Type 10 Conditional Coding & Decoding Practice Questions Answers Test With Solutions & More Shortcuts

Directions:

In each question below, is given a group of letters followed by four combinations of design/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 R E A U M D F P Q I O H N W Z B
Digit/Symbol code 7 # $ 6 % 8 5 * 4 9 @ © 3 d 1 2
  • If the first letter is a consonant and the fourth letter is a vowel, both are to be coded as the codes for the vowel.
  • If the first letter is a vowel and the fourth letter is a consonant, both are to be coded as the code for the vowel.
  • If the second and the third letters are consonants, both are to be coded as the code for the third letter.
[Allahabad Bank (PO) 2010]

Question : 1

Decode this AEIMVH using the above-given language?

a) 48@#78

b) #2@47$

c) 42@#7$

d) 42@47$

e) None of these

Answer: (a)

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

A → 4 E → 2 I → @ M → # V → 7 H → $.

But here the second letter is a vowel and the last letter is a consonant,

Therefore  condition (ii) is applied here.

As condition (ii) is applied here, both the second and the last letters are to be coded as 8.

A → 4 E → 8 I → @ M → # V → 7 H → 8

Hence, code for AEIMVH ⇒ 48@#78

Question : 2

Decode this WMELJU using the above-given language?

a) @#259© G4

b) 5#259©

c) 5#2@9©

d) @#2@9© E4+C8

e) None of the above

Answer: (d)

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

W → 5 M → # E → 2 I → @ J → 9 U → ©

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

therefore condition (i) is applied here.

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

W → @ M → # E → 2 I → @ J → 9 U → ©

Hence, code for WMEIJU ⇒ @#2@9©.

Question : 3

Decode this MBUVWE using the above-given language?

a) #*©#52

b) 7*©#52

c) #©*752

d) #©*7528

e) None of these

Answer: (e)

Here, none of the conditions is applied,

so the coding is done as follows. M → # B → * U → © V → 7 W → 5 E → 2

Hence, code for MBUVWE ⇒ #*©752

Question : 4

Decode this THAFIQ using the above-given language?

a) 1$48@3

b) 1$48@1

c) 3$48@1

d) 3$48@3

e) None  of these

Answer: (a)

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

T → 3 H → $ A → 4 F → 8 I → @ Q → 1

But here both the first and the last latter are consonants, therefore condition (iii) is applied here.

As condition (iii) as applied here, the codes for the first and the last letters are interchanged.

T → 1 H → $ A → 4 F → 8 I → @ Q → 3

Hence, code for THAFIQ ⇒ 1$48@3.

Question : 5

Decode this AJBMFU using the above-given language?

a) #9*#8©

b) ©9©#84

c) 49*48©

d) 49*#8©

e) None  of these

Answer: (d)

Here, none of the conditions is applied.

So the coding is done as follows. A → 4 J → 9 B → * M → # F → 8 U → ©

Hence, code for AJBMFU ⇒ 49*#8©.

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