Question

Use the formula for the compound interest with n compoundings each year to solve this problem. How long to the nearest tenth of a year will it take 12,500 to grow to 20000 at 6.5% annual interest compouned quarterly?

Answer 1

5.9 year

Explanation:

It takes 5.9 years to grow to 20000 at 6.5% annual interest compounded quartely!

Answer 2

7.3 years.

Explanation:

We are asked to find the time taken for 12,500 to grow to 20000 at 6.5% annual interest compounded quarterly. We will use compound interest formula to solve our given problem.

`A=P(1+(r)/(n))^(nt)`, where,

A = Final amount after t years,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years.

`6.5\%=(6.5)/(100)=0.065`

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

`20,000=12,500(1+(0.065)/(4))^(4\cdot t)`

`20,000=12,500(1+0.01625)^(4\cdot t)`

`20,000=12,500(1.01625)^(4\cdot t)`

`12,500(1.01625)^(4\cdot t)=20,000`

`(12,500(1.01625)^(4\cdot t))/(12,500)=(20,000)/(12,500)`

`(1.01625)^(4\cdot t)=1.6`

Now we will take natural log of both sides of equation as:

`\text{ln}((1.01625)^(4\cdot t))=\text{ln}(1.6)`

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

`4\cdot t\cdot \text{ln}(1.01625)=\text{ln}(1.6)`

`\frac{4\cdot t\cdot \text{ln}(1.01625)}{4\cdot \text{ln}(1.01625)}=\frac{\text{ln}(1.6)}{4\cdot \text{ln}(1.01625)}`

`t=(0.470003629245)/(4\cdot0.016119381879)`

`t=(0.470003629245)/(0.064477527516)`

`t=7.28941768`

Upon rounding to nearest tenth of year, we will get:

`t\approx 7.3`

Therefore, it will take approximately 7.3 years for 12,500 to grow to 20000.