Naomi invested $860 in an account paying an interest rate of 3, start fraction, 3, divided by, 4, end fraction3 43 % compounded monthly. Matthew invested $860 in an account paying an interest rate of 4, one quarter4 41 % compounded annually. After 10 years, how much more money would Matthew have in his account than Naomi, to the nearest dollar?
Question
Naomi invested 860 in an account paying an interest rate of 4, one quarter4 41 % compounded annually. After 10 years, how much more money would Matthew have in his account than Naomi, to the nearest dollar?
Solution
The problem involves calculating the future value of an investment using the formula for compound interest, which is:
A = P(1 + r/n)^(nt)
where: A = the amount of money accumulated after n years, including interest. P = the principal amount (the initial amount of money) r = annual interest rate (in decimal) n = number of times that interest is compounded per year t = time the money is invested for in years
For Naomi: P = $860 r = 3.75% = 0.0375 (in decimal) n = 12 (since it's compounded monthly) t = 10 years
Substitute these values into the formula:
A = 860(1 + 0.0375/12)^(12*10)
For Matthew: P = $860 r = 4.25% = 0.0425 (in decimal) n = 1 (since it's compounded annually) t = 10 years
Substitute these values into the formula:
A = 860(1 + 0.0425/1)^(1*10)
Calculate the future value for both Naomi and Matthew, then subtract Naomi's future value from Matthew's to find out how much more money Matthew would have. Round to the nearest dollar.
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