Linear Arrangement – MBA CET questions with solutions

Approach: Linear arrangement — build the row from the 'left of / right of' clues.

Solution:
D is left of C and A is right of C → D … C … A.
B is left of D → B … D … C … A.
B is right of E → E … B … D … C … A.
Five people, five seats: E B D C A, with E at a corner as given.
Middle (3rd) seat = D.

Why the other options are wrong:
A, E: A and E are at the two ends.
B, C: B and C are 2nd and 4th.

Cetking shortcut: E at one end and the chain E-B-D-C-A uses all five — the 3rd letter is the answer.

Answer: D

Approach: Count the slots in front of car 3 and use the 'first available slot' rule.

Solution:
Cars 1, 2, 3 parked first in slots from the start, so car 3 begins right after the slots of cars 1 and 2: s1 + s2 slots (each car takes 1 or 2).
Now cars 4, 5, 6 and one single vacant slot sit in front of car 3: s4 + s5 + s6 + 1 ≥ 4.
But s1 + s2 ≤ 4, so s1 + s2 = 4 and s4 = s5 = s6 = 1.
→ Car 1 and Car 2 are SUVs; cars 4, 5, 6 are compacts (A, B, C, D are certain).
Car 3's own size never affects the slots in front of it, so it may be a compact or an SUV → 'Car 3 is an SUV' is not necessarily true.

Why the other options are wrong:
A, C: cars 4 and 5 must be compact (one slot each).
B, D: cars 1 and 2 must have filled 4 slots, so both are SUVs.

Answer: E

Approach: Fix definite positions, then place blocks (8 seats, left to right).

Solution:
A is 3rd from left → seat 3.
B is 5th from right → 8 − 5 + 1 = seat 4.
C is immediately right of B → seat 5.
Free seats: 1, 2, 6, 7, 8. D is between F and G → a 3-seat block F–D–G, which only fits in 6–7–8.
H is left of E → H = 1, E = 2.
Seats 6 and 8 are F/G, so 6th from left is F or G; F is not offered → G.

Why the other options are wrong:
A A – seat 3.
B D – seat 7 (centre of the block).
D C – seat 5.
E B – seat 4.

Cetking shortcut: Only a 3-seat gap can hold the F–D–G block; seats 6–8 is the only one.

Answer: C

Approach: Linear order puzzle — build the fixed block first, then slide it.

Solution:
Q came immediately before P (no one between them, Q first) and V just before Q ⇒ block V Q P; S immediately after P ⇒ block V Q P S.
Three persons between R and S ⇒ R is 4 places before S (or after it).
Try S = 4: R would be at 0 or 8 — impossible.
S = 5 (V2 Q3 P4 S5): R = 1, then T (one between R and T) = 3 — taken. ✗
S = 6 (V3 Q4 P5 S6): R = 2, T = 4 — taken. ✗
S = 7 (V4 Q5 P6 S7): R = 3, T = 1 (5 is taken) ✓, U = 2.
Final order: T, U, R, V, Q, P, S.
First person: T.

Why the other options are wrong:
Q: Q is 5th. R: R is 3rd. S: S is last. P: P is 6th.

Cetking shortcut: Four people form one rigid block (V Q P S) in a line of 7 — only 4 positions to test.

Answer: D

Approach: Use the final order from the arrangement.

Solution:
Q came immediately before P (no one between them, Q first) and V just before Q ⇒ block V Q P; S immediately after P ⇒ block V Q P S.
Three persons between R and S ⇒ R is 4 places before S (or after it).
Try S = 4: R would be at 0 or 8 — impossible.
S = 5 (V2 Q3 P4 S5): R = 1, then T (one between R and T) = 3 — taken. ✗
S = 6 (V3 Q4 P5 S6): R = 2, T = 4 — taken. ✗
S = 7 (V4 Q5 P6 S7): R = 3, T = 1 (5 is taken) ✓, U = 2.
Final order: T, U, R, V, Q, P, S.
R is 3rd, Q is 5th ⇒ only V (4th) is between them ⇒ 1 person.

Why the other options are wrong:
4, 3, 2, 5: come from a wrong order; positions 3 and 5 have exactly one place between them.

Cetking shortcut: Persons between positions a and b = |a − b| − 1 = 5 − 3 − 1 = 1.

Answer: E

Approach: Use the final order from the arrangement.

Solution:
Q came immediately before P (no one between them, Q first) and V just before Q ⇒ block V Q P; S immediately after P ⇒ block V Q P S.
Three persons between R and S ⇒ R is 4 places before S (or after it).
Try S = 4: R would be at 0 or 8 — impossible.
S = 5 (V2 Q3 P4 S5): R = 1, then T (one between R and T) = 3 — taken. ✗
S = 6 (V3 Q4 P5 S6): R = 2, T = 4 — taken. ✗
S = 7 (V4 Q5 P6 S7): R = 3, T = 1 (5 is taken) ✓, U = 2.
Final order: T, U, R, V, Q, P, S.
R is 3rd ⇒ the person just before R (2nd) is U.

Why the other options are wrong:
P, Q, S: all come after R (positions 6, 5, 7). T: T is 1st, two places before R.

Answer: E

Approach: Linear seating (8, facing North): fix H, then test cases.

Solution:
H has fewer than 2 people to its left → H is 1st or 2nd. B is H's neighbour; C is 2nd to the right of B.
G is exactly midway between H and F, not adjacent to either → G = H + k, F = H + 2k, k ≥ 2.
Case H = 1: B = 2, C = 4 → G = 3, F = 5 → D = 6, A = 7 → E = 8 is next to A ✗.
Case H = 2, B = 1: C = 3 → (G 4, F 6, D 7, A 8, E 5 next to G ✗) or (G 5, F 8, D 7, A 6, E 4 next to G ✗).
Case H = 2, B = 3: C = 5 → G = 4, F = 6, D = 7, A = 8, E = 1 ✓.
Row (left→right): E H B G C F D A. People to the left of G: E, H, B = 3.

Cetking shortcut: Lock H and B first (only three seat options), then G–F 'midway' leaves very few cases – kill each with E's condition.

Answer: C

Approach: Linear order from left/right clues.

Solution:
Kajal is left of Aakansha; Divya is right of Aakansha; Anita is between Aakansha and Divya → Kajal, Aakansha, Anita, Divya.
Lila is left of Kajal → Lila, Kajal, Aakansha, Anita, Divya.
From the right: Divya (1), Anita (2), Aakansha (3).

Why the other options are wrong:

• A Lila – 1st from the left, i.e. 5th from the right.
• B Divya – 1st from right; D Anita – 2nd from right; E Kajal – 4th from right.

Cetking shortcut: With 5 people, 3rd from the right = 3rd from the left = the middle person.

Answer: C

Approach: Chain the 'immediate' clues into one block.

Solution:
Ajit is immediately left of Bhola and immediately right of Chetan → Chetan, Ajit, Bhola.
Chetan is immediately right of Firoj → Firoj, Chetan, Ajit, Bhola.
Dilip is immediately right of Eknath, and Eknath is left of Firoj → Eknath, Dilip, Firoj, Chetan, Ajit, Bhola.
Centre (3rd and 4th seats) = Firoj and Chetan.

Answer: C

Approach: Put the stops in positions 1–5 (upward) and test the gap condition: C is 2x stops before D while B is x stops after A.

Solution:
Terminals (1 and 5) can only be B, C or D (A and E are not terminals); C ≠ 3.
x = 1 → C is 2 before D, B is just after A.
• C = 1, D = 3 → A, B, E fill 2, 4, 5 with A–B adjacent → A = 4, B = 5, E = 2.
• C = 2, D = 4 → spots 1, 3, 5 have no adjacent pair for A–B ✗.
Upward: C E D A B → Downward (reverse): B A D E C.
(x = 2 gives C A E B D upward → D B E A C downward, which is not among the options.)

Why the other options are wrong: (reverse each to get the upward order)
A – upward D E C A B: C is the middle stop ✗.
B – upward C E B A D: B comes before A, not after ✗.
C – upward A B C E D: A is a terminal stop ✗.
E – upward C D A E B: C is 1 stop before D but B is 2 after A — the ratio is reversed ✗.

Answer: D

Approach: Two-row seating with rows facing each other — fix left/right for each row first.

Solution:
Facing each other: Row 1 faces South (its right = West), Row 2 faces North (its right = East). Number seats 1–6 from West to East.
Person in front of Aman is immediately left of Sunil → that person is at Sunil+1, so corner Sunil = seat 1, Aman = R2 seat 2.
Sunil in front of Raj → Raj = R2 seat 1. Sandeep diagonally opposite Raj → R1 seat 6; Rohit immediately right of Sandeep → R1 seat 5.
Ramesh 2nd right of Rajesh → Rajesh 4, Ramesh 2, R1 seat 3 vacant. Other vacant is a corner → R2 seat 6.
Aman (R2-2) → Rohit (R1-5) is 3 seats across; same for Ramesh (R1-2) → Omkar = R2-5. Suresh not next to Aman → Suresh 4, Yash 3.

Why the other options are wrong:
A, C: Aman is at seat 2, not beside Suresh (the puzzle says so).
D: Raj is at seat 1.
E: Sunil is in the other row.

Cetking shortcut: Start with the clue that ties two rows (front of Aman / left of Sunil) — it pins Sunil to a corner at once.

Answer: B

Approach: Use the final two-row arrangement.

Solution:
Facing each other: Row 1 faces South (its right = West), Row 2 faces North (its right = East). Number seats 1–6 from West to East.
Person in front of Aman is immediately left of Sunil → that person is at Sunil+1, so corner Sunil = seat 1, Aman = R2 seat 2.
Sunil in front of Raj → Raj = R2 seat 1. Sandeep diagonally opposite Raj → R1 seat 6; Rohit immediately right of Sandeep → R1 seat 5.
Ramesh 2nd right of Rajesh → Rajesh 4, Ramesh 2, R1 seat 3 vacant. Other vacant is a corner → R2 seat 6.
Aman (R2-2) → Rohit (R1-5) is 3 seats across; same for Ramesh (R1-2) → Omkar = R2-5. Suresh not next to Aman → Suresh 4, Yash 3.

Why the other options are wrong:
A, B, C: each is correct but incomplete.
E: Sandeep is at R1-6, next to Rohit only.

Answer: D

Approach: Use the final arrangement; 'in front of' = same seat number, other row.

Solution:
Facing each other: Row 1 faces South (its right = West), Row 2 faces North (its right = East). Number seats 1–6 from West to East.
Person in front of Aman is immediately left of Sunil → that person is at Sunil+1, so corner Sunil = seat 1, Aman = R2 seat 2.
Sunil in front of Raj → Raj = R2 seat 1. Sandeep diagonally opposite Raj → R1 seat 6; Rohit immediately right of Sandeep → R1 seat 5.
Ramesh 2nd right of Rajesh → Rajesh 4, Ramesh 2, R1 seat 3 vacant. Other vacant is a corner → R2 seat 6.
Aman (R2-2) → Rohit (R1-5) is 3 seats across; same for Ramesh (R1-2) → Omkar = R2-5. Suresh not next to Aman → Suresh 4, Yash 3.

Why the other options are wrong:
A: Rajesh faces Suresh.
B: Suresh faces Rajesh, not a vacant seat.
D: Raj faces Sunil.
E: Rohit faces Omkar.

Answer: C

Approach: Use the final arrangement.

Solution:
Facing each other: Row 1 faces South (its right = West), Row 2 faces North (its right = East). Number seats 1–6 from West to East.
Person in front of Aman is immediately left of Sunil → that person is at Sunil+1, so corner Sunil = seat 1, Aman = R2 seat 2.
Sunil in front of Raj → Raj = R2 seat 1. Sandeep diagonally opposite Raj → R1 seat 6; Rohit immediately right of Sandeep → R1 seat 5.
Ramesh 2nd right of Rajesh → Rajesh 4, Ramesh 2, R1 seat 3 vacant. Other vacant is a corner → R2 seat 6.
Aman (R2-2) → Rohit (R1-5) is 3 seats across; same for Ramesh (R1-2) → Omkar = R2-5. Suresh not next to Aman → Suresh 4, Yash 3.

Why the other options are wrong:
A, E: Yash and Suresh are in Row 2.
B: Ramesh is two seats away (vacant seat between).
D: Sandeep is at seat 6.

Answer: C

Approach: Use the final arrangement; opposite = same seat number in the other row.

Solution:
Facing each other: Row 1 faces South (its right = West), Row 2 faces North (its right = East). Number seats 1–6 from West to East.
Person in front of Aman is immediately left of Sunil → that person is at Sunil+1, so corner Sunil = seat 1, Aman = R2 seat 2.
Sunil in front of Raj → Raj = R2 seat 1. Sandeep diagonally opposite Raj → R1 seat 6; Rohit immediately right of Sandeep → R1 seat 5.
Ramesh 2nd right of Rajesh → Rajesh 4, Ramesh 2, R1 seat 3 vacant. Other vacant is a corner → R2 seat 6.
Aman (R2-2) → Rohit (R1-5) is 3 seats across; same for Ramesh (R1-2) → Omkar = R2-5. Suresh not next to Aman → Suresh 4, Yash 3.

Why the other options are wrong:
B, C: Sunil faces Raj, not Yash.
D: Sandeep and Raj are diagonally opposite, not directly opposite.
E: Rohit faces Omkar and Sunil faces Raj.

Answer: A

Approach: Two parallel rows facing North: fix positions by the 'not at end' and 'nth to the left/right' clues first.

Solution:
Blue line cars face the back of Yellow line cars → Yellow is the front (north) row. All face North, so left = West. Number positions 1–4 from West.
H3 is 2nd left of H4 and H4 is not at an end → H4 = 3, H3 = 1 (same row).
H5 3rd right of H2 → H2 = 1, H5 = 4 in the other row.
H1 is in a different row from H4 → H1 in the H2/H5 row (position 2 or 3).
H7 is opposite H1, so H7 is in H4's row and that row is Blue; H7 can only be at 2 → H1 = 2.
H8 not at an end → Yellow 3; H6 → Blue 4.

Cetking shortcut: H8's clue already says it is not at an end — every other option is a corner car.

Answer: D

Approach: Opposite = same position number in the other row.

Solution:
Blue line cars face the back of Yellow line cars → Yellow is the front (north) row. All face North, so left = West. Number positions 1–4 from West.
H3 is 2nd left of H4 and H4 is not at an end → H4 = 3, H3 = 1 (same row).
H5 3rd right of H2 → H2 = 1, H5 = 4 in the other row.
H1 is in a different row from H4 → H1 in the H2/H5 row (position 2 or 3).
H7 is opposite H1, so H7 is in H4's row and that row is Blue; H7 can only be at 2 → H1 = 2.
H8 not at an end → Yellow 3; H6 → Blue 4.

Why the other options are wrong:
A: H3 is in H4's own row.
C: H5 faces H6.
D: H1 faces H7.
E: H2 faces H3.

Answer: B

Approach: Read row membership from the final arrangement.

Solution:
Blue line cars face the back of Yellow line cars → Yellow is the front (north) row. All face North, so left = West. Number positions 1–4 from West.
H3 is 2nd left of H4 and H4 is not at an end → H4 = 3, H3 = 1 (same row).
H5 3rd right of H2 → H2 = 1, H5 = 4 in the other row.
H1 is in a different row from H4 → H1 in the H2/H5 row (position 2 or 3).
H7 is opposite H1, so H7 is in H4's row and that row is Blue; H7 can only be at 2 → H1 = 2.
H8 not at an end → Yellow 3; H6 → Blue 4.

Why the other options are wrong:
B, C, D, E: H2, H1, H5, H8 are all Yellow-line cars.

Answer: A

Approach: Same-row relative position; all cars face North.

Solution:
Blue line cars face the back of Yellow line cars → Yellow is the front (north) row. All face North, so left = West. Number positions 1–4 from West.
H3 is 2nd left of H4 and H4 is not at an end → H4 = 3, H3 = 1 (same row).
H5 3rd right of H2 → H2 = 1, H5 = 4 in the other row.
H1 is in a different row from H4 → H1 in the H2/H5 row (position 2 or 3).
H7 is opposite H1, so H7 is in H4's row and that row is Blue; H7 can only be at 2 → H1 = 2.
H8 not at an end → Yellow 3; H6 → Blue 4.

Why the other options are wrong:
A, D: there is a car (H7) between them.
B: that is H3 with respect to H4.
E: both are in the Blue line.

Answer: C

Approach: Opposite = same position in the other row.

Solution:
Blue line cars face the back of Yellow line cars → Yellow is the front (north) row. All face North, so left = West. Number positions 1–4 from West.
H3 is 2nd left of H4 and H4 is not at an end → H4 = 3, H3 = 1 (same row).
H5 3rd right of H2 → H2 = 1, H5 = 4 in the other row.
H1 is in a different row from H4 → H1 in the H2/H5 row (position 2 or 3).
H7 is opposite H1, so H7 is in H4's row and that row is Blue; H7 can only be at 2 → H1 = 2.
H8 not at an end → Yellow 3; H6 → Blue 4.

Why the other options are wrong:
A: H1 faces H7.
C, D: H7 and H4 are in H6's own row.
E: H2 faces H3.

Answer: B

Approach: Positional order — remove the players who are out and take the next closest.

Solution: Order behind the ball carrier: Amol (directly behind) → Jeet and Neeraj side by side → Nikhil.
Amol misses and falls; Neeraj trips.
The closest remaining defender is Jeet (he is ahead of Nikhil), so Jeet makes the tackle.

Why the other options are wrong: Neeraj tripped and Amol fell; Nikhil is behind Jeet; Rohit is not in the chase.

Answer: C

Approach: Seat numbering — right = higher number, left = lower number.

Solution: Hemanth = 253. Manoj is to his right in 254, so seat numbers increase to the right.
Gopal, to Hemanth's left = 252.
Iqbal, to Gopal's left = 251.

Why the other options are wrong: 252 is Gopal's seat; 254 and above are on the right side.

Cetking shortcut: Fix the direction from the one given pair (253 → 254 on the right), then step left one seat at a time.

Answer: A

Approach: Height ordering — build the full chain once, then read each question.

Solution: D > E > C > A (from clue 2).
F is between B and A in height: either B < F < A or A < F < B.
If A < F < B, everyone (B, F, C, E, D) is taller than A → A is the shortest, which clue 4 forbids. So B < F < A.
Order from shortest: B, F, A, C, E, D. Check clue 3: F is 2nd, E is 5th → two boys (A, C) between them. ✓

Why the other options are wrong:
A, B, D, E — E's neighbours are C and D only.

Cetking shortcut: Clue 4 ("A is not the shortest") is the key that decides B < F < A — apply it first.

Answer: C

Approach: Read the tallest from the fixed order.

Solution: D > E > C > A (from clue 2).
F is between B and A in height: either B < F < A or A < F < B.
If A < F < B, everyone (B, F, C, E, D) is taller than A → A is the shortest, which clue 4 forbids. So B < F < A.
Order from shortest: B, F, A, C, E, D. Check clue 3: F is 2nd, E is 5th → two boys (A, C) between them. ✓

Answer: D

Approach: Count positions from the shortest.

Solution: D > E > C > A (from clue 2).
F is between B and A in height: either B < F < A or A < F < B.
If A < F < B, everyone (B, F, C, E, D) is taller than A → A is the shortest, which clue 4 forbids. So B < F < A.
Order from shortest: B, F, A, C, E, D. Check clue 3: F is 2nd, E is 5th → two boys (A, C) between them. ✓

Answer: C

Approach: Read the shortest from the fixed order.

Solution: D > E > C > A (from clue 2).
F is between B and A in height: either B < F < A or A < F < B.
If A < F < B, everyone (B, F, C, E, D) is taller than A → A is the shortest, which clue 4 forbids. So B < F < A.
Order from shortest: B, F, A, C, E, D. Check clue 3: F is 2nd, E is 5th → two boys (A, C) between them. ✓

Answer: B

Approach: Linear arrangement of 5 with Monika fixed in the middle.

Solution:
Monika = position 3. Rashi is next to Monika → position 2 or 4.
Rashi is not next to Surekha or Abha, and Monika is on her other side, so her outer neighbour (1 or 5) must be Anuradha.
Case Rashi at 4: Anuradha at 5; Surekha and Abha at 1, 2. Anuradha is not next to Surekha ✓.
(Rashi at 2 is the mirror image.)
Anuradha's only neighbour = Rashi.

Why the other options are wrong:
B Surekha: banned next to Anuradha.
C Abha, E Surekha or Abha: they sit on the far side of Monika.
D Monika: Rashi sits between Monika and Anuradha.

Cetking shortcut: Rashi's outer neighbour cannot be Surekha or Abha → it is Anuradha; no full arrangement needed.

Answer: A

Approach: Linear arrangement of 5 with an extreme-end anchor.

Solution:
R is at the extreme left (1); K is next to R → K = 2.
T is not next to K → T is not 3, so T = 4 or 5.
If T = 4, S and P take 3 and 5 — not adjacent ✗.
So T = 5; S is not next to T → S = 3, P = 4.
Row: R, K, S, P, T → S sits between K and P.

Why the other options are wrong:
B R and P: R is at position 1.
C Only P: K is also next to S.
D P and T: S cannot be next to T.
E R and K: R is not next to S.

Answer: A

Approach: Linear arrangement — option elimination using fixed positions.

Solution:
D is 2nd from left; 5 people between D and B → B at position 8.
Check A (ADHECGFB): positions A1 D2 H3 E4 C5 G6 F7 B8.
C at 5 = 4th from right ✓; E (4) next to C ✓; A (1) and E (4) have two between ✓; H (3) and C (5) have one between ✓; G–F adjacent ✓.
All conditions hold, so A is correct.

Why the other options are wrong:
B ADCEHFGB — C is 3rd from left and 6th from right, not 4th from an end.
C GDAFCEHB — G and F are not adjacent.
D ADGHFCEB — C at 6 is 3rd from right.
E ADCEHFBG — B is at 7, so only 4 people sit between D and B.

Cetking shortcut: First fix D at 2 and B at 8 — this alone kills option E; then check C's position.

Answer: A

Approach: Linear arrangement, all facing the same side; seats 1–8 from left.

Solution:
Roshesh and Urus at seats 1 and 8; their neighbours (seats 2 and 7) are Quila and Tanaav.
Pravesh is 4th to the right of Tanaav → Tanaav cannot be at 7, so Tanaav = 2, Pravesh = 6, Quila = 7 (Pravesh is next to Quila ✓).
Wahaab is immediately left of Pravesh → Wahaab = 5.
Wahaab is 2nd to the right of Smitha → Smitha = 3; Vihaaz takes the last seat, 4.
Order: R/U, Tanaav, Smitha, Vihaaz, Wahaab, Pravesh, Quila, U/R → Wahaab's neighbours are Vihaaz and Pravesh.

Why the other options are wrong:
A, C, D, E — Urus is at an end and Tanaav/Quila sit at seats 2 and 7, far from Wahaab at seat 5.

Answer: B

Approach: Five in a row, all facing North.

Solution:
Dony = seat 3; Tony is next to Dony → seat 2 or 4.
Tony's other neighbour (seat 1 or 5) cannot be Bony or Mony → it is Sony.
So the row is Sony, Tony, Dony, _, _ (or its mirror); Bony and Mony take the other side.
Ends = Sony and one of Bony/Mony. Among the options only 'Sony, Mony' fits.

Why the other options are wrong:
A, B, C, D — Dony is in the middle and Tony at seat 2/4, so neither can be at an end.

Cetking shortcut: Dony (middle) and Tony (next to middle) can never be at an end — this removes A–D at once.

Answer: E