Answer:
a.$185,050.55
b.$191.11
c.$300.00
Step-by-step explanation:
a.Given that the initial value of the house was $95,000 with an annual appreciation rate of 4%.
-We get the difference between the two years to get the total years of appreciation,n:
[tex]n=2007-1990\\\\=17[/tex]
#The value after 17 yrs is calculated using the compound formula ;
[tex]V_f=V_o(1+i)^n, n=17, V_o=95000, i=0.04\\\\V_f=95000(1+0.04)^{17}\\\\=185050.55[/tex]
Hence, the value of the house in 2007 is $185,050.55
b. Given the invested amount is $500, n=10 and a rate of 3.25%(compounded 3.25%):
#First we calculate the effective annual rate:
[tex]i_m=(1+i/m)^m-1, i=0.0325, m=4\\\\i_m=(1+0.0325/4)^4-1=0.032898[/tex]
The amount of the investment after 10 yrs is:
[tex]A=P(1+i_m)^n, P=500,n=10,i_m=0.032898\\\\A=500(1.032898)^{10}\\\\=691.11[/tex]
The amount of interest earned is the final value minus the principal value:
[tex]I=A-P\\\\=691.11-500\\\\=191.11[/tex]
Hence, the interest earned on $500 is $191.11
c.Given the Final amount is $1131.73, n=20 yrs and the rate is 6.75% compounded semiannually.
#First we calculate the effective annual rate:
[tex]i_m=(1+i/m)^m-1, i=0.0675, m=2\\\\i_m=(1+0.0675/2)^2-1=0.068639[/tex]
#we use the compound interest formula to equate the principal to the final value to solve for principal;
[tex]A=P(1+i_m)^n, P=P,n=20,i_m=0.068639,A=1131.73\\\\1131.73=P(1.032898)^{20}\\\\P=300.00[/tex]
Hence, the initial investment was $300.00
