type 10 conditional coding & decoding Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (d)

When no condition is applied, the coding is done as follows.

E → # M → % I → 9 R → 7 D → 8 P → *

But here the first letter is a vowel and the fourth letter is a consonant,

therefore condition (ii) is applied.

As condition (ii) is applied here, both the first and the fourth letters are to be coded as the code for the vowel.

E → @ M → % I → 9 R → # D → 8 P → *

Hence, code for EMIRDP ⇒ #%9#8*.

Answer: (e)

When no condition is applied, coding is done as follows. P → * Q → 4 I → 9 M → % H → © Z © 1

But here the first letter is a consonant and the third letter is a vowel, therefore condition (i) is applied.

As condition (i) is applied here, the codes for first and third letters are to be interchanged.

P → 9 Q → 4 I → * M → % H → © Z → 1

Hence, code for PQIMHZ ⇒ 94*%©1.

Answer: (b)

When no condition is applied, the coding is done as follows.

F → 5 W → d Z → 1 E → # R → 7 A → $

But here the second and third letters are consonants, therefore condition (iii) is applied here.

As condition (iii) is applied here, both the second and third letters are to be coded as the code for the third letter.

F → 5 W → 1 Z → 1 E → # R → 7 A → $

Hence, code for FWZERA ⇒ 511#7$.

Answer: (e)

Here, none of the conditions is applied,

so the coding is done as follows.

N → 3, O → @, B → 2, O → @, Q → 4, E → #.

Hence, code for NABAQE ⇒3@2@4#.

Answer: (c)

Here, none of the condition is applied,

so the coding is done as follows. H → © U → 6 B → 2 D → 8 I → 9 N → 3

Hence, code for HUBDIN ⇒ ©62893.