Can anyone please explain why the interest rate (in the solution to the example at the bottom of this page) is converted into monthly while it already says 15% compounded monthly?
Example 2.7
. . .A principal of $50,000 is to be borrowed at an interest rate of 15% compounded monthly for 30 years. What will be the monthly payment to repay the loan?
. . .Solution: Monthly interest i = 0.15/12 = 0.0125, or 1.25%. Since Appendix A does not contain a table for that interest rate, one must use the formulas.
. . . . .\(\displaystyle A\, =\, P\, (A/P)_{i,n}\, =\, 50000\, (A/P)_{1.25, 360}\)
. . . .. . .\(\displaystyle =\, 50000\, \cdot\, \dfrac{(0.0125)\, (1\, +\, 0.0125)^{360}}{(1\, +\, 0.0125)\, -\, 1}\)
. . .. . . .\(\displaystyle =\, 50000\, (0.012644)\, =\, \$632\)
in my mind i have A = P(A/P,i,n); i = 15%, n = 360
but solution is different here
Example 2.7
. . .A principal of $50,000 is to be borrowed at an interest rate of 15% compounded monthly for 30 years. What will be the monthly payment to repay the loan?
. . .Solution: Monthly interest i = 0.15/12 = 0.0125, or 1.25%. Since Appendix A does not contain a table for that interest rate, one must use the formulas.
. . . . .\(\displaystyle A\, =\, P\, (A/P)_{i,n}\, =\, 50000\, (A/P)_{1.25, 360}\)
. . . .. . .\(\displaystyle =\, 50000\, \cdot\, \dfrac{(0.0125)\, (1\, +\, 0.0125)^{360}}{(1\, +\, 0.0125)\, -\, 1}\)
. . .. . . .\(\displaystyle =\, 50000\, (0.012644)\, =\, \$632\)
in my mind i have A = P(A/P,i,n); i = 15%, n = 360
but solution is different here
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