To solve this question, we'll first rearrange the letters in each word in alphabetical order:

(A) Code: cdeo
(B) Lack: ackl
(C) Meet: eemt
(D) Deaf: adef
(E) Road: ador

Now, let's rearrange the groups of letters formed alphabetically:

1. adef (from "Deaf")
2. ackl (from "Lack")
3. ador (from "Road")
4. cdeo (from "Code")
5. eemt (from "Meet")

Now, we see that there are five groups of letters. Since there is an odd number of groups, the middle group will be the third one when arranged alphabetically.

So, the word with its group of letters in the middle is "Road" (Option E).

To solve this problem, we first need to convert the given time (19:45) to the corresponding hour and minute positions on the clock where the digits have been replaced by alternate English alphabets beginning with D.

Given that D replaces 1, F replaces 2, and so on, we can create a mapping for the hour positions:

1 → D

2 → F

3 → H

4 → J

5 → L

6 → N

7 → P

8 → R

9 → T

10 → V

11 → X

12 → Z

So, at 19:45, the hour hand would correspond to position P (for 7) and the minute hand would correspond to position T (for 9).

The hour hand is positioned between P and R, and the minute hand is at position T. So, the correct answer is:

(C) Between P and R

To solve this problem, we need to select the first (P), fifth (L), ninth (A), and eleventh (E) letters from the word PENULTIMATE, and then form words using each letter only once.

The selected letters are P, L, A, and E.

Now, let's form words using these letters:

1. PALE
2. LEAP
3. PEAL

All three words are meaningful and fulfill the conditions given in the question.

Now, we need to find the third letter of these words:

1. PALE - The third letter is L.
2. LEAP - The third letter is A.
3. PEAL - The third letter is A.

As there are multiple possible words with different third letters, the answer is 'D'. So, the correct option is (D) D.

analyze the pairs of letters with only one letter between them in both the word and the alphabet:

1. CD: C directly precedes D in both the word and the alphabet.
2. DI: There is one letter (E) between D and I in both the word and the alphabet.
3. EF: There is one letter (G) between E and F in both the word and the alphabet.
4. IB: There is one letter (C) between I and B in both the word and the alphabet.
5. IL: There is one letter (M) between I and L in both the word and the alphabet.
6. TY: There is one letter (U) between T and Y in both the word and the alphabet.

Hence, there are two pairs of letters in the word "CREDIBILITY" that have only one letter between them in both the word and the alphabet: DI and IL.

So, the correct answer is indeed (C) Two.

After dropping every third letter, we get

To solve this problem, let's first identify the given word: EXPLOSION.

Now, we need to find pairs of letters in this word such that the number of letters between them in the word is the same as the number of letters between them in the English alphabet.

Let's go through the word letter by letter:

1. E and X: There are four letters (F, G, H, I) between E and X in the English alphabet and there are also four letters (P, L, O, S) between them in the word.
2. L and O: There are two letters (M, N) between L and O in the English alphabet and there are also two letters (P, S) between them in the word.
3. P and S: There are three letters (Q, R, S) between P and S in the English alphabet and there are also three letters (L, O, I) between them in the word.
4. S and I: There are eleven letters (T, U, V, W, X, Y, Z, A, B, C, D) between S and I in the English alphabet and there are also eleven letters (P, L, O, S, I, O, N, S, I, T, I) between them in the word.

So, there are three pairs of letters in the word "EXPLOSION" that have as many letters between them as in the English alphabet.

Hence, the correct answer is (C) Three.

The 3rd, 6th, 8th and 11th letters are S, N, U and H respectively. The word that can be made is SHUN.

To solve this problem, let's analyze each letter group based on their positions in the English alphabet:

(A) DKG:

• D is at position 4.
• K is at position 11.
• G is at position 7.

The difference between consecutive letters:

• K - D = 11 - 4 = 7
• G - K = 7 - 11 = -4

(B) FMI:

• F is at position 6.
• M is at position 13.
• I is at position 9.

The difference between consecutive letters:

• M - F = 13 - 6 = 7
• I - M = 9 - 13 = -4

(C) MTP:

• M is at position 13.
• T is at position 20.
• P is at position 16.

The difference between consecutive letters:

• T - M = 20 - 13 = 7
• P - T = 16 - 20 = -4

(D) JQM:

• J is at position 10.
• Q is at position 17.
• M is at position 13.

The difference between consecutive letters:

• Q - J = 17 - 10 = 7
• M - Q = 13 - 17 = -4

(E) HOL:

• H is at position 8.
• O is at position 15.
• L is at position 12.

The difference between consecutive letters:

• O - H = 15 - 8 = 7
• L - O = 12 - 15 = -3

Based on the analysis:

• Options (A), (B), (C), and (D) follow a pattern where the difference between consecutive letters alternates between +7 and -4.
• Option (E) does not follow this pattern. Although the difference between O and H is +7, the difference between L and O is -3 instead of -4.

Therefore, the letter group (E) HOL does not belong to the group formed by the other options.

The correct answer is (E) HOL.

To solve this problem, let's extract the fifth, seventh, eighth, ninth, and thirteenth letters from the word "EXTRAORDINARY":

Fifth letter: O
Seventh letter: R
Eighth letter: D
Ninth letter: I
Thirteenth letter: A

Now, let's form words using each letter only once:

1. DIARY
2. DAIRY

Both "DIARY" and "DAIRY" are meaningful words.

The second letter of "DIARY" is I.
The second letter of "DAIRY" is A.

As there are multiple possible words with different second letters, the answer is 'M'.

So, the correct option is (D) M.