(4) Present value of $1000 is invested:

a)     Future value is $3105.85 at 12% annual interest rate, compounding annually. How many years did it take?

b)    Future value is $1790.85 at 12% annual interest rate, compounding semi-annually. How many years did it take?

c)     Future value is $1608.44 at 12% annual interest rate, compounding every two months. How many years did it take?

(5) Beginning now, you make $1,000 deposits at the beginning of every 3 months for the next 4 years. The annual interest rate is 10% :

a)     Present value of these deposits

b)    Future value of these deposits in 4 years

4.a. = 10 years

b. = 5 years

c. = 4 years

 

5.a. = $13055

b = $19380.22

Step-by-step explanation

4.a. The number of years taken to reach $3105.85 at 12% annual interest rate is given as follows:

the formula to compute future value is given as follows:

FV = P * (1+i)ⁿ

where FV = $3105.85, P = $1000, i = 12%

 

therefore, the time n is given as follows:

FV = P * (1+i)ⁿ

$3105.85 = $1000 * (1+0.12)ⁿ

(1+0.12)ⁿ? = 3105.85/1000

(1.12)ⁿ = 3.10585

 

we introduce logarithms to solve:

(1.12)ⁿ = 3.10585

log (1.12)ⁿ = log 3.10585

n log 1.12 = log 3.10585

n = (log 3.10585)/log 1.12

n = 10.000000 years

therefore, it will take 10 years.

 

b. The number of years taken to reach $1790.85 at 12% annual interest rate is given as follows:

the formula to compute future value is given as follows:

FV = P * (1+i/m)^nm

where FV = $1790.85, P = $1000, i = 12%, m=2 (interest compounded semi-annually)

 

therefore, the time n is given as follows:

FV = P * (1+i/m)^nm

$1790.85 = $1000 * (1+0.12/2)²ⁿ

(1+0.12/2)²ⁿ = 1790.85/1000

(1.06)²ⁿ = 1.79085

 

we introduce logarithms to solve:

(1.06)²ⁿ = 1.79085

log (1.06)²ⁿ = log 1.79085

2n log 1.06 = log 1.79085

n = (log 1.79085)/2log 1.06

n = 5.0000 years

therefore, it will take 5 years.

 

 

c. The number of years taken to reach $1608.44 at 12% annual interest rate is given as follows:

the formula to compute future value is given as follows:

FV = P * (1+i/m)^nm

where FV = $1608.44, P = $1000, i = 12%, m=6 (interest compounded after 2 months)

 

therefore, the time n is given as follows:

FV = P * (1+i/m)^nm

$1608.44 = $1000 * (1+0.12/6)⁶ⁿ

(1+0.12/6)⁶ⁿ = $1608.44/1000

(1.02)⁶ⁿ = 1.60844

 

we introduce logarithms to solve:

(1.02)⁶ⁿ = 1.60844

log (1.02)⁶ⁿ = log 1.60844

6n log 1.02 = log 1.60844

n = (log 1.60844)/6log 1.02

n = 4.0000 years

therefore, it will take 4 years.

 

 

5.a. Present value of these deposits is given as follows:

the formula for present value of annuity is:

PV = P * [(1-(1+i/m)^-mn)/(i/m)]

where P = $1000, i = 10%, m = 4 (interest compounded after 3 months), n = 4 years

 

therefore,

PV = P * [(1-(1+i/m)^-mn)/(i/m)]

PV = 1000 * [(1-(1+0.10/4)^-4*4)/(0.10/4)]

PV = $13055.00266

therefore, the present value is $13055

 

 

b. Future value of these deposits in 4 years is given as follows:

the formula for future value of annuity is:

FV = P * [((1+i/m)^mn -1)/(i/m)]

where P = $1000, i = 10%, m = 4 (interest compounded after 3 months), n = 4 years

 

therefore,

FV = P * [((1+i/m)^mn -1)/(i/m)]

FV = 1000 * [((1+0.1/4)^4*4 -1)/(0.1/4)]

PV = $19380.22483

therefore, the present value is $19380.22