CAT 2023 Slot 2 Solved Paper

It is given that $\mathit{A}\mathit{B}=9\mathrm{cm},\mathit{B}\mathit{C}=6\mathrm{cm}$.
It is also known that the areas of the figures $\mathit{A}\mathit{B}\mathit{P},\mathit{A}\mathit{P}\mathit{Q}$, and $\mathit{A}\mathit{Q}\mathit{C}\mathit{D}$ are in geometric progression.
Hence, the area of the ABP, APQ, and AQCD are $\mathit{k},2\mathit{k}$, and $4\mathit{k}$ respectively.
The ratio of $\mathit{B}\mathit{P},\mathit{P}\mathit{Q},\mathit{Q}\mathit{C}$ will be the ratio of the respective triangles. Hence, we can draw a line from point $\mathit{A}$ to point C.Let the area of triangle $\mathit{A}\mathit{Q}\mathit{C}$ be $\mathit{x}$, which implies the area of triangle $\mathit{A}\mathit{D}\mathit{C}=\mathit{A}\mathit{D}\mathit{Q}\mathit{C}-\mathit{A}\mathit{Q}\mathit{C}=4\mathit{k}-\mathit{x}$, which is equal to the sum of the area of triangle APB, AQP, and ACQ, respectively.
Therefore, $4\mathrm{k}-\mathrm{x}=3\mathrm{k}+\mathrm{x}$
$\Rightarrow \mathrm{x}=\frac{\mathrm{k}}{2}$
Hence the ratio of $\mathit{B}\mathit{P}:\mathit{P}\mathit{Q}:\mathit{C}\mathit{Q}=\mathit{k}:2\mathit{k}:\frac{\mathit{k}}{2}=2:4:1$