type 10 conditional coding & decoding Detailed Explanation And More Example

Answer: (e)

When condition is not applied, the coding is done as follows.

3 → * 6 → F 4 → A 8 → K 1 → % 9 → M

But here the first and the last digits are odd, therefore condition (i) is applied here.

As condition (i) is applied here 3 and 9 will be coded as X.

3 → X 6 → F 4 → A 8 → K 1 → % 9 → X

Hence code for 364819 ⇒ XFAK%X

Answer: (a)

When condition is not applied, the coding is done as follows.

5 → B 4 → A 6 → F 8 → K 3 → * 9 → 1

But here the first and the last digits are odd, therefore condition (i) is applied here.

As condition (i) is applied here 5 and 9 will be coded as X. 5 → X 4 → A 6 → F 8 → K 3 → * 9 → X , X A F K * X.

Hence code for 546839 ⇒ XAFK*X

Answer: (c)

Here, none of the condition is applied,

so the coding will be done as follows.

7 → E 6 → F 5 → B 0 → R 8 → K 2 → @

Hence, code for 765082 ⇒ EFBRK@.

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: (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)

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.