Directions: In the following question, the symbols #, &, @ and $ are used with…

2017

Directions: In the following question, the symbols #, &, @ and $ are used with the meanings illustrated below. Study the information and answer the question. (The directions given indicate exact directions.)

  • P#Q → Q is to the South of P.

  • P@Q → Q is to the North of P.

  • P&Q → Q is to the East of P, at a distance of either 12 m or 6 m.

  • P$Q → Q is to the West of P, at a distance of either 15 m or 3 m.

  • P#&Q → Q (or P) is to the South-East of P (or Q).

  • P@&Q → P (or Q) is to the North-East of Q (or P).

If A&B#&C$D&E@F are related to each other such that D@&B, then what is the probable direction of A with respect to F?

  1. A.

    North-west

  2. B.

    West

  3. C.

    South-west

  4. D.

    East

  5. E.

    Can't be determined

Attempted by 1 students.

Show answer & explanation

Correct answer: E

Concept

In a direction-and-distance puzzle, a point's position is fixed only when both its bearing and its distance from a reference point are fixed. A relationship that gives a diagonal bearing (South-East, North-East) carries no magnitude, and a relationship that allows more than one distance leaves the point free along a range. If the net horizontal shift between two end points can come out positive in one valid layout and negative in another, and likewise for the vertical shift, then their relative direction is not unique — it cannot be determined.

Applying it to this chain

Take East as +x and North as +y, place A at (0, 0), and read A&B#&C$D&E@F link by link:

  1. A&B → B is East of A by 6 m or 12 m, so B = (a, 0) with a ∈ {6, 12}.

  2. B#&C → C is South-East of B by free amounts, so C = (a + s, −t) with s > 0, t > 0.

  3. C$D → D is West of C by 3 m or 15 m, so D = (a + s − w, −t) with w ∈ {3, 15}.

  4. D&E → E is East of D by 6 m or 12 m, so E = (a + s − w + e, −t) with e ∈ {6, 12}.

  5. E@F → F is North of E by a free amount r, so F = (a + s − w + e, −t + r).

Now impose D@&B: B must be North-East of D, i.e. B.x > D.x and B.y > D.y. The first gives a > a + s − w, i.e. w > s; the second gives 0 > −t, which always holds. So the only extra restriction is that the West step w exceeds the South-East horizontal shift s — every other quantity stays free.

Two valid layouts that disagree

Both layouts below satisfy w > s (so D@&B holds), yet they put A on opposite sides of F:

  • Layout 1: a = 6, s = 1, t = 1, w = 3, e = 6, r = 5 → F = (10, 4). A = (0, 0) is West of F and South of F, i.e. A lies South-West of F.

  • Layout 2: a = 6, s = 1, t = 5, w = 15, e = 6, r = 1 → F = (−2, −4). A = (0, 0) is East of F and North of F, i.e. A lies North-East of F.

Cross-check

The horizontal offset of F is a + s − w + e: with the West step as large as 15 m it can overshoot the eastward links and turn negative, but with the West step only 3 m it stays positive — so the East/West sense flips. The vertical offset is r − t, and since both the northward rise r and the South-East drop t are free, the North/South sense flips too. Because A switches from South-West of F to North-East of F across equally valid layouts, no single bearing holds for every case, so the probable direction of A with respect to F is Can't be determined.

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