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# Compound Interest Question

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The formula for an investment worth with interest compounded annually is A = P(1+i)^n, where P represents the initial investment, i is the interest rate, and A is the worth of the investment after n years.

a) Rearrange the formula for P. What was the initial investment of an investment worth \$1000 that compounded 10% interest for 10 years?
b) Rearrange the formula for i. What is the interest rate of an investment whose worth went from \$1000 to \$1200 in 2 years?
c) Explain a method with which you could estimate how many years it would take for an investment to reach a certain worth at a certain interest rate.
d) Estimate how many years would it take an investment of \$2100 at 20% interest to reach a worth of \$5225?

Sep 23, 2022

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a)  P = A (1 + i)^(-n)

P ==1000(1.10)^-10

P ==1000  x  0.3855432

P ==\$385.54 - initial investment.

b)  i = (A/P)^(1/n) - 1

i ==(1200 / 1000)^(1/2) - 1

i ==Sqrt(1.2) - 1

i ==1.095445 - 1  x 100 ==9.5445% - interest rate.

c) Use "Rule of 72".  \$100 at 10% will double to \$200 in: 72 / 10 [interest rate] ==~7.2 years.

d)  5225 ==2100 x [1 + 0.20]^N.  Note: We don't have to "estimate" but will solve it exactly!!!

5225 / 2100 ==1.20^N

2.488095238 ==1.20^N

Take the log of both sides

N ==log(2.488095238) / log(1.20)

N ==5 years for 2100 to grow to 5225 at 20% comp. annually.

Sep 23, 2022