[Bouncing cars, asymmetric start]

Take 9:00~a.m.\ as time $t=0$. Then $Y$ starts at $t=0.75$.

• $X$ reaches $Q$ at $t = D/50$, then turns back.
• $Y$ reaches $P$ at $t = 0.75 + D/40$, then turns back.

For the second meeting to occur with both cars having bounced once each:

$

\text{Position of } X = 2D - 50t,\qquad \text{Position of } Y = 40(t - 0.75) - D = 40t - 30 - D.

$

Setting them equal:

$

2D - 50t = 40t - 30 - D ;\Longrightarrow; 3D + 30 = 90t ;\Longrightarrow; D = 30t - 10.

$

At $t=3$: $D = 80$. Verification at $t=3$: $X$ is at $80 - 50(3-1.6) = 10$; $Y$ is at $40(3-2.75) = 10$. They coincide at $10$ km from $P$. Answer: (B) 80.

Why CAT-level: The standard ``combined distance $= 3D$ at second meeting'' shortcut breaks because $Y$ starts later. Method selection (bouncing analysis vs.\ symmetric shortcut) is the trap.