Question

A person invests 5500 dollars in a bank. The bank pays 6.75% interest compound monthly. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 13200 dollars?

Answer 1

We have been given that a person invests 5500 dollars in a bank. The bank pays 6.75% interest compound monthly. We are asked to find the time that will take the amount to 13,200 dollars.

We will use compound interest formula to solve our given problem.  

$A=P(1+\frac{r}{n})^{nt}$, where,

A = Final amount,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years.

$6.75\%=\frac{6.75}{100}=0.0675$

Upon substituting our given values in above formula, we will get:

$13200=5500(1+\frac{0.0675}{12})^{12\cdot t}$

$13200=5500(1+0.005625)^{12\cdot t}$

$13200=5500(1.005625)^{12\cdot t}$

$\frac{13200}{5500}=(1.005625)^{12\cdot t}$

$2.4=(1.005625)^{12\cdot t}$

Now we will take natural log on both sides.

$\text{ln}(2.4)=\text{ln}((1.005625)^{12\cdot t})$

Using natural log property $\text{ln}(a^b)=b\cdot \text{ln}(a)$, we will get:

$\text{ln}(2.4)=12t\cdot \text{ln}(1.005625)$

$t=\frac{\text{ln}(2.4)}{12\cdot \text{ln}(1.005625)}$

$t=13.00635$

Upon rounding to nearest tenth, we will get:

$t\approx 13.0$

Therefore, it will take approximately 13.0 years for the amount to reach $13200.