Algebra 3 Quiz: Growth And Decay Modeling

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Growth And Decay ModelingQuestion 1 of 20

A population of bacteria starts at 500 cells and triples every 90 minutes. Which equation can be used to find h, the number of hours needed for the population to reach 121,500 cells?

121{,}500=500(3)^{h/1.5}

121{,}500=500(3)^{1.5h}

121{,}500=500(3)^{h/90}

121{,}500=500(3)^{90h}

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Algebra 3 Quiz

Algebra 3 Quiz: Growth And Decay Modeling

Practice Growth And Decay Modeling in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Growth And Decay Modeling, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A population of bacteria starts at 500 cells and triples every 90 minutes. Which equation can be used to find h, the number of hours needed for the population to reach 121,500 cells?

$121{,}500=500(3)^{h/1.5}$ (correct answer)
$121{,}500=500(3)^{1.5h}$
$121{,}500=500(3)^{h/90}$
$121{,}500=500(3)^{90h}$

Explanation: When you see exponential growth, the key is to write the equation in the form

\text{final amount}=\text{initial amount}\cdot(\text{growth factor})^{\text{number of growth periods}}.

Here the population triples every 90 minutes, and 90 minutes is

1.5

hours. If

h

is the number of hours that have passed, then the number of 1.5-hour growth periods is

\frac{h}{1.5}.

So the correct equation is

121{,}500=500(3)^{h/1.5}.

This works because when

h=1.5

, the exponent is

1

, giving one tripling; when

h=3

, the exponent is

2

, giving two triplings, and so on. The choice

500(3)^{1.5h}

has the exponent inverted: it would mean the population triples

1.5

times per hour, i.e. every

2/3

hour, not every 1.5 hours. The choice

500(3)^{h/90}

mixes units:

h

is in hours but 90 is in minutes, so the exponent is not meaningful unless

h

were also in minutes. The choice

500(3)^{90h}

implies 90 triplings per hour, which would mean a tripling every

1/90

hour, far too fast. Strategy: identify the growth period, convert it to the same unit as the variable, and put the elapsed time over that period in the exponent. That single step avoids most exponential-model errors.

Question 2

A city's population is 100,000 and grows at a continuous rate of 3% per year. Which of the following is the population after 10 years, rounded to the nearest thousand?

130,000
134,000
135,000 (correct answer)
2,009,000

Explanation: Whenever you see a question about continuous growth, your first move is to write the exponential model

P(t)=P_0 e^{rt}.

Here

P_0=100{,}000

r=0.03

, and

t=10

, so

P(10)=100{,}000e^{0.03\cdot 10}=100{,}000e^{0.3}.

Since

e^{0.3}\approx 1.34986

, the population is about

134{,}986

, which rounds to the nearest thousand as

135{,}000

. The

130{,}000

option comes from computing

3\%\times 10=30\%

growth without compounding:

100{,}000+30{,}000

. The

134{,}000

option comes from using annual compounding instead of continuous compounding:

100{,}000(1.03)^{10}\approx 134{,}392

, which rounds to

134{,}000

. The very large

2{,}009{,}000

option appears if you double-count the 10 years: first multiply the rate by 10 to get

0.3

, then use that

0.3

as the annual rate in

e^{0.3\cdot 10}=e^3

, giving

100{,}000e^3\approx 2{,}008,500

. That is a percent-versus-rate error. On continuous-growth problems, keep the rate as a decimal per year, let time in years be the only place you multiply by 10, and round only at the very end.

Question 3

You want to have $10,000 in an account 5 years from now. The account pays 4% interest compounded monthly. Approximately how much must you deposit today?

$8,000
$8,190 (correct answer)
$8,219
$8,333

Explanation: When you see a question asking how much to deposit today to reach a target future amount, you're doing a present value problem. The core formula is:

P = FV \times \left(1+\frac{r}{n}\right)^{-nt}

Here

FV=10000

r=0.04

n=12

, and

t=5

. So:

P = 10000 \times \left(1+\frac{0.04}{12}\right)^{-60}

The monthly rate is about

0.003333

, and after 60 compoundings the factor is about

1.221

. Dividing

10000

1.221

gives approximately $8,190. Why the others are off:

$8,219 is the result of using annual compounding: $10000/(1.04)^5$ . It ignores the monthly compounding. More frequent compounding makes money grow faster, so the deposit needed is slightly lower than the annual-compounding version.
$8,333 comes from simple interest: $10000/(1 + 0.04 \times 5)$ . That treats the interest as if it never compounds, which understates how much the account will grow.
$8,000 comes from subtracting $20\%$ of the future value: $10000 - 2000$ . But $4\%$ for 5 years is $20\%$ of the principal, not of the future amount.

A good habit: always convert the annual rate to the per-month rate and multiply years by 12. If your answer looks like the annual-compounding answer, check whether you used

n=12

correctly.

Question 4

A town's population grew exponentially from 20,000 in 2000 to 30,000 in 2010. If the growth remained exponential, what was the population in 2005?

24,000
24,500 (correct answer)
25,000
25,500

Explanation: When you see exponential growth, remember that the population multiplies by the same factor in equal time intervals—it does not increase by the same number each year. Here, the town grew from 20,000 to 30,000 over 10 years, so its 10-year growth factor is

\frac{30000}{20000} = 1.5

. Since 2005 is halfway through that 10-year period, the growth factor over 5 years must be the square root of the 10-year factor:

P(2005)=20000\sqrt{1.5}\approx 20000(1.2247)=24{,}494 \approx 24{,}500.

So the correct choice is 24,500. The choice 25,000 is the classic linear-interpolation trap: it takes the halfway point between 20,000 and 30,000, but exponential growth is not a straight line. The choice 25,500 comes from assuming the annual growth rate is

5\%

because

50\%/10=5\%

, then compounding that for 5 years:

20000(1.05)^5\approx 25{,}526

. However, the actual annual growth rate is

1.5^{1/10}-1\approx 4.14\%

, not

5\%

. The choice 24,000 underestimates the population, likely by using a growth factor of

1.2

instead of

\sqrt{1.5}\approx 1.2247

. A useful habit: when an exponential growth question asks for the value at the midpoint of a known interval, take the square root of the growth factor. That instantly separates exponential reasoning from linear averaging.

Question 5

A 100 mg sample of a chemical decreases by 8% every 2 years. Which function models the amount A(t) remaining after t years?

$A(t)=100(0.92)^{t/2}$ (correct answer)
$A(t)=100(0.92)^{2t}$
$A(t)=100(0.08)^{t/2}$
$A(t)=100(0.08)^{2t}$

Explanation: When you see exponential decay, identify the decay factor per time unit and the exponent that counts how many time units have passed. Here the sample loses 8% every 2 years, so after any 2-year period, 92% remains. The multiplier is 0.92. If

t

years have passed, the number of 2-year periods is

t/2

, not

2t

. Therefore the function is

A(t)=100(0.92)^{t/2}

: start with 100 mg and apply the per-period factor once for every two years elapsed. The choice

100(0.92)^{2t}

uses the correct decay factor but the wrong exponent: multiplying by 0.92 every half-year would mean an 8% drop every 6 months, which is faster than the given rate. The choices using 0.08 treat 8% as the remaining amount, not the decrease.

100(0.08)^{t/2}

would leave only 8% after every two years, and

100(0.08)^{2t}

compounds that mistaken 8% twice per year. Both confuse the decay rate (8% lost) with the decay factor (92% retained). A quick check: plug in

t=2

. The correct formula gives

100(0.92)

, exactly one 8% decrease. The others give

100(0.92)^4

100(0.08)

, or

100(0.08)^4

. So always ask, "How many of my time periods fit into

t

years?" and remember that in exponential decay, the base is what remains, not what is lost.

Question 6

After 10 days, 60% of a radioactive sample remains. What is the half-life of the sample, to the nearest tenth of a day?

8.3 days
10.0 days
13.6 days (correct answer)
16.7 days

Explanation: Whenever you see a half-life question, the key is to remember that radioactive decay is exponential, not linear. The amount remaining follows

N = N_0\left(\frac12\right)^{t/T},

where

T

is the half-life. Here

N/N_0 = 0.60

after

t=10

days, so

0.60 = \left(\frac12\right)^{10/T}.

Taking natural logs gives

\ln(0.60) = \frac{10}{T}\ln(0.5),

T = \frac{10\ln(0.5)}{\ln(0.60)} \approx 13.57,

which rounds to 13.6 days. A quick sanity check: since 60% remains, more than half of the sample is still there, so the half-life must be longer than 10 days. That rules out 10.0 days and 8.3 days right away. The 16.7 days answer comes from dividing 10 by 0.6, treating the percentage as a linear rate — but decay is exponential, so that doesn't work. The 8.3 days answer may come from multiplying 10 by

0.5/0.6

, which confuses the fraction remaining with a time proportion. For half-life problems, first compare the remaining fraction to

1/2

: if more than half remains, the half-life is longer than the elapsed time; if less than half remains, it's shorter. Then set up the exponential equation and use logs to solve.

Question 7

An account pays 5% compounded quarterly. What is the effective annual rate, rounded to the nearest hundredth of a percent?

5.00%
5.09% (correct answer)
5.13%
21.55%

Explanation: When you see a nominal annual rate like 5% compounded quarterly, the question is really asking for the effective annual rate — the actual percentage your money grows in one year after compounding kicks in. The key formula is

\text{EAR} = \left(1 + \frac{r}{n}\right)^n - 1

where

r

is the stated annual rate and

n

is the number of compounding periods per year. Here,

r = 0.05

and

n = 4

because quarterly means four times a year. So compute:

\left(1 + \frac{0.05}{4}\right)^4 - 1 = (1.0125)^4 - 1 \approx 1.050945 - 1 = 0.050945

That is 5.09% after rounding to the nearest hundredth of a percent. Now, why not the others? 5.00% is just the nominal rate — it ignores compounding entirely, as if interest were paid only once at year-end. 5.13% looks tempting because it is what you would get from continuous compounding:

e^{0.05} - 1 \approx 5.13\%

, but the problem specifically says quarterly, not continuous. 21.55% comes from treating the 5% as a quarterly rate and raising

1.05

to the fourth power, which massively overstates growth; remember to divide the annual rate by 4 before compounding. Your memory hook: for effective annual rate questions, always divide the nominal rate by the number of periods first, then compound for the year. Misplacing that single step leads to all the wrong choices.

Question 8

A car purchased for $30,000 depreciates so that its value after t years is $ $V(t)=30000(0.85)^t$ $. Approximately how many years will it take for the car's value to fall to $12,000?

4.0 years
8.7 years
6.1 years
5.6 years (correct answer)

Explanation: When you see an exponential depreciation problem, think: the variable you're solving for is in the exponent, so logarithms are the right tool. Here, the car's value is

V(t)=30000(0.85)^t

, and you want it to equal

\

12{,}000$. Set up the equation:

12000=30000(0.85)^t

Divide both sides by 30000:

0.4=(0.85)^t

Take the natural log of both sides:

\ln(0.4)=t\ln(0.85)

t=\frac{\ln(0.4)}{\ln(0.85)} \approx \frac{-0.9163}{-0.1625} \approx 5.64

That rounds to 5.6 years. The 6.1-year choice comes from incorrectly treating the 15% loss as continuous decay, solving

e^{-0.15t}=0.4

instead of using the given

(0.85)^t

. The 4.0-year choice treats depreciation as linear: losing

\

4{,}500

each year means

$18{,}000/$4{,}500=4

. But exponential depreciation takes 15% of the *current* value, not the original value, each year. The **8.7-year** choice is roughly the time to fall to one-quarter of the original value (

$7{,}500

), not the 40% value (

$12{,}000$) the question asks for. A good habit: isolate the exponential term first, then take logs. Also notice that

\

12{,}000

is 40% of

$30{,}000$, which immediately helps you eliminate options that correspond to 50% or 25% remaining.

Question 9

Approximately how many years, to the nearest tenth, will it take a $2,000 investment to grow to $6,000 if it earns 6% compounded quarterly?

12.0 years
18.3 years
18.4 years (correct answer)
18.9 years

Explanation: When you see a question about how long it takes an investment to reach a target, you're working with compound interest. The key is to use the formula

A = P\left(1 + \frac{r}{n}\right)^{nt}

, where

P

is the principal,

r

is the annual rate,

n

is the number of compounding periods per year, and

t

is years. Here

P=2000

A=6000

r=0.06

n=4

. Dividing by 2000 gives

\left(1.015\right)^{4t}=3

. Taking natural logs:

4t\ln(1.015)=\ln(3)

, so

t=\frac{\ln(3)}{4\ln(1.015)} \approx \frac{1.0986}{0.05956} \approx 18.447

, which rounds to 18.4 years. Now the traps: 12.0 years comes from the Rule of 72 —

72/6=12

— but that gives doubling time, not tripling. 18.3 years is what you get if you incorrectly use continuous compounding:

\ln(3)/0.06 \approx 18.31

. 18.9 years results from compounding annually instead of quarterly:

\ln(3)/\ln(1.06) \approx 18.85

. Each wrong choice stems from misapplying a simpler formula or ignoring the compounding frequency. Your takeaway: always check that the compounding periods match the rate. Convert the annual rate to a per‑period rate and multiply the number of years by the periods per year. Also, remember that more frequent compounding shortens the time needed — so quarterly should give a slightly smaller time than annual. Practice converting between different compounding frequencies to avoid these classic errors.

Question 10

A deposit of $5,000 grows to $6,000 in 3 years in an account that compounds interest monthly. What is the annual nominal interest rate, to the nearest tenth of a percent?

6.0%
6.1% (correct answer)
6.3%
6.7%

Explanation: Whenever you see "compounds interest monthly," remember the formula is

A=P\left(1+\frac{r}{n}\right)^{nt}

where

n=12

. Here,

6000=5000\left(1+\frac{r}{12}\right)^{36}

. Divide by 5000:

1.2=\left(1+\frac{r}{12}\right)^{36}.

Take the natural log of both sides:

\ln(1.2)=36\ln\left(1+\frac{r}{12}\right).

Since

\ln(1.2)\approx0.18232

, divide by 36 to get

\ln(1+\frac{r}{12})\approx0.005064

. Exponentiating gives

1+\frac{r}{12}\approx1.005077

, so

r\approx0.06093

, or

6.1\%

when rounded to the nearest tenth. The 6.3% choice is the trap for using annual compounding instead of monthly: solving

(1+r)^3=1.2

gives about

6.3\%

, but that ignores the 12 compounding periods per year. The 6.7% choice comes from simply dividing the total 20% growth by 3 years, which is linear thinking and ignores compounding entirely. The 6.0% choice might come from misremembering the rate or using rough simple-interest logic; simple interest on $5,000 at 6% for 3 years would only produce $5,900, not $6,000. On exam day, when you see monthly compounding, always write

n=12

and

t=3

, then solve for

r

using logs. Checking your answer by plugging it back into the compound interest formula is a quick way to confirm it really grows $5,000 to $6,000.

Question 11

A radioactive isotope has a half-life of 24,000 years. If a sample initially contains 80 grams, how many grams remain after 60,000 years?

10 grams
40 grams
20 grams
14.1 grams (correct answer)

Explanation: Whenever you see a half-life question, you're dealing with exponential decay, not linear loss. The amount remaining is the initial amount multiplied by a power of one-half. Here, the half-life is 24,000 years, so 60,000 years means

\frac{60,000}{24,000} = 2.5

half-lives. Using the formula

A = 80 \cdot \left(\frac{1}{2}\right)^{2.5}

, compute

\left(\frac{1}{2}\right)^{2.5} = \frac{1}{4} \cdot \frac{1}{\sqrt{2}} \approx \frac{1}{5.656}

, giving

\frac{80}{5.656} \approx 14.14

grams. That's the correct answer: 14.1 grams. Now the wrong answers each come from a specific misconception. 40 grams is what you'd get after only one half-life (24,000 years) — you confused the elapsed time with the half-life itself. 20 grams is correct after two half-lives (48,000 years), but you still have 12,000 more years to go, so you must halve again partially. 10 grams would result if you incorrectly treated 60,000 as exactly three half-lives (72,000 years) or used

80/8

— a rounding or integer-half-life trap. Your study tip: always compute

\frac{t}{t_{1/2}}

first. If it's not an integer, keep the fractional exponent in your calculation — don't round to the nearest whole half-life. Also remember that after a fraction of a half-life, the remaining fraction is

\left(\frac{1}{2}\right)^{\text{fraction}}

, not a linear subtraction.

Question 12

An investment grows exponentially. After 2 years the balance is $5,000, and after 5 years the balance is $8,000. No deposits or withdrawals were made. What is the annual growth rate, to the nearest tenth of a percent?

16.0%
17.0% (correct answer)
18.0%
20.0%

Explanation: Whenever you see exponential growth with balances at two different times, write the model

A=P(1+r)^t

and set up a ratio—the unknown starting principal

P

cancels. Here the balance went from $5,000 at year 2 to $8,000 at year 5, so the elapsed time is 3 years: [ \frac{8000}{5000}=(1+r)^3 \quad\Rightarrow\quad 1.6=(1+r)^3. Taking the cube root: [ 1+r=\sqrt[3]{1.6}\approx1.1696 \quad\Rightarrow\quad r\approx0.1696=16.96%. Since the question asks for the nearest tenth of a percent, 16.96% rounds to 17.0%. The 20.0% option is the classic simple-interest trap: someone takes the

3,000 increase, divides by the

5,000 starting balance, and by 3 years: [ \frac{3000}{5000\cdot3}=0.20. ] But exponential growth isn't simple interest—the balance itself compounds, so you cannot just split the total percentage gain evenly. The 16.0% option often comes from using the continuous-compounding model

A=Pe^{rt}

; that gives

r=\frac{\ln(1.6)}{3}\approx0.1567,

about 15.7%, which rounds to 16.0% as a whole percent but is not the effective annual rate this problem intends. The 18.0% option comes from overestimating the cube root, such as treating

\sqrt[3]{1.6}

as 1.18 instead of 1.1696. Since

1.18^3\approx1.643

whereas

1.17^3\approx1.602

, the

1.17

factor—17.0%—is the one that actually reproduces the $8,000 balance.

Question 13

An account pays nominal annual rate $r=\frac{\ln2}{10}$ compounded continuously, so an initial deposit doubles in 10 years. A second account pays the same nominal annual rate $r$ compounded daily. Which statement best describes the second account's balance after 10 years?

It is exactly double the initial deposit because equal nominal rates produce equal effective yields over equal time.
It is slightly more than double the initial deposit because daily compounding applies interest 365 times per year.
It cannot be determined without knowing how the bank rounds daily interest.
It is slightly less than double the initial deposit because, for a fixed nominal rate, continuous compounding produces the largest possible effective yield. (correct answer)

Explanation: For a fixed nominal rate

r>0

(1+r/365)^{365}<e^r

. Therefore after 10 years the daily account has

P(1+r/365)^{3650}<Pe^{10r}=2P

. Equal nominal rates do not imply equal effective rates; daily compounding is not greater than continuous compounding; and the comparison is determined by the formula, independent of rounding conventions.

Question 14

An account earns 4.8% annual interest compounded quarterly. How many years will it take for an initial deposit to triple?

11.5
23.0 (correct answer)
46.1
92.2

Explanation: Solve

(1+0.048/4)^{4t}=3

, or

(1.012)^{4t}=3

. Thus

t=\ln3/(4\ln1.012)\approx23.0

years. 92.2 is the number of quarterly periods rather than years; 46.1 divides by 2 instead of 4; 11.5 divides by 8 instead of 4.

Question 15

Population $P$ is modeled by $P(t)=100(1.08)^t$ and population $Q$ by $Q(t)=100e^{0.08t}$ . Which expression gives the first time $t$ at which $Q(t)$ is 1% larger than $P(t)$ ?

$\frac{\ln 1.01}{\ln 1.08-0.08}$
$\frac{\ln 1.01}{0.08-\ln 1.08}$ (correct answer)
$\frac{\ln 1.01}{0.08-0.08}$
$\frac{\ln 1.01}{0.08-\ln 1.01}$

Explanation: Set

e^{0.08t}=1.01(1.08)^t

. Taking logs gives

0.08t=\ln1.01+t\ln1.08

, so

t=\ln1.01/(0.08-\ln1.08)

. This is positive because

0.08>\ln1.08

. Reversing the denominator gives a negative time; the other options use the wrong rates or omit the log of 1.08.

Question 16

An investment earns 5.5% per year. Inflation is 2% per year. To the nearest year, how many years will it take for the real value of the investment to double?

13
20
21 (correct answer)
26

Explanation: The real-value factor each year is

1.055/1.02\approx1.03431

. Solve

1.03431^t=2

, so

t=\ln2/\ln1.03431\approx20.5

, which rounds to 21. Thirteen years ignores inflation, and 20 years comes from treating

5.5\%-2\%

as a continuous rate; 26 comes from using a real rate of about 2.75%.

Question 17

A population triples in 18 years under continuous exponential growth. Once it has reached three times its original size, how many additional years are needed for it to reach five times its original size?

8.4 (correct answer)
11.4
13.3
26.4

Explanation: The continuous growth rate satisfies 18r = ln(3), so r = ln(3)/18. The additional time T from 3P0 to 5P0 satisfies 5P0 = 3P0 e^(rT), so T = ln(5/3)/r = 18 ln(5/3)/ln(3) ≈ 8.4 years. 11.4 is the doubling time 18 ln(2)/ln(3), which would take the population from 3P0 to 6P0; 13.3 is obtained by incorrectly using ln(2) in the denominator, treating the 18-year tripling time as a doubling time; 26.4 is the total time to reach 5P0 from the original size, not the additional time after tripling.

Question 18

An investment grows from $5,000 to $8,000 in 9 years with interest compounded monthly. What is the nominal annual interest rate, to the nearest hundredth of a percent?

5.22%
5.23% (correct answer)
5.36%
6.67%

Explanation: Solve

8000=5000(1+r/12)^{108}

. Then

r=12[(1.6)^{1/108}-1]\approx0.05234

, or 5.23%. 5.22% is the continuous-compounding approximation

\ln1.6/9

; 5.36% is the annual effective rate if compounding were yearly; 6.67% is the linear average percentage gain per year.

Question 19

An insect population was 2000 in 2015 and 3200 in 2020. Assuming continuous exponential growth, what was the population in 2018?

2480
2651 (correct answer)
2720
8192

Explanation: The model is

P(t)=2000e^{kt}

with

t

years after 2015. Since

3200=2000e^{5k}

k=\ln(1.6)/5

. In 2018,

t=3

, so

P(3)=2000e^{3k}=2000(1.6)^{3/5}\approx2651

. The 2480 and 2720 options are linear interpolations using 2/5 and 3/5 of the 5-year change; 8192 incorrectly treats the population as growing by 60% each year.

Question 20

A portfolio is modeled by $V(t)=1000(1.05)^t$ , where $t$ is in years. Which statement about its growth rate is correct?

Its value increases by exactly 5% each year, and its continuous growth rate is exactly 5%.
Its continuous growth rate is 5% per year, so its annual multiplier is $e^{0.05}\approx1.0513$ , not $1.05$ .
Its value increases by exactly 1.05% each year, so its continuous growth rate is approximately 1.04%.
Its value increases by exactly 5% each year, and its continuous growth rate is approximately 4.88%. (correct answer)

Explanation: Because

1.05=e^k

k=\ln1.05\approx0.04879

. The annual multiplier 1.05 is a 5% yearly increase; the continuous rate is about 4.88%. Setting the continuous rate to 5% would produce annual multiplier

e^{0.05}\approx1.0513

, a different model; reading 1.05 as 1.05% misstates the multiplier.