jmc

First, note that $P<Q<0<R<S<T$. Thus, $P$ and $Q$ are negative, while $R$, $S$, and $T$ are positive. Option $A$ is the difference of two numbers. If the first number is lower than the second number, the answer will be negative. In this case, $P<Q$, so $P-Q<0$, and $P-Q$ is thus indeed a negative number. So option $\boxed{P-Q}$ is correct. Option $B$ is the product of two negatives, which is always positive. Option $C$ starts with the quotient of a positive and a negative. This is negative. But this negative quotient is then multiplied by a negative number, which gives a positive number. Option $D$ has the product of two negatives in the denominator. Thus, the denominator is positive. Additionally, the numerator is (barely) positive. A positive over a positive will be positive. Option $E$ has the sum of two positives in the numerator. That will be positive. The denominator is also positive. Again, the quotient of two positives is positive.