CAT 2025 Slot 2 · Quant · Question 16

Let k be the common root, \alpha the other root of the first equation and \beta the other root of the second.

Sum of roots gives k+\alpha =\frac{5}{3} for the first and k+\beta =1 for the second.

Adding, \alpha +\beta =\frac{5}{3}+1-2k=\frac{8}{3}-2k, so everything now hinges on k.

Both equations hold at x=k: 3{k}^{2}-5k+p=0 and 2{k}^{2}-2k+q=0.

Multiplying the first by 2 and the second by 3 to match the {k}^{2} terms: 6{k}^{2}-10k+2p=0 and 6{k}^{2}-6k+3q=0.

Subtracting, -4k+2p-3q=0, so k=\frac{2p-3q}{4}.

Substituting back, \alpha +\beta =\frac{8}{3}-\frac{2p-3q}{2}=\frac{8}{3}-p+\frac{3}{2}q.

Shortcut. The constant \frac{8}{3} comes from the two sums of roots alone, so it is fixed before any work on k. Only option 2 and option 4 start with \frac{8}{3}, and a common root must make the answer decrease with p (larger p pushes k up, and \alpha +\beta =\frac{8}{3}-2k), so the -p version is the one.