2015-I-10

$\begin{array}{rcll}
10600 & = & k_1 + k_2(10) & \\
k_1 + 10k_2 & = & 10600 & \ldots \unicode{x2460}
\end{array}$

When $n=6$, $S=9\ 000$, we have

$\begin{array}{rcll}
9000 & = & k_1 + k_2(6) &\\
k_1 + 6k_2 & = & 9000 & \ldots \unicode{x2461}
\end{array}$

$\unicode{x2460} – \unicode{x2461}$, we have

$\begin{array}{rcl}
4k_2 & = & 1600 \\
k_2 & = & 400
\end{array}$

Sub. $k_2=400$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
k_1 + 10(400) & = & 10600 \\
k_1 & = & 6600
\end{array}$

Therefore, $S=6600+400n$. For $n=20$,

$\begin{array}{rcl}
S & = & 6600 + 400(20) \\
S & = & 14600
\end{array}$

Hence, her income in that month is $\$14\ 600$.

$\begin{array}{rcl}
18000 & = & 6600 + 400 n \\
400n & = & 11400 \\
n & = & 28.5
\end{array}$

Note that $n$ must be an integer. Hence, it is impossible that her income in a month is $\$18\ 000$.