If $$pq \lt 5$$, is $$p \lt 1$$?
>(1) $$p \lt 5$$
>(2) $$q \gt 5$$

Incorrect.
[[snippet]]
According to Stat. (1), $$p \lt 5$$ and $$pq \lt 5$$. Thus, $$p$$ can be 0, 1, 2, 3, or 4, and you will be able to find a $$q$$ that matches it. No definite answer, so **Stat.(1) → IS → BCE**.

Correct.
[[snippet]]
According to Stat. (1), $$p \lt 5$$ and $$pq \lt 5$$. Thus, $$p$$ can be 0, 1, 2, 3, or 4, and you will be able to find a $$q$$ that matches it. No definite answer, so **Stat.(1) → IS → BCE**.
According to Stat. (2), $$q \gt 5$$ and $$pq \lt 5$$.
* If $$q=6$$, then $$6p \lt 5$$, so $$p \lt \frac{5}{6}$$.
* If $$q=8$$, then $$8p \lt 5$$, so $$p \lt \frac{5}{8}$$.
* If $$q=10$$, then $$10p \lt 5$$, so $$p \lt \frac{1}{2}$$.
* If $$q=20$$, then $$20p \lt 5$$, so $$p \lt \frac{1}{4}$$.
It seems that for all values of $$q \gt 5$$, $$p \lt 1$$. That's a definite "Yes," which is sufficient, so **Stat.(2) → S → B**.
__Alternative method__:
Since $$q \gt 5$$, $$q$$ is definitely positive. Thus, you can divide the inequality $$pq \lt 5$$ by $$q$$, without changing the sign, getting $$p \lt \frac{5}{q}$$. Since $$q \gt 5$$, it follows that $$p$$ is smaller than $$\frac{5}{\text{greater than 5}}$$. Therefore, $$p$$ is smaller than a fraction, which means that $$p < 1$$.

Incorrect.
[[snippet]]
According to Stat. (2), $$q \gt 5$$ and $$pq \lt 5$$.
* If $$q=6$$, then $$6p \lt 5$$, so $$p \lt \frac{5}{6}$$.
* If $$q=8$$, then $$8p \lt 5$$, so $$p \lt \frac{5}{8}$$.
* If $$q=10$$, then $$10p \lt 5$$, so $$p \lt \frac{1}{2}$$.
* If $$q=20$$, then $$20p \lt 5$$, so $$p \lt \frac{1}{4}$$.
It seems that for all values of $$q \gt 5$$, $$p \lt 1$$. That's a definite "Yes," which is sufficient, so **Stat.(2) → S → BD**.

Incorrect.
[[snippet]]
According to Stat. (1), $$p \lt 5$$ and $$pq \lt 5$$. Thus, $$p$$ can be 0, 1, 2, 3, or 4, and you will be able to find a $$q$$ that matches it. No definite answer, so **Stat.(1) → IS → BCE**.

Incorrect.
[[snippet]]
According to Stat. (2), $$q \gt 5$$ and $$pq \lt 5$$.
* If $$q=6$$, then $$6p \lt 5$$, so $$p \lt \frac{5}{6}$$.
* If $$q=8$$, then $$8p \lt 5$$, so $$p \lt \frac{5}{8}$$.
* If $$q=10$$, then $$10p \lt 5$$, so $$p \lt \frac{1}{2}$$.
* If $$q=20$$, then $$20p \lt 5$$, so $$p \lt \frac{1}{4}$$.
It seems that for all values of $$q \gt 5$$, $$p \lt 1$$. That's a definite "Yes," which is sufficient, so **Stat.(2) → S → BD**.

Statement (1) ALONE is sufficient, but Statement (2) alone is not sufficient to answer the question asked.

Statement (2) ALONE is sufficient, but Statement (1) alone is not sufficient to answer the question asked.

BOTH Statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.

EACH statement ALONE is sufficient to answer the question asked.

Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.