Direction Sense – MBA CET questions with solutions

Approach: Direction tracking with coordinates (left of South = East, left of East = North).

Solution:
Start at O(0, 0).
25 m South → (0, −25).
Facing South, left turn = East; 20 m → (20, −25).
Facing East, left turn = North; 25 m → (20, 0).
Facing North, right turn = East; 15 m → (35, 0).
Final point is on the same horizontal line as O, 20 + 15 = 35 m away, towards East.

Why the other options are wrong:
B, E: he ends exactly back on the start's East–West line (the 25 m South and 25 m North cancel), so no North-East component.
C, D: 40 / 60 m add distances wrongly; only the East moves (20 + 15) count.

Cetking shortcut: The two 25 m legs (South and North) cancel; just add the East legs 20 + 15 = 35 m East.

Answer: A

Approach: Facing-direction tracking (right of South = West, left of West = South).

Solution:
Walks South → facing South.
Turns right: right of South is West; walks 3 km facing West.
Turns left: left of West is South; walks 5 km.
He is now facing South.

Why the other options are wrong:
A: West was the direction of the middle leg only.
C, D: no diagonal movement is involved.
E: East would need a right turn from South twice.

Cetking shortcut: A right turn followed by a left turn cancels out — he faces his original direction, South.

Answer: B

Approach: Rotated direction pole — find the rotation, then apply it to the pointer followed.

Solution:
The West pointer now points South. Going from West to South is a 90° anticlockwise turn.
So every pointer has turned 90° anticlockwise: North→West, East→North, South→East, West→South.
Harish follows the pointer marked 'East'.
That pointer now actually points North, so he is travelling North.

Why the other options are wrong:
A: South is where the West pointer points, not the East pointer.
C: West is where the North pointer now points.
D: East would mean the pole had not turned.
E: a 90° turn cannot produce a diagonal direction.

Cetking shortcut: Opposite pointers stay opposite: West→South means East→North.

Answer: B

Approach: Direction distance — opposite legs cancel.

Solution:
Start (0, 0) → 6 km North → (0, 6).
8 km West → (−8, 6).
6 km South → (−8, 0).
Distance from start = 8 km (due West).

Why the other options are wrong:
D: 10 km uses √(6² + 8²), ignoring the return 6 km South.
A, B, E: do not match the net displacement.

Cetking shortcut: North 6 and South 6 cancel — only the 8 km West remains.

Answer: C

Approach: Right-angled triangle from perpendicular directions (Pythagoras).

Solution:
B is North of A and C is East of A → angle BAC = 90°.
BC² = AB² + AC² = 5² + 12² = 25 + 144 = 169.
BC = 13 km.

Why the other options are wrong:
A: 60 is the product 5 × 12.
C: 17 is the path B → A → C, not the shortest distance.
D: 7 is the difference.
E: 14 has no basis.

Cetking shortcut: 5-12-13 is a standard Pythagorean triplet.

Answer: B

Approach: Track east-west and north-south separately.

Solution:
West 15 → turn left (facing west, left = south) 20 → east 25 → turn left (facing east, left = north) 20.
North-south: −20 + 20 = 0.
East-west: −15 + 25 = 10 km east.
Distance from house = 10 km.

Why the other options are wrong:
A 5 km: subtracts wrongly.
C 40 km: adds the two 20 km legs.
D 80 km: total distance walked.
E 60 km: no link.

Cetking shortcut: The two 20 km legs cancel; answer = 25 − 15 = 10.

Answer: B

Approach: Place B at the origin.

Solution:
A is south of B → A at (0, −y).
C is east of B → C at (x, 0).
From C, A is to the west and to the south → south-west.

Why the other options are wrong:
A North-east: direction of C from A (reversed).
B North-west, C South-east: one axis wrong.
E West: ignores the south component.

Cetking shortcut: A is left-and-down of C → south-west.

Answer: D

Approach: Place B at the origin.

Solution:
A: 40 m SW → (−a, −a). C: 40 m SE → (a, −a).
Same y-coordinate, C has larger x → C is due east of A.

Why the other options are wrong:
B West: A is west of C, not C of A.
C North-east, E North-west: both points are at the same level.
D South: they differ only east-west.

Cetking shortcut: Equal SW and SE legs are mirror images about the N-S line → the points are level.

Answer: A

Approach: Fix R and place the others.

Solution:
Partners sit opposite: P opposite R, S opposite Q.
R faces west → R's right side is north → S sits on the north side, facing south.
Q sits opposite S (south side) and faces the board → north.

Why the other options are wrong:
B South: that is S's facing direction.
C East: P's facing direction.
D West: R's direction.
E North-east: players face straight across the board.

Cetking shortcut: Facing west, right = north; Q opposite S faces the opposite way to S.

Answer: A

Approach: Let the first northward leg be x.

Solution:
North x → left (west) 1 km → left (south) 2 km.
Final position: (−1, x − 2). It is exactly 1 km west of start → x − 2 = 0.
x = 2 km.

Why the other options are wrong:
A 1 km: would leave him 1 km south-west.
C 3 km, D 5 km, E 11 km: leave him north/south of the start line.

Cetking shortcut: 'Exactly west' means no north-south gap: north leg = south leg = 2 km.

Answer: B

Approach: Track the net East–West and North–South movement.

Solution:
Start facing East: walk 10 m East.
Turn left → North: walk 12 m.
Turn right → East: walk 18 m.
Turn right → South: walk 12 m.
North–South: 12 − 12 = 0.
East: 10 + 18 = 28 m.
He ends 28 m due East of the start.

Cetking shortcut: Equal and opposite legs (12 N, 12 S) cancel — just add the parallel legs.

Answer: B

Approach: Place everyone on a grid with North at the top.

Solution:
All face North, so 'right' = East and 'behind' = South.
B is in the middle row. C is just to the right of B → C is East of B.
D is just behind C → D is South of C (back row).
A is North of B (front row).
From A: D is one column East and two rows South → South-East.

Why the other options are wrong:
A, D, E: D is below A, so no North component.
B: D is to the East of A, not West.

Answer: C

Approach: Find the fixed rotation that maps old directions to new ones.

Solution:
Measure clockwise from North: SE = 135°, N = 0°, NE = 45°, W = 270°.
SE (135°) → N (0°): change = −135° (135° anticlockwise).
NE (45°) → W (270° = −90°): change = −135° ✓.
So every direction turns 135° anticlockwise.
West = 270° → 270° − 135° = 135° = South-East.

Why the other options are wrong:
B (North-East) comes from rotating the wrong way (clockwise 135°).
A, C, D do not match a 135° anticlockwise shift.

Cetking shortcut: Count 45° steps on the compass: SE → N is 3 steps anticlockwise, so move West 3 steps anticlockwise: W → SW → S → SE.

Answer: E

Approach: Only movement along the main road reduces the gap between the cars.

Solution:
Car 1: 50 km along the road → right turn 30 km (away from the road) → left turn 50 km (parallel to the road) → turns back 30 km to rejoin the road.
Progress along the road = 50 + 50 = 100 km; the two 30 km legs cancel.
Car 2 covers 70 km along the road from the other end.
Distance = 300 − 100 − 70 = 130 km.

Why the other options are wrong:
E (150) and others come from counting the 30 km detours or dropping one car's progress.

Cetking shortcut: Detours perpendicular to the line of approach cancel out — add only the parallel legs.

Answer: A

Approach: Convert each turn into a bearing change.

Solution:
Start facing North (0°).
270° anticlockwise = 90° clockwise → facing East.
45° clockwise from East → facing South-East (135°).
The distances walked do not affect the facing direction.

Why the other options are wrong:
B (North-West): turning 270° clockwise instead of anticlockwise gives West, then 45° clockwise gives North-West.
A, C, D: do not follow the two turns.

Cetking shortcut: 270° one way = 90° the other way.

Answer: E

Approach: Track the facing direction after each turn.

Solution:
Walks South → facing South.
Turns right → facing West.
Turns left → facing South.

Cetking shortcut: One right + one left cancel out, so he faces his original direction, South. Distances are irrelevant.

Answer: B

Approach: Net rotation.

Solution:
+90° (clockwise) − 180° (anticlockwise) + 90° (clockwise) = 0°.
Net rotation 0° → still facing North.

Answer: A

Approach: Resolve each diagonal into components.

Solution:
10 m NE = 10/√2 m North + 10/√2 m East.
10 m SE = 10/√2 m South + 10/√2 m East.
North and South cancel; net = 10√2 m due East.

Answer: A

Approach: Net displacement along each axis.

Solution:
North–South: 5 N − 2 S = 3 km North.
East–West: 3 E − 3 W = 0.
Distance from start = 3 km.

Answer: A

Approach: Opposite directions.

Solution:
South-West was exactly behind her, so she was facing the opposite of South-West.
Opposite of South-West = North-East; she pointed straight ahead, i.e. North-East.

Answer: A

Approach: Track the facing direction after each turn.

Solution:
Walks North 10 m.
Turns left → faces West, walks 25 m.
Turns right → faces North, walks 5 m.
Turns right → faces East, walks 25 m.
He is now facing East.

Answer: A

Approach: Coordinates — North +y, East +x.

Solution:
Start (0, 0). North 20 → (0, 20).
Right → East 30 → (30, 20).
Right → South 35 → (30, –15).
Left → East 15 → (45, –15).
Left → North 15 → (45, 0).
She is 45 m East of the start.

Cetking shortcut: Net N–S: 20 – 35 + 15 = 0; net E–W: 30 + 15 = 45 East.

Answer: D

Approach: Direction sense — track only the facing direction through each turn.

Solution:
Start facing North; moves 20 km North.
Turns right ⇒ East (20 km).
Moves 10 km North-East ⇒ now facing North-East.
Turns right (90°) from NE ⇒ South-East (20 km).
Turns right from SE ⇒ South-West (20 km).
Turns left from SW ⇒ South-East (20 km).
Final facing direction: South-East.

Cetking shortcut: Only the turns matter for facing direction — ignore all distances.

Answer: A

Approach: Direction sense with coordinates; shortest distance by Pythagoras.

Solution:
Take office = (0, 0).
8 m South ⇒ park = (0, −8).
Facing South, a left turn = East; 15 m ⇒ traffic signal = (15, −8).
Office to signal: √(15² + 8²) = √(225 + 64) = √289 = 17 m.

Cetking shortcut: 8–15–17 is a Pythagorean triplet — spot it and answer instantly.

Answer: C

Approach: Direction sense — plot all points, then compare park and home.

Solution:
Office (0, 0) → park (0, −8) → signal (15, −8).
At the signal he faces East; 90° anticlockwise ⇒ North; 8 m ⇒ coffee shop (15, 0).
Facing North, left = West; 8 m ⇒ home (7, 0).
Park (0, −8) relative to home (7, 0): 7 m West and 8 m South ⇒ South-West.

Why the other options are wrong:
South/North/East: park is both west and south of home, so a pure direction is wrong. North-West: park is lower (south), not north.

Cetking shortcut: Direction of X w.r.t. Y = look from Y towards X; here both x and y decrease ⇒ South-West.

Answer: A

Approach: Direction tracing with right turns.

Solution:
Start O. 15 km North → (0, 15).
Turn right → East, 15 km → (15, 15).
Turn right → South, 15 km → (15, 0).
Final point is 15 km directly East of O.

Cetking shortcut: Two equal opposite legs (North 15, South 15) cancel – only the East leg remains.

Answer: E

Approach: Track facing direction as a bearing (N = 0°, clockwise positive).

Solution:
Facing West = 270°.
45° clockwise → 315° = North-West.
135° anticlockwise → 315° − 135° = 180° = South.
135° anticlockwise again → 180° − 135° = 45° = North-East.

Cetking shortcut: Net turn = +45 − 135 − 135 = −225° (i.e. 135° clockwise). West + 135° clockwise = North-East.

Answer: B

Approach: Same rotation applied to a new starting direction.

Solution:
South-West (225°) → East (90°): a turn of 135° anticlockwise (equivalently 225° clockwise).
North-East (45°) turned 135° anticlockwise → 45° − 135° = −90° = 270° = West.
(225° clockwise also gives 45° + 225° = 270° = West.)

Cetking shortcut: SW and NE are opposite, so the results are opposite too: opposite of East = West.

Answer: C

Approach: Convert turns into facing directions; distances are irrelevant to the question.

Solution:
Starts facing North, walks 10 km.
270° anticlockwise = 90° clockwise → faces East, walks 15 km.
45° clockwise from East → South-East, walks 25 km.
He is now facing South-East.

Cetking shortcut: 270° anticlockwise = 90° clockwise – always convert large turns to the smaller equivalent.

Answer: B

Approach: Coordinate tracing + Pythagoras.

Solution:
A = (0, 0). 3 km East → B = (3, 0).
Turn left → North, 3 × 3 = 9 km → C = (3, 9).
Turn left → West, 5 × 3 = 15 km → D = (3 − 15, 9) = (−12, 9).
AD = √(12² + 9²) = √(144 + 81) = √225 = 15 km.

Cetking shortcut: (12, 9) is 3 × (4, 3) – a 3-4-5 triangle, so the distance is 3 × 5 = 15.

Answer: A

Approach: Direction sense — net displacement using N–S and E–W components.

Solution:
Take start as (0, 0).
7 km North → (0, 7).
Turn right (facing East), 6 km → (6, 7).
Turn right again (facing South), 7 km → (6, 0).
North and South cancel; distance from start = 6 km (due East).

Why the other options are wrong:
B, C, D, E: come from adding paths or using Pythagoras wrongly — the 7 km legs cancel exactly.

Cetking shortcut: Equal and opposite legs cancel; only the 6 km East leg remains.

Answer: A

Approach: Direction sense — fix the facing direction from the 'right-hand' clue.

Solution:
West is exactly to the right of the direction he pointed.
If you face South, your right hand points West.
So he was pointing towards South (his home is South, not West).

Why the other options are wrong:
A North: right side would be East.
B North-East: right side would be South-East.
C West: right side would be North.
E East: right side would be South.

Cetking shortcut: Right of X = X turned 90° clockwise; so X = West turned 90° anticlockwise = South.

Answer: D

Approach: Direction sense — track each turn with coordinates.

Solution:
Start (0, 0), facing North; walk 20 m → (0, 20).
Turn right (East), 30 m → (30, 20).
Turn right (South), 35 m → (30, −15).
Turn left (from South, left = East), 15 m → (45, −15).
Turn left (from East, left = North), 15 m → (45, 0).
Final point is 45 m due East of the start.

Why the other options are wrong:
A: correct distance but wrong side (West).
C, D: 30 m ignores the extra 15 m East leg.
E: 15 m West mixes up the left turn from South.

Cetking shortcut: Add East moves: 30 + 15 = 45; North–South: +20 − 35 + 15 = 0 → 45 m East.

Answer: B

Approach: Clock-orientation direction — rotate the whole dial so 12 points to the given direction.

Solution:
At 2:00 p.m. the minute hand points at 12, so the '12' mark faces North-West.
So the '6' mark (opposite 12) faces South-East.
At 6:00 p.m. the hour hand points at 6.
Hence the hour hand points South-East.

Why the other options are wrong:
B North-West: that is where 12 points, not 6.
C North-East, D South-West: these are the 3 and 9 marks (90° away).
E East: not an angle consistent with the rotated dial.

Cetking shortcut: At any 'o'clock', the minute hand shows where 12 is; 6 is always directly opposite.

Answer: A

Approach: Direction sense with coordinates — place K at the origin.

Solution:
K = (0, 0). J is 1 km north of K → J = (0, 1).
Q is 2 km south of J → Q = (0, −1).
P is 1 km west of Q → P = (−1, −1).
KP = √(1² + 1²) = √2 ≈ 1.41 km.
(L, M, N, O, R are not needed for this question.)

Why the other options are wrong:
A 1.61, B 1.51, C 1.31, E 1.23: none equals √2 = 1.414.

Cetking shortcut: A 1 km × 1 km offset always gives √2 ≈ 1.41 km.

Answer: D

Approach: Place points on a grid with E at the origin.

Solution:
E = (0, 0). N is east of E → N = (1, 0).
G is south of E → G = (0, −1). L is west of G → L = (−1, −1).
From L to N: +2 east and +1 north → N is North-East of L.

Why the other options are wrong:
A, C, E: N is above L (north), not south.
B: N is to the east of L, not west.

Answer: D

Approach: Track facing direction after each turn.

Solution:
Start facing South, walk 5 km → (0, −5).
Right turn while facing South → now facing West; walk 6 km → (−6, −5).
Left turn while facing West → now facing South; walk 4 km → (−6, −9).
Final point is west and south of start → South-West.

Why the other options are wrong:
A: he is also 9 km south.
B: he is also 6 km west.
C, E: he is never north of the start.

Cetking shortcut: Right of South = West; left of West = South.

Answer: D

Approach: Find the rotation that maps North to South-East and apply it to East.

Solution:
North → South-East is a clockwise turn of 135° (N 0°, E 90°, SE 135°).
East (90°) + 135° = 225° = South-West.

Why the other options are wrong:
A: NE is a 45° turn.
B: SE is what North became.
D: NW is the opposite (anticlockwise) result.
E: South is a 90° turn.

Cetking shortcut: Each direction moves 3 steps clockwise on the 8-point compass (N→NE→E→SE); East moves E→SE→S→SW.

Answer: C

Approach: Rotated clock — find the rotation, then turn the hand by its angle.

Solution: At 12 noon the minute hand is on 12, and it points north-east. So the "12" mark faces NE: the clock is turned 45° clockwise from normal.
At 1:30 the hour hand is 1.5 × 30° = 45° clockwise from 12.
Direction = NE + 45° clockwise = East.

Why the other options are wrong:
A North — would need the hand 45° anticlockwise of 12.
B South, D West — 135°/225° away from 12 on this clock.
E North-East — that is the 12 position, hour hand is at 12 only at 12:00.

Cetking shortcut: Hour hand moves 30° per hour, so 1:30 = 45° — exactly one compass step (NE → E).

Answer: C

Approach: Net rotation.

Solution: Clockwise: 45° + 180° = 225°. Anticlockwise: 270°.
Net = 225° − 270° = −45°, i.e. 45° anticlockwise.
From West, 45° anticlockwise → South-West.

Why the other options are wrong:
A South — 90° anticlockwise of West.
C North — 90° clockwise of West.
D North-West — 45° clockwise of West.
E North-East — opposite of South-West.

Cetking shortcut: Net every turn to one number first; never trace turn by turn.

Answer: B

Approach: Direction by turning angles: find the net turn.

Solution:
Start facing East.
Turn 120° clockwise, then 210° anticlockwise.
Net turn = 210° − 120° = 90° anticlockwise.
90° anticlockwise from East = North.

Cetking shortcut: Cancel the turns first (net angle and its direction), then turn once from the starting direction.

Answer: A

Approach: Direction and distance: net displacement on N–S and E–W axes, then Pythagoras.

Solution:
12 km North; turns right → faces East, walks 4 km; turns right → faces South, walks 9 km.
North–South: 12 − 9 = 3 km North.
East–West: 4 km East.
Distance = √(3² + 4²) = √25 = 5 km, direction North-East.

Why the other options are wrong:
A, B: 10 km ignores that 9 km is walked back South.
C, E: the 4 km East leg is ignored.

Cetking shortcut: 3-4-5 triangle: spot it without calculating.

Answer: D

Approach: Direction and distance: track each leg on a grid.

Solution:
10 km North.
Turns left → faces West, 4 km West.
Turns right → faces North, 5 km North.
Turns right → faces East, 4 km East — this cancels the 4 km West.
Net: 10 + 5 = 15 km North of the start.

Why the other options are wrong:
A: she never moves South.
B: ignores the second 5 km northward leg.
C, E: the East and West legs cancel, so there is no East displacement.

Cetking shortcut: Equal and opposite legs (4 W, 4 E) cancel — add only the North legs.

Answer: D

Approach: Direction sense: track coordinates (East = +x, North = +y).

Solution:
Start (0, 0).
20 m North → (0, 20).
Right = East, 30 m → (30, 20).
Right = South, 35 m → (30, −15).
Left = East, 15 m → (45, −15).
Left = North, 15 m → (45, 0).
He is 45 m East of the start.

Why the other options are wrong:
A 15 m West, C 30 m West, E 45 m West: he only ever moves East horizontally.
B 30 m East: forgets the second 15 m eastward leg.

Cetking shortcut: North moves: +20 −35 +15 = 0; East moves: 30 + 15 = 45 → straight East 45 m.

Answer: D

Approach: Direction sense: net displacement and Pythagoras.

Solution:
Start (0, 0). 90 m East → (90, 0).
Right = South, 20 m → (90, −20).
Right = West, 30 m → (60, −20) (uncle's place).
100 m North → (60, 80).
Distance = √(60² + 80²) = √(3600 + 6400) = √10000 = 100 m.

Why the other options are wrong:
A 80 m: only the north–south part.
C 140 m: adds 60 + 80.
D 260 m: total path walked.
E 90 m: only the first leg.

Cetking shortcut: 60 and 80 form a 3-4-5 triple (×20) → 100 directly.

Answer: B

Approach: Circular arrangement mapped on the 8 compass points.

Solution:
8 people at equal distances = 8 compass points, 45° apart, clockwise order P, Q, R, S, T, U, V, W.
V = North; moving clockwise: W = NE, P = E, Q = SE, R = S, S = South-West, T = W, U = NW.

Why the other options are wrong:
A East: P.
B South-East: Q.
C South: R.
E North: V.

Cetking shortcut: S is 5 places clockwise after V (or 3 places anticlockwise): N → NW → W → SW.

Answer: D

Approach: Direction sense on a square table: partners sit opposite and face each other.

Solution:
R faces West → R sits on the East side.
R's right hand points North → S sits on the North side, facing South.
Q is S's partner → Q sits opposite S, on the South side, facing North.
(P sits on the West side facing East.)

Why the other options are wrong:
B South: that is the direction S faces.
C East: P's facing direction.
D West: R's facing direction.
E South-West: no player faces a diagonal.

Cetking shortcut: Partners always face opposite directions: S faces South, so Q faces North.

Answer: A

Approach: Direction sense with coordinates: T lies on line QP extended.

Solution:
Put P at (0, 0). Q is South-West → Q = (−1, −1).
R is East of Q and South-East of P → R = (1, −1).
T is North of R and on the line Q–P (the line y = x) → T = (1, 1).
(1, 1) from P(0, 0) is North-East.

Why the other options are wrong:
A South-East: that is R.
B North: T has an East component.
D East: T is above P's level.
E West: T is on the right of P.

Cetking shortcut: Line Q→P runs towards NE; extending it past P keeps going NE — T must be NE of P.

Answer: C

Approach: Track direction after each turn and add East–West and North–South moves separately.

Solution:
Walk 20 m North.
Turn right → East, 30 m.
Turn right → South, 35 m.
Turn left → East, 15 m.
Turn left → North, 15 m.
North–South: 20 − 35 + 15 = 0.
East–West: 30 + 15 = 45 m East.
He is 45 m East of the start.

Why the other options are wrong: West options reverse the right turns; 30 m East ignores the last East leg of 15 m.

Cetking shortcut: The vertical legs cancel (20 + 15 = 35), so only the two East legs count: 30 + 15.

Answer: C

Approach: Track the facing direction after each turn; use Pythagoras for the shortest distance.

Solution:
Facing East, turns left → North, walks 10 m.
Turns left → West, walks 5 m.
Turns right → North, walks 2 m.
Net: North 10 + 2 = 12 m, West 5 m.
Distance = √(12² + 5²) = √169 = 13 m; he is facing North.

Why the other options are wrong: 12 m counts only the northward legs; 11 m is a wrong sum; 'West' is his direction of displacement on one leg, not his final facing.

Cetking shortcut: 5–12–13 triangle.

Answer: A