Question

Ronisha invests $5000 in an account that earns 2.3% annual simple interest. After 3 years, Ronisha invests the interest earned in a second account. This account earns 3.2% annual interest. Ronisha leaves the money in this account until she has earned $55.20 in interest. Ronisha does not deposit or withdraw any money from the accounts. How long does Ronisha invest money into the second account?

Answer 1

Very complicated problem, but thank goodness I have too much time and am a nerd.

Basically model 1st account as so.
Fv=pv(1+I)^N
Fv future value
Pv present value
I interest
N number years
So first equation is interest earned after 3 years
Y=5000(1+0.023)^3
Y=5353 (rounded up)
So we know that the interest earned is 5353-5000 which is 353.

Now Ronisha (clearly the name of a future investment genius) invests these 353$ in a new account.
Now remember we’re not solving for FV we’re cause we’re given that: 55.2$
However this is the interest earned not the future value. So if interest earned is fv - pv and we know pv
Fv - 353 = 55.2
Fv = 408.2$
So now we reuse the formula
408.2 = 353(1+0.032)^x
Now just solve for x:
First divide both sides by 353
1.156 = 1.032^x
Remember the log rule that states x=b^y is same as y=logb(x)
So using the same logic:
X= log1.032(1.156)
Use some kind of calculator for that where you can adjust log base. But you basically get:
X= 4.602
So Ronisha has to basically invest 5 years or 4.6 years which is 4 years and 7 months.

Omg nevermind, wait it’s simple interest….
Sorry here’s the simple solution. My bad, but I worked so hard on the top part I don’t want to delete it.
I=prt
Interest = principal * rate * time
I= 5000(0.023)(3)=345
Then just do the same but plug 55.2 for I
55.2 = 345(0.032)t
Now solve for t
55.2 = 11.04t
t= 5

As you can see, similar logic where ultimately it takes 5 years. But this “genius” Ronisha should’ve just done compounding interest (my first calculation) and gotten it done in 4.6 years. Almost 5 months faster.