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Algebra II : Mathematical Relationships and Basic Graphs

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Example Questions

Example Question #1 : Radioactive Decay Equations

An animal population is dying out. There is a decrease in this number of animals by $8\%$ every year. In years, there will be of this animal left. What is the current population of this animal today?

Possible Answers:

143

101

Correct answer:

143

Explanation:

This is an exponential decay problem. Therefore, we can use this equation.

$\small P_{7} = P_{0}\cdot(1-d)^t$

$\small P_{7}$ is the animal population after the 7 years. $\small P_{0}$ is the animal population right now. $\small d$ is the decay of the animal population every year. $\small t$ is the time period of the animal populations decay.

From the problem we know after the 7 years the animal population will be 80, so

$\small P_{7} = 80$

The population is decreasing by 8% every year, therefore

$\small d$ = 8% = 0.08

$\small t$ is equal to the number of times a decay of 8% has occurred. Since an 8% decay happens every year, and the population is 7 years from now, the population has decayed 8%, 7 times. In other words,

$\small t = \frac{7 \text{ years}}{1 \text{ year}} = 7$

These values give us,

$\small 80 = P_{0}\cdot(1-0.08)^7$

Rearranging this equation we can solve for $\small P_{0}$

$\small P_{0} = \frac{80}{(1-0.08)^7} = \frac{80}{0.92^7} = 143$

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Example Question #5 : Radioactive Decay Equations

A school is losing a certain number of students each year. This year, the school has 242 students. Four years ago the school had 591 students. The yearly rate of the school losing students has been the same for the last four years. What is the school's yearly rate of losing students?

Possible Answers:

$80\%$

$12\%$

$20\%$

$67\%$

Correct answer:

$20\%$

Explanation:

This is an exponential decay problem, meaning that the decay of the school's students can be found using

$\small S_{now} = S_{past}\cdot(1-d)^t$

$\small S_{now}$ is the number of students currently at the school, which is 242

$\small S_{past}$ is the number of students that were at the school 4 years ago, which is 591

$\small t$ is the number of times the decay has occurred. Since, we are trying to find the yearly decay, the decay that happened to the school from one year to the next, and we have the number of students from 4 years ago, $\small t =4$ .

$\small d$ is the decay rate of the school that we are trying to find. Because we have every number except for $\small d$ , we can plug the values into the equation to solve for $\small d$ .

$\small 242 = 591\cdot(1-d)^4$

$\small (1-d)^4 = \frac{242}{591} \approx 0.41$

$\small 1-d = \sqrt[4]{0.41} \approx 0.80$

$\small d = 0.20$ = 20%

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Example Question #6 : Radioactive Decay Equations

Cells in a dish have started to decay. The cells are decaying by $3\%$ every minutes. When you left the cells there were 1000 cells in the dish. Now there are cells in the dish. Approximately how long did you leave the cells for?

Possible Answers:

$90 \text{ hours and } 40 \text{ minutes}$

$26 \text{ hours and } 6 \text{ minutes}$

$23 \text{ hours and } 4 \text{ minutes}$

$78 \text{ hours and } 20 \text{ minutes}$

Correct answer:

$26 \text{ hours and } 6 \text{ minutes}$

Explanation:

This is an exponential decay problem. That means that after 20 minutes 3% of the cells decay and 97% of the cells in the dish are left. To find the new amount of cells in the dish, we multiple the original number by the 97%.

$\small 1000 \cdot(0.97)$

The number of cells after every 20 minute interval can be calculated this way. Therefore, to find how long the cell were decaying we use,

$\small \small \small 1000 \cdot(0.97) \cdot(0.97)\cdot... = 92$

Which can be rewritten as,

$\small \small 1000(0.97)^t = 92$

Now we can solve for $\small t$ , which is the amount of time that you were gone

$\small (1-0.03)^t = \frac{92}{1000}$

To solve for $\small t$ , we must take the log of both sides to base 10. This will give us,

$\small t\cdotlog(0.97) = log\bigg(\frac{92}{1000}\bigg)$

$\small \small t = \frac{log(\frac{92}{1000})}{log(0.97)} \approx 78.333$

Remember! $\small t$ is the number of decays that the cells went through. Each decay took 20 minutes to get through, however.

Therefore, the time that you were gone is

$\small \small t = 78.333\cdot\frac{20}{60} \approx 26 \text{ hours and} \ 6 \text{ minutes}$

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Example Question #791 : Mathematical Relationships And Basic Graphs

Suppose 5 milligrams of element X decays to 3.2 milligrams after 48 hours. What is the decay rate on a day to day basis?

Possible Answers:

$34\%$

$15\%$

$30\%$

$28\%$

$22\%$

Correct answer:

$22\%$

Explanation:

Write the formula for radioactive decay.

$X=X_0e^{-\lambda T }$

$X= \textup{final amount} = 3.2$

$X_0=\textup{ initial amount }= 5$

$T=\textup{time in days} = \frac{48 \textup{ hours}}{ \frac{24\textup{ hours}}{1\textup{ day}}} = 2$

Substitute the values in the equation and solve for lambda.

$5=3.2e^{-2\lambda }$

Divide by 3.2 on both sides.

$1.5625=e^{-2\lambda }$

Take the natural log of both sides to eliminate the exponential.

$ln(1.5625)=ln(e^{-2\lambda })$

$ln(1.5625)=-2\lambda$

Divide by negative two on both sides.

$\frac{ln(1.5625)}{-2}=\frac{-2\lambda}{-2}$

$\lambda=\frac{ln(1.5625)}{-2}=-0.223144$

Convert this to a percentage.

This element's decay rate is approximately: $22\%$

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Example Question #3451 : Algebra Ii

An element has a half life of 6 days. What is the approximate amount remaining for a 50 mg sample of this element after 5 days?

Possible Answers:

1.5625

48.4375

$\textup{The answer is not given.}$

28.06155

21.93845

Correct answer:

28.06155

Explanation:

Write the formula for half life.

$y=a(\frac{1}{2})^x$

$a=\textup{ sample size} = 50$

$x=\textup{ time in days}$

Since the time requested is five out of the six day of the half life, the value of is:

$x=\frac{5}{6}$

Substitute all the known values into the equation.

$y=50(\frac{1}{2})^{\frac{5}{6}} = 28.06155$

The answer is: 28.06155

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Example Question #791 : Mathematical Relationships And Basic Graphs

The number of butterflies in an exhibit is decreasing at an exponential rate of decay. The number of butterflies is decreasing by $5\%$ every year. There are 500 butterflies in the exhibit right now. How many butterflies will be in the exhibit in years?

Possible Answers:

250

296

349

Correct answer:

349

Explanation:

Because the butterflies are decreasing exponentially, we can use this equation

$\small F = O\cdot(1-\text{decay})^{\text{time}}$

$\small F$ is the final value

$\small O$ is the original value

The decay for this problem is 5% or 0.05

The period of time is 7 years

Using this equation we can solve for $\small F$

$\small F = 500(1-0.05)^7 = 500\cdot0.95^7 \approx 349$

$\small F = 349$

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Example Question #1 : Other Exponent Applications

A biologist figures that the population of cane toads in a certain lake he is studying can be modeled by the equation

$P = 253 \cdot 1.04 ^{t}$ ,

where is the number of days elapsed in 2015. For example, t= 1 represents January 1, t = 2 represents January 2, and so forth.

If this model continues, in what month will the population of cane toads in the lagoon reach 5,000?

Possible Answers:

April

June

March

May

February

Correct answer:

March

Explanation:

Set P = 5,000 and solve for :

$5,000 = 253 \cdot 1.04^{t}$

$1.04 ^{t} = \frac{5,000 }{253 }$

$1.04^{t} \approx 19.7628$

$\ln 1.04^{t} \approx \ln 19.7628$

$t \ln 1.04 \approx \ln 19.7628$

$t \approx \frac{\ln 19.7628}{\ln 1.04 } \approx \frac{2.9838}{0.0392} \approx 76.1$

January and February have 59 days total; add March, and this is 90 days. The 76th day is in March.

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Example Question #2 : Other Exponent Applications

A biologist figures that the population of cane toads in a certain lake he is studying can be modeled by the equation

$P = 318\cdot 1.07 ^{t}$ ,

where is the number of days elapsed in 2015. For example, t= 1 represents January 1, t = 2 represents January 2, and so forth.

Assuming that this has been the model for their growth throughout the previous year as well, in what month did the population hit 100 cane toads?

Possible Answers:

August 2014

September 2014

October 2014

December 2014

November 2014

Correct answer:

December 2014

Explanation:

Set P = 100 and solve for :

$100 = 318\cdot 1.07 ^{t}$

$\frac{100}{318} = 1.07 ^{t}$

$1.07 ^{t} \approx 0.3145$

$\ln 1.07 ^{t} \approx \ln 0.3145$

$t \ln 1.07 \approx \ln 0.3145$

$t \approx \frac{\ln 0.3145}{ \ln 1.07} \approx \frac{-1.1568}{0.0677} \approx -17.1$

17 days before January 1 was in December of 2014.

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Example Question #3 : Other Exponent Applications

Lucia deposits $40,000 into a savings account that pays 5.5% annual interest compounded continuously. Assuming she neither deposits nor withdraws money, how long will it take for her to have $60,000 in the account?

Possible Answers:

Between 6 and 7 years

Between 9 and 10 years

Between 7 and 8 years

Between 8 and 9 years

Between 5 and 6 years

Correct answer:

Between 7 and 8 years

Explanation:

We set final principal A = 60,000 original principal $A_{0} = 40,000$ , and interest rate r = 0.055 . We solve for in the continuous compound interest formula:

$A = A_{0} e ^{rt}$

$60,000 =40,000 e ^{0.055t}$

$\frac{60,000 }{40,000 }=\frac{40,000 e ^{0.055t}}{40,000 }$

$1.5 = e ^{0.055t}$

$0.055t = \ln 1.5$

$t = \frac{\ln 1.5}{0.055} \approx \frac{0.4055}{0.055} \approx 7.37$

The correct response is therefore between 7 and 8 years.

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Example Question #2 : Other Exponent Applications

Ann deposits $30,000 into a savings account that pays 4.5% annual interest compounded quarterly. Assuming she neither deposits nor withdraws money, what is the amount of time it will take for her to have at least $50,000 in the account?

Possible Answers:

$11 \textup{ yrs } 6\textup{ mo}$

$13\textup{ yrs }$

$11 \textup{ yrs }$

$12\textup{ yrs }$

$12 \textup{ yrs } 6\textup{ mo}$

Correct answer:

$11 \textup{ yrs } 6\textup{ mo}$

Explanation:

Apply the compound interest formula:

$A = A_{0} \left ( 1 + \frac{r}{n}\right ) ^{nt}$ .

We set final principal A = 50,000 original principal $A_{0} = 30,000$ , interest rate r = 0.045 , number of periods per year n = 4 (quarterly). We solve for in the equation

$50,000=30,000 \left ( 1 + \frac{0.045}{4}\right ) ^{4t}$

$50,000=30,000 \left ( 1 .01125 \right ) ^{4t}$

$\frac{50,000}{30,000}=\frac{30,000 \left ( 1 .01125 \right ) ^{4t}}{30,000}$

$1.6667= \left ( 1 .01125 \right ) ^{4t}$

$\log 1.6667= \log \left ( 1 .01125 \right ) ^{4t}$

$\log 1.6667= 4t \log 1 .01125$

$t = \frac{\log 1.6667}{4\log 1 .01125} \approx \frac{0.2219}{4 \cdot 0.004859} \approx 11.4$

This is rounded up to the next quarter of a year, so the correct response is 11.5 years, or 11 years 6 months.

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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Radioactive Decay Equations

Example Question #5 : Radioactive Decay Equations

Example Question #6 : Radioactive Decay Equations

Example Question #791 : Mathematical Relationships And Basic Graphs

Example Question #3451 : Algebra Ii

Example Question #791 : Mathematical Relationships And Basic Graphs

Example Question #1 : Other Exponent Applications

Example Question #2 : Other Exponent Applications

Example Question #3 : Other Exponent Applications

Example Question #2 : Other Exponent Applications

All Algebra II Resources

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