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College Algebra : Applications

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Example Questions

Example Question #1 : Applications

On the day of a child's birth, a sum of money is to be invested into a certificate of deposit (CD) that draws $6.2\%$ annual interest compounded continuously. The plan is for the value of the CD to be at least $\$20,000$ on the child's $18^{th}$ birthday.

If the amount of money invested is to be a multiple of $\$1,000$ , what is the minimum that should be invested initially, assuming that there are no further deposits or withdrawals?

Possible Answers:

$\$9,000$

$\$10,000$

$\$6,000$

$\$7,000$

$\$8,000$

Correct answer:

$\$7,000$

Explanation:

If we let be the initial amount invested and be the annual interest rate of the CD expressed as a decimal, then at the end of years, the amount of money that the CD will be worth can be determined by the formula

$A = Pe^{rt}$

Substitute A = 20,000 , r = 0.062 , t = 18 , and solve for .

$20,000 = Pe^{0.062 \cdot 18}$

$P = \frac{20,000 }{e^{0.062 \cdot 18}} \approx \frac{20,000 }{e^{1.116}} \approx \frac{20,000 }{3.0526} \approx 6,551$

The minimum principal to be invested initially is $6,551. However, since we are looking for the multiple of $1,000 that guarantees a minimum final balance of $20,000, we round up to the nearest such multiple, which is $7,000 - the correct response.

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Example Question #3311 : Algebra Ii

Twelve years ago, your grandma put money into a savings account for you that earns $7.5\%$ interest annually and is continuously compounded. How much money is currently in your account if she initially deposited $\$10,000$ and you have not taken any money out?

Possible Answers:

$8,103

$24,596

$81,030

$10,778

$21,170

Correct answer:

$24,596

Explanation:

1. Use $x=Pe^{rt}$ where is the current amount, is the interest rate, is the amount of time in years since the initial deposit, and is the amount initially deposited.

r=0.075

P=$10,000

t=12

2. Solve for

$x=Pe^{rt}$

$x=10,000e^{(0.075\cdot12)}$

x=24,596.03

You currently have $24,596 in your account.

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Example Question #76 : College Algebra

Jeffrey has won a lottery and has elected to take a $10,000 per month payment.

At the beginning of the year, Jeffrey deposits the first payment of $10,000 in an account that pays 7.6% interest annually, compounded continuously. At the very beginning of each month, he deposits another $10,000. How much will he have at the very end of the year?

Possible Answers:

$\$125,072.98$

$\$124,269.51$

$\$123,561.85$

$\$124,283.35$

$\$125.056.56$

Correct answer:

$\$125,072.98$

Explanation:

The continuous compound interest formula is

$A= Pe^{rt}$ ,

where is the amount of money in the account at the end of the period, is the principal at the beginning, is the annual interest rate in decimal form, and is the number of years over which the interest accumulates. Since Jeff deposits $1,000 per month, we apply this formula twelve times, with equal to the principal at the beginning of each successive month, r = 0.076 , and $t = \frac{1}{12} \approx 0.08333$ .

We can go ahead and calculate $e^{rt}$ , since and do not change:

$e^{rt}= e^{0.076 \cdot \frac{1}{12}} \approx e^{0.00633 } \approx 1.00635$

The formula can be rewritten as

A= P ( 1.006353)

At the beginning of January, Jeffrey deposits $10,000. At the end of January, there is

$A= \$10,000 ( 1.006353) = \$10,063.53$

in the account.

At the beginning of February, he again deposits $10,000, so there is now

$\$10,063.53+ \$10,000 = \$20,063.53$

in the account.

At the end of February, there is

$A= \$20,063.53 ( 1.006353) = \$ 20,191.01$

in the account.

Repeat addition of $10,000, then multiplication by 1.006353, ten more times to get the amount of money in the account at the end of December. This will be $125,072.98.

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Example Question #81 : Exponential And Logarithmic Functions

Sheila has won a lottery and has elected to take a $10,000 per month payment.

At the beginning of the year, Sheila deposits the first payment of $10,000 in an account that pays 7.6% interest annually, compounded monthly. At the very beginning of each month, she deposits another $10,000. How much will she have at the very end of the year?

Possible Answers:

$\$125.056.56$

$\$124,283.35$

$\$123,561.85$

$\$124,269.51$

Correct answer:

$\$125.056.56$

Explanation:

The periodic compound interest formula is

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

where is the amount of money in the account at the end of the period, is the principal at the beginning, is the annual interest rate in decimal form, is the number of periods per year at which the interest is compounded, and is the number of years over which the interest accumulates.

Since Sheila deposits $1,000 per month, we apply this formula twelve times, with equal to the principal at the beginning of each successive month, r = 0.076 , n = 12 (monthly interest), and $t = \frac{1}{12} \approx 0.08333$ .

We can go ahead and calculate $\left (1+ \frac{r}{n} \right ) ^ {nt}$ , since , , and remain constant:

$\left (1+ \frac{r}{n} \right ) ^ {t} = \left (1+ \frac{0.076}{12} \right ) ^ {12 \cdot \frac{1}{12}} \approx 1.006333$ .

The formula can be rewritten as

A= P( 1.006333)

At the beginning of January, Sheila deposits $10,000. At the end of January, there is

$A= \$10,000 ( 1.006333) = \$10,063.33$

in the account.

At the beginning of February, she again deposits $10,000, so there is now

$\$10,063.33+ \$10,000 = \$20,063.33$

in the account.

At the end of February, there is

$A= \$20,063.33 ( 1.006333) = \$ 20,190.40$

in the account.

Repeat addition of $10,000, then multiplication by 1.006333, ten more times to get the amount of money in the account at the end of December. This will be $125.056.56

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Example Question #82 : Exponential And Logarithmic Functions

An investor places $5,000 into an account that has a 5% interest rate. If the he keeps his money in the account for 57 months, and it is compounded quarterly, how much interest will he earn?

Round your answer to the nearest dollar, if needed

Possible Answers:

$\$6,500$

$\$6,330$

$\$6,000$

$\$7,330$

Correct answer:

$\$6,330$

Explanation:

Use the formula for compound interest to solve: $A=P(1+\frac{r}{m})^{tm}$

"A" is the amount of interest, "P" is the initial amount invested, "r" is the interest rate, "m" is the number of times per year the interest is compounded, and "t" is the number of years the money is invested.

Plug in the appropriate values for the equation: $A=5,000(1+\frac{0.05}{4})^{(4.75)(4)}$

Because "t" is measure in years, 57 months needs to be converted to years (i.e. 57 months=4.75 years) to solve this equation

Simplify the equation: $A=5,000(1.0125)^{19}=(5,000)(1.266)=6,330$

Solution: $A=\$6,330$

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College Algebra : Applications

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #1 : Applications

Example Question #3311 : Algebra Ii

Example Question #76 : College Algebra

Example Question #81 : Exponential And Logarithmic Functions

Example Question #82 : Exponential And Logarithmic Functions

All College Algebra Resources

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