Question

If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years for the following compounding methods, to the nearest cent.

(a) Annual: $______
(b) Semiannual: $ _____
(c) Monthly: $______
(d) Daily: $_______

Answer 1

Part A) Annual
`\$66,480.95`

Part B) Semiannual
`\$66,862.38`

Part C) Monthly
`\$67,195.44`

Part D) Daily
`\$67,261.54`

Explanation:

we know that

The compound interest formula is equal to

`A=P(1+(r)/(n))^(nt)`

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

Part A)

Annual

in this problem we have

`t=5\ years\\ P=\$47,400\\ r=0.07\\n=1`

substitute in the formula above

`A=47,400(1+(0.07)/(1))^(1*5)`

`A=47,400(1.07)^(5)`

`A=\$66,480.95`

Part B)

Semiannual

in this problem we have

`t=5\ years\\ P=\$47,400\\ r=0.07\\n=2`

substitute in the formula above

`A=47,400(1+(0.07)/(2))^(2*5)`

`A=47,400(1.035)^(10)`

`A=\$66,862.38`

Part C)

Monthly

in this problem we have

`t=5\ years\\ P=\$47,400\\ r=0.07\\n=12`

substitute in the formula above

`A=47,400(1+(0.07)/(12))^(12*5)`

`A=47,400(1.0058)^(60)`

`A=\$67,195.44`

Part D)

Daily

in this problem we have

`t=5\ years\\ P=\$47,400\\ r=0.07\\n=365`

substitute in the formula above

`A=47,400(1+(0.07)/(365))^(365*5)`

`A=47,400(1.0002)^(1,825)`

`A=\$67,261.54`