60 * 242.81 = 17.568.60
17,568.60 - 12,000 = 5,568.60
I = PRT
5,568.60 = 12,000 * r * 6
5,568.6/72,000 = r
The owner of an automobile repair center purchased new electronic diagnostic equipment for $12,000. He paid down and then paid 60 monthly payments of $242.81. Determine the APR of the loan to the nearest one-half of a percent.
3 answers
Ms Sue has 2 typos in that she used 6 years instead of 5, and 242.81*60 = 14568.86
Her rate would have been
R = 2568.60/(12000*5) = .043 or appr 4.3%
Using compound interest, since the time > 1 year
let the real monthly rate be x
242.81(1 - (1+x)^-60)/x = 12000
1 - (1+x)^-60 = 49.421x
this type of equation is very messy to solve, so let's go to Wolfram
https://www.wolframalpha.com/input/?i=solve+1+-+(1%2Bx)%5E-60+%3D+49.421x
and it said: x = .00659335
or per year : 12(.00659335) = .07912 or appr 7.9%
Her rate would have been
R = 2568.60/(12000*5) = .043 or appr 4.3%
Using compound interest, since the time > 1 year
let the real monthly rate be x
242.81(1 - (1+x)^-60)/x = 12000
1 - (1+x)^-60 = 49.421x
this type of equation is very messy to solve, so let's go to Wolfram
https://www.wolframalpha.com/input/?i=solve+1+-+(1%2Bx)%5E-60+%3D+49.421x
and it said: x = .00659335
or per year : 12(.00659335) = .07912 or appr 7.9%
I sure messed up! Thank you, Reiny!!