Q: A $15,000 investment is made for 5 years at a 7% annual interest rate. What is the total interest earned if the interest is compounded continuously?

$$15,000 \left( e$$^{0.07 \times 5}$$ - 1 \right) = 6$$,105.15

Q: Chris invests $$200 in a money market account. What quarterly interest rate is needed for the investment to grow to 1$$,800 in 14 years?

$$r = \frac{4}{14} \ln \left( \frac{1800}{200} \right) = 0.1516 \text{ or } 15.16\% $$

Q: A $50 item is subject to continuous inflation at 4% per year. What will it cost in 3 years?

$$50 \times e$$^{0.04 \times 3}$$ = 56.41 $$

Q: If $15,000 is invested for 5 years at 7% interest compounded monthly, what is the total interest earned?

$$15,000 \left( 1 + \frac{0.07}{12} \$$right)^{12 \times 5}$$ - 15$$,000 = $6,089.36

Q: A business wants to know how much a $10,000 investment will be worth in 10 years if compounded continuously at 6%. What is the correct formula and value?

$$A = 10,000 \times e$$^{0.06 \times 10}$$ = 18$$,221.19

Q: If inflation is continuously compounded at 5% per year, what will a $50 item cost in 3 years?

$$50 \times e$$^{0.05 \times 3}$$ = 57.97 $$

Q: A $15,000 investment is made for 5 years at 7% interest compounded quarterly. What is the total interest earned?

$$15,000 \left( 1 + \frac{0.07}{4} \$$right)^{4 \times 5}$$ - 15$$,000 = $6,073.12

Question 1

A $15,000 investment is made for 5 years at a 7% annual interest rate. What is the total interest earned if the interest is compounded continuously?

Accepted Answer

$$15,000 \left( e$$^{0.07 	imes 5}$$ - 1 ight) = 6$$,105.15

Question 2

Chris invests $$200 in a money market account. What quarterly interest rate is needed for the investment to grow to 1$$,800 in 14 years?

Accepted Answer

$$r = \frac{4}{14} \ln \left( \frac{1800}{200} ight) = 0.1516 	ext{ or } 15.16\% $$

Question 3

If $$5,000 is invested for 8 years and grows to 6$$,500 with interest compounded semiannually, what is the required annual interest rate?

Accepted Answer

$$r = 2 \left[ \left( \frac{6500}{5000} \$$right)^{\frac{1}$${16}} - 1 ight] = 0.0143 	ext{ or } 1.43\% $$

Question 4

A $50 item is subject to continuous inflation at 4% per year. What will it cost in 3 years?

Accepted Answer

$$50 	imes e$$^{0.04 	imes 3}$$ = 56.41 $$

Question 5

Which formula correctly calculates the future value $$A of a $$principal $$P$$ invested at an annual rate $$r$$ compounded $$n$$ times per year for $$t$$ years?

Accepted Answer

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

Question 6

If $15,000 is invested for 5 years at 7% interest compounded monthly, what is the total interest earned?

Accepted Answer

$$15,000 \left( 1 + \frac{0.07}{12} \$$right)^{12 	imes 5}$$ - 15$$,000 = $6,089.36

Question 7

A business wants to know how much a $10,000 investment will be worth in 10 years if compounded continuously at 6%. What is the correct formula and value?

Accepted Answer

$$A = 10,000 	imes e$$^{0.06 	imes 10}$$ = 18$$,221.19

Question 8

If inflation is continuously compounded at 5% per year, what will a $50 item cost in 3 years?

Accepted Answer

$$50 	imes e$$^{0.05 	imes 3}$$ = 57.97 $$

Question 9

A $15,000 investment is made for 5 years at 7% interest compounded quarterly. What is the total interest earned?

Accepted Answer

$$15,000 \left( 1 + \frac{0.07}{4} \$$right)^{4 	imes 5}$$ - 15$$,000 = $6,073.12

Question 10

Which scenario best illustrates the advantage of continuous compounding over annual compounding for a long-term investment?

Accepted Answer

Continuous compounding yields a slightly higher amount than annual compounding due to the exponential nature of $$A = P e$$^{rt}$$.

Question 11

A $$5,000 investment grows to 6$$,500 in 8 years with interest compounded annually. What is the required annual interest rate?

Accepted Answer

$$r = \left( \frac{6500}{5000} \$$right)^{\frac{1}$${8}} - 1 = 0.0341 	ext{ or } 3.41\% $$

Question 12

If a $15,000 investment is compounded semiannually at 7% for 5 years, what is the total interest earned?

Accepted Answer

$$15,000 \left( 1 + \frac{0.07}{2} \$$right)^{2 	imes 5}$$ - 15$$,000 = $6,045.13

Question 13

A $50 item is subject to continuous inflation at 3% per year. What will it cost in 3 years?

Accepted Answer

$$50 	imes e$$^{0.03 	imes 3}$$ = 54.77 $$

Question 14

Which of the following best describes the effect of increasing the compounding frequency on the total interest earned?

Accepted Answer

Increasing compounding frequency increases total interest earned, approaching the maximum as compounding becomes continuous.

Question 15

A $$200 investment grows to 1$$,800 in 14 years with interest compounded quarterly. Which formula should be used to find the required interest rate?

Accepted Answer

$$r = \frac{4}{14} \ln \left( \frac{1800}{200} \right) $$

Question 16

If a $15,000 investment is compounded annually at 7% for 5 years, what is the total interest earned?

Accepted Answer

$$15,000 \left( 1 + 0.07 \$$right)^5$$ - 15$$,000 = $6,002.13

Question 17

A $$5,000 investment grows to 6$$,500 in 8 years with interest compounded continuously. What is the required annual interest rate?

Accepted Answer

$$r = \frac{1}{8} \ln \left( \frac{6500}{5000} ight) = 0.0341 	ext{ or } 3.41\% $$

Business Calculus: Compound Interest and Exponential Growth Study Guide

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