92 = 4600, which is exactly 92% of 5000.

Why each other choice fails

Choice A: Reads the multiplier as the percent lost. A 92% decrease would leave a multiplier of 0.08 and a first-year population of 400, not 4600. The percent change is the distance from 1: |0.92 − 1| = 0.08, an 8% decrease.
Choice C: Treats 0.92 as an additive rate of people per year. The model multiplies; it does not subtract 0.92 residents annually. A linear reading would be P(t) = 5000 − 0.92t, which is a different function.
Choice D: Gets the 8% magnitude right but the direction wrong. A base below 1 is a decrease; growth of 8% would require the base 1.08.

Question 6 Harder

A stock is worth $500 at the start of year 1. During year 1 its value increases by 20%. During year 2 its value decreases by 10% of the value it has at the start of year 2. What is the stock's value at the end of year 2?

Show the answer Choice A

Why it is right

Each change needs its own multiplier, applied to the current value. Up 20% is ×1.20, so after year 1 the stock is 500 × 1.20 = 600. Down 10% is ×0.90, so after year 2 it is 600 × 0.90 = 540. The overall factor is 1.20 × 0.90 = 1.08, an 8% net gain on the original 500, which also gives 540.

Why each other choice fails

Choice B: Adds the percents: +20% − 10% = +10%, then 500 × 1.10 = 550. The second percent applies to a different whole (the year-1 value), so the only legal combination is to multiply the factors.
Choice C: Stops after year 1 (500 × 1.20 = 600) and ignores the second year's 10% decrease.
Choice D: Applies a 10% decrease to the original $500 and ignores the first-year gain: 500 × 0.90 = 450.

Question 7 Harder

A savings balance of $900 grows by 6% each year. Which expression gives the balance after m months?

Show the answer Choice C

Why it is right

The stated growth is 6% per year, so one full multiplication by 1.06 happens once per year. If m counts months, the number of years elapsed is m/12, and the balance is 900(1.06)^{m/12}. Check a full year: when m = 12 the exponent is 1 and the expression returns 900 × 1.06, exactly one year's growth.

Why each other choice fails

Choice A: Treats each month as a full year of 6% growth. After 12 months this multiplies by 1.06^12 ≈ 2.01 — roughly a doubling — instead of a single 6% rise.
Choice B: Divides the annual rate by 12 and uses 0.5% per month as if that were equivalent. The monthly model (1.005)^m is a different process; translating '6% per year' keeps the factor 1.06 and puts months in the exponent.
Choice D: Uses the percent 0.06 as the base (and still adjusts the exponent). A 6% rise needs the multiplier 1.06, not 0.06.

Question 8 Harder Student-produced response

A museum had 2,500 members at the start of 2022. Membership increases by 4% each year. How many members does the museum have at the start of 2024, according to this model?

Your answer

Show the answer 2704

Why it is right

From the start of 2022 to the start of 2024 is exactly 2 years of growth, so the membership is 2500 × (1.04)^2. Because 1.04^2 = 1.0816, the count is 2500 × 1.0816 = 2704. Stepping year by year: 2500 → 2600 at the start of 2023 → 2704 at the start of 2024.

Answers students type instead

100: Computes 4% of 2500 and reports the increase alone, or uses 0.04 as a base and collapses the product. The question asks for the membership count after two multiplications by 1.04.
2600: Applies only one year of 4% growth (2500 × 1.04 = 2600). The span from the start of 2022 to the start of 2024 is two full years.
2700: Adds the percents across periods: 2 × 4% = 8%, then 2500 × 1.08 = 2700. Growth compounds, so the correct factor is 1.04^2 = 1.0816, not 1.08.

Question 9 Harder

A colony of insects doubles in size every 4 weeks. There are currently 150 insects in the colony. How many insects will be in the colony after 12 weeks, according to this model?

Show the answer Choice D

Why it is right

Twelve weeks contain 12/4 = 3 doubling periods, so the population is multiplied by 2 three times: 150 × 2^3 = 150 × 8 = 1200. Stepping in 4-week jumps confirms it: 150 → 300 → 600 → 1200. The model form is N(w) = 150 · 2^{w/4}; at w = 12 the exponent is 3.

Why each other choice fails

Choice A: Treats three doublings as tripling once, or multiplies 150 by 3 (the number of periods) instead of by 2^3. Doubling three times multiplies by 8, not by 3.
Choice B: Stops after two doublings (the count at 8 weeks): 150 → 300 → 600. One more 4-week period fits inside 12 weeks, and doubling 600 gives the correct 1200.
Choice C: Uses the exponent 4 × 3 or treats the period as a multiplier in the base: for example 150 · 2^{4} after misplacing the 4, which is 2400 — the population after 16 weeks, not 12.

Question 10 Hardest

The number of active users of an app is 800 at t = 0 and grows by 24% each year. Which expression gives the number of active users after m months?

Show the answer Choice A

Why it is right

The growth is 24% per year, so the yearly multiplier is 1.24. When time is measured in months, the number of years that have passed is m/12, and the user count is 800(1.24)^{m/12}. At m = 12 the exponent is 1 and the expression returns exactly one year of 24% growth, matching the sentence.

Why each other choice fails

Choice B: Divides the annual percent by 12 and builds a monthly multiplier of 1.02. That is a different compounding schedule; translating '24% each year' keeps the factor 1.24 and places months in the exponent, not in a rewritten rate.
Choice C: Treats each month as a full year of 24% growth. After one year (m = 12) this multiplies by 1.24^12, far more than a single 24% rise.
Choice D: Uses the percent 0.24 as the base. A 24% rise needs the multiplier 1.24; a base of 0.24 would be a 76% collapse each year.

Question 11 Hardest Student-produced response

A phone is purchased for $800. Each year its resale value is 15% less than its resale value the year before. What is the resale value, in dollars, of the phone 2 years after the purchase?

Your answer

Show the answer 578

Why it is right

A 15% loss each year means the multiplier is 0.85. After 2 years the value is 800 × (0.85)^2. Because 0.85^2 = 0.7225, the resale value is 800 × 0.7225 = 578. Stepping year by year: 800 → 680 after one year → 578 after two years.

Answers students type instead

120: Reports the first year's dollar loss (15% of 800) instead of the value remaining after two multiplications by 0.85.
560: Subtracts a fixed 15% of the original price twice: 800 − 120 − 120 = 560. That is the linear trap (800 − 120t at t = 2). Percent decay multiplies by 0.85 each year, so the second loss is 15% of 680, not of 800.
680: Applies only one year of decay (800 × 0.85 = 680). The stem asks for 2 years after purchase.

Question 12 Hardest

The value of a machine decreases by the same percent each year. The machine was worth $10,000 when new and $6,400 after 2 years. By what percent does the value of the machine decrease each year?

Show the answer Choice B

Why it is right

After 2 years of the same annual multiplier b, the value satisfies 10000 · b^2 = 6400, so b^2 = 0.64 and b = 0.8 (positive root). A multiplier of 0.8 means the machine keeps 80% of its value each year and therefore loses 20% each year. Check: 10000 × 0.8 = 8000 after one year; 8000 × 0.8 = 6400 after two years.

Why each other choice fails

Choice A: Reports the total two-year percent decrease: (10000 − 6400)/10000 = 36%. That is the overall change, not the per-year rate. The annual rate is the factor that compounds to 0.64 in two steps, which is 0.8 (a 20% decrease).
Choice C: Halves the total 40% drop in a linear way (as if 20 percentage points were wrong-sized) or treats each year as losing 40% of the original. Losing 40% per year would leave 10000 × (0.6)^2 = 3600, not 6400.
Choice D: Reads the remaining fraction 0.64 as a 64% annual decrease (or reports the two-year retention as if it were the rate). The annual retention is 0.8, so the annual decrease is 20%.

Common mistakes

Using the percent as the base — 8% growth is $1.08$ ; 8% decay is $0.92$ . A base of $0.08$ is a 92% collapse.
Reading “decays 8%” as ×0.08 — same trap on the decay side; you keep 92%, you do not keep 8%.
Adding percents across periods — three years of 10% is not +30%. Multiply $1.10^{3}$ , or step year by year.
Dividing an annual rate by 12 for a monthly multiplier — 6% a year is $(1.06)^{m/12}$ , not $(1.005)^{m}$ .
Treating a percent story as linear — subtracting 8% of the original every year builds a line that agrees once and then drifts.
Putting the period in the base instead of the exponent — “doubles every 4 weeks” is $2^{w/4}$ , not a factor of $2/4 = 0.5$ per week.
Swapping growth and decay — $1 + r$ when the story loses value, or $1 - r$ when it gains.
Answering the factor when the question wants a count (or the reverse) — $1.05$ , “5%”, and “353 seedlings” are three different answers.

FAQ

Is this the same as exponential functions? Same formula family, different job. Here you translate a percent story into a multiplier and a time unit. There you manipulate $a \cdot b^{x}$ when the model is already on the page or in a table/graph.

When is growth linear instead of exponential? When each period adds (or subtracts) a fixed amount, not a percent of the current value. “Gains 40 members a year” is linear; “gains 40% a year” is exponential.

Do I need the compound-interest formula with $n$ ? No. The Digital SAT does not require $A = P(1 + r/n)^{nt}$ . Translate the percent and the time unit you are given; keep $b = 1 \pm r$ for the stated period.

How do I enter a decimal on an SPR? Type the number the last line asks for — no percent sign, no dollar sign. If the stem wants a whole number of people, round only as instructed.

Desmos? Yes for evaluating a model you already built. No for choosing between $0.92$ and $0.08$ — that decision is verbal.

Editorial team · last reviewed 2026-08-12

Percent growth/decay in context

On the test

What this page owns

Pattern recognition

Method

Worked example 1 — percent becomes a multiplier

Worked example 2 — decay is 1−r1 - r1−r, not rrr

Worked example 3 — linear vs exponential, and a period in the exponent

Practice

Question 1 Warm-up

Why it is right

Why each other choice fails

Question 2 Standard

Why it is right

Why each other choice fails

Question 3 Standard

Why it is right

Why each other choice fails

Question 4 Standard

Why it is right

Why each other choice fails

Question 5 Standard

Why it is right

Why each other choice fails

Question 6 Harder

Why it is right

Why each other choice fails

Question 7 Harder

Why it is right

Why each other choice fails

Question 8 Harder Student-produced response

Why it is right

Answers students type instead

Question 9 Harder

Why it is right

Why each other choice fails

Question 10 Hardest

Why it is right

Why each other choice fails

Question 11 Hardest Student-produced response

Why it is right

Answers students type instead

Question 12 Hardest

Why it is right

Why each other choice fails

Common mistakes

FAQ

Related skills

Worked example 2 — decay is $1 - r$ , not $r$