Anathi will trade in the old grain tank to offset
some of the cost of the new one. Therefore, we
need to determine the current value of the old
grain tank:

\begin{align*}
A & = P(1 - i)^{n} \\
& = \text{196 000}(1 - \text{0,121}
)^{\text{14}} \\
& = \text{196 000}( \text{0,16437} \ldots ) \\
&= \text{32 218,12946} \ldots
\end{align*}

The value of the old grain tank for the trade in is
\(\text{R}\,\text{32 218,13}\).

Therefore, the loan amount for the new tank is:

\[\text{R}\,\text{219 450} - \text{R}\,\text{32 218,13}
= \text{R}\,\text{187 231,87}\]

Determine which interest rate to use: the loan amount
is more than \(\text{R}\,\text{170 000,00}\),
therefore Anathi gets the lower interest rate of
\(\text{9,31}\%\).

Calculate how much the loan will be worth at the end
of the grace period:

\begin{align*}
A & = P(1 + i)^{n} \\
& = \text{187 231,87} \left( 1 +
\frac{\text{0,0931}}{12} \right)^{\text{6}} \\
&= \text{196 118,31962} \ldots
\end{align*}

After the first six months of the loan period, the
amount she owes increases to
\(\text{R}\,\text{196 118,32}\).

Now we can use the present value formula to solve for
the value of \(x\). Remember that the time
period for this calculation is \(\text{29,5}\)
years.

\begin{align*}
\text{196 118,32} & = \frac{x \left[ 1 -
\left(1 + \frac{\text{0,0931}}{12}\right)
^{-(\text{29,5} \times 12)} \right]}
{\left(\frac{\text{0,0931}}{12} \right)} \\
x & = \frac{\text{196 118,32} \times
\frac{\text{0,0931}}{12} } { \left[ 1 - \left(1 +
\frac{\text{0,0931}}{12}\right) ^{-(\text{29,5} \times
12)} \right]} \\
x &= \text{1 627,0471} \ldots
\end{align*}

Therefore, Anathi must pay \(\text{R}\,\text{1
627,05}\) each month.