Question

A) Mr and Mrs Greenland bought a house for $95000 in 1990. Real estate values in their area increase approximately 4% each year. What was the value of the house in 2007?

B) Determine the amount of interest earned if $500 is invested interest rate of 3.25% compounded quarterly for 10 years.

C) Determine the initial investment with an ending balance of $1131.73 invested at an interest rate of 6.75% compounded semiannually for 20 years.

Answer 1

a.$185,050.55

b.$191.11

c.$300.00

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:

`n=2007-1990\\\\=17`

#The value after 17 yrs is calculated using the compound formula ;

`V_f=V_o(1+i)^n, n=17, V_o=95000, i=0.04\\\\V_f=95000(1+0.04)^(17)\\\\=185050.55`

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:

`i_m=(1+i/m)^m-1, i=0.0325, m=4\\\\i_m=(1+0.0325/4)^4-1=0.032898`

The amount of the investment after 10 yrs is:

`A=P(1+i_m)^n, P=500,n=10,i_m=0.032898\\\\A=500(1.032898)^(10)\\\\=691.11`

The amount of interest earned is the final value minus the principal value:

`I=A-P\\\\=691.11-500\\\\=191.11`

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:

`i_m=(1+i/m)^m-1, i=0.0675, m=2\\\\i_m=(1+0.0675/2)^2-1=0.068639`

#we use the compound interest formula to equate the principal to the final value to solve for principal;

`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`

Hence, the initial investment was $300.00