Alphabet Test Questions in English

(D) To find the number of pairs of letters in the word $DELUSION$ that have as many letters between them as in the English alphabet,we check the sequence forward and backward:
Word: $D, E, L, U, S, I, O, N$
Positions: $D(4), E(5), L(12), U(21), S(19), I(9), O(15), N(14)$
Checking forward:
- $D$ to $E$: $D(4), E(5)$ (Adjacent,$0$ letters between,matches alphabet sequence $D-E$)
- $D$ to $L$: $4$ to $12$ ($7$ letters between,alphabet has $7$ letters between $D$ and $L$)
- $E$ to $L$: $5$ to $12$ ($6$ letters between,alphabet has $6$ letters between $E$ and $L$)
- $I$ to $N$: $9$ to $14$ ($4$ letters between,alphabet has $4$ letters between $I$ and $N$)
Pairs found:
$1$. $(D, E)$
$2$. $(D, L)$
$3$. $(E, L)$
$4$. $(I, N)$
Since there are $4$ pairs,and the options provided are None,One,Two,None of these,the correct choice is $D$ (None of these).

(B) The letters available in the word $TEAR$ are $T, E, A, R$.
We need to form three-letter meaningful words starting with $A$.
Since we cannot repeat letters,we need to choose two more letters from the remaining set ${T, E, R}$.
The possible combinations of three letters starting with $A$ are:
$1$. $A + R + E = ARE$
$2$. $A + R + T = ART$
$3$. $A + T + E = ATE$
Thus,there are $3$ meaningful words that can be formed.

(D) The word is $ARROGANCE$. The positions are:
$1: A, 2: R, 3: R, 4: O, 5: G, 6: A, 7: N, 8: C, 9: E$.
According to the given condition:
$1st$ $(A)$ swaps with $5th$ $(G)$: $G$ comes to $1st$,$A$ comes to $5th$.
$2nd$ $(R)$ swaps with $6th$ $(A)$: $A$ comes to $2nd$,$R$ comes to $6th$.
$3rd$ $(R)$ swaps with $7th$ $(N)$: $N$ comes to $3rd$,$R$ comes to $7th$.
$4th$ $(O)$ swaps with $8th$ $(C)$: $C$ comes to $4th$,$O$ comes to $8th$.
$9th$ $(E)$ remains unchanged.
Combining these,the new arrangement is $GANCARROE$.

(C) To find the number of pairs of letters in the word $CHILDREN$ that have the same number of letters between them as in the English alphabet,we check the sequence:
$C(3), H(8), I(9), L(12), D(4), R(18), E(5), N(14)$
$1$. $H$ and $I$: $H$ is $8$th and $I$ is $9$th letter. In the word,they are adjacent (zero letters between them),and in the alphabet,there are zero letters between them. (Pair: $HI$)
$2$. $E$ and $I$: $E$ is $5$th and $I$ is $9$th letter. In the word,there are $3$ letters ($D, R, E$ is not correct,let's re-evaluate: $I(9), L(12), D(4), R(18), E(5)$). Between $I$ and $E$,there are $3$ letters $(L, D, R)$. In the alphabet,there are $3$ letters $(F, G, H)$. (Pair: $EI$)
$3$. $H$ and $N$: $H$ is $8$th and $N$ is $14$th. Between them in the word are $I, L, D, R, E$ ($5$ letters). In the alphabet,there are $5$ letters $(I, J, K, L, M)$. (Pair: $HN$)
$4$. $I$ and $N$: $I$ is $9$th and $N$ is $14$th. Between them in the word are $L, D, R, E$ ($4$ letters). In the alphabet,there are $4$ letters $(J, K, L, M)$. (Pair: $IN$)
Thus,there are $4$ such pairs: $(HI, EI, HN, IN)$.

(D) The given word is $OUTRAGEOUS$.
The letters at the specified positions are:
$3^{rd}$ letter: $T$
$5^{th}$ letter: $A$
$7^{th}$ letter: $E$
$10^{th}$ letter: $S$
The letters are $T, A, E, S$.
Using these letters,we can form the following meaningful words:
$1$. $SEAT$
$2$. $EAST$
$3$. $SATE$
Since more than one meaningful word can be formed,the answer is $X$.

(C) The original word is $PRINCE$.
The positions of the letters are:
$P(1), R(2), I(3), N(4), C(5), E(6)$.
Arranging the letters of $PRINCE$ in alphabetical order:
$C, E, I, N, P, R$.
Comparing the original and rearranged positions:
Original: $P, R, I, N, C, E$
Alphabetical: $C, E, I, N, P, R$
Comparing index by index:
$1: P \neq C$
$2: R \neq E$
$3: I = I$
$4: N = N$
$5: C \neq P$
$6: E \neq R$
Thus,the letters $I$ and $N$ remain in their original positions. Therefore,two letters remain unchanged.

(D) The letters provided are $E, S, R, O$.
Using each letter only once,the meaningful English words that can be formed are:
$1$. $ROSE$ (a flower)
$2$. $SORE$ (painful)
$3$. $EROS$ (the Greek god of love)
$4$. $ORES$ (plural of ore,a naturally occurring solid material)
Since there are $4$ meaningful words,the correct answer is 'More than three'.

(B) To find the number of such pairs,we compare the positions of letters in the word $CONSTABLE$ with their positions in the English alphabet.
Word: $C, O, N, S, T, A, B, L, E$
Positions: $3, 15, 14, 19, 20, 1, 2, 12, 5$
Checking pairs:
$1$. $C$ to $E$: $C(3), D(4), E(5)$. There is one letter $(D)$ between them in the alphabet,and in the word $CONSTABLE$,there are $7$ letters between them. (No match)
$2$. $A$ to $B$: $A(1), B(2)$. There are $0$ letters between them in the alphabet,and in the word $CONSTABLE$,there are $0$ letters between them. (Match: $1$ pair)
$3$. $L$ to $E$: $E(5), F(6), G(7), H(8), I(9), J(10), K(11), L(12)$. There are $6$ letters between them in the alphabet,and in the word $CONSTABLE$,there are $3$ letters between them. (No match)
Checking all forward and backward sequences,the only pair that satisfies the condition is $(A, B)$.
Therefore,there is only $1$ such pair.

(D) The letters provided are $A, C, E, R$.
By rearranging these letters,we can form the following meaningful words:
$1. \text{RACE}$
$2. \text{CARE}$
$3. \text{ACRE}$
Since there are $3$ meaningful words that can be formed,the correct option is $D$.

(B) $1$. The word is $FLOURISH$. The vowels are $O, U, I$ and consonants are $F, L, R, S, H$.
$2$. Arrange vowels alphabetically: $I, O, U$.
$3$. Arrange consonants alphabetically: $F, H, L, R, S$.
$4$. Combine them: $I, O, U, F, H, L, R, S$.
$5$. Replace vowels with the previous letter: $I \rightarrow H, O \rightarrow N, U \rightarrow T$.
$6$. Replace consonants with the next letter: $F \rightarrow G, H \rightarrow I, L \rightarrow M, R \rightarrow S, S \rightarrow T$.
$7$. The new sequence is $H, N, T, G, I, M, S, T$.
$8$. The third letter from the right end is $S$.

(B) To find the number of pairs of letters in the word $PHYSICAL$ that have the same number of letters between them as in the English alphabet,we check both forward and backward directions:
$1$. Word: $P, H, Y, S, I, C, A, L$
$2$. Forward direction:
- $P$ to $H, Y, S, I, C, A, L$: No pairs found.
- $H$ to $Y, S, I, C, A, L$: No pairs found.
- $Y$ to $S, I, C, A, L$: No pairs found.
- $S$ to $I, C, A, L$: No pairs found.
- $I$ to $C, A, L$: No pairs found.
- $C$ to $A, L$: No pairs found.
- $A$ to $L$: No pairs found.
$3$. Backward direction:
- $L$ to $A, C, I, S, Y, H, P$: $L$ to $A$ (no),$L$ to $C$ (no),$L$ to $I$ (no),$L$ to $S$ (no),$L$ to $Y$ (no),$L$ to $H$ (no),$L$ to $P$ (no).
- $A$ to $C, I, S, Y, H, P$: $A$ to $C$ (one letter $B$ in alphabet,but none here),$A$ to $I$ (no).
- $C$ to $I, S, Y, H, P$: $C$ to $I$ (no).
- $I$ to $S, Y, H, P$: $I$ to $S$ (no).
- $S$ to $Y, H, P$: $S$ to $Y$ (no).
- $Y$ to $H, P$: $Y$ to $H$ (no).
- $H$ to $P$: $H$ to $P$ (in alphabet $H, I, J, K, L, M, N, O, P$ - $7$ letters between them; in word $H, Y, S, I, C, A, L, P$ - $6$ letters between them,no).
Wait,let's re-check: $P(16), H(8), Y(25), S(19), I(9), C(3), A(1), L(12)$.
Checking pairs: $(I, L)$ - $I(9)$ and $L(12)$. Between them are $J, K$ ($2$ letters). In the word $PHYSICAL$,between $I$ and $L$ are $C, A$ ($2$ letters). Thus,$(I, L)$ is one pair.
Therefore,there is only one such pair.

(D) To find the number of pairs of letters in the word $SHIFTED$ that have the same number of letters between them as in the English alphabet,we check both forward and backward directions:
$1$. Word: $S, H, I, F, T, E, D$
$2$. Forward direction:
- $S$ to $H$: $S(19), H(8)$ (No)
- $S$ to $I$: $S(19), I(9)$ (No)
- $S$ to $F$: $S(19), F(6)$ (No)
- $S$ to $T$: $S(19), T(20)$ (One letter between them: $S, T$ are adjacent in alphabet,but here $H, I, F$ are between them. No.)
- $H$ to $I$: $H(8), I(9)$ (Adjacent in word,adjacent in alphabet. Pair: $(H, I)$)
- $H$ to $F$: $H(8), F(6)$ (No)
- $I$ to $F$: $I(9), F(6)$ (No)
- $F$ to $E$: $F(6), E(5)$ (Adjacent in word,adjacent in alphabet. Pair: $(F, E)$)
- $E$ to $D$: $E(5), D(4)$ (Adjacent in word,adjacent in alphabet. Pair: $(E, D)$)
$3$. Backward direction:
- $D$ to $E$: $(D, E)$ (Already counted as $E, D$)
- $D$ to $F$: $D(4), F(6)$ (One letter $E$ between them in word,one letter $E$ between them in alphabet. Pair: $(D, F)$)
- $D$ to $T$: No
- $E$ to $F$: $(E, F)$ (Already counted)
- $F$ to $I$: No
- $I$ to $H$: $(I, H)$ (Already counted)
- $I$ to $S$: No
- $H$ to $S$: No
Checking the pairs: $(H, I), (F, E), (E, D), (D, F)$.
Wait,let's re-verify:
$S(19), H(8), I(9), F(6), T(20), E(5), D(4)$
Pairs:
$1$. $(H, I)$ - $8, 9$ (Yes)
$2$. $(F, E)$ - $6, 5$ (Yes)
$3$. $(E, D)$ - $5, 4$ (Yes)
$4$. $(D, F)$ - $4, 6$ (Yes,$E$ is between them)
Total pairs = $4$. Since $4$ is not in the options,let's re-read the word $SHIFTED$.
$S(19), H(8), I(9), F(6), T(20), E(5), D(4)$.
Pairs found: $(H, I), (F, E), (E, D), (D, F)$.
Given the options,if the question implies only forward or specific constraints,but standard logic yields $4$. Since $4$ is not an option,we select the closest logical interpretation or verify if $SHIFTED$ has a typo. Assuming the question is correct,the answer is $3$ if we exclude one,but mathematically it is $4$.

(B) The letters available in the word $GOT$ are $G, O, T$.
Possible two-letter combinations are $GO, GT, OG, OT, TG, TO$.
Among these,the meaningful English words are $GO$ and $TO$.
Therefore,a total of $2$ meaningful words can be formed.

(C) To find the number of pairs of letters in the word $KNIGHT$ that have the same number of letters between them as in the English alphabet,we check the alphabetical order:
$1$. $K$ $(11)$ to $N$ $(14)$: There are $2$ letters $(L, M)$ between them in the alphabet,but $1$ letter $(N)$ in the word. (No)
$2$. $K$ $(11)$ to $I$ $(9)$: No.
$3$. $K$ $(11)$ to $G$ $(7)$: No.
$4$. $K$ $(11)$ to $H$ $(8)$: No.
$5$. $K$ $(11)$ to $T$ $(20)$: No.
$6$. $N$ $(14)$ to $I$ $(9)$: No.
$7$. $N$ $(14)$ to $G$ $(7)$: No.
$8$. $N$ $(14)$ to $H$ $(8)$: No.
$9$. $N$ $(14)$ to $T$ $(20)$: No.
$10$. $I$ $(9)$ to $G$ $(7)$: There is $1$ letter $(H)$ between them in the alphabet,and $1$ letter $(G)$ in the word. (Yes,$I$ and $G$ form a pair).
$11$. $I$ $(9)$ to $H$ $(8)$: No.
$12$. $I$ $(9)$ to $T$ $(20)$: No.
$13$. $G$ $(7)$ to $H$ $(8)$: There are $0$ letters between them in the alphabet,and $0$ letters in the word. (Yes,$G$ and $H$ form a pair).
$14$. $G$ $(7)$ to $T$ $(20)$: No.
$15$. $H$ $(8)$ to $T$ $(20)$: No.
Thus,there are $2$ such pairs: $(I, G)$ and $(G, H)$.

(C) The given word is $STABLE$.
$1$. Identify the consonants and vowels: $S$ (consonant),$T$ (consonant),$A$ (vowel),$B$ (consonant),$L$ (consonant),$E$ (vowel).
$2$. Replace consonants with the previous letter in the alphabet:
$S \rightarrow R$
$T \rightarrow S$
$B \rightarrow A$
$L \rightarrow K$
$3$. Replace vowels with the next letter in the alphabet:
$A \rightarrow B$
$E \rightarrow F$
$4$. The new word formed is $RSAKBF$.
$5$. The third letter from the left end is $A$.

(C) The given letters are $A, E, H, T$.
By using each letter only once,we can form the following meaningful words:
$1$. $HATE$
$2$. $HEAT$
Thus,a total of $2$ meaningful words can be formed.

(D) The word is $PASSIONATELY$.
The letters are:
1st: $P$
2nd: $A$
3rd: $S$
4th: $S$
5th: $I$
6th: $O$
7th: $N$
8th: $A$
9th: $T$
10th: $E$
11th: $L$
12th: $Y$
The 2nd,3rd,10th,and 11th letters are $A, S, E, L$.
Using these letters $(A, S, E, L)$,we can form the following meaningful words:
$1$. $SALE$
$2$. $SEAL$
Since more than one word can be formed,the answer is $Y$.

(A) The letters available in the word $NOTE$ are $N, O, T, E$.
We need to form three-letter words starting with $T$.
The remaining two letters must be chosen from ${N, O, E}$.
The possible combinations are:
$1. T + O + N = TON$
$2. T + O + E = TOE$
$3. T + E + N = TEN$
All these are meaningful English words.
Thus,there are $3$ such words.

(D) The word is $OBSERVANT$. The positions are: $O(1), B(2), S(3), E(4), R(5), V(6), A(7), N(8), T(9)$.
According to the rule:
$1st \leftrightarrow 9th$: $T$ moves to $1st$,$O$ moves to $9th$.
$2nd \leftrightarrow 8th$: $N$ moves to $2nd$,$B$ moves to $8th$.
$3rd \leftrightarrow 7th$: $A$ moves to $3rd$,$S$ moves to $7th$.
$4th \leftrightarrow 6th$: $V$ moves to $4th$,$E$ moves to $6th$.
$5th$ remains unchanged: $R$ stays at $5th$.
Combining these,the new arrangement is $TNAVRSEBO$.

(B) To find the number of pairs of letters in the word $ANSWER$ that have the same number of letters between them as in the English alphabet,we check both forward and backward directions:
$1$. Word: $A, N, S, W, E, R$
$2$. Forward direction:
- $A$ to $N$: $1$ letter between ($N$ is at position $2$,$A$ is at $1$,$2-1-1 = 0$ letters between). In English,$A$ and $N$ have $12$ letters between them.
- $A$ to $S$: $A(1), N(2), S(3)$. $2$ letters between. In English,$A$ and $S$ have $17$ letters between them.
- $N$ to $S$: $N(14), S(19)$. $4$ letters between $(O, P, Q, R)$. In the word,there is $1$ letter $(S)$.
- $R$ to $E$: $R(18), E(5)$. Backward: $E, F, G, H, I, J, K, L, M, N, O, P, Q, R$. There are $12$ letters between $E$ and $R$ in the alphabet. In the word $ANSWER$,the letters between $E$ and $R$ are none. Wait,let's check alphabetical order:
- $A(1), N(14), S(19), W(23), E(5), R(18)$.
- Pair $(A, E)$: $A(1)$ and $E(5)$. Letters between in word: $N, S, W$ ($3$ letters). Letters between in alphabet: $B, C, D$ ($3$ letters). This is a match.
- Pair $(N, R)$: $N(14)$ and $R(18)$. Letters between in word: $S, W, E$ ($3$ letters). Letters between in alphabet: $O, P, Q$ ($3$ letters). This is a match.
There are $2$ such pairs.

(B) Step $1$: Identify the letters in the word $MERCIFUL$: $M, E, R, C, I, F, U, L$.
Step $2$: Arrange these letters in alphabetical order: $C, E, F, I, L, M, R, U$.
Step $3$: Substitute each letter with the alphabet immediately preceding it in the English alphabet:
$C \rightarrow B$
$E \rightarrow D$
$F \rightarrow E$
$I \rightarrow H$
$L \rightarrow K$
$M \rightarrow L$
$R \rightarrow Q$
$U \rightarrow T$
Step $4$: The new arrangement is $BDEHKLQT$.

(B) The word is $CERTIFICATE$. The total number of letters is $11$.
The letters are: $1:C, 2:E, 3:R, 4:T, 5:I, 6:F, 7:I, 8:C, 9:A, 10:T, 11:E$.
According to the rule,we interchange the pairs: $(1,7), (2,8), (3,9), (4,10), (5,11)$.
Original: $C, E, R, T, I, F, I, C, A, T, E$
New positions:
$1 leftrightarrow 7: I, E, R, T, I, F, C, C, A, T, E$
$2 leftrightarrow 8: I, C, R, T, I, F, C, E, A, T, E$
$3 leftrightarrow 9: I, C, A, T, I, F, C, E, R, T, E$
$4 leftrightarrow 10: I, C, A, T, I, F, C, E, R, T, E ightarrow I, C, A, T, I, F, C, E, R, T, E$
Wait,let's re-map carefully:
Original: $C(1), E(2), R(3), T(4), I(5), F(6), I(7), C(8), A(9), T(10), E(11)$
After swapping $(1,7), (2,8), (3,9), (4,10), (5,11)$:
New word: $I, C, A, T, E, F, C, E, R, T, T$
Wait,let's re-index: $I(1), C(2), A(3), T(4), E(5), F(6), C(7), E(8), R(9), T(10), T(11)$.
We need the third to the right of the sixth from the right end.
Sixth from the right end is the $6^{th}$ letter from the end: $11, 10, 9, 8, 7, 6$. The $6^{th}$ letter is $F$.
Third to the right of $F$ is the $3^{rd}$ letter to the right of $F$ (towards the end): $F(6), C(7), E(8), R(9)$.
The letter is $R$.

(D) To find the number of pairs of letters in the word $DOCUMENT$ that have the same number of letters between them as in the English alphabet,we assign the position of each letter: $D(4), O(15), C(3), U(21), M(13), E(5), N(14), T(20)$.
Checking forward and backward:
$1$. $D$ to $E$: $D(4)$ and $E(5)$ have no letters between them in the word,but in the alphabet,there is no letter between them. This is a pair.
$2$. $M$ to $N$: $M(13)$ and $N(14)$ have no letters between them in the word,and in the alphabet,they are consecutive. This is a pair.
$3$. $C$ to $E$: $C(3), U(21), M(13), E(5)$. Between $C$ and $E$ in the word,there are $2$ letters $(U, M)$. In the alphabet,there is $1$ letter $(D)$. Not a match.
$4$. $O$ to $M$: $O(15)$ and $M(13)$ have $1$ letter $(C)$ between them in the word. In the alphabet,there is $1$ letter $(N)$. This is a pair.
The pairs are $(D, E)$,$(M, N)$,and $(O, M)$.
Thus,there are $3$ such pairs.

(D) The word is $ENTERPRISE$. The letters at the first,third,fifth,and eighth positions are $E, T, R, I$ respectively.
Using these letters $(E, T, R, I)$,we can form the following meaningful words: $TIRE, TIER, RITE$.
Since more than one meaningful word can be formed,the answer is $'X'$.

(A) $1$. Identify the letters in the word $DISTANCE$: $D, I, S, T, A, N, C, E$.
$2$. Apply the transformation rules:
- Vowels $(I, A, E)$ are replaced by the next letter: $I \rightarrow J, A \rightarrow B, E \rightarrow F$.
- Consonants $(D, S, T, N, C)$ are replaced by the previous letter: $D \rightarrow C, S \rightarrow R, T \rightarrow S, N \rightarrow M, C \rightarrow B$.
$3$. The new set of letters is: $C, J, R, S, B, M, B, F$.
$4$. Arrange these letters in alphabetical order: $B, B, C, F, J, M, R, S$.
$5$. The third letter from the right is the sixth letter from the left,which is $M$.

(D) Step $1$: Arrange the letters of the word $FOLK$ in alphabetical order.
The alphabetical order of the letters $F, O, L, K$ is $F, K, L, O$.
Step $2$: Substitute each letter with the letter immediately preceding it in the English alphabet.
$F$ becomes $E$ (since $E$ precedes $F$).
$K$ becomes $J$ (since $J$ precedes $K$).
$L$ becomes $K$ (since $K$ precedes $L$).
$O$ becomes $N$ (since $N$ precedes $O$).
Thus,the resultant word is $EJKN$.

(C) The pattern follows a cycle of $1, 2, 3$ for the alphabet sequence.
Assigning values to each letter:
$M$ is the $13^{th}$ letter. $13 \div 3 = 4$ remainder $1$,so $M = 1$.
$U$ is the $21^{st}$ letter. $21 \div 3 = 7$ remainder $0$,so $U = 3$.
$L$ is the $12^{th}$ letter. $12 \div 3 = 4$ remainder $0$,so $L = 3$.
$E$ is the $5^{th}$ letter. $5 \div 3 = 1$ remainder $2$,so $E = 2$.
Sum $= M + U + L + E = 1 + 3 + 3 + 2 = 9$.

(D) The series consists of two parts: the first letter and the second letter of each group.
For the first letter: $W \xrightarrow{-5} R \xrightarrow{-4} N \xrightarrow{-3} K \xrightarrow{-2} I$.
For the second letter: $C \xrightarrow{+2} E \xrightarrow{+4} I \xrightarrow{+6} O \xrightarrow{+8} W$.
Combining these,the next term is $IW$.

(D) The letters provided are $A, R, T, S, E$.
By rearranging these letters,we can form the following meaningful English words:
$1. TEARS$
$2. STARE$
$3. RATES$
$4. ASTER$
Since there are $4$ such words,the correct answer is 'More than three'.

(C) To find the pairs,we check the alphabetical positions of letters in $PROFITABLE$:
$P(16), R(18), O(15), F(6), I(9), T(20), A(1), B(2), L(12), E(5)$.
$1$. Checking forward:
- $P(16)$ to $T(20)$: $16, 17, 18, 19, 20$ (No)
- $A(1)$ to $B(2)$: $1, 2$ (Yes,one pair: $A-B$)
$2$. Checking backward:
- $E(5)$ to $I(9)$: $5, 6, 7, 8, 9$ (No)
- $B(2)$ to $F(6)$: $2, 3, 4, 5, 6$ (Yes,one pair: $B-F$)
Thus,there are two pairs: $(A, B)$ and $(B, F)$.

(D) To find the pairs,we check the alphabetical sequence in the word $ELEVATION$:
$1$. $E$ to $I$: $E(F, G, H)I$ ($3$ letters between them in both).
$2$. $E$ to $A$: $A(B, C, D)E$ ($3$ letters between them in both).
$3$. $V$ to $T$: $T(U)V$ ($1$ letter between them in both).
$4$. $O$ to $N$: $N(O)$ ($0$ letters between them in both).
Thus,there are four such pairs: $(EA, EI, VT, ON)$.

(B) The given word is $WEBPAGE$.
$1$. Identify vowels and consonants:
- $W$ (consonant) $\rightarrow$ preceding letter is $V$.
- $E$ (vowel) $\rightarrow$ next letter is $F$.
- $B$ (consonant) $\rightarrow$ preceding letter is $A$.
- $P$ (consonant) $\rightarrow$ preceding letter is $O$.
- $A$ (vowel) $\rightarrow$ next letter is $B$.
- $G$ (consonant) $\rightarrow$ preceding letter is $F$.
- $E$ (vowel) $\rightarrow$ next letter is $F$.
$2$. The new word formed is $VFAOBFF$.
$3$. Counting the frequency of each letter:
- $V$: $1$ time
- $F$: $3$ times
- $A$: $1$ time
- $O$: $1$ time
- $B$: $1$ time
Therefore,the letter $F$ appears thrice.

(C) To find the pairs,we check the number of letters between each pair in the word $DOCUMENTARY$ and compare it with the English alphabet:
$1$. $D$ to $O$: $D(E, F, G, H, I, J, K, L, M, N)O$ ($10$ letters in word,$10$ in alphabet) - Pair found: $(D, O)$.
$2$. $M$ to $R$: $M(N, O, P, Q)R$ ($4$ letters in word,$4$ in alphabet) - Pair found: $(M, R)$.
$3$. $T$ to $R$: $T(S)R$ ($1$ letter in word,$1$ in alphabet) - Pair found: $(T, R)$.
Wait,let's re-verify:
$D$ to $O$: $D, O$ (Positions $1$ and $3$). Letters between: $C, U, M, E, N, T, A, R, Y$ ($9$ letters). Alphabet: $E, F, G, H, I, J, K, L, M, N$ ($10$ letters). No.
Let's check forward and backward:
$D(4) O(15) C(3) U(21) M(13) E(5) N(14) T(20) A(1) R(18) Y(25)$
Pairs:
- $M(13)$ and $R(18)$: Between them in word is $E, N, T, A$ ($4$ letters). In alphabet: $N, O, P, Q$ ($4$ letters). Yes.
- $T(20)$ and $R(18)$: Between them in word is $A$ ($1$ letter). In alphabet: $S$ ($1$ letter). Yes.
- $D(4)$ and $A(1)$: Between them in word is $O, C, U, M, E, N, T$ ($7$ letters). In alphabet: $B, C$ ($2$ letters). No.
There are two pairs: $(M, R)$ and $(T, R)$.

(C) The given letters are $T, P, S, I$.
By rearranging these letters,we can form the following meaningful English words:
$1$. $TIPS$ (plural of tip)
$2$. $SPIT$ (to eject saliva)
$3$. $PITS$ (plural of pit)
Thus,a total of $3$ meaningful words can be formed.

(D) To find the pairs,we count the letters in the word $HORIZONTAL$ and compare their positions with the English alphabet:
$H(8), O(15), R(18), I(9), Z(26), O(15), N(14), T(20), A(1), L(12)$.
Checking forward and backward:
$1$. $H$ to $L$: $H(8), I(9), J(10), K(11), L(12)$. There are $3$ letters between them in the word $(O, R, I, Z, O, N, T, A)$ - No.
$2$. $H$ to $N$: $H(8)$ to $N(14)$ has $5$ letters $(I, J, K, L, M)$ in the alphabet. In the word,$H$ to $N$ has $O, R, I, Z, O$ ($5$ letters). This is a pair.
$3$. $R$ to $N$: $R(18)$ to $N(14)$ has $3$ letters $(O, P, Q)$ in the alphabet. In the word,$R$ to $N$ has $I, Z, O$ ($3$ letters). This is a pair.
$4$. $O$ to $N$: $O(15)$ to $N(14)$ has $0$ letters in the alphabet. In the word,$O$ and $N$ are adjacent. This is a pair.
$5$. $R$ to $O$: $R(18)$ to $O(15)$ has $2$ letters $(P, Q)$ in the alphabet. In the word,$R$ to $O$ has $I, Z$ ($2$ letters). This is a pair.
Thus,there are four pairs: $(H, N), (R, N), (O, N), (R, O)$.
Since there are four pairs,the correct option is 'More than three'.

(B) The given letters are $S, L, I, K, L$.
To form a five-letter word using each letter exactly once,we need to arrange these letters.
The meaningful words that can be formed are:
$1. SKILL$
$2. KILLS$
Since there are two meaningful words,the correct option is $B$.

(C) The original word is $HABITUAL$.
$1$. Identify vowels and consonants:
- $H$ (consonant),$A$ (vowel),$B$ (consonant),$I$ (vowel),$T$ (consonant),$U$ (vowel),$A$ (vowel),$L$ (consonant).
$2$. Apply the transformation rules:
- For consonants,change to the previous letter: $H \rightarrow G$,$B \rightarrow A$,$T \rightarrow S$,$L \rightarrow K$.
- For vowels,change to the next letter: $A \rightarrow B$,$I \rightarrow J$,$U \rightarrow V$,$A \rightarrow B$.
$3$. Construct the new word:
- $H \rightarrow G$
- $A \rightarrow B$
- $B \rightarrow A$
- $I \rightarrow J$
- $T \rightarrow S$
- $U \rightarrow V$
- $A \rightarrow B$
- $L \rightarrow K$
The new word is $GBAJSVBK$.
$4$. Identify the fourth letter from the left:
- The sequence is $G(1), B(2), A(3), J(4), S(5), V(6), B(7), K(8)$.
- The fourth letter is $J$.

(B) The sequence follows a pattern of $6$ letters in each group.
Group $1$: $b, w, x, y, z, a$ (The sequence is $b, w, x, y, z, a$ where $x$ is the missing letter).
Group $2$: $d, s, t, u, v, c$ (The sequence is $d, s, t, u, v, c$ where $c$ is the missing letter).
Group $3$: $f, o, p, q, r, e$ (The sequence is $f, o, p, q, r, e$ where $f$ is the missing letter).
Thus,the missing letters are $x, c, f$.

(C) Original word: $W, A, L, K, I, N, G$
Positions: $1, 2, 3, 4, 5, 6, 7$
Arranging the letters of the word $WALKING$ in alphabetical order:
Sorted letters: $A, G, I, K, L, N, W$
Positions: $1, 2, 3, 4, 5, 6, 7$
Comparing the original and sorted positions:
Original: $W, A, L, K, I, N, G$
Sorted: $A, G, I, K, L, N, W$
Comparing index by index:
Position $1: W
eq A$
Position $2: A
eq G$
Position $3: L
eq I$
Position $4: K = K$ (Unchanged)
Position $5: I
eq L$
Position $6: N = N$ (Unchanged)
Position $7: G
eq W$
The letters $K$ and $N$ remain in their original positions.
Therefore,the number of unchanged positions is $2$.

(D) To find the pairs,we compare the positions of letters in the word $REGIONAL$ with their positions in the English alphabet:
$1$. $R$ to $G$: $R(18), E(5), G(7) \rightarrow$ No.
$2$. $G$ to $L$: $G(7), I(9), O(15), N(14), A(1), L(12)$. Between $G$ and $L$ in the word,there are $4$ letters $(I, O, N, A)$. In the alphabet,between $G$ and $L$ are $H, I, J, K$ ($4$ letters). This is a pair $(G, L)$.
$3$. $O$ to $N$: $O(15), N(14)$. Adjacent in the word,adjacent in the alphabet. This is a pair $(O, N)$.
$4$. $N$ to $L$: $N(14), A(1), L(12)$. Between $N$ and $L$ in the word is $A$ ($1$ letter). In the alphabet,between $L$ and $N$ is $M$ ($1$ letter). This is a pair $(N, L)$.
$5$. $O$ to $L$: $O(15), N(14), A(1), L(12)$. Between $O$ and $L$ in the word are $N, A$ ($2$ letters). In the alphabet,between $L$ and $O$ are $M, N$ ($2$ letters). This is a pair $(O, L)$.
Thus,there are four pairs: $(G, L), (O, N), (N, L), (O, L)$.

(A) The word is $'AUTOMOBILE'$.
The $1^{st}$ letter is $A$.
The $5^{th}$ letter is $M$.
The $9^{th}$ letter is $L$.
The $10^{th}$ letter is $E$.
The letters are $A, M, L, E$.
Possible meaningful words that can be formed using these letters are $MALE, MEAL, LAME$.
Since more than one word can be formed,the answer is $'Y'$.

(D) To find the pairs of letters,we check the sequence in the word $CORPORATE$ against the English alphabet:
$1$. $C$ to $O$: $C(3), D, E, F, G, H, I, J, K, L, M, N, O(15)$. (Difference is $11$ letters,not matching).
$2$. $P$ to $R$: $P, Q, R$. (There is $1$ letter between them in the word,and $1$ letter in the alphabet).
$3$. $R$ to $T$: $R, S, T$. (There is $1$ letter between them in the word,and $1$ letter in the alphabet).
$4$. $O$ to $R$: $O, P, Q, R$. (There are $2$ letters between them in the word,and $2$ letters in the alphabet).
However,the condition specifies the same sequence as the English alphabet. Checking forward:
- $(P, R)$: $P$ is $16^{th}$,$R$ is $18^{th}$. In the word,$P$ is at index $3$ and $R$ is at index $5$. Gap is $1$ letter $(O)$. This matches.
- $(R, T)$: $R$ is $18^{th}$,$T$ is $20^{th}$. In the word,$R$ is at index $5$ and $T$ is at index $7$. Gap is $1$ letter $(A)$. This matches.
- $(O, R)$: $O$ is $15^{th}$,$R$ is $18^{th}$. In the word,$O$ is at index $2$ and $R$ is at index $5$. Gap is $2$ letters $(R, P)$. This matches.
Thus,there are $3$ such pairs: $(P, R), (R, T),$ and $(O, R)$.

(A) The word is $COMPREHENSION$. The total number of letters is $14$.
Step $1$: Reverse the first three letters $(COM \rightarrow MOC)$.
Step $2$: Add the last three letters $(ION)$.
Step $3$: Reverse the remaining letters $(PREHENS \rightarrow SNEHERP)$.
Combining these parts: $MOC + ION + SNEHERP = MOCIONSNEHERP$.
The new word is $MOCIONSNEHERP$.
The total number of letters is $14$. The middle position is between the $7^{th}$ and $8^{th}$ letter. However,the question asks for the letter exactly in the middle of the sequence. In a $14$-letter word,there is no single middle letter. Re-evaluating the sequence: $M(1), O(2), C(3), I(4), O(5), N(6), S(7), N(8), E(9), H(10), E(11), R(12), P(13)$. Wait,the word $COMPREHENSION$ has $13$ letters $(C-O-M-P-R-E-H-E-N-S-I-O-N)$.
Let's re-count: $C(1), O(2), M(3), P(4), R(5), E(6), H(7), E(8), N(9), S(10), I(11), O(12), N(13)$.
Step $1$: Reverse first three $(COM \rightarrow MOC)$.
Step $2$: Add last three $(ION)$.
Step $3$: Reverse remaining $(PREHENS \rightarrow SNEHERP)$.
Result: $MOC + SNEHERP + ION = MOCSNEHERPION$.
The middle letter of a $13$-letter word is the $7^{th}$ letter.
$M(1), O(2), C(3), S(4), N(5), E(6), H(7), E(8), R(9), P(10), I(11), O(12), N(13)$.
The $7^{th}$ letter is $H$.

(D) The original word is $DEPRESSION$.
There are $10$ letters in the word.
We interchange the letters in pairs: $(D, E), (P, R), (E, S), (S, I), (O, N)$.
After interchanging,the new word becomes $EDRPESISON$.
The positions are:
$1: E, 2: D, 3: R, 4: P, 5: S, 6: E, 7: I, 8: S, 9: O, 10: N$.
The seventh letter from the right is the letter at position $10 - 7 + 1 = 4$ from the left.
The letter at the $4^{th}$ position is $P$.

(C) The word is $DOCUMENTATION$. The total number of letters is $13$.
Original positions: $1(D), 2(O), 3(C), 4(U), 5(M), 6(E), 7(N), 8(T), 9(A), 10(T), 11(I), 12(O), 13(N)$.
Applying the changes:
$1$. Interchange $3rd$ $(C)$ and $10th$ $(T)$: $D O T U M E N T A C I O N$
$2$. Interchange $4th$ $(U)$ and $7th$ $(N)$: $D O T N M E U T A C I O N$
$3$. Interchange $2nd$ $(O)$ and $6th$ $(E)$: $D E T N M O U T A C I O N$
The new word is $D E T N M O U T A C I O N$.
We need the $11th$ letter from the right end.
Counting from the right: $1(N), 2(O), 3(I), 4(C), 5(A), 6(T), 7(U), 8(O), 9(M), 10(N), 11(T)$.
The $11th$ letter from the right is $T$.

(A) To arrange the words in alphabetical order,we compare them letter by letter:
$1$. Random $(R-a-n...)$
$2$. Restaurant $(R-e-s-t-a...)$
$3$. Restrict $(R-e-s-t-r...)$
$4$. Robber $(R-o-b...)$
$5$. Rocket $(R-o-c...)$
The ordered sequence is: Random,Restaurant,Restrict,Robber,Rocket.
There are $5$ words in total. The word at the middle position ($3^{rd}$ position) is Restrict.