Question 724057: How much money must be deposited now in an account paying 8% annual interest, compounded quarterly, to have a balance of $1,000 after 10 years?
Answer by Jstrasner(112)   (Show Source):
You can put this solution on YOUR website! Hey,
So for this one, we need to create a formula for calculating the amount. A normal compounded interest formula looks like this:
A = B ( 1 + r ) ^n
Where A is the ending amount of money, in this case $1000. B is the beginning amount of money, in this case x because we don't know. Little r is the rate, in this case .08 (or 8%). Little n is the number of years the money is in the bank, in this case 10 years. This would work, HOWEVER, the interest is compounded quarterly, which means 4 times a year. Therefore the formula is change to look like this:
A = B ( 1 + r/t ) ^ (n)(t)
Where everything is the same except that we divide the rate (r) by the number of times a year the interest is made (t) and then instead of just taking that amount to the exponent of n, we need to first multiply n by the number of times per year (t). So this is what this problem would look like:
$1,000 = x ( 1 + .08/4 ) ^ (10)(4)
First we solve what is in the parentheses:
1 + .08/4 => 1 + .02 => 1.02
Then we multiply the exponents together:
(10)(4) => 40
Then we take 1.02 ^ 40 => 2.208...
Now we have to isolate the x so we can solve for it. In order to do this, we need to get rid of the 2.208 by dividing both sides by 2.208 :
$1000/2.208 = x (2.208)/2.208 => 452.8985..... = x
And now we have the answer.
If we deposit around $453 into a bank with compound interest of 8% quarterly, we will have $1000 after 10 years.
Again the answer is $453.
I hope this helps!
|
|
|
