Ans: A
Note that according to the stem-and-leaf diagram, $0 \le a, b \le 9$.
Note that according to the stem-and-leaf diagram, $0 \le a, b \le 9$.
Note that the lower and upper quartiles are $30+a$ and $60+b$ respectively.
Since the inter-quartile range of the distribution is at most $25$, then we have
$\begin{array}{rcl}
(60 + b) – (30 + a) & \le & 25 \\
b – a & \le & -5 \\
a – b & \ge & 5
\end{array}$
If $a = 9$, then $0 \le b \le 4$.
Since the minimum value of $b$ is $0$, then the minimum value of $a$ is $5$.
Therefore, $5 \le a \le 9$ and $0 \le b \le 4$.
Hence, I and II must be true.
III may not be true. If we takes $a = 9$ and $b = 0$, then $a-b = 9 \ge 6$.
