Osmanian

2.1 Reversing Entire Alphabet

2.1 - Reversing Entire Alphabet - Alphabetical Reasoning Problems

Below are 10 example problems for the Reversing the Entire Alphabet type of Alphabetical Reasoning questions, as described. Each problem includes the question, the process of solving it, the solution, and an explanation. The problems cover the common variants: finding the letter in the reversed alphabet, finding the original letter, and identifying patterns or sequences in the reversed alphabet. The reversed alphabet maps A to Z, B to Y, C to X, ..., Z to A, based on position (1st → 26th, 2nd → 25th, ..., 26th → 1

Problem 1: Find the letter in the reversed alphabet

Question: If the alphabet is reversed, what letter replaces 'F'?

Process:

The normal alphabet is A, B, C, ..., Z (26 letters).
The reversed alphabet is Z, Y, X, ..., A.
F is the 6th letter in the normal alphabet (A=1, B=2, ..., F=6).
In the reversed alphabet, the 6th letter from the start corresponds to the 6th letter from the end in the normal alphabet.
Calculate: 26 - 6 + 1 = 21. The 21st letter in the normal alphabet is U (A=1, B=2, ..., U=21).
Alternatively, map directly: A→Z, B→Y, C→X, D→W, E→V, F→U.

Solution: U

Explanation: Since F is the 6th letter in the forward alphabet, we find the 6th letter in the reversed alphabet (Z, Y, X, W, V, U, ...), which is U. The formula for the reversed letter is: for the nth letter, the reversed letter is the (27-n)th letter in the normal alphabet. Here, n=6, so 27-6=21, and the 21st letter is U.

Problem 2: Find the letter in the reversed alphabet

Question: What letter corresponds to 'K' in the reversed alphabet?

Process:

K is the 11th letter in the normal alphabet (A=1, B=2, ..., K=11).
In the reversed alphabet, the 11th letter corresponds to the (26 - 11 + 1) = 16th letter in the normal alphabet.
The 16th letter is P (A=1, B=2, ..., P=16).
Alternatively: A→Z, B→Y, ..., K→P (pairing: Z-A, Y-B, ..., P-K).

Solution: P

Explanation: K, being the 11th letter, maps to the 11th letter in the reversed alphabet (Z, Y, X, ..., P, O, ...), which is P. Using the position formula (27-11=16), the 16th letter in the normal alphabet is P.

Problem 3: Find the original letter

Question: In the reversed alphabet, the letter 'S' is given. What was the original letter in the normal alphabet?

Process:

In the reversed alphabet (Z, Y, X, ..., A), S is the 8th letter (Z=1, Y=2, X=3, W=4, V=5, U=6, T=7, S=8).
The 8th letter in the reversed alphabet corresponds to the 8th letter from the end in the normal alphabet.
Calculate: 26 - 8 + 1 = 19. The 19th letter in the normal alphabet is S (A=1, ..., S=19).
Alternatively: S in reversed alphabet pairs with I in normal alphabet (Z-A, Y-B, ..., S-I).

Solution: I

Explanation: S is the 8th letter in the reversed alphabet, so we find the 8th letter from the end in the normal alphabet (or 19th from the start: 26-8+1=19). The 19th letter is I. The pairing confirms S→I.

Problem 4: Find the original letter

Question: If 'M' is a letter in the reversed alphabet, what is its corresponding letter in the normal alphabet?

Process:

In the reversed alphabet (Z, Y, X, ..., A), M is the 14th letter (Z=1, Y=2, ..., M=14).
The 14th letter in the reversed alphabet corresponds to the 14th letter from the end in the normal alphabet: 26 - 14 + 1 = 13.
The 13th letter in the normal alphabet is M (A=1, ..., M=13).
Alternatively: M in reversed alphabet pairs with N (Z-A, Y-B, ..., M-N).

Solution: N

Explanation: M is the 14th letter in the reversed alphabet, mapping to the 14th letter from the end in the normal alphabet (26-14+1=13), which is N. The pairing M→N confirms this.

Problem 5: Identify a pattern in the reversed alphabet

Question: In the reversed alphabet, what are the first 5 letters of the sequence starting from 'Z'?

Process:

The reversed alphabet is Z, Y, X, W, V, U, T, ..., A.
The sequence starts at Z (1st letter in reversed alphabet).
The first 5 letters are: Z (1st), Y (2nd), X (3rd), W (4th), V (5th).

Solution: Z, Y, X, W, V

Explanation: The reversed alphabet begins with Z, Y, X, W, V, ..., so the first 5 letters are simply the first 5 in this order. This tests understanding of the reversed alphabet’s sequence.

Problem 6: Identify a sequence in the reversed alphabet

Question: In the reversed alphabet, what is the sequence of letters at positions 3, 6, and 9?

Process:

Reversed alphabet: Z, Y, X, W, V, U, T, S, R, ...
Position 3: X
Position 6: U
Position 9: R
Sequence: X, U, R.

Solution: X, U, R

Explanation: We identify the letters at the specified positions in the reversed alphabet. Position 3 is X, position 6 is U, and position 9 is R, forming the sequence X, U, R.

Problem 7: Find the letter in the reversed alphabet (complex)

Question: If the alphabet is reversed, what letter replaces the 10th letter of the normal alphabet?

Process:

The 10th letter in the normal alphabet is J (A=1, B=2, ..., J=10).
In the reversed alphabet, the 10th letter from the start corresponds to the 10th letter from the end in the normal alphabet.
Calculate: 26 - 10 + 1 = 17. The 17th letter is Q (A=1, ..., Q=17).
Alternatively: J→Q (Z-A, Y-B, ..., Q-J).

Solution: Q

Explanation: The 10th letter (J) maps to the 10th letter in the reversed alphabet, which is Q (position 17 in the normal alphabet: 26-10+1=17). The pairing J→Q confirms.

Problem 8: Find the original letter (complex)

Question: In the reversed alphabet, the 20th letter is given. What is the original letter in the normal alphabet?

Process:

Reversed alphabet: Z, Y, X, ..., G, F, E, D, C, B, A.
The 20th letter: Count Z=1, Y=2, ..., I=18, H=19, G=20. So, the 20th letter is G.
The 20th letter in the reversed alphabet corresponds to the 20th letter from the end in the normal alphabet: 26 - 20 + 1 = 7.
The 7th letter in the normal alphabet is G (A=1, ..., G=7).
Alternatively: G in reversed alphabet pairs with U (Z-A, Y-B, ..., G-U).

Solution: U

Explanation: The 20th letter in the reversed alphabet is G, which maps to the 20th letter from the end in the normal alphabet (26-20+1=7), giving U. The pairing G→U confirms.

Problem 9: Identify a pattern in the reversed alphabet

Question: In the reversed alphabet, what are the letters that correspond to the vowels (A, E, I, O, U) in the normal alphabet?

Process:

Vowels in normal alphabet: A (1st), E (5th), I (9th), O (15th), U (21st).
Reversed alphabet mapping:
- A (1st) → 26th = Z (26-1+1=26).
- E (5th) → 22nd = V (26-5+1=22).
- I (9th) → 18th = R (26-9+1=18).
- O (15th) → 12th = L (26-15+1=12).
- U (21st) → 6th = U (26-21+1=6).
Letters: Z, V, R, L, U.

Solution: Z, V, R, L, U

Explanation: Each vowel’s position in the normal alphabet is mapped to its corresponding position in the reversed alphabet using the formula (26-n+1). The resulting letters are Z, V, R, L, U.

Problem 10: Identify a sequence in the reversed alphabet

Question: In the reversed alphabet, what is the sequence formed by every 4th letter starting from the 1st letter?

Process:

Reversed alphabet: Z, Y, X, W, V, U, T, S, R, Q, P, ...
Start at 1st letter (Z), then take every 4th letter:
- 1st: Z
- 5th: V (Z, Y, X, W, V)
- 9th: R (V, U, T, S, R)
- 13th: N (R, Q, P, O, N)
- Stop at a reasonable sequence length (4 letters).
Sequence: Z, V, R, N.

Solution: Z, V, R, N

Explanation: Starting at Z (1st), we select every 4th letter in the reversed alphabet: 1st (Z), 5th (V), 9th (R), 13th (N). This forms the sequence Z, V, R, N.

Summary of Problems

Problems 1-2, 7: Find the letter in the reversed alphabet (direct mapping).
Problems 3-4, 8: Find the original letter given a letter in the reversed alphabet.
Problems 5-6, 9-10: Identify patterns or sequences in the reversed alphabet (sequences, specific positions, or properties like vowels).
Each solution uses the reversed alphabet’s positional logic (nth letter → (27-n)th letter) or direct pairing (A-Z, B-Y, ..., Z-A).

If you need more examples, specific variants, or deeper explanations for any problem, let me know!

Major Types of Patterns in Alphabetical Reasoning

Fixed Positional Shifts (Forward or Backward) - Practice
Reversing Letters - Practice
Alternating Shifts - Practice
Positional Relationships (Letter Positions in Alphabet) - Practice
Non-Alphabetical Relationships (Pattern-Based) - Practice
Mirroring or Complementary Letters - Practice
Skipping Letters - Practice
Grouping or Clustering - Practice
Cyclic or Modular Patterns - Practice
Letter Substitution or Replacement - Practice
Interleaving or Alternating Letters - Practice
Anagram-Based Patterns - Practice

Fixed Positional Shifts variations and example problems for practice

In alphabetical reasoning, the Fixed Positional Shifts category involves transforming letters by shifting them a fixed number of positions forward or backward in the alphabet (A=1, B=2, ..., Z=26). This is one of the most common pattern types in aptitude tests, as it is straightforward yet allows for variations in complexity. Below, I’ll outline the types of questions that can be asked within this category, identify the distinct varieties, and provide 10 Practice questions for each variety, complete with solutions and explanations.

Types of Questions in Fixed Positional Shifts

The Fixed Positional Shifts category can be broken down into distinct varieties based on the nature of the shift and the structure of the question. The key variations include:

Uniform Forward Shift: All letters in a pair or word are shifted forward by a fixed number of positions (e.g., +2, +3). – View
Uniform Backward Shift: All letters in a pair or word are shifted backward by a fixed number of positions (e.g., -2, -3). – View
Mixed Forward and Backward Shifts (Same Question): Different parts of the question (e.g., different letters in a word or different pairs) use forward and backward shifts, but each shift is fixed. – View
Wrap-Around Shifts (Modular Arithmetic): Shifts that go beyond Z or before A, wrapping around the alphabet (e.g., Z+1=A, A-1=Z). – View
Shifts Applied to Sequences: The shift is applied to a sequence of letters or pairs, where the pattern continues across multiple terms. – View

Mixed Forward and Backward Shifts - Alphabetical Reasoning Problems

3. Mixed Forward and Backward Shifts - Alphabetical Reasoning Problems

Description: Different letters or parts of a pair/word use fixed forward and backward shifts (e.g., first letter +2, second letter -2). This adds complexity by requiring identification of multiple fixed shifts.

10 Practice Questions

AB : CE :: PQ : ?

Options: SU, TV, RU, QT
Solution: SU
Explanation: In AB to CE, the first letter shifts forward by 2 (A+2=C), and the second shifts forward by 3 (B+3=E). For PQ (P+2=S, Q+3=U), the answer is SU.

CAT : AFU :: DOG : ?

Options: FJY, EKX, GLZ, HMU
Solution: FJY
Explanation: In CAT to AFU, the first letter shifts backward by 2 (C-2=A), the second forward by 2 (A+2=C), and the third forward by 1 (T+1=U). For DOG (D-2=B, O+2=Q, G+1=I), the answer is FJY.

XY : UW :: BC : ?

Options: ZA, YB, AC, IE
Solution: ZA
Explanation: In XY to UW, the first letter shifts backward by 1 (X-1=W), and the second backward by 2 (Y-2=W). For BC (B-1=A, C-2=A), the answer is ZA.

LMN : KOP :: DEF : ?

Options: AFI, BGH, CGJ, DIK
Solution: AFI
Explanation: In LMN to KOP, the first letter shifts backward by 2 (L-2=J), the second forward by 1 (M+1=N), and the third forward by 2 (N+2=P). For DEF (D-2=B, E+1=F, F+2=I), the answer is AFI.

PQ : OS :: UV : ?

Options: SW, TX, RY, QZ
Solution: SW
Explanation: In PQ to OS, the first letter shifts backward by 1 (P-1=O), and the second forward by 2 (Q+2=S). For UV (U-1=T, V+2=X), the answer is SW.

BIG : AJI :: FUN : ?

Options: DWP, EVQ, FXS, GYT
Solution: EVQ
Explanation: In BIG to AJI, the first letter shifts backward by 1 (B-1=A), the second forward by 1 (I+1=J), and the third forward by 2 (G+2=I). For FUN (F-1=E, U+1=V, N+2=P), the answer is EVQ.

EF : IG :: KL : ?

Options: NJ, OK, PI, QM
Solution: NJ
Explanation: In EF to IG, the first letter shifts forward by 3 (E+3=I), and the second forward by 1 (F+1=G). For KL (K+3=N, L+1=M), the answer is NJ.

SUN : QRM :: MOON : ?

Options: JHLK, KIMP, LJNQ, MKOS
Solution: KIMP
Explanation: In SUN to QRM, the first letter shifts backward by 2 (S-2=Q), the second backward by 1 (U-1=T), and the third forward by 1 (N+1=O). For MOON (M-2=K, O-1=N, O+1=P, N+1=O), the answer is KIMP.

CD : AE :: JK : ?

Options: IL, HM, GN, FO
Solution: IL
Explanation: In CD to AE, the first letter shifts backward by 1 (C-1=B), and the second forward by 1 (D+1=E). For JK (J-1=I, K+1=L), the answer is IL.

HAT : JCV :: BAG : ?

Options: AFI, BGH, CGJ, DIK
Solution: AFI
Explanation: In HAT to JCV, the first letter shifts forward by 1 (H+1=I), the second forward by 2 (A+2=C), and the third backward by 1 (T-1=S). For BAG (B+1=C, A+2=C, G-1=F), the answer is AFI.