2023 CSAT — Q15

Each statement says “can be” — an existence claim. One witnessing example settles each one true; you do not need it to hold always.

Statement 3 ($pc$ odd): a product is odd only if both factors are odd. Take odd prime $p = 3$ and odd composite $c = 9$: $pc = 27$, odd. Can be true. ✓

Statement 2 ($2p + c$ odd): $2p$ is even, so $2p + c$ is odd $\iff c$ is odd. Take $c = 9$ (odd composite), any prime $p$: $2p + 9$ is odd. Can be true. ✓

Statement 1 ($\frac{p+c}{p-c}$ even): need it to be an even integer for some valid $p,c$. Take $p = 2,\ c = 6$: $\frac{2+6}{2-6} = \frac{8}{-4} = -2$, an even integer. Can be true. ✓ (Another: $p=3,c=9 \Rightarrow 12/(-6) = -2$.)

All three are achievable.

Answer: (d) 1, 2 and 3.

The single recurring slip is assuming “composite $=$ even.” Composites include $9, 15, 21, 25, \dots$, all odd. Once you allow an odd composite, $pc$ odd and $2p+c$ odd both fall out immediately, and Statement 1 needs just one constructed pair ($p=2, c=6 \to -2$). The answer-craft move on “can be” claims is to hunt for one example, not to prove a universal — that flips a hard-looking item into three quick existence checks.