Example: A $100 investment now in an account that pays compound interest annually will be worth $250 at a point exactly 31 years from now. What annual interest rate does this account pay?
Solving the equation for i:
250 = 100 (1 + i)31
1 + i = (250/100)1/31= 1.0299
yields an answer of 3%.
Or, using interest tables, note that i = 3%, because (F/P,3%,31) = 2.500.
1. How many years are required for an investment to double in value at 10% interest?
a. n = 2 years, because (F/A,10%,2)>2 and (F/A,10%,1) < 2
b. n = 3 years, because (P/G,10%,3)>2 and (P/G,10%,2) < 2
c. n = 8 years, because (F/P,10%,8)>2 and (F/P,10%,7) < 2
d. n = 10 years, because interest is 10% per year
2. At what annual compound interest rate is $100 today equivalent to $370, seventeen years from now?
a. i = 6%
b. i = 7%
c. i = 8%
d. i = 9%
Example: A $100 investment now in an account that pays compound interest annually will be worth $250 at a point exactly 31 years from now. What annual interest rate does this account pay?
Solving the equation for i:
250 = 100 (1 + i)31
1 + i = (250/100)1/31= 1.0299
yields an answer of 3%.
Or, using interest tables, note that i = 3%, because (F/P,3%,31) = 2.500.
1. How many years are required for an investment to double in value at 10% interest?
a. n = 2 years, because (F/A,10%,2)>2 and (F/A,10%,1) < 2
b. n = 3 years, because (P/G,10%,3)>2 and (P/G,10%,2) < 2
c. n = 8 years, because (F/P,10%,8)>2 and (F/P,10%,7) < 2
d. n = 10 years, because interest is 10% per year
2. At what annual compound interest rate is $100 today equivalent to $370, seventeen years from now?
a. i = 6%
b. i = 7%
c. i = 8%
d. i = 9%
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