Coding and Decoding Questions For SSC GD PDF
SSC GD Constable Coding and Decoding Question and Answers download PDF based on previous year question paper of SSC GD exam. 40 Very important Coding and Decoding questions for GD Constable.
CODING AND DECODING QUESTIONS FOR SSC GD PDF
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Instructions
In a certain language, “England won against Colombia” is coded as “ta ja je tu”.
”Rivals are playing against Colombia” is coded as “zu ta ja ga xu”.
”England and India are rivals” is coded as “yu je xu zu ha”.
”India lost against Australia” is coded as “li ta du yu”.
Question 1: If the code for the statement ‘Australia likes rivals’ is ‘fa du xu’ then find out the code for the statement ‘Colombia and India are lost’?
a) tu zu ja li ha
b) zu ja yu li ha
c) yu li ha zu ta
d) yu li ha xu ta
e) ha xu ja zu li
Question 2: Which of the following can be the code for “Colombia and Australia are friends”?
a) xu ja zu po ha
b) du po li ha ja
c) ha po du zu ja
d) li ja po ha fu
e) du zu ja ha po
Question 3: Which of the following can be the code for “Playing against Austria”?
a) du ta ga
b) ga ta li
c) du li ga
d) ta so ga
e) ta tu so
Question 4: Which of the following must be the code for “and”?
a) yu
b) je
c) xu
d) ha
e) ja
Question 5: Which of the following must be the code for “playing”?
a) ga
b) ta
c) ja
d) xu
e) zu
Question 6: In a certain coding language, ‘Brian gets up early’ is coded as ‘Pa ta na ma’, ‘Adam gets up late’ is coded as ‘Ba ta na ka’, ‘Brian is moving up the ladder’ is coded as ‘Pa su la ta va za’. What is the code word for ‘gets’ in this language?
a) pa
b) ta
c) na
d) ma
e) Cannot be determined
Instructions
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 combinations correctly represents the group of letters based on the following coding system and the conditions and mark the numbers of that combination as your answer. If none of the four combinations correctly represents the group of letters, give e: i.e. ‘None of these’ as your answer.
Letters K E T J H I F A L U B M O R P
Digits/Symbol 3 7 % $ 4 * 1 9 8 6 # @ 2 5 ©
Conditions:
(i) If the first as well as the last letter is a vowel their codes are to be swapped.
(ii) If the first as well as the last letter is a consonant both are to be coded by £.
(iii) If the first letter is a vowel and the last letter is a consonant, the vowel is to be coded by ∆ and the consonant is to be coded by ↑.
Question 7: IJLTPU
a) *$8%©6
b) *$8%©*
c) 6$8%©6
d) 6$8%©*
e) None of these
Question 8: KEOMPA
a) ↑72@©∆
b) ∆72@©↑
c) 372@©9
d) 972@©3
e) None of these
Question 9: ORBETH
a) ↑5#7%∆
b) ∆5#7%↑
c) 25#7%∆
d) 45#7%2
e) None of these
Question 10: AJTKLU
a) 9$%386
b) £$%38£
c) ∆$%38↑
d) ↑$%38∆
e) None of these
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Question 11: TARIFM
a) £95*1£
b) %95*1!@
c) %95* 1 %
d) @95*1@
e) None of these
Instructions
The following rules are followed while coding a word.
Letters from A to Z are assigned numbers from 26 to 1 in that order. (A corresponds to 26, B corresponds to 25 and so on).
Question 12: The fourth letter represented by the word 261216723 cannot be
a) Y
b) O
c) T
d) Z
e) U
Question 13: Which of the following cannot be the word represented by 11131415?
a) ZZZXWZV
b) ZZZXZWL
c) PNZWL
d) PZXMZV
e) More than one of the above
Question 14: If it is known for sure that the word represented by 271213 has 4 letters, then the word is
a) YKOX
b) YTYX
c) YTON
d) YTFX
e) YTYN
Question 15: Which of the following conditions will be sufficient to uniquely determine the word represented by 261712?
a) The word coded is 4 lettered
b) The word coded is 5 lettered
c) The word coded is 6 lettered
d) The word coded is 3 lettered
e) More than one of the above.
Question 16: Which of the following words can be represented by 212312?
a) YOXO
b) FYXO
c) YZDO
d) FDZY
e) More than one of the above
Instructions
Following is given a set of digits and the corresponding letter code of each digit followed by certain conditions for coding. In each question one number consisting six digits followed by four combinations of letter codes. You have to find out which of the combination of letter codes represents the set of digits based on the above codes and the conditions given below and mark your answer accordingly. Otherwise give answer e:, i.e., None of these.
Conditions :
(I) If both the first and the last digits of the number are odd digits then both are to be coded as I.
(II) If both the first and last digits of the number are even digits then both are to be coded as Y.
Question 17: 831795
a) INJTWR
b) BNJTWR
c) BNJTWY
d) YNJYWY
e) None of these
Question 18: 615824
a) IJRBMI
b) IJRBMY
c) YJRBMY
d) DJRBMK
e) None of these
Question 19: 591248
a) IWJMKB
b) RWJMKY
c) YWJMKY
d) RWJMKB
e) None of these
Question 20: 263847
a) IDNBKI
b) YDNBKY
c) IDNBKY
d) MDNBKY
e) None of these
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Question 21: 726395
a) IMDNWI
b) YMDNWY,,
c) TMDNWR
d) IMDNWR
e) None of these
Instructions
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 combinations correctly represents the group of letters based on the following coding system and mark the number of that combination as the answer. If none of the four combinations correctly represents the group of letters, mark e: i.e. ‘None of these’ as the answer. Symbol
Conditions :
(i) If both the first and the last letters of the group are vowels, their codes are to be interchanged.
(ii) If the first letter is a consonant and the last letter is a vowel, both are to be coded by the code for the consonant.
Question 22: QRLGHM
a) 219#@2
b) 619#@6
c) 619#@2
d) 6190#2
e) None of these
Question 23: BQRLHA
a) $21908
b) 851908
c) 821908
d) $219@$
e) None of these
Question 24: IGCHRE
a) %#*@13
b) 3#*@1%
c) 3#*@13
d) %#*01%
e) None of these
Question 25: EBHRMT
a) %0$16O
b) %$@16©
c) 6$@1©%
d) @$@16%
e) None of these
Question 26: JQGALI
a) 32#893
b) 52#893
c) 52#895
d) 32#895
e) None of these
Instructions
In each questions below is given a group of digits followed by four combinations of letters or symbols numbered a, b, c and d. You have to find out which of the combinations correctly represents the group of digits based on the coding system and the conditions given below and mark the number of that combination as your answer. If none of the combinations correctly represents the group of digits, mark e:, i.e. None of these’ as your answer.
Conditions :
(i) If the first digit is odd and the last digit is even, the codes for the first and last digits are to be interchanged.
(ii) If both the first and the last digits are even, both are to be coded as *.
(iii) If both the first and the last digits are odd, both are to be coded as $.
Question 27: 591426
a) @RA%©P
b) PRA%©@
c) @AR%©P
d) $RA%©*
e) None of these
Question 28: 794821
a) MR%D©A
b) AR%D©M
c) M%RD©A
d) $R%D@$
e) None of these
Question 29: 813469
a) RAK%@D
b) DAK%@R
c) DAP%@R
d) *AK%@*
e) None of these
Question 30: 671254
a) ©MA©P%
b) $MA©P$
c) *MA©P*
d) %MA©P©
e) None of these
Question 31: 215349
a) RAPK%©
b) *APK%$
c) $APK%$
d) ©PAK%R
e) None of these
Instructions
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 combinations correctly represents the group of letters based on the following coding system and mark the number of that combination as the answer. If none of the four combinations correctly represents the group of letters. marks e: i.e ‘None of these’ as the answer
Conditions :
(i) If both the first and the last letters of the group are consonants, both are to be coded as the code for the last letter.
(ii) If the first letter is a consonant and the last letter is a vowel, the codes are to be interchanged.
Question 32: RKUMFP
a) 7 % C * # 4 3
b) 3 * % # 4 7
c) 3 % * # 4 3
d) 3 % * # 4 7
e) None of these
Question 33: MDEAJI
a) 1 $ @ 8 2 #
b) # $ @ 8 2 1
c) 1 $@ 2 1
d) # $ @ 8 2 #
e) None of these
Question 34: EMNTKU
a) * # © 1 % @
b) @ # © 1 4 *
c) @ # © 1 % *
d) # @ © 1 % *
e) None of these
Question 35: AWBRND
a) $563©8
b) 8563©$
c) 8365©$
d) 8536©$
e) None of these
Question 36: BDAIFE
a) 6$8146
b) 68814@
c) @$814@
d) @$8146
e) None of these
Question 37: In a certain coding system, PAPER is written as PERPA and SUBJECT is written as JECTSUB, what should be the code for COUNCIL ?
a) NCILCOU
b) LICNOUC
c) NCOUCIL
d) NLICUOC
e) None of these
Instructions
In each question below a group of letters is given followed by four combinations of digits and/or symbols numbered a:, b:, c: and d:. You have to find out which of the combinations correctly represents the group of letters based on the following coding system and the conditions those follow and give the number of that combination as the answer. If none of the combinations correctly represents, give e:, i.e. None of these as the answer.
Letter P M A J E T K I R B U F H
Code : 5 6 1 # 9 2 8 $ 3 @ 7 © 4
Conditions : (i) If the first letter is a vowel and the last letter is a consonant, both are to be coded as *.
(ii) If the first letter is consonant and the last letter is a vowel, both are to be coded as %.
Question 38: TMRBFJ
a) *6 3@ ©*
b) %63@ ©%
c) 236@ ©#
d) 263 @ ©#
e) None of these
Question 39: RFHKJA
a) 3 ©48#1
b) 483 ©#1
c) * ©48#*
d) %©48#%
e) None of these
Question 40: IPAUHM
a) *5174*
b) %5 I 74%
c) $51746
d) $51476
e) None of these
Answers & Solutions:
1) Answer (B)
We are given that
(1)“England won against Colombia” is coded as “ta ja je tu”.
(2)”Rivals are playing against Colombia” is coded as “zu ta ja ga xu”.
(3)”England and India are rivals” is coded as “yu je xu zu ha”.
(4)”India lost against Australia” is coded as “li ta du yu”.
From statement (1) and (4) we can see that only word ‘against’ is common hence code for the word ‘against’ is ‘ta’.
From statement (1) and (2) we can see that words ‘against’ and ‘Colombia’ are common hence code for word ‘Colombia’ is ‘ja’.
From statement (3) and (4) we can see that only word ‘India’ is common hence code for word ‘India’ is ‘yu’.
From statement (1) and (3) we can see that only word ‘England’ is common hence code for word ‘England’ is ‘je’.
From statement (1) we can say that code for word ‘won’ is ‘tu’.
In statement (3) we know that ‘yu’ and ‘je’ are codes for words ‘India’ and ‘England’ respectively. 2 codes out of remaining 3 codes ‘xu’, ‘zu’ and ‘ha’ should be common with codes available in statement (2) because of the words ‘Rivals’ and ‘are’. Those two common codes are ‘xu’ and ‘zu’ and hence we can say that ‘ha’ is code for the word ‘and’ in statement 3 and ‘ga’ is code for the word ‘playing’.
For words ‘lost’ and ‘Australia’ codes are ‘li’ and ‘du’ in any order.
For words ‘Rivals’ and ‘are’ codes are ‘xu’ and ‘zu’ in any order. Tabulating all available data
‘Australia likes rivals’ is ‘fa du xu’ that means codes for words ‘Australia’ and ‘rivals’ are ‘du’ and ‘xu’ respectively. Consequently codes for words ‘lost’ and ‘are’ will be ‘li’ and ‘zu’ respectively.
Hence we can say that code for statement ‘Colombia and India are lost’ will be ‘ja’, ‘ha’, ‘yu’, ‘zu’ and ‘li’ in any order. We can see that all codes are available in option B. Hence, option B is the correct answer.
2) Answer (E)
We are given that
(1)“England won against Colombia” is coded as “ta ja je tu”.
(2)”Rivals are playing against Colombia” is coded as “zu ta ja ga xu”.
(3)”England and India are rivals” is coded as “yu je xu zu ha”.
(4)”India lost against Australia” is coded as “li ta du yu”.
From statement (1) and (4) we can see that only word ‘against’ is common hence code for the word ‘against’ is ‘ta’.
From statement (1) and (2) we can see that words ‘against’ and ‘Colombia’ are common hence code for word ‘Colombia’ is ‘ja’.
From statement (3) and (4) we can see that only word ‘India’ is common hence code for word ‘India’ is ‘yu’.
From statement (1) and (3) we can see that only word ‘England’ is common hence code for word ‘England’ is ‘je’.
From statement (1) we can say that code for word ‘won’ is ‘tu’.
In statement (3) we know that ‘yu’ and ‘je’ are codes for words ‘India’ and ‘England’ respectively. 2 codes out of remaining 3 codes ‘xu’, ‘zu’ and ‘ha’ should be common with codes available in statement (2) because of the words ‘Rivals’ and ‘are’. Those two common codes are ‘xu’ and ‘zu’ and hence we can say that ‘ha’ is code for the word ‘and’ in statement 3 and ‘ga’ is code for the word ‘playing’.
For words ‘lost’ and ‘Australia’ codes are ‘li’ and ‘du’ in any order.
For words ‘Rivals’ and ‘are’ codes are ‘xu’ and ‘zu’ in any order. Tabulating all available data
We can see that codes for words “Columbia” “and” “Australia” and “are” are ‘ja’ ‘ha’, ‘du/li’ and ‘xu/zu’ respectively.Code for word ‘friends‘ is not known to us.
Option (A) xu ja zu po ha: – Here both codes ‘xu/zu’ are used. Hence, this option is incorrect.
Option (B) du po li ha ja: – Here both codes ‘du/li’ are used. Hence, this option is incorrect.
Option (C) hi po du zu ja: – Code for word ‘and‘, ‘ha’ is not available. Hence this option is incorrect.
Option (D) li ja po ha fu: – Code for word ‘are‘, ‘xu/zu‘ is not available. Hence this option is incorrect.
Option (E) du zu ja ha po: – Code for word ‘friends‘ may be ‘po‘. This option is correct.
3) Answer (D)
We are given that
(1)“England won against Colombia” is coded as “ta ja je tu”.
(2)”Rivals are playing against Colombia” is coded as “zu ta ja ga xu”.
(3)”England and India are rivals” is coded as “yu je xu zu ha”.
(4)”India lost against Australia” is coded as “li ta du yu”.
From statement (1) and (4) we can see that only word ‘against’ is common hence code for the word ‘against’ is ‘ta’.
From statement (1) and (2) we can see that words ‘against’ and ‘Colombia’ are common hence code for word ‘Colombia’ is ‘ja’.
From statement (3) and (4) we can see that only word ‘India’ is common hence code for word ‘India’ is ‘yu’.
From statement (1) and (3) we can see that only word ‘England’ is common hence code for word ‘England’ is ‘je’.
From statement (1) we can say that code for word ‘won’ is ‘tu’.
In statement (3) we know that ‘yu’ and ‘je’ are codes for words ‘India’ and ‘England’ respectively. 2 codes out of remaining 3 codes ‘xu’, ‘zu’ and ‘ha’ should be common with codes available in statement (2) because of the words ‘Rivals’ and ‘are’. Those two common codes are ‘xu’ and ‘zu’ and hence we can say that ‘ha’ is code for the word ‘and’ in statement 3 and ‘ga’ is code for the word ‘playing’.
For words ‘lost’ and ‘Australia’ codes are ‘li’ and ‘du’ in any order.
For words ‘Rivals’ and ‘are’ codes are ‘xu’ and ‘zu’ in any order. Tabulating all available data
We can see that codes for words ‘playing’ and ‘against’ are ‘ga’ and ‘ta’ respectively.Code for word ‘Austria’ is not known to us.
Option (A) du ta ga: – du is one of two possible codes for word ‘Australia’ not ‘Austria’. Hence this option is incorrect.
Option (B) ga ta li: – li is one of two possible codes for word ‘Australia’ not ‘Austria’.Hence this option is incorrect.
Option (C) du li ga: – Code for word ‘against’ is not available. Hence this option is incorrect.
Option (E) ta tu so: – Code for word ‘playing’ is not available. Hence this option is incorrect.
Option (D) ta so ga: – Code for word ‘Austria’ may be ‘so’. Hence, this option is correct.
4) Answer (D)
We are given that
(1)“England won against Colombia” is coded as “ta ja je tu”.
(2)”Rivals are playing against Colombia” is coded as “zu ta ja ga xu”.
(3)”England and India are rivals” is coded as “yu je xu zu ha”.
(4)”India lost against Australia” is coded as “li ta du yu”.
From statement (1) and (4) we can see that only word ‘against’ is common hence code for the word ‘against’ is ‘ta’.
From statement (1) and (2) we can see that words ‘against’ and ‘Colombia’ are common hence code for word ‘Colombia’ is ‘ja’.
From statement (3) and (4) we can see that only word ‘India’ is common hence code for word ‘India’ is ‘yu’.
From statement (1) and (3) we can see that only word ‘England’ is common hence code for word ‘England’ is ‘je’.
From statement (1) we can say that code for word ‘won’ is ‘tu’.
In statement (3) we know that ‘yu’ and ‘je’ are codes for words ‘India’ and ‘England’ respectively. 2 codes out of remaining 3 codes ‘xu’, ‘zu’ and ‘ha’ should be common with codes available in statement (2) because of the words ‘Rivals’ and ‘are’. Those two common codes are ‘xu’ and ‘zu’ and hence we can say that ‘ha’ is code for the word ‘and’ in statement 3 and ‘ga’ is code for the word ‘playing’.
For words ‘lost’ and ‘Australia’ codes are ‘li’ and ‘du’ in any order.
For words ‘Rivals’ and ‘are’ codes are ‘xu’ and ‘zu’ in any order. Tabulating all available data
From the table we can see that code for word “and” is ‘ha’. Hence, option D is the correct answer.
5) Answer (A)
We are given that
(1)“England won against Colombia” is coded as “ta ja je tu”.
(2)”Rivals are playing against Colombia” is coded as “zu ta ja ga xu”.
(3)”England and India are rivals” is coded as “yu je xu zu ha”.
(4)”India lost against Australia” is coded as “li ta du yu”.
From statement (1) and (4) we can see that only word ‘against’ is common hence code for the word ‘against’ is ‘ta’.
From statement (1) and (2) we can see that words ‘against’ and ‘Colombia’ are common hence code for word ‘Colombia’ is ‘ja’.
From statement (3) and (4) we can see that only word ‘India’ is common hence code for word ‘India’ is ‘yu’.
From statement (1) and (3) we can see that only word ‘England’ is common hence code for word ‘England’ is ‘je’.
From statement (1) we can say that code for word ‘won’ is ‘tu’.
In statement (3) we know that ‘yu’ and ‘je’ are codes for words ‘India’ and ‘England’ respectively. 2 codes out of remaining 3 codes ‘xu’, ‘zu’ and ‘ha’ should be common with codes available in statement (2) because of the words ‘Rivals’ and ‘are’. Those two common codes are ‘xu’ and ‘zu’ and hence we can say that ‘ha’ is code for the word ‘and’ in statement 3 and ‘ga’ is code for the word ‘playing’.
For words ‘lost’ and ‘Australia’ codes are ‘li’ and ‘du’ in any order.
For words ‘Rivals’ and ‘are’ codes are ‘xu’ and ‘zu’ in any order. Tabulating all available data
From the table we can see that code for word “playing” is ‘ga’. Hence, option A is the correct answer.
6) Answer (C)
We can see that ‘gets up’ is common between the first and second coded sentences. Hence, the code word for these words must be ‘ta na’ in no particular order. ‘up’ is common between the second and third coded sentences. Hence, ‘up’ must be coded as ‘ta’. Thus, the code word for ‘gets’ must be ‘na’. Hence, option C is the correct answer.
7) Answer (D)
Expression : IJLTPU
Since, the first and the last letters are vowels, rule (i) is applied and codes for ‘I’ and ‘U’ are swapped.
=> Code for IJLTPU = 6$8%©*
=> Ans – (D)
8) Answer (C)
Expression : KEOMPA
None of the rules is applied, and the codes are written as it is.
Thus, code for KEOMPA = 372@©9
=> Ans – (C)
9) Answer (B)
Expression : ORBETH
Since, the first letter is a vowel and the last is a consonant, thus rule (iii) is applied, the vowel is to be coded by ∆ and the consonant is to be coded by ↑.
Code for ORBETH = ∆5#7%↑
=> Ans – (B)
10) Answer (E)
Expression : AJTKLU
Since, the first and the last letters are vowels, rule (i) is applied and codes for ‘A’ and ‘U’ are swapped.
Code for AJTKLU = 6$%389, which is not given in the options.
=> Ans – (E) : none of these
11) Answer (A)
Expression : TARIFM
Since, both the first and last letters are consonant, thus by applying rule (ii), both are to be coded by £.
Code for TARIFM = £95*1£
=> Ans – (A)
12) Answer (B)
261216723 can be YUZYZUTYX, AOKTD, AOZUTYX, YUOZUTYX etc. The fourth letter cannot be O since O is represented by 12 and we cannot split 261216723 with 12 as the fourth letter.
13) Answer (A)
Let us write the code of the words to find out.
ZZZXWZV – 1113415
ZZZXZWL – 11131415
PNZWL – 11131415
PZXMZV – 11131415
Hence, option A is the right answer.
14) Answer (C)
271213 – There is no letter represented by 27. Hence, the first 2 letters are represented by 2 and 7 for sure. The first and second letters are YT. Now, the third and fourth letters are represented by 1213. The only way of splitting 1213 is as 12 and 13. Hence, the last 2 letters are O and N. Hence, the word is YTON. Option C is the right answer.
15) Answer (E)
261712 – To determine all the letters uniquely, we must know what each number represents. If it is provided that the word is 6 lettered, we can determine the letter represented by each number. If it is given that the word is 3 lettered, then 2 digits in the number must represent 1 alphabet. Hence, for more than one case, we can determine the letters uniquely. Hence, option E is the right answer.
16) Answer (E)
YOXO – 212312
FYXO – 212312
YZDO – 212312
FDZY – 212312
Hence, option E is the right answer.
17) Answer (B)
18) Answer (C)
19) Answer (D)
20) Answer (E)
21) Answer (A)
22) Answer (E)
Expression : QRLGHM
Since both the first and last letters are consonant. Thus, the code for this expression will be written as it is.
=> Code for QRLGHM = 219#@6, which is not in the options.
=> Ans – (E)
23) Answer (D)
Expression : BQRLHA
Since, the first letter is a consonant and the last letter is a vowel, both are to be coded as the code for consonant, i.e. ‘$’
Thus, code for BQRLHA = $219@$
=> Ans – (D)
24) Answer (A)
Expression : IGCHRE
Since, the first and last letters are both vowels, their codes are interchanged.
Thus, code for IGCHRE = %# *@13
=> Ans – (A)
25) Answer (B)
Expression : EBHRMT
Since both the first and last letters are consonant. Thus, the code for this expression will be written as it is.
=> Code for EBHRMT = %$@16©, which is not in the options.
=> Ans – (B)
26) Answer (C)
Expression : JQGALI
Since, the first letter is a consonant and the last letter is a vowel, both are to be coded as the code for consonant, i.e. ‘5’
Thus, code for JQGALI = 52#895
=> Ans – (C)
27) Answer (A)
Expression : 591426
Since, the first digit is odd and last digit is even, thus rule (i) is applied and codes for ‘5’ and ‘6’ are interchanged.
=> Code for 591426 = @RA%©P
=> Ans – (A)
28) Answer (D)
Expression : 794821
Since, the first and last digit are both odds, thus rule (iii) is applied and both are to be coded as $
=> Code for 794821 = $R%D@$
=> Ans – (D)
29) Answer (B)
Expression : 813469
Since, first digit is even and last digit is odd, thus none of the rules is applied and codes are written as it is.
=> Code for 813469 = DAK%@R
=> Ans – (B)
30) Answer (C)
Expression : 671254
Since, the first and last digit are both even, thus rule (ii) is applied and both are to be coded as *
=> Code for 671254 = *MA©P*
=> Ans – (C)
31) Answer (E)
Expression : 215349
Since, first digit is even and last digit is odd, thus none of the rules is applied and codes are written as it is.
=> Code for 215349 = ©APK%R, which is not given in the options
=> Ans – (E)
32) Answer (E)
Expression : RKUMFP
Since, both the first and last letters are consonant, thus rule (i) is applied and both are to be coded as code for ‘P’
=> Code for RKUMFP = 7% *#47, which is not in the options
=> Ans – (E)
33) Answer (E)
Expression : MDEAJI
Since, first letter is consonant and last letter is vowel, thus rule (ii) is applied and their codes are interchanged.
=> Code for MDEAJI = 9 $ @ 8 2 #, which is not in the options
=> Ans – (E)
34) Answer (C)
Expression : EMNTKU
Since both the first and last letters are vowels, thus none of the rules is applied and the codes are written as it is.
=> Code for EMNTKU = @ # © 1 % *
=> Ans – (C)
35) Answer (B)
Expression : AWBRND
Since the first letter is vowel and last letter is consonant, thus none of the rules is applied and the codes are written as it is.
=> Code for AWBRND = 8563©$
=> Ans – (B)
36) Answer (D)
Expression : BDAIFE
Since, first letter is consonant and last letter is vowel, thus rule (ii) is applied and their codes are interchanged.
=> Code for BDAIFE = @$8146
=> Ans – (D)
37) Answer (A)
PAPER is written as PERPA
The pattern followed is that the first half of the word is written at the end of the code and the second half is written at the start of the code.
Thus, code for SUB JECT is = JECT SUB
Similarly, for COUNCIL, ‘NCIL’ will be written at the beginning, and ‘COU’ at the end.
=> COUNCIL : NCILCOU
=> Ans – (A)
38) Answer (D)
Word – TMRBFJ
Since, both the first and last letters are consonants, thus none of the conditions is applied.
Thus, TMRBFJ : 263 @ ©#
=> Ans – (D)
39) Answer (D)
Word – RFHKJA
Since, the first letter is a consonant and the last letter is a vowel, thus condition (ii) is applied, both are coded as = %
Thus, RFHKJA : %©48#%
=> Ans – (D)
40) Answer (A)
Word – IPAUHM
Since, the first letter is a vowel and the last letter is a consonant, thus condition (i) is applied, both are coded as = *
Thus, IPAUHM : *5174*
=> Ans – (A)
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We hope this Coding and Decoding questions for SSC GD will be highly useful for your preparation.
