Let $B$, $G$ be the original numbers of boys and girls. Original average:

$

\frac{68B + 80G}{B+G} = 74 ;\Longrightarrow; 68B + 80G = 74B + 74G ;\Longrightarrow; 6G = 6B ;\Longrightarrow; B=G.

$

Let $n$ new boys (average $62$) join. New overall average:

$

\frac{74(B+G) + 62n}{B+G+n} = 71 ;\Longrightarrow; 74\cdot 2B + 62n = 71(2B+n) ;\Longrightarrow; 148B + 62n = 142B + 71n.

$

$6B = 9n \;\Longrightarrow\; B/n = 9/6 = 3/2.$

Original boys : new boys $= 3:2$.

Why this is CAT-level: The two-stage weighted-average computation forces students to derive $B=G$ as an interim result, then use it. A direct alligation attempt on the new group vs the original mean (gap $74-62=12$, gap $74-71=3$) gives ratio of (original total) : (new) $= 12:3 = 4:1$, i.e., $2B:n = 4:1 \Rightarrow B = 2n$ -- a tempting trap that ignores the original boy-girl split. Careful weighted averaging recovers $B = 1.5n$, i.e., $3:2$.

Answer: (B) $3:2$.