CSAT Solved Papers/ 2021/Q18

2021 CSAT — Q18

Quant Logical & quantitative reasoning 2.5 marks Medium

A bank employee drives 1010 km towards South from her house and turns to her left and drives another 2020 km. She again turns left and drives 4040 km, then she turns to her right and drives for another 55 km. She again turns to her right and drives another 3030 km to reach her bank where she works. What is the shortest distance between her bank and her house?

  1. A 20 km
  2. B 25 km Answer
  3. C 30 km
  4. D 35 km

Worked rationale

Track coordinates from the house at (0,0)(0,0) with East =+x=+x, North =+y=+y. A left turn rotates the heading 90∘90^\circ anticlockwise, a right turn 90∘90^\circ clockwise.

  • Start facing South, drive 1010: (0,−10)(0,-10).
  • Left (South →\to East), drive 2020: (20,−10)(20,-10).
  • Left (East →\to North), drive 4040: (20,30)(20,30).
  • Right (North →\to East), drive 55: (25,30)(25,30).
  • Right (East →\to South), drive 3030: (25,0)(25,0).

The bank is at (25,0)(25,0), the house at (0,0)(0,0). They share the same yy, so the shortest (straight-line) distance is

(25−0)2+(0−30+30)2=252=25 km.\sqrt{(25-0)^2+(0-30+30)^2}=\sqrt{25^2}=25 \text{ km}.

Answer: (b) 25 km.

Visual solution

The same solve, worked by hand — read it, then trace it.

Hand-drawn worked solution for UPSC 2021 CSAT Q18 — Logical & quantitative reasoning
Tap the drawing to open it full size for the fine detail.

Why the other options miss

  • A
    drops a leg: loses the +5+5 km eastward leg, leaving the east offset at 2020.
  • C
    turns confused: mixes up turn directions so the vertical legs (1010 S, 4040 N, 3030 S) fail to cancel, leaving a spurious north/south gap.
  • D
    sums instead of resolving: adds east legs 20+5+…20+5+\ldots or naively sums perpendicular legs instead of taking net displacement.

Specialist insight

A turn-by-turn coordinate ledger beats mental visualisation every time on direction problems — write the heading after each turn and update one coordinate per leg. Here the elegance is that the north/south legs cancel exactly: −10+40−30=0-10+40-30 = 0, so the displacement is purely east, 20+5=2520+5 = 25. When the vertical net is 00, no Pythagoras is even needed — the answer is just the horizontal sum.

The trap, in one line

North/south legs cancel (−10+40−30=0)(-10+40-30=0); east legs sum 20+5=25⇒20+5=25 \Rightarrow (b).

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