Nolan began a savings account three years ago. He invested 100 dollars at a 2 percent interest rate according to the equation cap v sub n is equal to 100 times 1 point 0 2 to the x th power , where cap v sub n is the value of his account after x years. Anias started an account today. She invested 100 dollars at a 2 percent interest rate according to the equation cap v sub A is equal to 100 times 1 point 0 2 raised to the x minus 3 power , where cap v sub A is the value of her account. Let’s say Anias started saving at the same time Nolan did, three years ago. Approximately how much money would she have had to invest to have the same amount of money she has now?
At the present time, x = 3, so Anias’s account is worth 100 × 1.02^(3 − 3) = $100. If she had invested an amount P three years ago, it would grow to P × 1.02^3. Set that equal to $100: P = 100 ÷ 1.02^3 ≈ $94.23. So she would have needed to invest about $94.23, choice A.
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