Alphabet Series – MBA CET questions with solutions

Approach: Letter series — alternate terms move +2; each even term is the reverse of the term before it.

Solution:
Odd terms: FAG → HAI → ?: outer letters +2 each (F→H, G→I), A stays.
Even terms: GAF = FAG reversed, IAH = HAI reversed.
The 5th term is the next odd term: H+2 = J, A, I+2 = K → JAK.

Why the other options are wrong:
A, B: HAK / HAL keep H as first letter — no +2 move.
D: JAI moves only the first letter.
E: KAH does not follow either pattern.

Cetking shortcut: Pairs (FAG, GAF), (HAI, IAH): new word then its mirror — the 5th term starts a new pair: JAK.

Answer: C

Approach: Letter series — position-wise patterns in 5-letter groups.

Solution:
Position 1: A, D, G, J → +3 → M
Position 2: Y, W, U, S → −2 → Q
Position 3: B, E, H, K → +3 → N
Position 4: Z, X, V, T → −2 → R
Position 5: C, F, I, L → +3 → O
Next group: MQNRO.

Why the other options are wrong:
A: MQORN swaps the 3rd and 5th letters.
C, D, E: wrong first letter (must be M = J + 3).

Cetking shortcut: Positions 1, 3, 5 just run through the alphabet (ABC, DEF, GHI, JKL → MNO); slot Q and R into positions 2 and 4.

Answer: B

Approach: Letter series: treat each letter position as its own series.

Solution:
1st letter: W (−3) T (−3) Q (−3) N.
2nd letter: F (+1) G (+1) H (+1) I.
3rd letter: B (+2) D (+3) G (+4) K.
Next group: NIK.

Why the other options are wrong:
A (NIJ): 3rd letter uses +3 again instead of +4.
C (NJK): 2nd letter jumps +2.
D (NOK): 2nd letter wrong.
E (PJK): 1st and 2nd letters wrong.

Cetking shortcut: Third letter has increasing gaps (+2, +3, +4): G + 4 = K. Last letter K rules out A; first letter N rules out E; second letter I (not J or O) leaves only B.

Answer: B

Approach: Track each letter position separately.

Solution:
1st letters: S, R, Q, P → O (−1 each step)
2nd letters: H, I, J, K → L (+1)
3rd letters: G, F, E, D → C (−1)
Next term = OLC

Why the other options are wrong:
A NME, B NLB: first letter should be O, not N.
C OLE: third letter should be C, not E.
E SRQ: repeats a different pattern entirely.

Cetking shortcut: First letter must be O (eliminates A, B, E); third must be C → D.

Answer: D

Approach: Break the series into groups.

Solution:
m n o | n o p q | o p q r s | ...
Groups start at m, n, o → next group starts at p.
The next group therefore starts at p; among the options, the five-letter block that begins correctly is p q r s t.

Why the other options are wrong:
A mnopq, E mnono: restart from m — the start letter moves forward each group.
B oqrst: skips p and repeats o.
D qrstu: starts at q, one step too far.

Cetking shortcut: Each group starts one letter later than the previous one (m, n, o) — only C starts with p.

Answer: C

Approach: Write the new series: A–M in odd positions, N–Z in the gaps.

Solution:
Arrangement: A N B O C P D Q E R F S G T H U I V J W K X L Y M Z (positions 1–26).
7th from left = D.
Third to the right of D (position 7) = position 10 = R.

Why the other options are wrong:
A C: position 5.
B O: position 4 (counting left).
D S: position 12.
E A: position 1.

Cetking shortcut: Odd positions hold A–M: position 2k − 1 = k-th letter; even position 2k = (13 + k)-th letter. Position 10 → 13 + 5 = 18th = R.

Answer: C

Approach: Write the new series: A–M in odd positions, N–Z in the gaps.

Solution:
Arrangement: A N B O C P D Q E R F S G T H U I V J W K X L Y M Z (positions 1–26).
D is at position 7, K at position 21.
Middle position = (7 + 21) ÷ 2 = 14 → letter at 14 = T.

Why the other options are wrong:
A S: position 12.
C V: position 18.
D W: position 20.
E B: position 3.

Cetking shortcut: Even position 14 → (13 + 7)th letter = T.

Answer: B

Approach: Compare the given arrangement with alphabetical positions.

Solution:
Series positions: 1 H, 2 D, 3 Y, 4 S, 5 M, 6 W, 7 N, 8 B, 9 Q, 10 P, 11 O, 12 C, 13 R, 14 T, 15 B, 16 L, 17 Z, 18 V, 19 E, 20 G, 21 U, 22 F.
Option pairs and gaps: B–Q = 15, D–Y = 25 − 4 = 21, U–F = 15, V–E = 17, Q–O = 2.
Largest gap = D and Y (21).

Why the other options are wrong:
A B–Q: 15.
C U–F: 15.
D V–E: 17.
E Q–O: 2 (and they are not neighbours).

Cetking shortcut: Only check option pairs — D (4) vs Y (25) is the obvious extreme.

Answer: B

Approach: Compare the given arrangement with alphabetical positions.

Solution:
Series positions: 1 H, 2 D, 3 Y, 4 S, 5 M, 6 W, 7 N, 8 B, 9 Q, 10 P, 11 O, 12 C, 13 R, 14 T, 15 B, 16 L, 17 Z, 18 V, 19 E, 20 G, 21 U, 22 F.
P (position 10) has neighbours Q and O.
In the alphabet P lies between O and Q — same neighbours, sides swapped → P.

Why the other options are wrong:
A M: neighbours S, W.
B N: neighbours W, B.
C O: neighbours P, C.
E D: neighbours H, Y.

Cetking shortcut: Look for a run of three consecutive letters reversed — Q P O.

Answer: D

Approach: Compare the given arrangement with alphabetical positions.

Solution:
Series positions: 1 H, 2 D, 3 Y, 4 S, 5 M, 6 W, 7 N, 8 B, 9 Q, 10 P, 11 O, 12 C, 13 R, 14 T, 15 B, 16 L, 17 Z, 18 V, 19 E, 20 G, 21 U, 22 F.
Y, L, F are at positions 3, 16, 22 → gaps 13 and 6.
In the alphabet F = 6, L = 12, Y = 25 → gaps 6 and 13.
Same distances, order reversed → YLF.

Why the other options are wrong:
A HMP: series gaps 4, 5; alphabet gaps 5, 3.
B NQZ: series gaps 2, 8; alphabet 3, 9.
C QOE: series gaps 2, 8; alphabet 10, 2.
E HDY: series gaps 1, 1; alphabet 4, 17.

Cetking shortcut: Compute only the first gap for each option and match it to an alphabet gap — Y to L is 13 in both.

Answer: D

Approach: Mark vowels (A, E, I, O, U) and consonants, then count carefully.

Solution:
Drop A, E, I, E, U, O → C B T X F M K V N X D (11 letters).
9th from the right = (11 − 9 + 1) = 3rd from the left = T.

Why the other options are wrong:
B X: 4th from left (8th from right).
C B: 2nd from left (10th from right).
D F: 5th from left.
E C: 1st from left.

Cetking shortcut: Position from right = n − k + 1 from left: 11 − 9 + 1 = 3.

Answer: A

Approach: Mark vowels (A, E, I, O, U) and consonants, then count carefully.

Solution:
Positions: 1 C, 2 B, 3 T, 4 A, 5 E, 6 X, 7 F, 8 I, 9 M, 10 K, 11 E, 12 U, 13 V, 14 N, 15 X, 16 O, 17 D.
Pattern needed: vowel, consonant, consonant.
E X F → X ✓; I M K → M ✓; U V N → V ✓.
Others fail (e.g. A E X: E is a vowel; O D: D has nothing after it). Count = 3.

Why the other options are wrong:
A Seven, E Four: count consonants without checking both neighbours.
B One, C Two: miss a match such as U V N.

Cetking shortcut: Start from each vowel that is followed by a consonant and check just the next letter.

Answer: D

Approach: Compare position difference in the word with position difference in the alphabet.

Solution:
S E P A R A T I O N = positions 1–10; letter values S19 E5 P16 A1 R18 A1 T20 I9 O15 N14.
E(2)–A(6): word gap 4, alphabet gap 5 − 1 = 4 ✓
P(3)–R(5): gap 2, 18 − 16 = 2 ✓
P(3)–T(7): gap 4, 20 − 16 = 4 ✓
R(5)–T(7): gap 2, 20 − 18 = 2 ✓
O(9)–N(10): gap 1, 15 − 14 = 1 ✓
Total = 5 pairs.

Cetking shortcut: For each letter, scan right only as far as the word allows and check both A→Z and Z→A directions.

Answer: E

Approach: Sort and compare position-wise.

Solution:
Original: S A T E L L I T E
Sorted: A E E I L L S T T
Position 5: L = L ✓
Position 6: L = L ✓
Position 8: T = T ✓
All other positions differ → 3 letters unchanged.

Why the other options are wrong:
E 0 – overlooks positions 5, 6 and 8.
A 2 / B 1 / D 4 – miscounts.

Answer: C

Approach: Letter positions with increasing gaps (cyclic alphabet).

Solution: C(3) → H(8): +5; H(8) → P(16): +8; next gap +11: 16 + 11 = 27 → 27 − 26 = 1 = A.
Check: A(1) + 14 = 15 = O ✔; O(15) + 17 = 32 − 26 = 6 = F ✔.
Gaps +5, +8, +11, +14, +17.

Cetking shortcut: Write positions under letters; gaps increasing by 3 is a classic CET pattern. Wrap past Z by subtracting 26.

Answer: D

Approach: Letter positions with decreasing steps (cyclic alphabet).

Solution: Z(26) → T(20): −6; T → M(13): −7; next −8: 13 − 8 = 5 = E.
Check: E(5) − 9 = −4 → −4 + 26 = 22 = V ✔; V(22) − 10 = 12 = L ✔.

Cetking shortcut: Backward gaps −6, −7, −8, −9, −10; add 26 when you go below A.

Answer: A

Approach: Letter positions: third column = first + second in each row.

Solution: Row 1: F(6) + I(9) = 15 = O ✔.
Row 2: A(1) + J(10) = 11 = K ✔.
Row 3: E(5) + M(13) = 18 = R.

Cetking shortcut: Convert to numbers immediately; row-wise sums are the first test.

Answer: B

Approach: Letter-position shift in each slot of a 4-letter group.

Solution:
1st letter: O, Q, S, ?, W → +2 each → U.
2nd letter: L constant.
3rd letter: P, R, T, ?, X → +2 each → V.
4th letter: A constant.
Missing term = U L V A.

Why the other options are wrong: OLPA repeats the first term; KLMA and LLMA move backwards; ULAV swaps the last two letters.

Answer: B

Approach: Sort the letters alphabetically and compare position by position.

Solution:
Original: C R E A T I O N
Sorted:   A C E I N O R T
Position 3: E = E — the only match.
All other positions differ.

Answer: B

Approach: For each pair, letters between them in the word must equal letters between them in the alphabet (count both forward and backward pairs).

Solution: B(1) E(2) A(3) U(4) T(5) I(6) F(7) Y(8) I(9) N(10) G(11)
E–I: 3 letters between in word (A,U,T), alphabet E→I has F,G,H = 3 ✓
U–T: adjacent in word, adjacent in alphabet (T,U) ✓
U–Y: 3 between in word (T,I,F), alphabet U→Y has V,W,X = 3 ✓
I–G (positions 9, 11): 1 between (N), alphabet G→I has H = 1 ✓
Total = 4 pairs.

Cetking shortcut: Check each letter only against the next 4–5 letters; far-apart pairs rarely match.

Answer: D

Approach: Column rule with alphabet positions: row 1 – row 2 = row 3.

Solution:
Column 1: E(5) – A(1) = 4 = D ✓
Column 2: J(10) – C(3) = 7 = G ✓
Column 3: O(15) – B(2) = 13 = M.

Why the other options are wrong: N(14), L(12), T(20), Q(17) do not give 15 – 2.

Answer: A

Approach: Fill the blanks so the series becomes a repeating block.

Solution:
Fill with a, a, n, a, n:
ananna | ananna | ananna
= ananna | ananna | ananna — the block 'ananna' repeats three times.

Why the other options are wrong: A has only 4 letters for 5 gaps; C, D, E break the repeating block.

Cetking shortcut: Total length 18 with 5 gaps → try a 6-letter repeating block.

Answer: B

Approach: Track first and second letters separately.

Solution:
First letters: A(1), G(7), M(13), ?, Y(25) → +6 each → S(19).
Second letters: Z(26), T(20), N(14), ?, B(2) → –6 each → H(8).
Missing term = SH.

Why the other options are wrong: HS reverses the pair; PW, ST, RX fail one of the ±6 rules.

Cetking shortcut: Each pair is a letter and its opposite (A–Z, G–T, M–N) — S's opposite is H.

Answer: D

Approach: Alphabet manipulation — reverse, delete vowels, then count position.

Solution:
Reversed: Z Y X W V U T S R Q P O N M L K J I H G F E D C B A.
Delete A, E, I, O, U ⇒ Z Y X W V T S R Q P N M L K J H G F D C B (21 letters).
15th from the right = (21 − 15 + 1) = 7th from the left.
Z(1) Y(2) X(3) W(4) V(5) T(6) S(7).

Why the other options are wrong:
L, M, T, V are at other positions (T is 6th, V 5th from left).

Cetking shortcut: n-th from right = (total − n + 1)-th from left; 21 − 15 + 1 = 7.

Answer: B

Approach: Apply the three operations in order, then count letters between two letters in the alphabet.

Solution:
ANALYTICS: 1 A, 2 N, 3 A, 4 L, 5 Y, 6 T, 7 I, 8 C, 9 S.
Odd positions +2: A→C, A→C, Y→A, I→K, S→U. Even positions −1: N→M, L→K, T→S, C→B.
Word: C M C K A S K B U.
Perfect-square positions 1, 4, 9 swap with the next letter: 1↔2 → M C; 4↔5 → A K; 9 has no 10th letter.
New word: M C C A K S K B U.
3rd letter = C, 7th letter = K. Letters between C and K in the alphabet: D, E, F, G, H, I, J = 7.

Cetking shortcut: Between two letters = difference of positions − 1: K(11) − C(3) − 1 = 7.

Answer: C

Approach: Sort the letters alphabetically and compare position by position.

Solution:
UNIDENTIFIED: U N I D E N T I F I E D (12 letters).
Alphabetical: D D E E F I I I N N T U.
Compare: U/D, N/D, I/E, D/E, E/F, N/I, T/I, I/I, F/N, I/N, E/T, D/U.
Only the 8th letter (I) stays in place → 1.

Cetking shortcut: Write the sorted word directly under the original – a match jumps out.

Answer: C

Approach: Rewrite the word per the rule and compare positions.

Solution:
SEPTUAGENARIAN: S E P T U A G E N A R I A N (14 letters).
First three reversed: SEP → P E S. Next three reversed: TUA → A U T.
Remaining G E N A R I A N in alphabetical order: A A E G I N N R.
New word: P E S A U T A A E G I N N R.
Compare with S E P T U A G E N A R I A N: only positions 2 (E) and 5 (U) are unchanged.
Letters whose positions change = 14 − 2 = 12.

Cetking shortcut: Reversing a group of 3 keeps its middle letter fixed – so check positions 2 and 5 first, then the sorted tail.

Answer: E

Approach: Transform each letter, then sort.

Solution:
EDUCATION: E D U C A T I O N.
Consonants +1: D→E, C→D, T→U, N→O. Vowels −1: E→D, U→T, A→Z, I→H, O→N.
New letters: D E T D Z U H N O.
Alphabetical: D D E H N O T U Z.
4th from the left = H.

Why the other options are wrong:

• D E is 3rd, not 4th (D, D, E, H…).
• A N, B U, E O – later positions.

Answer: C

Approach: For every pair of letters, compare the gap in the word with the gap in the alphabet.

Solution:
Word positions: P1 H2 Y3 S4 I5 C6 A7 L8.
Alphabet values: P16 H8 Y25 S19 I9 C3 A1 L12.
A pair qualifies when |word-position difference| = |alphabet-value difference|.
P–S: word gap 4−1 = 3, alphabet gap 19−16 = 3 ✓
I–L: word gap 8−5 = 3, alphabet gap 12−9 = 3 ✓
Every other pair fails (e.g. H–I: word gap 3, alphabet gap 1; C–A: 1 vs 2).
Total qualifying pairs = 2.

Cetking shortcut: Write the alphabet values under the word and scan each letter only against letters within a few places; large value jumps (Y=25, A=1) can be ignored quickly.

Answer: D

Approach: Rewrite only the changed part of the alphabet, then count positions.

Solution:
A–G swap places with T–Z, with A taking T's place: positions 20–26 now read A B C D E F G (and T–Z move to positions 1–7).
18th letter from the left is still R (positions 8–19 are unchanged).
Third letter to the right of the 18th = position 21.
Position 21 now holds B.

Why the other options are wrong: C, D, E, F sit at positions 22–25 after the swap — they come from counting 4–7 places instead of 3, or from starting at the wrong letter.

Cetking shortcut: Positions 8–19 never change, so only map position 21 → A at 20, B at 21.

Answer: E

Approach: Write the word and its alphabetical arrangement position by position and count matches.

Solution:

Cetking shortcut: Sort by counting letters (A2, E1, I2, L2, N1, O1, R1, T2) and fill positions; compare only where the same letter could land.

Answer: D

Approach: Write alphabet positions; a pair qualifies when |difference in values| = gap in positions in the word.

Solution:
S19 T20 R18 E5 N14 G7 T20 H8 E5 N14 (positions 1–10).
S–T (pos 1, 2): values differ by 1, positions by 1 → no letter between in either ✓
E–G (pos 4, 6): values differ by 2 (F between), positions differ by 2 (N between) ✓
Every other pair fails, e.g. R–T (2 vs 4), G–H (1 vs 2), H–E (3 vs 1), N–T (6 vs 2).
Total = 2 pairs.

Cetking shortcut: Check each letter only against letters within ~8 places and with close alphabet values; long gaps rarely match.

Answer: B

Approach: Apply the letter changes, sort, then count letters between in the alphabet.

Solution:
TREATMENT: vowels +1 (E→F, A→B, E→F), consonants −1 (T→S, R→Q, T→S, M→L, N→M, T→S).
New letters: S Q F B S L F M S.
Alphabetical: B F F L M Q S S S.
3rd from left = F; 4th from right = 6th from left = Q.
Letters between F and Q in the alphabet: G to P = 10.

Cetking shortcut: Between letters at positions a and b = |a − b| − 1 → F(6), Q(17): 17 − 6 − 1 = 10.

Answer: D

Approach: Two letter pairs moving in opposite directions.

Solution: First pair: WX → UV → ST → QR (each letter −2) → next OP.
Second pair: CD → EF → GH → IJ (each letter +2) → next KL.
Next term = OPKL.

Why the other options are wrong:
B OPQR — second pair should be KL.
C, D, E — first pair is not OP.

Cetking shortcut: First pair alone (OP) cuts to A or B; second pair KL settles it.

Answer: A

Approach: Letter matrix: convert letters to alphabet positions.

Solution: Row 1: F(6) + I(9) = 15 = O ✓
Row 2: A(1) + J(10) = 11 = K ✓
Row 3: E(5) + M(13) = 18 = R.

Why the other options are wrong:
A Q = 17, C S = 19, D B = 2, E T = 20 – none equals 5 + 13.

Cetking shortcut: Memorise EJOTY (5, 10, 15, 20, 25) to convert letters instantly.

Answer: B

Approach: Letter series — each letter moves by a fixed step.

Solution:
First letters: C, E, G, __, K → +2 each → I.
Second & third letters (same in each term): M, O, Q, __, U → +2 each → S.
Missing term = ISS.

Why the other options are wrong:
A (GRR), B (GSS): first letter G repeats instead of moving to I.
D (ITT): T is +3 from Q.
E (HSS): H is +1 from G.

Cetking shortcut: Check the first letter only: C→E→G→I→K. Only C and D start with I; the Q→S step then picks C.

Answer: C

Approach: Letter groups grow by one letter; starting positions jump by +2, +4, +6, +8.

Solution:
Group sizes: 1, 2, 3, ?, 5 → the missing group has 4 letters.
Starting letters: A(1), C(3), G(7), ?, U(21).
Jumps: +2, +4, so next +6 → 13 = M, then 13 + 8 = 21 = U ✓
Missing group = MNOP.
Equivalently, letters skipped between groups: B (1), EF (2), JKL (3), QRST (4).

Why the other options are wrong:
A LMNO: starts at L (12), breaks the +6 jump.
B MNO: only 3 letters.
D NOPQ: starts at N (14).
E OPQZ: not consecutive letters.

Cetking shortcut: Count skipped letters: 1, 2, 3, 4 — after GHI skip JKL, so the group starts at M.

Answer: C

Approach: Reverse the alphabet, then count positions to the right of the given letter.

Solution:
Reversed order: Z Y X W V U T S R Q P O N M L K J I H G F E D C B A.
In this order, the letters to the right of O are N, M, L, K, J, I, H, G ...
Counting 8 places: N(1), M(2), L(3), K(4), J(5), I(6), H(7), G(8).

Why the other options are wrong:
A (F): this is the 9th letter to the right.
C (V), D (W): these are to the LEFT of O in the reversed order.
E (A): this is the last letter of the reversed series, not the 8th from O.

Cetking shortcut: In the reversed series, moving right from a letter means going backwards in the normal alphabet. O = 15, so 15 − 8 = 7, which is G.

Answer: B

Approach: Reverse the alphabet, locate the reference letter from the right end, then count to its left.

Solution:
Reversed order: Z Y X W V U T S R Q P O N M L K J I H G F E D C B A.
Ninth letter from the right end (A is 1st): A, B, C, D, E, F, G, H, I.
Fifth letter to the left of I: J(1), K(2), L(3), M(4), N(5).

Why the other options are wrong:
A (P): this is the 7th letter to the left of I.
C (D): this is to the right of I, not the left.
D (W): far to the left; it is the 4th letter from the left end.
E (M): this is only the 4th letter to the left of I.

Cetking shortcut: In the reversed series, the k-th letter from the right is simply the k-th letter of the normal alphabet (9th = I). Left in the reversed series = forward in the normal alphabet: I + 5 = N.

Answer: B

Approach: Letter-deletion series: remove the first and last letters alternately.

Solution:
ADVENTURE → DVENTURE: first letter A removed.
DVENTURE → DVENTUR: last letter E removed.
DVENTUR → ? : remove the first letter D, giving VENTUR.
Check: VENTUR → VENTU (last letter R removed), which matches the next term.

Why the other options are wrong:
A (DVENT): two letters are dropped from the end at once.
B (VENTURE): brings back the E that was already removed.
D (DVENTU): drops the last letter twice in a row and breaks the alternation.
E (ADVENT): brings back the A that was already removed.

Cetking shortcut: Each term is one letter shorter than the one before. The missing term must have 6 letters and lie inside both DVENTUR and VENTU, so it is VENTUR.

Answer: C

Approach: Split the series into blocks of four consecutive letters.

Solution:
Block 1: m n o p
Block 2: n o p q (starts one letter later)
Block 3: o p q r
Block 4 must start at p: p q r s.

Why the other options are wrong: q r s t is block 5, not block 4; m n o p repeats block 1; o q r s and m o p q break the consecutive-letter rule.

Cetking shortcut: The first letters of the blocks (m, n, o, …) give the start of the missing block directly.

Answer: E

Approach: In the reversed alphabet, position from left = 27 − normal position.

Solution:
Reversed: Z Y X W V U T S R Q P O N M L K J I H G F E D C B A.
O (15th letter) is at 27 − 15 = 12th from the left.
Eighth to its right = 12 + 8 = 20th from left.
Letter = 27 − 20 = 7th letter = G.

Why the other options are wrong: I and H are the 18th/19th; Q and S are to the left of O in the reversed series.

Cetking shortcut: In the reversed series 'right' means going back in the alphabet: O − 8 = G.

Answer: B

Approach: In the reversed alphabet Z … A, the k-th letter from the right is the k-th letter of the normal alphabet.

Solution:
Reversed: Z Y X … C B A. Counting from the right: A(1), B(2), … → 9th from right = I.
I is at 27 − 9 = 18th from the left.
Fifth to its left = 18 − 5 = 13th from left = 27 − 13 = 14th letter = N.

Why the other options are wrong: M is 4th to the left of I; O is 6th; L and W are wrong counts/directions.

Cetking shortcut: Left in the reversed series = forward in the normal alphabet: I + 5 = N.

Answer: D

Approach: Write the new series and count positions.

Solution:
New series: A B C … O P | Z Y X W V U T S R Q.
Positions 1–16 unchanged; 13th from left = M.
Sixth to its right = 13 + 6 = 19th position.
Positions 17, 18, 19 = Z, Y, X.

Why the other options are wrong: S would be the 19th letter of the normal alphabet (not offered); W, V are the 20th/21st; Y is the 18th; G is to the left of M.

Answer: A

Approach: The n-th letter appears n times, so positions end at triangular numbers n(n + 1)/2.

Solution:
Last position of the n-th letter = n(n + 1)/2.
n = 23: 23 × 24/2 = 276 (W ends at 276).
n = 24: 24 × 25/2 = 300 (X occupies 277 to 300).
288 lies between 277 and 300 → 24th letter = X.

Cetking shortcut: Solve n²/2 ≈ 288 → n ≈ 24; check 276 < 288 ≤ 300.

Answer: B

Approach: Column-wise alphabet series.

Solution:
1st letters: I, L, O, R → +3 each → next U (21).
2nd letters: C, E, G, I → +2 each → next K (11).
3rd letters: F, I, L, O → +3 each → next R (18).
Next group = U K R.
The opening group UNR does not follow this progression (the series runs from ICF onward).

Why the other options are wrong:
E UKS, C ULS — third letter should be R.
D UJR — second letter should be K (I + 2).
A WTR — first letter should be U.

Cetking shortcut: The answer repeats the 1st and 3rd letters of the first term (U _ R), so only the middle letter needs work.

Answer: B

Approach: The n-th letter appears n times, so the last position of the n-th letter is n(n+1)/2.

Solution:
Find n with n(n+1)/2 ≥ 288.
n = 23: 23 × 24 / 2 = 276; n = 24: 24 × 25 / 2 = 300.
Positions 277–300 are the 24th letter = X (288 is the 12th 'x').

Why the other options are wrong:
W: 23rd letter, ends at position 276.
Z / V / U: 26th, 22nd, 21st letters.

Cetking shortcut: √(2 × 288) = √576 = 24 → the 24th letter, X (then confirm 276 < 288 ≤ 300).

Answer: A