$S_{A_{1} B C_{1}}=1,2 S_{A_{1} B C}=(1,2 \cdot 0,85) \cdot S_{A B C}=1,02 S_{A B C}$.

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Answer

$102 \%$.
$[$ The ratio of the areas of triangles with a common base or common height $]$

On side $A C$ of triangle $A B C$, a point $A_{1}$ is taken, and on the extension of side $B C$ beyond point $C$, a point $C_{1}$ is taken, the length of segment $A_{1} C$ is $85 \%$ of the length of side $A C$, and the length of segment $B C_{1}$ is $120 \%$ of the length of side $B C$. What percentage of the area of triangle $A B C$ is the area of triangle $A_{1} B C_{1}$?