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Algebra II : Applying Exponents

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Example Questions

Example Question #312 : Exponents

If a person deposits 300 dollars to a savings account, which earns one percent interest that is compounded annually, what is the balance after 60 years?

Possible Answers:

$\$5680$

$\$480$

$\$2540$

$\$545$

$\$385$

Correct answer:

$\$545$

Explanation:

Write the formula for compound interest.

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

$A= \textup{balance}$

$P=\textup{starting investment or principal} = \$300$

$r=\textup{ rate} = 1\% = 0.01$

$n= \textup{number of times compounded per year} =1$

$t= \textup{time in years} = 60$

Substitute all the known values into the formula.

$A=300(1+\frac{0.01}{1})^{(1)(60)} = 300(1.01)^{60} = \$545$

The answer is: $\$545$

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Example Question #313 : Exponents

Suppose Billy's has $\$300$ , invests the money at a bank at $1\%$ , compounded monthly. About how much will Billy have after 36 months?

Possible Answers:

$\$323$

$\$318$

$\$309$

$\$303$

$\$333$

Correct answer:

$\$309$

Explanation:

Write the compound interest formula.

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

where is the total, is the principal, is the rate, is the number of times compounded annually, and is the time in years.

$P=\$300$

$r=1\% = 0.01$

n=12

$t=\frac{36}{12}=3$

Substitute all into the equation.

$A=300(1+\frac{0.01}{12})^{12\times 3}$

$A=309.132 = \$309$

The answer is: $\$309$

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Example Question #314 : Exponents

Peter opens a savings account on his $21^{st}$ t birthday. He makes a deposit of $\$5,000$ . The account earns percent interest, compounded annually. Peter plans to take the money out when he is years old. If he doesn't make any deposits or withdrawals until then, how much money will be in the account?

Possible Answers:

$\$10,150.00$

$\$ 147,285.13$

$\$ 35,571.29$

$\$15,150.00$

Correct answer:

$\$ 35,571.29$

Explanation:

The formula for calculating compount interest is as follows:

$FV = PV\cdot (1+r)^n$

where

= future value

= present value

= interest rate

= number of times the interest is compounded

In this problem, the present value of the money is $5000, and the interest rate is 7%. If Peter takes the money out when he is 50, it would have been compounded 29 times (once per year). Therefore:

$FV = \$5000\cdot (1+.07)^{29}$

${\color{Red} FV = \$35,571.29}$

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Example Question #1 : Radioactive Decay Equations

Over the past few years, the number of students enrolled at a certain university has been decreasing. Each year there is a 12% decrease in student enrollement. Currently, 14,286 students are enrolled. If this trend continues, how many students will be enrolled in 5 years?

Possible Answers:

9,756

12,426

3,571

7,539

5,714

Correct answer:

7,539

Explanation:

This is an exponential decay problem. The formula for exponential decay is:

FV = PV(1- d)^n

Where

= future value

= present value

= rate of decay

= number of periods

This problem requests the number of students five years in the future. The rate of decay is twelve percent. Therefore:

FV = 14,286(1-.12)^5

FV = 7,539

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Example Question #1 : Radioactive Decay Equations

Sceintists recently discovered a new type of metal compound. They have roughly 15 grams of this compound, which has a half life of 16 hours. Approximately how much of this substance will the scientists have in 24 hours?

Possible Answers:

0.005 grams

grams

0.50 grams

5.00 grams

0.06 grams

Correct answer:

0.06 grams

Explanation:

Recall the radioactive decay formula:

$Q_{(t)}=Q_{(o)} (^{-r\cdot t})$

The half life formula is:

$r=\frac{ln2}{(h)}$ , where is the half life.

Plug in the given half life:

$r=\frac{ln2}{(16)}\approx 0.0433216$

Plug this value into the radioactive decay formula:

$Q_{(24)}=15 (^{-0.0433216\cdot 24})\approx 0.06$ grams

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Example Question #1 : Radioactive Decay Equations

The equation for radioactive decay is,

$A=A_{0}\cdot \left (\frac{1}{2} \right )^{t/h}$ .

Where $A_{0}$ is the original amount of a radioactive substance, is the final amount, is the half life of the substance, and is time.

The half life of Carbon-14 is about 5730 years. If a fossile contains grams of Carbon-14 at time t = 0 , how much Carbon-14 remains at time t = 17190 years?

Possible Answers:

$16 \text{ grams}$

$4 \text{ grams}$

$8 \text{ grams}$

None of the other answers.

$96 \text{ grams}$

Correct answer:

$8 \text{ grams}$

Explanation:

Using the equation for radioactive decay, we get:

$A=64\cdot \left (\frac{1}{2} \right )^{17190/5730}$ .

$A = 64\cdot \left ( \frac{1}{2} \right )^{3} = 64 \cdot \frac{1}{8} = 8 \text{ grams}$

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Example Question #3 : Radioactive Decay Equations

The number of fish in an aquarium is decreasing with exponential decay. The population of fish is decreasing by $7\%$ each year. There are 1500 fish in the aquarium today. If the decay continues how many fish will be in the aquarium in years?

Possible Answers:

990

None of these answers are correct

1122

1200

Correct answer:

1122

Explanation:

Every year the population of fish losses 7%. In other words, every year 93% of the fish remain from the previous year. Knowing this, we can use the original number of fish to find the number of fish for the next year. Since we want to know the number of fish 4 years from now, we multiply 1500 by 93% four times.

$\small \small 1500\cdot0.93\cdot0.93\cdot0.93\cdot0.93 = 1500\cdot(0.93)^4 \approx 1122$

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Example Question #319 : Exponents

The population of a city is decreasing. The city has a population of , 529 people today, but the population decreases by $10.5\%$ every year. What will be the population of the city in years if this continues?

Possible Answers:

6565

7769

7312

6953

Correct answer:

6953

Explanation:

Because the population of the city is decreasing every year at 10.5% we can find the population after each year by using

$\small 13,529\cdot(1-0.105) = 13,529\cdot(0.895)$

Because this decrease will continue every year for the 6 years, we can continue to multiply the population by the decay for every year.

$\small \small \small 13,529\cdot0.895\cdot0.895\cdot0.895\cdot0.895\cdot0.895\cdot0.895 \Rightarrow$

$\small \small \Rightarrow 13,529\cdot0.895^6 \approx 6953$

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Example Question #4 : Radioactive Decay Equations

There is water leaking out of a cup. $12\%$ of the water is leaking out every minute. How many kilograms of water will be left in minutes and seconds, if there are kilograms of water , $\small W$ , in the cup right now?

Possible Answers:

None of these answers are correct

5.40 kilograms

2.32 kilograms

4.68 kilograms

Correct answer:

4.68 kilograms

Explanation:

Because the water is leaking at a continuous rate, we can use the exponential decay equation.

$\small W_{new} = W\cdot(1-D)^t$

$\small D$ is the decay of the problem, 12% or 0.12. $\small t$ is equal to how many times the water will have a 12% decay. This can be calculated as

$\small t = \frac{4min 12secs}{1min}$

To calculate this we must first convert both time to seconds

$\small t = \frac{4\cdot60+12}{1\cdot60} = \frac{240+12}{60} = \frac{252}{60}$

Our equation is then

$\small \large \large W_{new} = 8\cdot(1-0.12)^{\frac{252}{60}} \approx 4.68$

$\small W_{new} = 4.68$

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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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Algebra II : Applying Exponents

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #312 : Exponents

Example Question #313 : Exponents

Example Question #314 : Exponents

Example Question #1 : Radioactive Decay Equations

Example Question #1 : Radioactive Decay Equations

Example Question #1 : Radioactive Decay Equations

Example Question #3 : Radioactive Decay Equations

Example Question #319 : Exponents

Example Question #4 : Radioactive Decay Equations

Example Question #1 : Radioactive Decay Equations

All Algebra II Resources

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