IIM-K (BMSAT) 2025 (QA) PYQs

ABCD is a quadrilateral whose diagonals AC and BD intersect at O. If triangles AOB and COD have areas 4 and 9 respectively, then the minimum area that ABCD can have is

If one angle of a parallelogram is $24^{\circ}$ less than twice its adjacent angle, then the largest angle of the parallelogram is

Points $P$, $Q$, $R$, and $S$ are taken on sides $AB$, $BC$, $CD$, and $DA$ of square $ABCD$ respectively, so that $\frac{AP}{PB} = \frac{BQ}{QC} = \frac{CR}{RD} = \frac{DS}{SA} = \frac{1}{n}$. Then the ratio of the area of $PQRS$ to the area of $ABCD$ is