Question

How much money must be deposited now in an

account paying 7.25% annual interest, compounded
quarterly, to have a balance of $1000 after 10
years?

Answer 1

`\$487.48`

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

in this problem we have

`t=10\ years\\ A=\$1,000\\ r=7.25\%=7.25/100=0.0725\\n=4`

substitute in the formula above

`1,000=P(1+(0.0725)/(4))^(4*10)`

`1,000=P((4.0725)/(4))^(40)`

`P=1,000/((4.0725)/(4))^(40)`

`P=\$487.48`