Question

Suppose that $4000 is placed in a savings account at an annual rate of 6.4%, compounded quarterly. Assuming that no withdrawals are made, how long will it
take for the account to grow to $4916?
Do not round any intermediate computations, and round your answer to the nearest hundredth.

Answer 1

Answer:

3.25 years (nearest hundredth)

Step-by-step explanation:

Compound Interest Formula

$\large \sf A=P\left(1+\frac{r}{n}\right)^{nt}$

where:

• A = final amount
• P = principal amount
• r = interest rate (in decimal form)
• n = number of times interest applied per time period
• t = number of time periods elapsed

Given:

• A = $4916
• P = $4000
• r = 6.4% = 0.064
• n = 4 (quarterly)

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

$\implies \sf 4916=4000\left(1+\frac{0.064}{4}\right)^{4t}$

$\implies \sf 4916=4000\left(1.016\right)^{4t}$

$\implies \sf \dfrac{4916}{4000}=\left(1.016\right)^{4t}$

$\implies \sf \dfrac{1229}{1000}=\left(1.016\right)^{4t}$

$\implies \sf \ln \left(\dfrac{1229}{1000}\right)=\ln \left(1.016\right)^{4t}$

$\implies \sf \ln \left(\dfrac{1229}{1000}\right)=4t \ln \left(1.016\right)$

$\implies \sf t=\dfrac{\ln \left(\frac{1229}{1000}\right)}{4\ln \left(1.016\right)}$

$\implies \sf t=3.247594892...$

Therefore, it will take 3.25 years (nearest hundredth) for the account to grow to $4916.

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