The scores of a large group in an aptitude test have a first quartile of 40, a median of 50 and a…

Step 1. Before the conversion. $\text{IQR} = 62 - 40 = 22$. The upper fence is $62 + 1.5 \times 22 = 95$. The highest score, $98$, is above $95$, so it is an outlier.

Step 2. The new quartiles. $Q_1' = 1.5 \times 40 + 5 = 65$ and $Q_3' = 1.5 \times 62 + 5 = 98$. The new IQR is $98 - 65 = 33 = 1.5 \times 22$.

Step 3. The new fence. $98 + 1.5 \times 33 = 147.5$.

Step 4. The highest score after. $1.5 \times 98 + 5 = 152$, which is above $147.5$. It is still an outlier.

Why. A conversion $y = ax + b$ with $a > 0$ keeps the order and keeps every distance in the same proportion, so the fences move with the data and an outlier stays an outlier.