This question is based on the words given below.
IRE AWE DAY FUR
In each of the words, each vowel is changed to the second letter preceding it in the English alphabetical order, and each consonant is changed to the letter immediately following it in the English alphabetical order. In how many letter-clusters thus formed will no vowel appear?

Let us analyze the given words step-by-step to find the correct answer.

The original words provided are:
IRE, AWE, DAY, and FUR.

Let us recall the rules for transforming the letters:
1. Each vowel (A, E, I, O, U) is changed to the second letter preceding it in the English alphabetical order.
- A (1st letter) ⇒ Y (25th letter) [wrapping around alphabetically: A - 1 = Z, Z - 1 = Y]
- E (5th letter) ⇒ C (3rd letter)
- I (9th letter) ⇒ G (7th letter)
- O (15th letter) ⇒ M (13th letter)
- U (21st letter) ⇒ S (19th letter)
2. Each consonant is changed to the letter immediately following it in the English alphabetical order.

Let us apply these rules to each word:

1. IRE
- I (vowel) ⇒ G
- R (consonant) ⇒ S
- E (vowel) ⇒ C
Formed cluster: GSC (Contains no vowels: Yes)

2. AWE
- A (vowel) ⇒ Y
- W (consonant) ⇒ X
- E (vowel) ⇒ C
Formed cluster: YXC (Contains no vowels: Yes, as Y is treated as a consonant here)

3. DAY
- D (consonant) ⇒ E
- A (vowel) ⇒ Y
- Y (consonant) ⇒ Z
Formed cluster: EYZ (Contains no vowels: No, it contains the vowel E)

4. FUR
- F (consonant) ⇒ G
- U (vowel) ⇒ S
- R (consonant) ⇒ S
Formed cluster: GSS (Contains no vowels: Yes)

Therefore, the letter-clusters with no vowels are GSC, YXC, and GSS. There are 3 such clusters in total.

Thus, the correct option is Three.