2024 CSAT — Q56

Don’t guess the start direction — express the net displacement as a vector in terms of the unknown start direction, then force it to point North-East.

Let the initial heading be the unit vector $D$, with “turn left” $=$ a $90^\circ$ counter-clockwise rotation $R$. The four legs are walked in headings $D,\ R(D),\ R^2(D),\ R(D)$ (the final right turn from heading $R^2(D)$ returns to $R(D)$):

using $R^2(D) = -D$. So $\text{net} = 200\big(R(D) - D\big)$, which must point North-East, i.e. have equal positive North and East components.

Write $D = (d_x, d_y)$; then $R(D) = (-d_y, d_x)$ and $R(D) - D = (-d_y - d_x,\ d_x - d_y)$. Setting the two components equal: $-d_y - d_x = d_x - d_y \Rightarrow d_x = 0$. The components then become $(-d_y, -d_y)$, both positive iff $d_y < 0$ — that is, $D$ points South.

Check with $D = (0,-1)$ (South): legs are $(0,-100)$, then East $(100,0)$, then North $(0,300)$, then East $(100,0)$; sum $= (200, 200)$ — exactly North-East. ✓

Answer: (c) South.

The whole item collapses to one identity: $\text{net} = 200\big(R(D) - D\big)$. Once you see that the two $100$ m legs along $R(D)$ add while the outbound $D$ and the $300$ m along $-D$ leave a net $-200\,D$, you never have to draw a careful path — you solve a two-line vector equation. The “equal North and East components” condition for North-East is the lever: it forces $d_x = 0$ and the sign of $d_y$ fixes South. Under the clock, set the office direction as the target vector and back-solve the start; it is faster and trap-proof than tracing arrows.