Solve 120r = 10.1(1 - 1/(1 + r)^12) for r. This is the rate on a loan of 120 repaid by 12 payments of 10.1.

Step-by-step solution

1. 1

   Recognise the annuity present-value equation. Dividing both sides by $r$ recovers the standard form

   $120=10.1\cdot\frac{1-(1+r)^{-12}}{r},$

   i.e. principal $=$ payment $\times$ annuity factor with $n=12$. Identifying the finance behind the algebra immediately supplies a sanity check the pure algebra does not.

2. 2

   Estimate the answer before computing it. Total repayments are $12\times10.1=121.20$ against a principal of $120$, so the borrower pays just $1.20$ in interest over twelve periods — roughly $1\%$ of the principal spread across a whole year of periods. The rate must therefore be a small fraction of $1\%$ per period, not close to $1\%$. Any answer near $0.9\%$ can be rejected on sight: at $r=0.00893$ the annuity factor prices the loan at only $114.45$, not $120$.

3. 3

   Set up the root-finding problem. Define

   $f(r)=120r-10.1\left(1-(1+r)^{-12}\right).$

   Note $f(0)=0$ is a trivial root — at zero interest both sides vanish — so the meaningful root is the other sign change. Evaluating: $f(0.001)=-0.000416$ and $f(0.005)=+0.013244$, so a root lies in $(0.001,0.005)$.

4. 4

   Bisect to convergence. Halving that bracket repeatedly gives

   $r=0.0015341499\ldots\approx0.0015341,$

   that is $0.15341\%$ per period. A sweep across $(-0.5,1)$ finds no further sign changes, so this is the only economically meaningful solution.

5. 5

   Verify through the annuity factor. With $r=0.0015341499$,

   $\frac{1-(1+r)^{-12}}{r}=11.8811881,\qquad 10.1\times11.8811881=120.000000 ✓,$

   and indeed $\frac{120}{10.1}=11.8811881$ ✓. If the periods are months, the nominal annual rate is $12r\approx1.84\%$ and the effective annual rate is $(1+r)^{12}-1\approx1.857\%$.