Consider two decision problems $Q_1, Q_2$ such that $Q_1$ reduces in polynomial time to 3-SAT and 3-SAT reduces in polynomial time to $Q_2$. Then which one of the following is consistent with the above statement?

We are given two polynomial-time reductions:

1. $Q_1 \le_p \text{3-SAT}$

2. $\text{3-SAT} \le_p Q_2$

Recall that:

• $\text{3-SAT}$ is an $\text{NP}$-complete problem (hence it is both in $\text{NP}$ and $\text{NP}$-hard).

• If a problem $A$ reduces in polynomial time to a problem $B \in \text{NP}$ ($A \le_p B$), then $A \in \text{NP}$. Since $\text{3-SAT} \in \text{NP}$, it follows that $Q_1 \in \text{NP}$.

• If a known $\text{NP}$-hard (or $\text{NP}$-complete) problem reduces in polynomial time to a problem $C$ ($\text{3-SAT} \le_p C$), then $C$ is $\text{NP}$-hard. Hence, $Q_2$ is $\text{NP}$-hard.

Thus, $Q_1$ is in $\text{NP}$, and $Q_2$ is $\text{NP}$-hard.

Therefore, A is the correct option.