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## Single Payment Compound Interest Formulas (annual)

### Interest and Equivalence

#### Single payment compound interest formulas (annual)

Go to questions covering topic below

Given a present dollar amount P, interest rate i% per year, compounded annually, and a future amount F that occurs n years after the present, the relationship between these terms is F = P (1 + i) n

In equations, the interest rate i must be in decimal form, not percent.

Example: If \$100 is invested at 6% interest per year, compounded annually, then the future value of this investment after 4 years is

F = P (1 + i) n = \$100 (1 + 0.06) 4

= \$100 (1.06) 4 = \$100 (1.2625) = \$126.25

Solving the above equation for P yields: P = F (1 + i) -n

The factors F/P and P/F are available in interest tables, simplifying somewhat the calculations. The common notation for these factors is

(F/P,i%,n) = (1 + i) n

(P/F,i%,n) = (1 + i) – n

### Interest and Equivalence

#### Single payment compound interest formulas (annual)

Question 1

Question 2

Question 3

Question 4

Question 1.

Suppose that \$1,000 is invested for six years at an interest rate of 10% per year, compounded annually. How much will be in the account at the end of six years?

Choose an answer by clicking on one of the letters below, or click on "Review topic" if needed.

A F = \$1,000 + \$600 = \$1,600

B F = \$1,000 (1.06)10 = \$1,791

C F = \$1,000 (1.10) 6 = \$1,772

D F = \$1,000 (1.06) -10 = \$558

Review topic

Question 2.

Suppose that an investor wishes to deposit an amount now so that in 30 years \$1,000,000 will be in an account that pays 10% interest per year, compounded annually. What amount must be deposited now?

Choose an answer by clicking on one of the letters below, or click on "Review topic" if needed.

A P = \$1,000,000 (1.10) -30 = \$57,309

B P = \$1,000,000 / 30 = \$33,333

C P = \$1,000,000 (0.10) = \$100,000

D P = \$1,000,000 (1.10) 30 = \$17,449,402

Review topic

Question 3.

This question deals with the use of interest tables instead of equations. Use the interest tables to answer the question.

Suppose that \$100 is invested for five years at an interest rate of 8% per year, compounded annually. How much will be in the account at the end of five years?

Choose an answer by clicking on one of the letters below, or click on "Review topic" if needed.

A F = \$100 (P/F,8%,5) = \$100 (0.6806) = \$68.06

B F = \$100 (F/P,8%,5) = \$100 (1.469) = \$146.90

C P = \$100 (P/A,8%,5) = \$100 (3.993) = \$399.30

D F = \$100 (F/A,8%,5) = \$100 (5.867) = \$586.70

Review topic

Question 4.

This question deals with the use of interest tables instead of equations. Use the interest tables to answer the question.

Suppose that an investor wishes to deposit an amount now so that in 20 years there will be \$50,000 in an account that pays 7% interest, compounded annually. What amount must be deposited now?

Choose an answer by clicking on one of the letters below, or click on "Review topic" if needed.

A F = \$50,000 (F/P,7%,20) = \$50,000 (3.870) = \$193,500

B P = \$50,000 (P/F,20%,7) = \$50,000 (0.2791) = \$13,955

C P = \$50,000 (P/F,7%,20) = \$50,000 (0.2584) = \$12,920

D P = \$50,000 (P/F,7%,20) = \$50,000 (0.2415) = \$12,075

Review topic