{"id":24604,"date":"2019-01-28T13:18:00","date_gmt":"2019-01-28T07:48:00","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=24604"},"modified":"2019-01-28T14:55:55","modified_gmt":"2019-01-28T09:25:55","slug":"coding-and-decoding-questions-for-ssc-gd-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/coding-and-decoding-questions-for-ssc-gd-pdf\/","title":{"rendered":"Coding and Decoding Questions For SSC GD PDF"},"content":{"rendered":"

Coding and Decoding Questions For SSC GD PDF<\/h1>\n

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.<\/p>\n

CODING AND DECODING QUESTIONS FOR SSC GD PDF<\/a><\/p>\n

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SSC GD Free Mock Test<\/a> (Latest pattern)<\/p>\n

SSC GD Previous Papers<\/a><\/p>\n

Instructions<\/b><\/p>\n

In a certain language, \u201cEngland won against Colombia\u201d is coded as \u201cta ja je tu\u201d.
\n\u201dRivals are playing against Colombia\u201d is coded as \u201czu ta ja ga xu\u201d.
\n\u201dEngland and India are rivals\u201d is coded as \u201cyu je xu zu ha\u201d.
\n\u201dIndia lost against Australia\u201d is coded as \u201cli ta du yu\u201d.<\/p>\n

Question 1:\u00a0<\/b>If the code for the statement \u2018Australia likes rivals\u2019 is \u2018fa du xu\u2019 then find out the code for the statement \u2018Colombia and India are lost\u2019?<\/p>\n

a)\u00a0tu zu ja li ha<\/p>\n

b)\u00a0zu ja yu li ha<\/p>\n

c)\u00a0yu li ha zu ta<\/p>\n

d)\u00a0yu li ha xu ta<\/p>\n

e)\u00a0ha xu ja zu li<\/p>\n

Question 2:\u00a0<\/b>Which of the following can be the code for \u201cColombia and Australia are friends\u201d?<\/p>\n

a)\u00a0xu ja zu po ha<\/p>\n

b)\u00a0du po li ha ja<\/p>\n

c)\u00a0ha po du zu ja<\/p>\n

d)\u00a0li ja po ha fu<\/p>\n

e)\u00a0du zu ja ha po<\/p>\n

Question 3:\u00a0<\/b>Which of the following can be the code for \u201cPlaying against Austria\u201d?<\/p>\n

a)\u00a0du ta ga<\/p>\n

b)\u00a0ga ta li<\/p>\n

c)\u00a0du li ga<\/p>\n

d)\u00a0ta so ga<\/p>\n

e)\u00a0ta tu so<\/p>\n

Question 4:\u00a0<\/b>Which of the following must be the code for \u201cand\u201d?<\/p>\n

a)\u00a0yu<\/p>\n

b)\u00a0je<\/p>\n

c)\u00a0xu<\/p>\n

d)\u00a0ha<\/p>\n

e)\u00a0ja<\/p>\n

Question 5:\u00a0<\/b>Which of the following must be the code for \u201cplaying\u201d?<\/p>\n

a)\u00a0ga<\/p>\n

b)\u00a0ta<\/p>\n

c)\u00a0ja<\/p>\n

d)\u00a0xu<\/p>\n

e)\u00a0zu<\/p>\n

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Question 6:\u00a0<\/b>In a certain coding language, \u2018Brian gets up early\u2019 is coded as \u2018Pa ta na ma\u2019, \u2018Adam gets up late\u2019 is coded as \u2018Ba ta na ka\u2019, \u2018Brian is moving up the ladder\u2019 is coded as \u2018Pa su la ta va za\u2019. What is the code word for \u2018gets\u2019 in this language?<\/p>\n

a)\u00a0pa<\/p>\n

b)\u00a0ta<\/p>\n

c)\u00a0na<\/p>\n

d)\u00a0ma<\/p>\n

e)\u00a0Cannot be determined<\/p>\n

Instructions<\/b><\/p>\n

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. \u2018None of these\u2019 as your answer.<\/p>\n

Letters K E T J H I F A L U B M O R P
\nDigits\/Symbol 3 7 % $ 4 * 1 9 8 6 # @ 2 5 \u00a9<\/p>\n

Conditions:
\n(i) If the first as well as the last letter is a vowel their codes are to be swapped.
\n(ii) If the first as well as the last letter is a consonant both are to be coded by \u00a3.
\n(iii) If the first letter is a vowel and the last letter is a consonant, the vowel is to be coded by \u2206 and the consonant is to be coded by \u2191.<\/p>\n

Question 7:\u00a0<\/b>IJLTPU<\/p>\n

a)\u00a0*$8%\u00a96<\/p>\n

b)\u00a0*$8%\u00a9*<\/p>\n

c)\u00a06$8%\u00a96<\/p>\n

d)\u00a06$8%\u00a9*<\/p>\n

e)\u00a0None of these<\/p>\n

Question 8:\u00a0<\/b>KEOMPA<\/p>\n

a)\u00a0\u219172@\u00a9\u2206<\/p>\n

b)\u00a0\u220672@\u00a9\u2191<\/p>\n

c)\u00a0372@\u00a99<\/p>\n

d)\u00a0972@\u00a93<\/p>\n

e)\u00a0None of these<\/p>\n

Question 9:\u00a0<\/b>ORBETH<\/p>\n

a)\u00a0\u21915#7%\u2206<\/p>\n

b)\u00a0\u22065#7%\u2191<\/p>\n

c)\u00a025#7%\u2206<\/p>\n

d)\u00a045#7%2<\/p>\n

e)\u00a0None of these<\/p>\n

Question 10:\u00a0<\/b>AJTKLU<\/p>\n

a)\u00a09$%386<\/p>\n

b)\u00a0\u00a3$%38\u00a3<\/p>\n

c)\u00a0\u2206$%38\u2191<\/p>\n

d)\u00a0\u2191$%38\u2206<\/p>\n

e)\u00a0None of these<\/p>\n

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Question 11:\u00a0<\/b>TARIFM<\/p>\n

a)\u00a0\u00a395*1\u00a3<\/p>\n

b)\u00a0%95*1!@<\/p>\n

c)\u00a0%95* 1 %<\/p>\n

d)\u00a0@95*1@<\/p>\n

e)\u00a0None of these<\/p>\n

Instructions<\/b><\/p>\n

The following rules are followed while coding a word.
\nLetters 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).<\/p>\n

Question 12:\u00a0<\/b>The fourth letter represented by the word 261216723 cannot be<\/p>\n

a)\u00a0Y<\/p>\n

b)\u00a0O<\/p>\n

c)\u00a0T<\/p>\n

d)\u00a0Z<\/p>\n

e)\u00a0U<\/p>\n

Question 13:\u00a0<\/b>Which of the following cannot be the word represented by 11131415?<\/p>\n

a)\u00a0ZZZXWZV<\/p>\n

b)\u00a0ZZZXZWL<\/p>\n

c)\u00a0PNZWL<\/p>\n

d)\u00a0PZXMZV<\/p>\n

e)\u00a0More than one of the above<\/p>\n

Question 14:\u00a0<\/b>If it is known for sure that the word represented by 271213 has 4 letters, then the word is<\/p>\n

a)\u00a0YKOX<\/p>\n

b)\u00a0YTYX<\/p>\n

c)\u00a0YTON<\/p>\n

d)\u00a0YTFX<\/p>\n

e)\u00a0YTYN<\/p>\n

Question 15:\u00a0<\/b>Which of the following conditions will be sufficient to uniquely determine the word represented by 261712?<\/p>\n

a)\u00a0The word coded is 4 lettered<\/p>\n

b)\u00a0The word coded is 5 lettered<\/p>\n

c)\u00a0The word coded is 6 lettered<\/p>\n

d)\u00a0The word coded is 3 lettered<\/p>\n

e)\u00a0More than one of the above.<\/p>\n

Question 16:\u00a0<\/b>Which of the following words can be represented by 212312?<\/p>\n

a)\u00a0YOXO<\/p>\n

b)\u00a0FYXO<\/p>\n

c)\u00a0YZDO<\/p>\n

d)\u00a0FDZY<\/p>\n

e)\u00a0More than one of the above<\/p>\n

Instructions<\/b><\/p>\n

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.
\nConditions :
\n(I) If both the first and the last digits of the number are odd digits then both are to be coded as I.
\n(II) If both the first and last digits of the number are even digits then both are to be coded as Y.<\/p>\n

Question 17:\u00a0<\/b>831795<\/p>\n

a)\u00a0INJTWR<\/p>\n

b)\u00a0BNJTWR<\/p>\n

c)\u00a0BNJTWY<\/p>\n

d)\u00a0YNJYWY<\/p>\n

e)\u00a0None of these<\/p>\n

Question 18:\u00a0<\/b>615824<\/p>\n

a)\u00a0IJRBMI<\/p>\n

b)\u00a0IJRBMY<\/p>\n

c)\u00a0YJRBMY<\/p>\n

d)\u00a0DJRBMK<\/p>\n

e)\u00a0None of these<\/p>\n

Question 19:\u00a0<\/b>591248<\/p>\n

a)\u00a0IWJMKB<\/p>\n

b)\u00a0RWJMKY<\/p>\n

c)\u00a0YWJMKY<\/p>\n

d)\u00a0RWJMKB<\/p>\n

e)\u00a0None of these<\/p>\n

Question 20:\u00a0<\/b>263847<\/p>\n

a)\u00a0IDNBKI<\/p>\n

b)\u00a0YDNBKY<\/p>\n

c)\u00a0IDNBKY<\/p>\n

d)\u00a0MDNBKY<\/p>\n

e)\u00a0None of these<\/p>\n

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Question 21:\u00a0<\/b>726395<\/p>\n

a)\u00a0IMDNWI<\/p>\n

b)\u00a0YMDNWY,,<\/p>\n

c)\u00a0TMDNWR<\/p>\n

d)\u00a0IMDNWR<\/p>\n

e)\u00a0None of these<\/p>\n

Instructions<\/b><\/p>\n

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. \u2018None of these\u2019 as the answer. Symbol
\nConditions :
\n(i) If both the first and the last letters of the group are vowels, their codes are to be interchanged.
\n(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.<\/p>\n

Question 22:\u00a0<\/b>QRLGHM<\/p>\n

a)\u00a0219#@2<\/p>\n

b)\u00a0619#@6<\/p>\n

c)\u00a0619#@2<\/p>\n

d)\u00a06190#2<\/p>\n

e)\u00a0None of these<\/p>\n

Question 23:\u00a0<\/b>BQRLHA<\/p>\n

a)\u00a0$21908<\/p>\n

b)\u00a0851908<\/p>\n

c)\u00a0821908<\/p>\n

d)\u00a0$219@$<\/p>\n

e)\u00a0None of these<\/p>\n

Question 24:\u00a0<\/b>IGCHRE<\/p>\n

a)\u00a0%#*@13<\/p>\n

b)\u00a03#*@1%<\/p>\n

c)\u00a03#*@13<\/p>\n

d)\u00a0%#*01%<\/p>\n

e)\u00a0None of these<\/p>\n

Question 25:\u00a0<\/b>EBHRMT<\/p>\n

a)\u00a0%0$16O<\/p>\n

b)\u00a0%$@16\u00a9<\/p>\n

c)\u00a06$@1\u00a9%<\/p>\n

d)\u00a0@$@16%<\/p>\n

e)\u00a0None of these<\/p>\n

Question 26:\u00a0<\/b>JQGALI<\/p>\n

a)\u00a032#893<\/p>\n

b)\u00a052#893<\/p>\n

c)\u00a052#895<\/p>\n

d)\u00a032#895<\/p>\n

e)\u00a0None of these<\/p>\n

Instructions<\/b><\/p>\n

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\u2019 as your answer.<\/p>\n

Conditions :
\n(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.
\n(ii) If both the first and the last digits are even, both are to be coded as *.
\n(iii) If both the first and the last digits are odd, both are to be coded as $.<\/p>\n

Question 27:\u00a0<\/b>591426<\/p>\n

a)\u00a0@RA%\u00a9P<\/p>\n

b)\u00a0PRA%\u00a9@<\/p>\n

c)\u00a0@AR%\u00a9P<\/p>\n

d)\u00a0$RA%\u00a9*<\/p>\n

e)\u00a0None of these<\/p>\n

Question 28:\u00a0<\/b>794821<\/p>\n

a)\u00a0MR%D\u00a9A<\/p>\n

b)\u00a0AR%D\u00a9M<\/p>\n

c)\u00a0M%RD\u00a9A<\/p>\n

d)\u00a0$R%D@$<\/p>\n

e)\u00a0None of these<\/p>\n

Question 29:\u00a0<\/b>813469<\/p>\n

a)\u00a0RAK%@D<\/p>\n

b)\u00a0DAK%@R<\/p>\n

c)\u00a0DAP%@R<\/p>\n

d)\u00a0*AK%@*<\/p>\n

e)\u00a0None of these<\/p>\n

Question 30:\u00a0<\/b>671254<\/p>\n

a)\u00a0\u00a9MA\u00a9P%<\/p>\n

b)\u00a0$MA\u00a9P$<\/p>\n

c)\u00a0*MA\u00a9P*<\/p>\n

d)\u00a0%MA\u00a9P\u00a9<\/p>\n

e)\u00a0None of these<\/p>\n

Question 31:\u00a0<\/b>215349<\/p>\n

a)\u00a0RAPK%\u00a9<\/p>\n

b)\u00a0*APK%$<\/p>\n

c)\u00a0$APK%$<\/p>\n

d)\u00a0\u00a9PAK%R<\/p>\n

e)\u00a0None of these<\/p>\n

Instructions<\/b><\/p>\n

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 \u2018None of these\u2019 as the answer<\/p>\n

Conditions :
\n(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.
\n(ii) If the first letter is a consonant and the last letter is a vowel, the codes are to be interchanged.<\/p>\n

Question 32:\u00a0<\/b>RKUMFP<\/p>\n

a)\u00a07 % C * # 4 3<\/p>\n

b)\u00a03 * % # 4 7<\/p>\n

c)\u00a03 % * # 4 3<\/p>\n

d)\u00a03 % * # 4 7<\/p>\n

e)\u00a0None of these<\/p>\n

Question 33:\u00a0<\/b>MDEAJI<\/p>\n

a)\u00a01 $ @ 8 2 #<\/p>\n

b)\u00a0# $ @ 8 2 1<\/p>\n

c)\u00a01 $@ 2 1<\/p>\n

d)\u00a0# $ @ 8 2 #<\/p>\n

e)\u00a0None of these<\/p>\n

Question 34:\u00a0<\/b>EMNTKU<\/p>\n

a)\u00a0* # \u00a9 1 % @<\/p>\n

b)\u00a0@ # \u00a9 1 4 *<\/p>\n

c)\u00a0@ # \u00a9 1 % *<\/p>\n

d)\u00a0# @ \u00a9 1 % *<\/p>\n

e)\u00a0None of these<\/p>\n

Question 35:\u00a0<\/b>AWBRND<\/p>\n

a)\u00a0$563\u00a98<\/p>\n

b)\u00a08563\u00a9$<\/p>\n

c)\u00a08365\u00a9$<\/p>\n

d)\u00a08536\u00a9$<\/p>\n

e)\u00a0None of these<\/p>\n

Question 36:\u00a0<\/b>BDAIFE<\/p>\n

a)\u00a06$8146<\/p>\n

b)\u00a068814@<\/p>\n

c)\u00a0@$814@<\/p>\n

d)\u00a0@$8146<\/p>\n

e)\u00a0None of these<\/p>\n

Question 37:\u00a0<\/b>In a certain coding system, PAPER is written as PERPA and SUBJECT is written as JECTSUB, what should be the code for COUNCIL ?<\/p>\n

a)\u00a0NCILCOU<\/p>\n

b)\u00a0LICNOUC<\/p>\n

c)\u00a0NCOUCIL<\/p>\n

d)\u00a0NLICUOC<\/p>\n

e)\u00a0None of these<\/p>\n

Instructions<\/b><\/p>\n

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.<\/p>\n

Letter P M A J E T K I R B U F H
\nCode : 5 6 1 # 9 2 8 $ 3 @ 7 \u00a9 4
\nConditions : (i) If the first letter is a vowel and the last letter is a consonant, both are to be coded as *.
\n(ii) If the first letter is consonant and the last letter is a vowel, both are to be coded as %.<\/p>\n

Question 38:\u00a0<\/b>TMRBFJ<\/p>\n

a)\u00a0*6 3@ \u00a9*<\/p>\n

b)\u00a0%63@ \u00a9%<\/p>\n

c)\u00a0236@ \u00a9#<\/p>\n

d)\u00a0263 @ \u00a9#<\/p>\n

e)\u00a0None of these<\/p>\n

Question 39:\u00a0<\/b>RFHKJA<\/p>\n

a)\u00a03 \u00a948#1<\/p>\n

b)\u00a0483 \u00a9#1<\/p>\n

c)\u00a0* \u00a948#*<\/p>\n

d)\u00a0%\u00a948#%<\/p>\n

e)\u00a0None of these<\/p>\n

Question 40:\u00a0<\/b>IPAUHM<\/p>\n

a)\u00a0*5174*<\/p>\n

b)\u00a0%5 I 74%<\/p>\n

c)\u00a0$51746<\/p>\n

d)\u00a0$51476<\/p>\n

e)\u00a0None of these<\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

We are given that<\/p>\n

(1)\u201cEngland won against Colombia\u201d is coded as \u201cta ja je tu\u201d.<\/p>\n

(2)\u201dRivals are playing against Colombia\u201d is coded as \u201czu ta ja ga xu\u201d.<\/p>\n

(3)\u201dEngland and India are rivals\u201d is coded as \u201cyu je xu zu ha\u201d.<\/p>\n

(4)\u201dIndia lost against Australia\u201d is coded as \u201cli ta du yu\u201d.<\/p>\n

From statement (1) and (4) we can see that only word \u2018against\u2019 is common hence code for the word \u2018against\u2019 is \u2018ta\u2019.<\/p>\n

From statement (1) and (2) we can see that words \u2018against\u2019 and \u2018Colombia\u2019 are common hence code for word \u2018Colombia\u2019 is \u2018ja\u2019.<\/p>\n

From statement (3) and (4) we can see that only word \u2018India\u2019 is common hence code for word \u2018India\u2019 is \u2018yu\u2019.<\/p>\n

From statement (1) and (3) we can see that only word \u2018England\u2019 is common hence code for word \u2018England\u2019 is \u2018je\u2019.<\/p>\n

From statement (1) we can say that code for word \u2018won\u2019 is \u2018tu\u2019.<\/p>\n

In statement (3) we know that \u2018yu\u2019 and \u2018je\u2019 are codes for words \u2018India\u2019 and \u2018England\u2019 respectively. 2 codes out of remaining 3 codes \u2018xu\u2019, \u2018zu\u2019 and \u2018ha\u2019 should be common with codes available in statement (2) because of the words \u2018Rivals\u2019 and \u2018are\u2019. Those two common codes are \u2018xu\u2019 and \u2018zu\u2019 and hence we can say that \u2018ha\u2019 is code for the word \u2018and\u2019 in statement 3 and \u2018ga\u2019 is code for the word \u2018playing\u2019.<\/p>\n

For words \u2018lost\u2019 and \u2018Australia\u2019 codes are \u2018li\u2019 and \u2018du\u2019 in any order.<\/p>\n

For words \u2018Rivals\u2019 and \u2018are\u2019 codes are \u2018xu\u2019 and \u2018zu\u2019 in any order. Tabulating all available data<\/p>\n

<\/figure>\n

\u2018Australia likes rivals\u2019 is \u2018fa du xu\u2019 that means codes for words ‘Australia’<\/strong> and ‘rivals’<\/strong> are ‘du’<\/strong> and ‘xu’<\/strong> respectively. Consequently codes for words ‘lost’<\/strong> and ‘are’<\/strong> will be ‘li’<\/strong> and ‘zu’<\/strong> respectively.<\/p>\n

Hence we can say that code for statement ‘Colombia and India are lost’ <\/span><\/strong>will be ‘ja’, ‘ha’, ‘yu’, ‘zu’<\/strong> and ‘li’<\/strong> in any order. We can see that all codes are available in option B. Hence, option B is the correct answer.<\/p>\n

2)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

We are given that<\/p>\n

(1)\u201cEngland won against Colombia\u201d is coded as \u201cta ja je tu\u201d.<\/p>\n

(2)\u201dRivals are playing against Colombia\u201d is coded as \u201czu ta ja ga xu\u201d.<\/p>\n

(3)\u201dEngland and India are rivals\u201d is coded as \u201cyu je xu zu ha\u201d.<\/p>\n

(4)\u201dIndia lost against Australia\u201d is coded as \u201cli ta du yu\u201d.<\/p>\n

From statement (1) and (4) we can see that only word \u2018against\u2019 is common hence code for the word \u2018against\u2019 is \u2018ta\u2019.<\/p>\n

From statement (1) and (2) we can see that words \u2018against\u2019 and \u2018Colombia\u2019 are common hence code for word \u2018Colombia\u2019 is \u2018ja\u2019.<\/p>\n

From statement (3) and (4) we can see that only word \u2018India\u2019 is common hence code for word \u2018India\u2019 is \u2018yu\u2019.<\/p>\n

From statement (1) and (3) we can see that only word \u2018England\u2019 is common hence code for word \u2018England\u2019 is \u2018je\u2019.<\/p>\n

From statement (1) we can say that code for word \u2018won\u2019 is \u2018tu\u2019.<\/p>\n

In statement (3) we know that \u2018yu\u2019 and \u2018je\u2019 are codes for words \u2018India\u2019 and \u2018England\u2019 respectively. 2 codes out of remaining 3 codes \u2018xu\u2019, \u2018zu\u2019 and \u2018ha\u2019 should be common with codes available in statement (2) because of the words \u2018Rivals\u2019 and \u2018are\u2019. Those two common codes are \u2018xu\u2019 and \u2018zu\u2019 and hence we can say that \u2018ha\u2019 is code for the word \u2018and\u2019 in statement 3 and \u2018ga\u2019 is code for the word \u2018playing\u2019.<\/p>\n

For words \u2018lost\u2019 and \u2018Australia\u2019 codes are \u2018li\u2019 and \u2018du\u2019 in any order.<\/p>\n

For words \u2018Rivals\u2019 and \u2018are\u2019 codes are \u2018xu\u2019 and \u2018zu\u2019 in any order. Tabulating all available data<\/p>\n

<\/figure>\n

We can see that codes for words “Columbia”<\/strong> “and”<\/strong> “Australia”<\/strong> and “are”<\/strong> are ‘ja’<\/span> ‘ha’, ‘du\/li’ <\/strong>and ‘xu\/zu’<\/strong> respectively.Code for word ‘friends<\/strong>‘ is not known to us.<\/p>\n

Option (A) xu ja zu po ha: – Here both codes ‘xu\/zu’<\/strong> are used. Hence, this option is incorrect.<\/p>\n

Option (B) du po li ha ja: – Here both codes ‘du\/li’<\/strong> are used. Hence, this option is incorrect.<\/p>\n

Option (C) hi po du zu ja: – Code for word ‘and<\/strong>‘, ‘ha’ is not available. Hence this option is incorrect.<\/p>\n

Option (D) li ja po ha fu: – Code for word ‘are<\/strong>‘, ‘xu\/zu<\/strong>‘ is not available. Hence this option is incorrect.<\/p>\n

Option (E) du zu ja ha po: – Code for word ‘friends<\/strong>‘ may be ‘po<\/strong>‘. This option is correct.<\/p>\n

3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

We are given that<\/p>\n

(1)\u201cEngland won against Colombia\u201d is coded as \u201cta ja je tu\u201d.<\/p>\n

(2)\u201dRivals are playing against Colombia\u201d is coded as \u201czu ta ja ga xu\u201d.<\/p>\n

(3)\u201dEngland and India are rivals\u201d is coded as \u201cyu je xu zu ha\u201d.<\/p>\n

(4)\u201dIndia lost against Australia\u201d is coded as \u201cli ta du yu\u201d.<\/p>\n

From statement (1) and (4) we can see that only word \u2018against\u2019 is common hence code for the word \u2018against\u2019 is \u2018ta\u2019.<\/p>\n

From statement (1) and (2) we can see that words \u2018against\u2019 and \u2018Colombia\u2019 are common hence code for word \u2018Colombia\u2019 is \u2018ja\u2019.<\/p>\n

From statement (3) and (4) we can see that only word \u2018India\u2019 is common hence code for word \u2018India\u2019 is \u2018yu\u2019.<\/p>\n

From statement (1) and (3) we can see that only word \u2018England\u2019 is common hence code for word \u2018England\u2019 is \u2018je\u2019.<\/p>\n

From statement (1) we can say that code for word \u2018won\u2019 is \u2018tu\u2019.<\/p>\n

In statement (3) we know that \u2018yu\u2019 and \u2018je\u2019 are codes for words \u2018India\u2019 and \u2018England\u2019 respectively. 2 codes out of remaining 3 codes \u2018xu\u2019, \u2018zu\u2019 and \u2018ha\u2019 should be common with codes available in statement (2) because of the words \u2018Rivals\u2019 and \u2018are\u2019. Those two common codes are \u2018xu\u2019 and \u2018zu\u2019 and hence we can say that \u2018ha\u2019 is code for the word \u2018and\u2019 in statement 3 and \u2018ga\u2019 is code for the word \u2018playing\u2019.<\/p>\n

For words \u2018lost\u2019 and \u2018Australia\u2019 codes are \u2018li\u2019 and \u2018du\u2019 in any order.<\/p>\n

For words \u2018Rivals\u2019 and \u2018are\u2019 codes are \u2018xu\u2019 and \u2018zu\u2019 in any order. Tabulating all available data<\/p>\n

<\/figure>\n

We can see that codes for words ‘playing’ and ‘against’ are ‘ga’ and ‘ta’ respectively.Code for word ‘Austria’ is not known to us.<\/p>\n

Option (A) du ta ga: – du is one of two possible codes for word ‘Australia’ not ‘Austria’. Hence this option is incorrect.<\/p>\n

Option (B) ga ta li: – li is one of two possible codes for word ‘Australia’ not ‘Austria’.Hence this option is incorrect.<\/p>\n

Option (C) du li ga: – Code for word ‘against’ is not available. Hence this option is incorrect.<\/p>\n

Option (E) ta tu so: – Code for word ‘playing’ is not available. Hence this option is incorrect.<\/p>\n

Option (D) ta so ga: – Code for word ‘Austria’ may be ‘so’. Hence, this option is correct.<\/p>\n

4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

We are given that<\/p>\n

(1)\u201cEngland won against Colombia\u201d is coded as \u201cta ja je tu\u201d.<\/p>\n

(2)\u201dRivals are playing against Colombia\u201d is coded as \u201czu ta ja ga xu\u201d.<\/p>\n

(3)\u201dEngland and India are rivals\u201d is coded as \u201cyu je xu zu ha\u201d.<\/p>\n

(4)\u201dIndia lost against Australia\u201d is coded as \u201cli ta du yu\u201d.<\/p>\n

From statement (1) and (4) we can see that only word \u2018against\u2019 is common hence code for the word \u2018against\u2019 is \u2018ta\u2019.<\/p>\n

From statement (1) and (2) we can see that words \u2018against\u2019 and \u2018Colombia\u2019 are common hence code for word \u2018Colombia\u2019 is \u2018ja\u2019.<\/p>\n

From statement (3) and (4) we can see that only word \u2018India\u2019 is common hence code for word \u2018India\u2019 is \u2018yu\u2019.<\/p>\n

From statement (1) and (3) we can see that only word \u2018England\u2019 is common hence code for word \u2018England\u2019 is \u2018je\u2019.<\/p>\n

From statement (1) we can say that code for word \u2018won\u2019 is \u2018tu\u2019.<\/p>\n

In statement (3) we know that \u2018yu\u2019 and \u2018je\u2019 are codes for words \u2018India\u2019 and \u2018England\u2019 respectively. 2 codes out of remaining 3 codes \u2018xu\u2019, \u2018zu\u2019 and \u2018ha\u2019 should be common with codes available in statement (2) because of the words \u2018Rivals\u2019 and \u2018are\u2019. Those two common codes are \u2018xu\u2019 and \u2018zu\u2019 and hence we can say that \u2018ha\u2019 is code for the word \u2018and\u2019 in statement 3 and \u2018ga\u2019 is code for the word \u2018playing\u2019.<\/p>\n

For words \u2018lost\u2019 and \u2018Australia\u2019 codes are \u2018li\u2019 and \u2018du\u2019 in any order.<\/p>\n

For words \u2018Rivals\u2019 and \u2018are\u2019 codes are \u2018xu\u2019 and \u2018zu\u2019 in any order. Tabulating all available data<\/p>\n

<\/figure>\n

From the table we can see that code for word “and” is ‘ha’. Hence, option D is the correct answer.<\/p>\n

5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

We are given that<\/p>\n

(1)\u201cEngland won against Colombia\u201d is coded as \u201cta ja je tu\u201d.<\/p>\n

(2)\u201dRivals are playing against Colombia\u201d is coded as \u201czu ta ja ga xu\u201d.<\/p>\n

(3)\u201dEngland and India are rivals\u201d is coded as \u201cyu je xu zu ha\u201d.<\/p>\n

(4)\u201dIndia lost against Australia\u201d is coded as \u201cli ta du yu\u201d.<\/p>\n

From statement (1) and (4) we can see that only word \u2018against\u2019 is common hence code for the word \u2018against\u2019 is \u2018ta\u2019.<\/p>\n

From statement (1) and (2) we can see that words \u2018against\u2019 and \u2018Colombia\u2019 are common hence code for word \u2018Colombia\u2019 is \u2018ja\u2019.<\/p>\n

From statement (3) and (4) we can see that only word \u2018India\u2019 is common hence code for word \u2018India\u2019 is \u2018yu\u2019.<\/p>\n

From statement (1) and (3) we can see that only word \u2018England\u2019 is common hence code for word \u2018England\u2019 is \u2018je\u2019.<\/p>\n

From statement (1) we can say that code for word \u2018won\u2019 is \u2018tu\u2019.<\/p>\n

In statement (3) we know that \u2018yu\u2019 and \u2018je\u2019 are codes for words \u2018India\u2019 and \u2018England\u2019 respectively. 2 codes out of remaining 3 codes \u2018xu\u2019, \u2018zu\u2019 and \u2018ha\u2019 should be common with codes available in statement (2) because of the words \u2018Rivals\u2019 and \u2018are\u2019. Those two common codes are \u2018xu\u2019 and \u2018zu\u2019 and hence we can say that \u2018ha\u2019 is code for the word \u2018and\u2019 in statement 3 and \u2018ga\u2019 is code for the word \u2018playing\u2019.<\/p>\n

For words \u2018lost\u2019 and \u2018Australia\u2019 codes are \u2018li\u2019 and \u2018du\u2019 in any order.<\/p>\n

For words \u2018Rivals\u2019 and \u2018are\u2019 codes are \u2018xu\u2019 and \u2018zu\u2019 in any order. Tabulating all available data<\/p>\n

<\/figure>\n

From the table we can see that code for word “playing” is ‘ga’. Hence, option A is the correct answer.<\/p>\n

SSC GD Free Study Material<\/a><\/p>\n

6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

We can see that \u2018gets up\u2019 is common between the first and second coded sentences. Hence, the code word for these words must be \u2018ta na\u2019 in no particular order. \u2018up\u2019 is common between the second and third coded sentences. Hence, \u2018up\u2019 must be coded as \u2018ta\u2019. Thus, the code word for \u2018gets\u2019 must be \u2018na\u2019. Hence, option C is the correct answer.<\/p>\n

7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Expression : IJLTPU<\/p>\n

Since, the first and the last letters are vowels, rule (i) is applied and codes for ‘I’ and ‘U’ are swapped.<\/p>\n

=> Code for IJLTPU = 6$8%\u00a9*<\/p>\n

=> Ans – (D)<\/p>\n

8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Expression : KEOMPA<\/p>\n

None of the rules is applied, and the codes are written as it is.<\/p>\n

Thus, code for KEOMPA = 372@\u00a99<\/p>\n

=> Ans – (C)<\/p>\n

9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Expression : ORBETH<\/p>\n

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 \u2206 and the consonant is to be coded by \u2191.<\/p>\n

Code for ORBETH = \u22065#7%\u2191<\/p>\n

=> Ans – (B)<\/p>\n

10)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Expression : AJTKLU<\/p>\n

Since, the first and the last letters are vowels, rule (i) is applied and codes for ‘A’ and ‘U’ are swapped.<\/p>\n

Code for AJTKLU = 6$%389, which is not given in the options.<\/p>\n

=> Ans – (E) : none of these<\/p>\n

11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Expression : TARIFM<\/p>\n

Since, both the first and last letters are consonant, thus by applying rule (ii), both are to be coded by \u00a3.<\/p>\n

Code for TARIFM = \u00a395*1\u00a3<\/p>\n

=> Ans – (A)<\/p>\n

12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

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.<\/p>\n

13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let us write the code of the words to find out.
\nZZZXWZV – 1113415
\nZZZXZWL – 11131415
\nPNZWL – 11131415
\nPZXMZV – 11131415
\nHence, option A is the right answer.<\/p>\n

14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

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.<\/p>\n

15)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

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.<\/p>\n

16)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

YOXO – 212312
\nFYXO – 212312
\nYZDO – 212312
\nFDZY – 212312
\nHence, option E is the right answer.<\/p>\n

17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

20)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

21)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

22)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Expression : QRLGHM<\/p>\n

Since both the first and last letters are consonant. Thus, the code for this expression will be written as it is.<\/p>\n

=> Code for QRLGHM = 219#@6, which is not in the options.<\/p>\n

=> Ans – (E)<\/p>\n

23)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Expression : BQRLHA<\/p>\n

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. ‘$’<\/p>\n

Thus, code for BQRLHA = $219@$<\/p>\n

=> Ans – (D)<\/p>\n

24)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Expression : IGCHRE<\/p>\n

Since, the first and last letters are both vowels, their codes are interchanged.<\/p>\n

Thus, code for IGCHRE = %# *@13<\/p>\n

=> Ans – (A)<\/p>\n

25)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Expression : EBHRMT<\/p>\n

Since both the first and last letters are consonant. Thus, the code for this expression will be written as it is.<\/p>\n

=> Code for EBHRMT = %$@16\u00a9, which is not in the options.<\/p>\n

=> Ans – (B)<\/p>\n

26)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Expression : JQGALI<\/p>\n

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’<\/p>\n

Thus, code for JQGALI = 52#895<\/p>\n

=> Ans – (C)<\/p>\n

27)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Expression : 591426<\/p>\n

Since, the first digit is odd and last digit is even, thus rule (i) is applied and codes for ‘5’ and ‘6’ are interchanged.<\/p>\n

=> Code for 591426 = @RA%\u00a9P<\/p>\n

=> Ans – (A)<\/p>\n

28)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Expression : 794821<\/p>\n

Since, the first and last digit are both odds, thus rule (iii) is applied and both are to be coded as $<\/p>\n

=> Code for 794821 = $R%D@$<\/p>\n

=> Ans – (D)<\/p>\n

29)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Expression : 813469<\/p>\n

Since, first digit is even and last digit is odd, thus none of the rules is applied and codes are written as it is.<\/p>\n

=> Code for 813469 = DAK%@R<\/p>\n

=> Ans – (B)<\/p>\n

30)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Expression : 671254<\/p>\n

Since, the first and last digit are both even, thus rule (ii) is applied and both are to be coded as *<\/p>\n

=> Code for 671254 = *MA\u00a9P*<\/p>\n

=> Ans – (C)<\/p>\n

31)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Expression : 215349<\/p>\n

Since, first digit is even and last digit is odd, thus none of the rules is applied and codes are written as it is.<\/p>\n

=> Code for 215349 = \u00a9APK%R, which is not given in the options<\/p>\n

=> Ans – (E)<\/p>\n

32)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Expression : RKUMFP<\/p>\n

Since, both the first and last letters are consonant, thus rule (i) is applied and both are to be coded as code for ‘P’<\/p>\n

=> Code for RKUMFP = 7% *#47, which is not in the options<\/p>\n

=> Ans – (E)<\/p>\n

33)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Expression : MDEAJI<\/p>\n

Since, first letter is consonant and last letter is vowel, thus rule (ii) is applied and their codes are interchanged.<\/p>\n

=> Code for MDEAJI = 9 $ @ 8 2 #, which is not in the options<\/p>\n

=> Ans – (E)<\/p>\n

34)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Expression : EMNTKU<\/p>\n

Since both the first and last letters are vowels, thus none of the rules is applied and the codes are written as it is.<\/p>\n

=> Code for EMNTKU = @ # \u00a9 1 % *<\/p>\n

=> Ans – (C)<\/p>\n

35)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Expression : AWBRND<\/p>\n

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.<\/p>\n

=> Code for AWBRND = 8563\u00a9$<\/p>\n

=> Ans – (B)<\/p>\n

36)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Expression : BDAIFE<\/p>\n

Since, first letter is consonant and last letter is vowel, thus rule (ii) is applied and their codes are interchanged.<\/p>\n

=> Code for BDAIFE = @$8146<\/p>\n

=> Ans – (D)<\/p>\n

37)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

PAPER is written as PERPA<\/p>\n

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.<\/p>\n

Thus, code for SUB JECT is = JECT SUB<\/p>\n

Similarly, for COUNCIL, ‘NCIL’ will be written at the beginning, and ‘COU’ at the end.<\/p>\n

=> COUNCIL : NCILCOU<\/strong><\/p>\n

=> Ans – (A)<\/p>\n

38)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Word – TMRBFJ<\/p>\n

Since, both the first and last letters are consonants, thus none of the conditions is applied.<\/p>\n

Thus, TMRBFJ : 263 @ \u00a9#<\/strong><\/p>\n

=> Ans – (D)<\/p>\n

39)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Word – RFHKJA<\/p>\n

Since, the first letter is a consonant and the last letter is a vowel, thus condition (ii) is applied, both are coded as = %<\/p>\n

Thus, RFHKJA : %\u00a948#%<\/strong><\/p>\n

=> Ans – (D)<\/p>\n

40)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Word – IPAUHM<\/p>\n

Since, the first letter is a vowel and the last letter is a consonant, thus condition (i) is applied, both are coded as = *<\/p>\n

Thus, IPAUHM : *5174*<\/strong><\/p>\n

=> Ans – (A)<\/p>\n

DOWNLOAD APP FOR SSC FREE MOCKS<\/a><\/p>\n

We hope this Coding and Decoding questions for SSC GD will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

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. SSC GD Free Mock Test (Latest pattern) SSC GD Previous Papers Instructions In a certain language, […]<\/p>\n","protected":false},"author":32,"featured_media":24605,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[9,1493,1459,1268],"tags":[1461],"class_list":{"0":"post-24604","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-ssc","8":"category-ssc-cpo","9":"category-ssc-gd","10":"category-ssc-stenographer","11":"tag-ssc-gd"},"better_featured_image":{"id":24605,"alt_text":"coding and decoding questions for ssc gd pdf","caption":"coding and decoding questions for ssc gd pdf","description":"coding and decoding questions for ssc gd 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