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A person invests $1000 in an account that earns 5.5% annual interest. Find when
the value of the investment has doubled.

Sagot :

Answer:

P: the principal, the amount invested

A: the new balance

t: the time

r: the rate, (in decimal form)

Ex1: If $1000 is invested now with simple interest of 8% per year. Find the new amount after two years.

P = $1000, t = 2 years, r = 0.08.

A = 1000(1+0.08(2)) = 1000(1.16) = 1160

Compound Interest:

P: the principal, amount invested

A: the new balance

t: the time

r: the rate, (in decimal form)

n: the number of times it is compounded.

Ex2:Suppose that $5000 is deposited in a saving account at the rate of 6% per year. Find the total amount on deposit at the end of 4 years if the interest is:

P =$5000, r = 6% , t = 4 years

a) simple : A = P(1+rt)

A = 5000(1+(0.06)(4)) = 5000(1.24) = $6200

b) compounded annually, n = 1:

A = 5000(1 + 0.06/1)(1)(4) = 5000(1.06)(4) = $6312.38

c) compounded semiannually, n =2:

A = 5000(1 + 0.06/2)(2)(4) = 5000(1.03)(8) = $6333.85

d) compounded quarterly, n = 4:

A = 5000(1 + 0.06/4)(4)(4) = 5000(1.015)(16) = $6344.93

e) compounded monthly, n =12:

A = 5000(1 + 0.06/12)(12)(4) = 5000(1.005)(48) = $6352.44

f) compounded daily, n =365:

A = 5000(1 + 0.06/365)(365)(4) = 5000(1.00016)(1460) = $6356.12

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