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+(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\%=(6.75)/(100)=0.0675`

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

`13200=5500(1+(0.0675)/(12))^(12\cdot t)`

`13200=5500(1+0.005625)^(12\cdot t)`

`13200=5500(1.005625)^(12\cdot t)`

`(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.