Question

Noah currently has an account balance of $2,976.45. He opened the account 11 years ago with a deposit of $1,963.45. If the interest compounds quarterly, what is the interest rate on the account?

Answer 1

What we know:
Acct. balance A (11)= $2,976.45
Pricipal=$1,963.45
Compounded (n)=quarterly (4)
rate=r
Compound interest formula: A (t)=p (1+r/n)^nt

What we need toi find:
If the interest compounds quarterly, what is the interest rate on the account? Find r.
A (t)=p (1+r/n)^nt
2976.45=1963.45 (1+r/4)^(4×11)
2976.45=1963.45 (1+r/4)^44
2976.45/1963.45=1963.45/1963.45 (1+r/4)^44
1.5159=(1+r/4)^44
(1.5159)^(1/44)=((1+r/4)^44)^(1/44)

1.0095=1+r/4
1.0095-1=1-1+r/4
.0095=r/4
.0095(4)=r/4 (4)
.038=r

Rate = 3.8%

Answer 2

The interest rate is 3.79% or 3.8%

Explanation:

The formula to find how much money A we have at the end of t years if we start with a principal or amount of money P today and the interest is compounded n times in the years is:

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

Here A = $2,976.45 the final amount

t = 11 years

P = $1,963.45.

and n = 4 because the interest is compounded quarterly or 4 times in the year.

We the need to simplify the interest rate r in (1) so:

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

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

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

(A/P)^(1/nt) - 1 = r/n

((A/P)^(1/nt) - 1 ) * n = r

r = ((A/P)^(1/nt) - 1 ) * n

We replace values and solve doing parenthesis exponents first

r = ((2,976.45/1,963.45.)^(1/(4*11)) - 1 ) * 4

r= ((1.51^(1/44)) -1)*4

((1.51^(1/44)) -1)*4

r= (1.009499443 - 1)*4

r= 0.009499443 *4 = 0.037997773

That expressed in percentage multiplying by 100 is 3.79% or 3.8% interest