Question

Carter invested $70,000 in an account paying an interest rate of 1.125% compounded continuously. Savannah invested $70,000 in an account paying an interest rate of 1.375% compounded quarterly. To the nearest dollar, how much money would Carter have in his account when Savannah's money has doubled in value?

Answer 1

Step 1: Figure out when Savannah will double her account value:

We need to solve $140000 = 70000\left(1+\frac{0.01375}{4}\right)^{4t}$.

Divide by 70000:

       $2 = \left(1+\frac{0.01375}{4}\right)^{4t}$

Take the natural log of both sides and bring the exponent down:

      $\ln(2) = 4t\cdot\ln\left(1+\frac{0.01375}{4}\right)$

Divide by that mess multiplied by t:

      $\dfrac{\ln(2)}{4\ln\left(1+\frac{0.01375}{4}\right)} = t$

Throw that into a calculator and you get about 50.497297884.

Depending on how picky your teacher is, we'd need to round this to the next time the interest is compounded, since it's only compounded each quarter, so t = 50.5.  (The fact is that Savannah's account will never exactly double in value.)

Step 2:

Now, we need to evaluate Carter's continuously compounded investment for 50.5 years:

    $B = 70000e^{0.01125\cdot(50.5)}\approx123546.825994$

    B ≈ 123547 to the nearest dollar.

As a comparison, here's the other calculation with the more precise t value.

    $B = 70000e^{0.01125\cdot(50.497297884)}\approx123543.070375$

    B ≈ 123543 to the nearest dollar.

Again, I would say the 50.5 calculation is actually more correct, since Savannah's account only compounds the interest each quarter, but you'll have to decide what your teacher would say.

   

Answer 2

Answer:

$123,543 (nearest dollar)

Step-by-step explanation:

Find the length of time Savannah has to invest her money for it to double.

$\boxed{\begin{minipage}{8.5 cm}\underline{Compound Interest Formula}\\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \\\\where:\\\\ \phantom{ww} \bullet A = final amount \\ \phantom{ww} \bullet P = principal amount \\ \phantom{ww} \bullet r = interest rate (in decimal form) \\ \phantom{ww} \bullet n = number of times interest is applied per year \\ \phantom{ww} \bullet t = time (in years) \\ \end{minipage}}$

Given:

• A = $140,000
• P = $70,000
• r = 1.375% = 0.01375
• n = 4 (quarterly)

Substitute the given values into the formula and solve for t:

$\implies 140000=70000\left(1+\dfrac{0.01375}{4}\right)^{4t}$

$\implies 140000=70000\left(1.0034375\right)^{4t}$

$\implies2=\left(1.0034375\right)^{4t}$

$\implies \ln2=\ln\left(1.0034375\right)^{4t}$

$\implies \ln2=4t\ln\left(1.0034375\right)$

$\implies t=\dfrac{\ln 2}{4 \ln\left(1.0034375\right)}$

$\implies t=50.49729788$

Therefore, it would take approximately 50.5 years for Savannah's principal investment to double.

$\boxed{\begin{minipage}{8.5 cm}\underline{Continuous Compounding Formula}\\\\ A=Pe^{rt} \\\\where:\\\\ \phantom{ww} \bullet A = final amount \\ \phantom{ww} \bullet P = principal amount \\ \phantom{ww} \bullet e = Euler's number (constant) \\ \phantom{ww} \bullet r = annual interest rate (in decimal form) \\ \phantom{ww} \bullet t = time (in years) \\ \end{minipage}}$

Given:

• P = $70,000
• r = 1.125% = 0.01125
• t = 50.49729788...

Substitute the given values into the formula along with the found value of t and solve for A:

$\implies A=70000e^{0.01125 \times 50.4972...}$

$\implies A=123543.0704...$

Therefore, the amount of money that Carter would have in his account when Savannah's money has doubled in value is $123,543 (nearest dollar).