How many years would it take for a deposit of $10,000 to become $200,000, if interest of 9% per year is compounded annually?
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A. B. C. D. E.D
On the BAII Plus, press 10000 PV, 0 PMT, 200000 +/- FV, 9 I/Y, CPT N. On the HP12C, press 10000 PV, 200000 CHS FV, 9 i, then press n. Note that the HP12C shows 35 as the answer.
To determine the number of years it would take for a deposit of $10,000 to grow to $200,000 at an interest rate of 9% per year compounded annually, we can use the compound interest formula:
Future Value (FV) = Present Value (PV) * (1 + interest rate)^number of periods
In this case, the present value (PV) is $10,000, the future value (FV) is $200,000, and the interest rate is 9% per year. We need to find the number of periods (years).
Let's plug in the given values into the formula and solve for the number of periods:
$200,000 = $10,000 * (1 + 0.09)^number of periods
Dividing both sides of the equation by $10,000, we have:
20 = (1 + 0.09)^number of periods
To isolate the exponent, we can take the natural logarithm (ln) of both sides of the equation:
ln(20) = ln((1 + 0.09)^number of periods)
Using the property of logarithms, we can bring down the exponent:
ln(20) = number of periods * ln(1 + 0.09)
Now we can solve for the number of periods:
number of periods = ln(20) / ln(1 + 0.09)
Using a calculator, the approximate value of ln(20) is 2.9957, and ln(1 + 0.09) is 0.0862.
Plugging in these values, we get:
number of periods = 2.9957 / 0.0862 ≈ 34.76
Therefore, it would take approximately 34.76 years for a deposit of $10,000 to grow to $200,000 at an interest rate of 9% per year compounded annually.
The correct answer is D. 34.76.