type 10 conditional coding & decoding Detailed Explanation And More Example

MOST IMPORTANT verbal reasoning - 12 EXERCISES

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Directions:

In each question below, is given a group of letters followed by four combinations of digits/symbols numbered (a), (b), (c) and (d).

You have to find out which of the four combinations correctly represents the group of letters based on the following coding system and the conditions that follow and mark the number of that combination as your answer. If none of the combinations correctly represents the group of letters, mark (e), i.e., 'None of these', as your answer.

Letter B A D E F H J K M I U O W F P
Digit/symbol code 6 $ 7 8 # 1 2 * % 3 © 4 9 @ 5
  • If the first letter is a vowel and the last letter is a consonant, their codes are to be interchanged.
  • If both the first and the last letters are consonants, both are to be coded as 'δ'.
  • If the first letter is a consonant and the last letter is a vowel, both are to be coded as the code for the vowel.
[Punjab & Sind Bank (PO) 2010]

The following question based on coding & decoding topic of verbal reasoning

Questions : Decode this UKPDMI using the above-given language?

(a) ©5*7%3

(b) δ*57%δ

(c) 3*57%©

(d) ©*5%73

e) None of these

The correct answers to the above question in:

Answer: (e)

Here, none of the condition is applied,

so the coding is done as follows

U → ©, K → *, P → 5, D → 7, M → %, I → 3.

Hence, code for UKPDMI ⇒ ©*57%3.

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Read more conditional coding decoding Based Verbal Reasoning Questions and Answers

Question : 1

Decode this EKFUDH using the above-given language?

a) δ*#©78

b) 1*#©71

c) 1*#©78

d) 8*#©78

e) None of these

Answer: (c)

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

E → 8, K →*, F → #, U → ©, D → 7, H → 1.

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

As condition (i) is applied here, the codes for the first and the last letter are interchanged.

E → 1 K → * F → # U → © D → 7 H → 8

Hence, code for EKFUDH ⇒ 1*#©78.

Question : 2

Decode this IBHWPO using the above-given language?

a) 361953

b) 461954

c) 461953

d) 361954

e) None of these

Answer: (d)

Here, none of the condition is applied,

so the coding is done as follows.

I → 3, B → 6, H → 1,  W → 9, P → 5, O → 4.

Hence, code for IBHWPO ⇒ 361954.

Question : 3

Decode this DMEAKJ using the above-given language?

a) 2%8$*7

b) δ%8$*δ

c) 7%8$*7

d) 7%8$*7

e) None of these

Answer: (b)

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

D → 7, M → %, E → 8, A → $, K →*, J → 2.

But here both the first and the last letter are consonants,

therefore condition (ii) is applied here.

As condition (ii) is applied here, both the first and the last letters are to be coded as δ

D → δ, M → %, E → 8, A → $, K → *, J → δ.

Hence, code for DMEAKJ ⇒ δ%8#*δ.

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GET coding & decoding PRACTICE TEST EXERCISES

type 1 opposite letter coding & decoding

type 2 left/right letter coding & decoding

type 3 letter rearrangement coding & decoding

type 4 alphabet replacement coding & decoding

type 5 word substitution coding & decoding

type 6 letter number coding & decoding

type 7 symbol coding & decoding

type 8 fictitious word, number, symbol coding & decoding

type 9 jumbled coding & decoding

type 10 conditional coding & decoding

type 11 matrix coding & decoding

type 12 comparison coding & decoding
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