Math II

<p>A sum of $10,000 is invested at a rate of 10 percent, with interest compounded semiannually. The value, in dollars, of this investment after t years is given by V(t)=10,000(1.05)^2t. Approximately how much greater is the value of this investment at the end of 2 years than the same amount invested at the rate of 10 percent compounded annually?</p>

<p>The answer is $55. I thought all I had to do was plug 2 into that equation and half the product to find the answer, but I'm apparently wrong. Can someone explain how to do this?</p>

<p>For the initial way it has been done:
x=10,000(1.05)^4 ->(here you were correct about plugging in)
ie x=12100.
For the second method of calculating the amount,</p>

<p>y= 10,000(1 + 10/100)^2 (because the interest is 10% and because you’re calculating for 2 years,so we square the value.)</p>

<p>ie y=10,000(1.1)^2
=12155</p>

<p>Answer = y-x =12155-12100 =55</p>