Direction Sense for Punjab Police SI

Answer: 2 — 30 km

Take the house as the origin, east as +x and north as +y, and add the legs as coordinates. ```puzzle:direction start: A A > E 30 > B B > N 30 > C C > W 60 > D D > S 40 > F F > E 30 > G G > S 20 > H ``` East–west: +30 − 60 + 30 = 0, so he ends on the same north–south line as his house. North–south: +30 − 40 − 20 = −30, i.e. 30 km due south of the start. With no sideways offset there is no Pythagoras to do — the straight-line distance is simply 30 km. Option 3 (32 km) comes from mis-adding the southward legs, and options 1 and 4 drop one leg altogether.

Answer: 2 — 20 m

Here the directions are named outright, so just add them as signed quantities. Take the office at (0, 0), east as +x and north as +y. Leg  ·  Direction  ·  Distance  ·  Position after 1  ·  South  ·  30 m  ·  (0, −30) 2  ·  East  ·  40 m  ·  (40, −30) 3  ·  North  ·  30 m  ·  (40, 0) 4  ·  West  ·  20 m  ·  (20, 0) North-south: 30 m down then 30 m up, net 0. East-west: 40 m east then 20 m west, net 20 m east. So he ends at (20, 0) and the distance from the office is 20 m, due east of it. The tempting wrong answer is 30 m, picked by candidates who notice the two 30 m legs and assume one of them survives; in fact they cancel exactly, and the answer comes only from 40 − 20. There is no right-angled triangle to solve here, which is why no square root appears.

Answer: 2 — North-West

Track the facing and the coordinates, with the start at (0, 0), east as +x and north as +y. Leg  ·  Facing  ·  Distance  ·  Position after 1  ·  North  ·  50 km  ·  (0, 50) 2 (right of north = east)  ·  East  ·  24 km  ·  (24, 50) 3 (left of east = north)  ·  North  ·  80 km  ·  (24, 130) 4 (left of north = west)  ·  West  ·  20 km  ·  (4, 130) 5 (stated)  ·  South  ·  150 km  ·  (4, −20) He finishes at (4, −20). Note the question asks the reverse of the usual: the direction of the start as seen from the finish. Start − finish = (0 − 4, 0 − (−20)) = (−4, +20): 4 km to the west and 20 km to the north. West and north together give North-West. The classic error is to answer South-East, which is the direction of the finish from the start — the exact opposite. Reading the last sentence carefully is the whole difficulty of this item.

Answer: 2 — 8 km towards south

Take A at (0, 0) with east as +x and north as +y, and track the heading after each turn. - East 12 km → (12, 0), facing east. - Left turn faces north; 4 km → (12, 4). - Left turn faces west; 8 km → (4, 4). - Right turn faces north; 4 km → (4, 8). - Left turn faces west; 4 km → (0, 8) = B. B is at (0, 8) and A is at (0, 0), so B is exactly 8 km due north of A. To get back he must walk 8 km towards the south. Option 4 (6 km towards west) tempts anyone who nets the east–west legs wrongly (12 − 8 − 4 = 0, not 6).

Answer: 1 — 22 km

Put the office at the origin and let Arun walk north 10 km, so Varun walks south 12 km. Arun, facing north, turns right, which is east, and covers 2 km — he is now at (2 east, 10 north). Varun, facing south, turns left, which is also east, and covers 2 km — he is now at (2 east, 12 south). Both men have shifted east by the same 2 km, so those legs cancel. The gap between them is purely north–south: 10 + 12 = 22 km. Option 2 (26 km) comes from wrongly adding both 2 km legs to the total, and option 4 (24 km) from adding only one.

Answer: 4 — South-West

Take the start as origin, with north up and east to the right. - He walks 5 km south — he is 5 km south of the start, facing south. - Facing south, a right turn makes him face west; 3 km west takes him 3 km west of the line he came down. - Facing west, a left turn makes him face south again; another 5 km south. Net displacement = 10 km south and 3 km west of the start. A point that is both south and west of the origin lies in the south-west direction. Option 2 (South) would need the two side moves to cancel, which they do not — there is only one westward leg.

Answer: 4 — 12 m

Put the start at the origin and track the turns. Facing south, a left turn faces east; from east a left turn faces north; from north a left turn faces west; from west a left turn faces south again. Leg 1: 18 m south, to (0, −18). Leg 2: 18 m east, to (18, −18). Leg 3: 28 m north, to (18, 10). Leg 4: 30 m west, to (−12, 10). Leg 5: 10 m south, to (−12, 0). The finish is level with the start, 12 m to its west, so the distance is 12 m. Note the north-south legs cancel exactly (18 + 10 = 28), which is why the answer is a plain east-west gap and no square root is needed. Option 3 (10 m) is the trap - it is the length of the last leg, not the displacement.

Answer: 4 — 15 m

Put the start at (0, 0), with east as +x and north as +y, and track the facing after every turn. Leg  ·  Facing  ·  Distance  ·  Position after Start  ·  —  ·  —  ·  (0, 0) 1  ·  North  ·  12 m  ·  (0, 12) 2 (right of north = east)  ·  East  ·  15 m  ·  (15, 12) 3 (left of east = north)  ·  North  ·  15 m  ·  (15, 27) 4 (left of north = west)  ·  West  ·  30 m  ·  (−15, 27) 5 (left of west = south)  ·  South  ·  27 m  ·  (−15, 0) He finishes at (−15, 0). The north-south movements cancel exactly: 12 + 15 = 27 up and 27 down. Left-right he is 15 m east then 30 m west, a net 15 m west. Distance from start = sqrt((−15)^2 + 0^2) = 15 m. The common error is to forget that the last two left turns reverse him completely and to add 15 and 30 as if they were in the same direction, giving 45, or to take the diagonal and get 25 or 20 — but with the vertical displacement zero, no diagonal arises at all.

Answer: 1 — West

Track her heading and her coordinates, taking east as +x and north as +y, starting at (0, 0). - 6 m north → (0, 6), facing north. - Left turn faces west; 3 m → (−3, 6). - Left turn faces south; 2 m → (−3, 4). - Right turn faces west; 4 m → (−7, 4). - Left turn faces south; 4 m → (−7, 0). The north–south movements cancel exactly (6 up, 2 + 4 = 6 down), leaving her 7 m due west of the start. Option 4 (South) is the trap for anyone who stops at the last leg's heading instead of comparing positions.

Answer: 2 — 50 metres

Take AB as north, so he is at B facing north. Turn right = east, 50 m. Turn left = north, 50 m. Turn left = west, 50 m. The 50 m east and the 50 m west cancel, leaving only 50 m north of B. So his distance from B is 50 m. Option 4 (200 m) is the A-to-B leg, not the distance from B, and 100 m would need the two 50 m north-south legs to add instead of the east-west pair cancelling.

Answer: 2 — 1 km

Put A at (0, 0) and let the first leg go north. A to B: 3 km north, so B = (0, 3). Right turn faces east; 5 km to C = (5, 3). Right turn faces south; 10 km to D = (5, −7). Right turn faces west; 5 km to E = (0, −7). Right turn faces north; 6 km ends at (0, −1). The east and west legs are both 5 km, so he returns to A's line, and along that line he went 3 + (−10) + 6 = −1, i.e. 1 km short of A. The distance from A is therefore 1 km. Option 3 (0 km) would need the last leg to be 7 km, not 6.

Answer: 3 — 20 km

Put the house at (0, 0), east positive along x and north positive along y. 6 km west → (−6, 0). Facing west, a left turn is south; 14 km south → (−6, −14). 26 km east → (20, −14). Facing east, a left turn is north; 14 km north → (20, 0). The two 14 km legs cancel, so he ends on the same east-west line as the house, at x = 20. Straight-line distance = 20 km. 26 km and 32 km tempt you into adding the eastward legs without subtracting the first 6 km westward.

Answer: 3 — 6 km

Track the headings. Facing south, a right turn points west; the next right turn points north; the next points east. Leg 1: 20 km south. Leg 2: 19 km west. Leg 3: 20 km north — this cancels leg 1 exactly. Leg 4: 13 km east. North-south displacement: 20 − 20 = 0. East-west displacement: 19 west − 13 east = 6 km west. Since the north-south part cancels, the straight-line distance is just 6 km (no Pythagoras needed).

Answer: 3 — 5 km

Facing east, a left turn points north; a second left turn points west. Leg 1: 5 km east. Leg 2: 5 km north. Leg 3: 5 km west — this cancels leg 1. East-west displacement: 5 − 5 = 0. North-south displacement: 5 km north. So he finishes due north of the start, 5 km away. The decoy 0 km would need a fourth leg southwards to close the square.

Answer: 1 — 2 km, south

Track the two axes separately. Facing east, a left turn gives north, the next left gives west, the last left gives south. East-west: 2 km east then 2 km west cancel, so E lies on the same north-south line as A. North-south: 5 km north then 3 km south leaves a net 2 km north. So E is 2 km due north of A, and Rita must drive 2 km south to return. Option 2 (2 km, north) is the displacement of E from A, not the correction needed, which is the usual trap.

Answer: 1 — West

Put the start at (0, 0) with north as +y and east as +x. Facing north, 3 m takes her to (0, 3). A left turn faces her west; 7 m takes her to (−7, 3). A left turn faces her south; 7 m takes her to (−7, −4). A left turn faces her east; 4 m takes her to (−3, −4). A left turn faces her north; 4 m takes her to (−3, 0). She ends level with the start in the north–south direction and 3 m to its left, i.e. due West. Option 2 (South) comes from stopping one leg early at (−3, −4).

Answer: 1 — 120 m

Track the facings. Walking west, a left turn faces south: 60 m south. Facing south, a right turn faces west: 40 m west. Facing west, a right turn faces north: 60 m north. The 60 m south and 60 m north cancel, so he ends up purely west of the start by 80 + 40 = 120 m. Because nothing is left over in the north-south direction there is no Pythagoras step, which is where 100 m (option 2) comes from.

Answer: 4 — South

Only the turns matter for this question, not the distances. He starts facing east. A left turn from east faces him north. A right turn from north faces him east again. A left turn from east faces him north. A left turn from north faces him west. A final left turn from west faces him south, and he walks the last 20 m in that direction. So he is moving south. ```puzzle:direction start: R0 R0 > E 20 > R1 R1 > N 20 > R2 R2 > E 20 > R3 R3 > N 20 > R4 R4 > W 40 > R5 R5 > S 20 > R6 ``` Option 3 (East) is the trap for anyone who loses track at the single right turn in the middle and ends up one quarter-turn out.

Answer: 2 — North

Only the turns matter; the distances are there to slow you down. - Start facing West. - Right from West → North (15 m). - Left from North → West (5 m). - Left from West → South (20 m). - Left from South → East (5 m). - Left from East → North. He will now walk north. Option 4 (East) is the trap — it is the direction one step before the last turn.

Answer: 2 — East

Only the turns decide the final facing; the distances are there to distract. Start facing north. A left turn from north puts him facing west. A right turn from west puts him facing north again. A second right turn, now from north, puts him facing east, and that is the direction in which he walks the last 12 m. Option 1 (west) is the facing after the first turn only, and option 3 (north) the facing after the second, so both are answers to an earlier stage of the walk.

Answer: 4 — 10 m

Every turn is to the left, so the facing runs South → East → North → West → South → East. Put the coaching at (0, 0), with east positive on the x-axis and north positive on the y-axis. South 20 m → (0, −20) East 25 m → (25, −20) North 40 m → (25, 20) West 50 m → (−25, 20) South 10 m → (−25, 10) East 25 m → (0, 10) He ends at (0, 10): the east–west movements cancel exactly (25 east, 50 west, 25 east) and the north–south movements leave him 10 m north of the start. The straight-line distance is therefore 10 m. ```puzzle:direction start: P0 P0 > S 20 > P1 P1 > E 25 > P2 P2 > N 40 > P3 P3 > W 50 > P4 P4 > S 10 > P5 P5 > E 25 > P6 ``` Option 2 (70 m) is what you get by adding leg lengths instead of resolving them into directions.

Answer: 4 — 8 m

Each left turn is 90° anticlockwise, so the headings run north, west, south, east. ```puzzle:direction start: A A > N 12 > B B > W 20 > C C > S 20 > D D > E 20 > F ``` With the start as the origin: 12 m north puts him at (0, 12); 20 m west at (−20, 12); 20 m south at (−20, −8); 20 m east at (0, −8). The two 20 m legs west and east cancel, and the two legs along the north–south axis leave 12 − 20 = −8. He is back on the starting line, 8 m due south of the start, so the distance is 8 m. Option 3 (20 m) is the length of one leg rather than the displacement, and option 2 (10 m) would need the first leg to be 10 m.

Answer: 3 — North

A left turn is 90° anticlockwise for the walker. Facing south, a left turn makes him face east. Facing east, another left turn makes him face north. The distances (3 km, 4 km, 1 km) do not affect the facing direction at all — they are there to distract. Option 1 (West) is what you get by turning right twice instead of left.

Answer: 2 — South

Each turn is a quarter turn, so track the facing one step at a time. He starts facing East. A left turn from East faces him North. A right turn from North faces him East again. A second right turn from East faces him South. The distances walked are never given and never matter; only the turns do. The trap is to cancel the left turn against one right turn and stop at East.

Answer: 4 — 16 m

Track the facing after every turn. He starts facing west. Leg  ·  Turn  ·  Facing  ·  Distance  ·  Effect 1  ·  —  ·  West  ·  7 m  ·  7 m west 2  ·  right  ·  North  ·  4 m  ·  4 m north 3  ·  left  ·  West  ·  9 m  ·  9 m west 4  ·  left  ·  South  ·  4 m  ·  4 m south The 4 m north on leg 2 is exactly cancelled by the 4 m south on leg 4, so B ends up on the same east–west line as A. Westward travel totals 7 + 9 = 16 m, so AB = 16 m and no Pythagoras is needed. 13 m is the trap for anyone who forgets the second westward leg is 9 and not 6; 2 m comes from subtracting instead of adding the west legs.

Answer: 3 — 50 metres

Track the turns. Facing north, a left turn points west; facing west, the next left turn points south. So the walk is 20 m north, 40 m west, 50 m south. ```puzzle:direction start: A A > N 20 > B B > W 40 > C C > S 50 > D ``` Taking the start as the origin with north as +y: after 20 m north he is at (0, 20); after 40 m west at (−40, 20); after 50 m south at (−40, −30). He is therefore 40 m west and 30 m south of the start, and the straight-line distance is sqrt(40^2 + 30^2) = sqrt(1600 + 900) = sqrt(2500) = 50 metres — the familiar 3-4-5 triangle. Option 1 (110 m) is the total path walked, 20 + 40 + 50, not the displacement; option 2 (70 m) adds 40 and 30 instead of using Pythagoras.

Answer: 1 — East

Rule: in the morning the sun is in the east, so every shadow is thrown towards the west. Ram's shadow therefore points west. The phrase "in front of Shyam" names the direction Shyam is facing — the same convention as the standard 'shadow fell to the left of X' puzzles — so Shyam faces west. The two men face each other, so Ram faces the opposite of Shyam: east. The trap is to read "in front of Shyam" as a position (the shadow lying between them) and answer west — which is exactly what the candidate marked. What the clue fixes is Shyam's facing, not where he stands.

Answer: 3 — If the data given in both statements, statement I alone and statement II alone are independently sufficient to answer the question

Early in the morning the sun is in the east, so every shadow points west. Statement I: the puppet's shadow (pointing west) lies to Shekhar's right, so west is on his right hand. A person whose right hand points west is facing south. Determined — I alone is sufficient. Statement II: his own shadow also points west. After he turns left the shadow is behind him, so after the turn he faces east; undoing one left turn, he was facing south before it. Either way the direction is fixed. Determined — II alone is sufficient. Both statements are independently sufficient (and they agree on south), so the answer is not 1, 2 or 4 — those all require at least one statement to fail.

Answer: 3 — North

At sunrise the sun is in the east, so every shadow is cast towards the west. Sanjay's shadow lies to Ravi's right, so Ravi's right hand points west. A person whose right hand points west is facing south — check it: face south, and west is on your right. The two men face each other, so Sanjay faces the opposite of south, i.e. north. Option 4 (south) is the trap for anyone who stops at Ravi's direction and forgets to turn it around for Sanjay.

Answer: 3 — South - west

Only three of the five clues touch M and P, so use those. - O is east of M, which is the same as saying M is west of O. - P is north of O, which is the same as saying O is south of P. So starting at P you go south to reach O, then west to reach M. M is therefore both south of P and west of P — south-west. The clues about N, Q and R are padding and change nothing. Option 1 (south-east) is the trap for anyone who reads "O is to the east of M" as putting M east.

Answer: 1 — North-east

Put G at the origin. I is south of G, so I is directly below G. K is south-west of I, so K is further down and to the left of I — that is, below G and west of the G–I line. H is east of G, so H is on G's own level but to the right. Now compare H with K: H is higher up than K (K is below I, and I is below G, while H is level with G), so H is north of K; and H is to the right of the G–I line while K is to the left of it, so H is east of K. North plus east gives north-east. Option 4 (south-east) tempts anyone who reads the statements in the reverse order, i.e. K with respect to H. The fourth statement about J is only padding and changes nothing.

Answer: 4 — North-East

Put P at the origin, with north up and east to the right. P is south of O, so O is due north of P — O is at (0, +a). M is north-east of O, so M is further north and also further east — M is at (+b, +a+c) with b and c positive. Relative to P, M therefore has a positive east component and a positive north component, which is exactly north-east. L and N are decoys: they sit on the west side and never touch the P–O–M chain. South-west and south are impossible because M is north of O, which is already north of P.

Answer: 3 — 16 m

Put the common start at (0, 0), with east positive on the x-axis and north positive on the y-axis. Jack: north 11 m → (0, 11). Facing north, a right turn is east; 7 m east → (7, 11). Bob: west 9 m → (−9, 0). Facing west, a right turn is north; 11 m north → (−9, 11). Both end at the same height, y = 11, so the straight-line gap is purely horizontal: 7 − (−9) = 16 m. Option 4 (20 m) comes from adding 9 + 11 as if the two paths did not overlap in the north-south direction; option 1 (9 m) is just Bob's first leg.