16 Jan 2016

PrepHelp Educators

Coding of Words

CODING OF WORDS:
Question based on coding of words are always asked in exams.In this type of questions two rows of letters and digits are given, which could be used as mutual codes. On the basis of the given information you are to solve the given set of questions. So let us Take some examples and try to solve them.

Ex. In each question below a group of letters is given followed by four combinations of digits 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 combination correctly represents give option (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

Condition : (1) If the first letter is a vowel and the last letter is a consonant, both are to be coded as *


Q(1) IKBUTA
(a) $8@721       (b) *8@72*       (c) %8@72%       (d) *8@72%        (e) None of these

Solution : I    K    B    U    T    A

$    8    @    7    2    1

Here first and last both letters are vowels, So option (a) is correct




Q(2). EMPRJH
(a) 9653#4       (b) *9653#       (c) %653#%       (d) *653#*        (e) None of these

Solution : E    M    P    R    J    H

*    6    5    3    #    *

Here first letter is vowel and last is consonant, so condition (1) is applied




Q(3) IPAUHM
(a) *5174*       (b) %5174%       (c) $51746       (d) $51476        (e) None of these

Solution : I    P    A    U    H    M

*    5    1    7    4    *

Here condition (1) is applied




Q(4) RFHKJA
(a) 3©48       (b) 483©#1       (c) *©48#*       (d) %©48#%        (e) None of these

Solution : R    F    H    K    J    A

%    ©    4    8    #    %

Here first letter is consonant and the last is vowel, so condition (2) is applied




Q(5) TMRBFJ
(a) *63@©*       (b) %63@©%       (c) 236@©#       (d) 263@©#        (e) None of these

Solution : T    M    R    B    F    J

2    6    3    @    ©   #

Both (1) and (2) not get full filled so simple coding is to be done.




Q(6) KTJUFA
(a) 82#7©1       (b) *2#7©*       (c) %2#7©%      (d) $2#7©1        (e) None of these

Solution : K    T    J    U    F    A

%    2    #    7    ©   %

Here condition (2) is satisfied.




Q(7) UMBKPE
(a) *6@85*       (b) 76@85©       (c) *6@85©      (d) 76@85%        (e) None of these

Solution : U    M    B    K    P    E

7    6    @    8    5   9

Here first and lst both words are vowels so none of the conditions get satisfied , so normal coding is to be done

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