Conditional Matrix MCQ Quiz - Objective Question with Answer for Conditional Matrix - Download Free PDF
Last updated on Jul 8, 2025
Latest Conditional Matrix MCQ Objective Questions
Conditional Matrix Question 1:
In this question, a group of number/symbol is coded using letter codes as per the codes given below and the conditions which follow. The correct combination of codes (following only the condition/s that apply) is your answer.
Conditions:
(i) If the first element is a symbol and the last a number, the codes for these two (the first and the last elements) are to be interchanged.
(ii) If the first element is an even number and the last an odd number, the first and last elements are to be coded as ©
(iii) If both the second and third elements are perfect squares, the third element is to be coded as the code for the second element
2 # @ 7 9
Answer (Detailed Solution Below)
Conditional Matrix Question 1 Detailed Solution
Given:
| Number/Symbol | 2 | # | @ | 7 | 9 |
| Code | M | Y | T | P | V |
| Conditions | Coded As |
| If the first element is a symbol and the last a number, the codes for these two (the first and the last elements) are to be interchanged. | Codes for first and last elements are interchanged |
| If the first element is an even number and the last an odd number, the first and last elements are to be coded as © | First and last elements are coded as © |
| If both the second and third elements are perfect squares, the third element is to be coded as the code for the second element | Third element takes the code for the second element |
Now, the group of numbers and symbols is: 2 # @ 7 9
Here, 2nd condition applies (If the first element is an even number and the last an odd number, the first and last elements are to be coded as ©
Hence, the code for 2 # @ 7 9 is → © P Z L ©
Therefore, the correct answer is "Option 3".
Conditional Matrix Question 2:
A group of numbers and symbols is coded using letter codes as per the codes given below and the conditions that follow. Study the given codes and conditions and answer the question that follows.
Note: If none of the conditions apply, then codes for the respective number/symbol are to be followed directly as given in the table.
| Number/symbol | 2 | @ | 9 | 5 | $ | & | 3 | % | # | 7 | + | 4 | 8 | 6 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Code | T | F | A | J | L | E | W | Q | D | P | R | B | U | S |
If the first element is a symbol and the last a number, the codes for these two (the first and the last elements) are to be interchanged. If the first element is an odd number and the last an even number, the first and last elements are to be coded as ©. If both second and third elements are perfect squares, the third element is to be coded as the code for the second element. What will be the code for the following group?
7 2 4 @ 5
Answer (Detailed Solution Below)
Conditional Matrix Question 2 Detailed Solution
| Number/symbol | 2 | @ | 9 | 5 | $ | & | 3 | % | # | 7 | + | 4 | 8 | 6 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Code | T | F | A | J | L | E | W | Q | D | P | R | B | U | S |
Conditions:
(i) If the first element is a symbol and the last a number, the codes for these two (the first and the last elements) are to be interchanged.
(ii) If the first element is an odd number and the last an even number, the first and last elements are to be coded as ©
(iii) If both the second and third elements are perfect squares, the third element is to be coded as the code for the second element.
Given- 7 2 4 @ 5
The first element is 7 (an odd number).
The last element is 5 (an odd number).
The second element is 2 (not a perfect square).
The third element is 4 (a perfect square, 2*2=4).
Rule No. (i): This condition does not apply because the first element (7) is a number, not a symbol.
Rule No. (ii): This condition does not apply because the last element (5) is an odd number, not an even number.
Rule No. (iii): This condition does not apply because the second element (2) is not a perfect square, even though the third element (4) is.
Since none of the given conditions apply, the codes for the respective number/symbol are to be followed directly as given in the table.
The code for 7 is P.
The code for 2 is T.
The code for 4 is B.
The code for @ is F.
The code for 5 is J.
Thus, '7 2 4 @ 5' is coded as → P T B F J.
Hence, the correct answer is "Option 3".
Conditional Matrix Question 3:
A group of numbers/symbols is coded using letter codes as per the codes given below and the conditions that follow. If none of the conditions apply, then codes for the respective number/symbol to be followed directly as given in the table.
| Number/Symbol | 9 | 2 | @ | & | 3 | 7 | 5 | % | 1 | 8 | Δ | 4 | 6 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Code | L | C | Z | M | D | W | A | P | R | T | F | B | K |
(i) If the first element is a symbol and the last a number, the codes for these two (the first and the last elements) are to be interchanged. (ii) If the first element is an odd number and the last an even number, the first and last elements are to be coded as ©.
(iii) If both the second and third elements are perfect squares, the third element is to be coded as the code for the second element
What will be the code for the following group?
5 & 3 @ 7 % 2
Answer (Detailed Solution Below)
Conditional Matrix Question 3 Detailed Solution
| Number/Symbol | 9 | 2 | @ | & | 3 | 7 | 5 | % | 1 | 8 | Δ | 4 | 6 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Code | L | C | Z | M | D | W | A | P | R | T | F | B | K |
Conditions:
(i) If the first element is a symbol and the last a number, the codes for these two (the first and the last elements) are to be interchanged.(ii) If the first element is an odd number and the last an even number, the first and last elements are to be coded as ©
(iii) If both the second and third elements are perfect squares, the third element is to be coded as the code for the second element.
Given- 5 & 3 @ 7 % 2
The first element is 5 (an odd number).
The last element is 2 (an even number).
The second element is & (a symbol, not a perfect square).
The third element is 3 (not a perfect square).
Rule No. (i): This condition does not apply because the first element (5) is a number, not a symbol.
Rule No. (ii): This condition applies because the first element (5) is an odd number and the last element (2) is an even number. So, the first and last elements will be coded as ©.
Rule No. (iii): This condition does not apply because neither the second element (&) nor the third element (3) are perfect squares.
Based on the applicable rule (ii), the first and last elements are coded as ©.
The code for & is M.
The code for 3 is D.
The code for @ is Z.
The code for 7 is W.
The code for % is P.
Thus, '5 & 3 @ 7 % 2' is coded as → © M D Z W P ©.
Hence, the correct answer is "Option 4".
Conditional Matrix Question 4:
A group of numbers/symbols is coded using letter codes as per the codes given below and the conditions that follow. If none of the conditions apply, then codes for the respective number/symbol to be followed directly as given in the table.
| Number/Symbol | 8 | 9 | @ | & | 5 | 7 | 2 | % | 1 | 3 | Δ | 4 | 6 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Code | L | C | Z | M | D | W | A | P | R | T | F | B | K |
(i) If the first element is a symbol and the last a number, the codes for these two (the first and the last elements) are to be interchanged. (ii) If the first element is an odd number and the last an even number, the first and last elements are to be coded as ©
(iii) If both the second and third elements are perfect squares, the third element is to be coded as the code for the second element. What will be the code for the following group?
7 2 @ 1 3 &
Answer (Detailed Solution Below)
Conditional Matrix Question 4 Detailed Solution
Given:
| Number/Symbol | 8 | 9 | @ | & | 5 | 7 | 2 | % | 1 | 3 | Δ | 4 | 6 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Code | L | C | Z | M | D | W | A | P | R | T | F | B | K |
Conditions:
(i) If the first element is a symbol and the last a number, the codes for these two (the first and the last elements) are to be interchanged.
(ii) If the first element is an odd number and the last an even number, the first and last elements are to be coded as ©
(iii) If both the second and third elements are perfect squares, the third element is to be coded as the code for the second element.
Given- 7 2 @ 1 3 &
Rule No. (i) This condition does not apply because the first element (7) is a number, not a symbol.
Rule No. (ii) This condition does not apply because while the first element (7) is an odd number, the last element (&) is a symbol, not an even number.
Rule No. (iii) This condition does not apply because neither the second element (2) nor the third element (@) are perfect squares.
So, the code for 7 2 @ 1 3 & is
The code for 7 is W.
The code for 2 is A.
The code for @ is Z.
The code for 1 is R.
The code for 3 is T.
The code for & is M.
Thus, '7 2 @ 1 3 &' is code 'W A Z R T M'.
Hence, the correct answer is "Option 2".
Conditional Matrix Question 5:
Read all information carefully and answer the following question?
Digits/Symbols: * $ % + # ∆ © 3 5 4 7 2 1 6 8 9
Codes: A B C D E F G H I J K L M N O P
(I) If first place is even number and last place is odd number then both places are to be coded by first digit/symbol.
(II) If first place is odd number and last place is even number then both place codes will be interchanged.
5 1 6 9 *
Answer (Detailed Solution Below)
Conditional Matrix Question 5 Detailed Solution
Given:
| Digits/symbols | * | $ | % | + | # | ∆ | © | 3 | 5 | 4 | 7 | 2 | 1 | 6 | 8 | 9 |
| Codes | A | B | C | D | E | F | G | H | I | J | K | L | M | N | O | P |
According to the given condition:
|
Rule no. |
Condition |
Result |
|
I |
First place is even number and last place is odd number |
Both places are to be coded by first digit/symbol |
|
II |
First place is odd number and last place is even number |
Codes will be interchanged |
Given: 5 1 6 9 *
Here, no rule is applicable.
5 → coded as I
1 → coded as M
6 → coded as N
9 → coded as P
* → coded as A
So, 5 1 6 9 * → Coded as IMNPA
Hence, "Option 3" is the correct answer.
Top Conditional Matrix MCQ Objective Questions
In this question, a group of numbers/symbols is coded using letters as per the table given below and the conditions which follow. The correct combination of codes following the conditions is your answer.
|
Number/Symbol |
6 |
8 |
4 |
+ |
5 |
# |
% |
2 |
$ |
1 |
^ |
7 |
9 |
* |
@ |
|
Code |
S |
U |
T |
W |
N |
P |
V |
D |
A |
Z |
J |
M |
O |
M |
Y |
Conditions:
(i) If the first and last elements are numbers, then the last element will be replaced by the symbol @.
(ii) If the first elements is a symbol, then the first element will be coded as ©.
(iii) If both the first and second elements are numbers, then the third element will be coded as *.
What will be the code for the following?
6 % $ 8 5
Answer (Detailed Solution Below)
Conditional Matrix Question 6 Detailed Solution
Download Solution PDFWe follow the given table and code the given combination:
|
Number/Symbol |
6 |
8 |
4 |
+ |
5 |
# |
% |
2 |
$ |
1 |
^ |
7 |
9 |
* |
@ |
|
Code |
S |
U |
T |
W |
N |
P |
V |
D |
A |
Z |
J |
M |
O |
M |
Y |
Given combination: 6 % $ 8 5
Since first and last elements are numbers, first condition will be applied which says 'If the first and last elements are numbers, then the last element will be replaced by the symbol @.' so the code will contain '@' at last.
Now none of the other conditions apply so the code becomes:
Hence, the correct answer is option 3).
In this question, a group of letters is coded using symbols as per the table given below and the conditions which follow. The correct combination of codes following the conditions is your answer.
|
Letters |
D |
T |
V |
O |
I |
A |
P |
E |
|
Codes |
@ |
> |
^ |
% |
= |
# |
$ |
+ |
Conditions:
i. If both the first and the last letters of the word are vowels, then both these letters will be coded as ‘&’.
ii. If both the first and the last letters of the word are consonants, then both these letters will be coded as ‘*'.
What will be the code for the word 'ADOPTIVE'?
Answer (Detailed Solution Below)
Conditional Matrix Question 7 Detailed Solution
Download Solution PDF|
Letters |
D |
T |
V |
O |
I |
A |
P |
E |
|
Codes |
@ |
> |
^ |
% |
= |
# |
$ |
+ |
Given word: 'ADOPTIVE'
Conditions:
i. If both the first and the last letters of the word are vowels, then both these letters will be coded as '&'.
ii. If both the first and last letters of the word are consonants, then both these letters will be coded as '*'.
As condition (i) is followed by the given word because the first and the last letters of the word are vowels. So both these letters will be coded as '&'.
So, ADOPTIVE is coded as '&@%$>=^&'
Hence, the correct answer is '&@%$>=^&'.
A word is represented by only one set of numbers as given in any one of the alternatives. The sets of numbers given in the alternatives are represented by two classes of alphabets as shown in the given two matrices. The columns and rows of Matrix-I are numbered from 0 to 4 and that of matrix-II are numbered form 5 to 9. A letter from these matrices can be represented first by its row and next by its column, for example, ‘Q’ can be represented by 23, 59, etc., and ‘S’ can be represented by 34, 99, etc. Similarly, you have to identify the set for the word “FAINT”
|
Matrix – I |
|
Matrix – II |
||||||||||
|
0 |
1 |
2 |
3 |
4 |
|
5 |
6 |
7 |
8 |
9 |
||
|
0 |
B |
M |
U |
A |
F |
5 |
D |
N |
X |
G |
Q |
|
|
1 |
H |
E |
O |
W |
D |
6 |
L |
V |
E |
O |
Y |
|
|
2 |
P |
J |
G |
Q |
Y |
7 |
S |
T |
C |
M |
W |
|
|
3 |
V |
R |
L |
I |
S |
8 |
H |
R |
B |
K |
U |
|
|
4 |
Z |
X |
T |
N |
K |
9 |
F |
P |
A |
I |
S |
|
Answer (Detailed Solution Below)
Conditional Matrix Question 8 Detailed Solution
Download Solution PDFLet’s check all the options,
1) 95, 03, 34, 43, 76 → FASNT
2) 95, 97, 33, 40, 42 → FAIZT
3) 04, 97, 99, 56, 34 → FASNS
4) 04, 97, 33, 56, 42 → FAINT
Thus the correct option is 04, 97, 33, 56, 42.Each of the questions below begins with a collection of numbers or letters, followed by four different symbol combinations with the letters and numbers A, B, C, and D. Based on the symbol codes and the requirements listed below, you must determine which of the following four combinations accurately depicts the set of digits or letters.
| Letter | Q | D | I | P | S | E | H | R | C | U | M | W | N | A | J | B | O |
| Code | 7 | @ | 5 | 4 | 1 | # | 6 | $ | % | 9 | & | 3 | 2 | ? | 0 | 8 | ! |
Situation 1:
If a word contains more than two vowels, the code will be written backward.
Situation 2:
The first three letters of the code will be written in reverse if the word has no vowels.
Situation 3:
The last two letters of the code will be modified if any two letters in the word are the same.
Note: All conditions will be applied if more than one is required.
What would the word "PMIMN" mean in code?
Answer (Detailed Solution Below)
Conditional Matrix Question 9 Detailed Solution
Download Solution PDFThe word given is "PMIMN".
Checking all the situation one by one.
Situation 1: Incorrect, because the word does not contain more than 2 vowels.
Situation 2: Incorrect, because the word contains 1 vowel.
Situation 3: Correct, because 2 letters in the word are same.
So on following this:
PMI will be coded as '4&5'
& 2 letters are same so last two letters will be modified and coded as "2&"
So the word 'PMIMN' will be coded as '4&52&'.
Hence the correct answer is "4&52&".
The letters (a), (b), (c), and (d) are followed by four sets of digits or symbols in each of the examples below (d). Determine which of the following coding combinations best depicts the set of letters, and then mark that combination's number as the correct response. Mark (e), which means "None of these," as the correct response if none of the four combinations accurately describes the set of letters.
Note: The conditions must be applied in the order listed below if more than one applies.
| Letter | P | D | E | R | A | S | K | T | O | N | I | H | C | L | Y |
| Symbol | * | % | 9 | 7 | @ | ? | $ | 5 | 4 | + | ~ | 8 | # | 3 | ! |
Conditions :
(i) Both letters are coded using the consonant's code if the first letter is a vowel and the last letter is a consonant.
(ii) If the second and last letters are both vowels, their codes must be switched.
(iii) Both letters must be coded as the vowel if the second letter is a consonant and the second to last letter is a vowel.
(iv) If the first letter and the fifth letter are both consonants, the third letter is denoted by both of them.
(v) If a term is given but does not fit any of the criteria listed above, the first letter's code is switched for the second letter, the third letter's code for the fourth letter, and so on.
Where can I find the "ADDICTION" code?
Answer (Detailed Solution Below)
Conditional Matrix Question 10 Detailed Solution
Download Solution PDFThe word given is "ADDICTION".
Checking all the conditions one by one.
Condition 1: Correct, because in the given word first letter is vowel and last letter is consonant. So A & N both will be coded as '+'.
Condition 2: Incorrect, because in the given word second and last letters are not vowels.
Condition 3: Correct, because in the given word the second letter is a consonant and the second to last letter is a vowel.
So D at 2nd position and O will be coded as '4'.
Condition 4: Incorrect, because 1st and 5th both letters are not consonant.
Condition 5: Incorrect, because three conditions among 4 are satisfying.
On following the conditions 'ADDICTION' will be coded as
'+ 4 % ~ # 5 ~ 4 +'.
Hence the correct answer is "+ 4 % ~ # 5 ~ 4 +".
Comprehension:
Comprehension:
Read the following passage and answer the questions that follows.
The codes for some letters are given below.
| Letters | O | T | D | U | A | C | I | Y | R |
| Codes | $ | @ | > | ^ | % | = | # | + | * |
If the first letter of the word is a consonant and the last letter is a vowel, then both these letters would be coded as '&'. If the fifth letter of the word is a vowel and the third letter is a consonant, then both these letters will be coded as 'K'.
What is the code for the word 'COSTARICA'?
Answer (Detailed Solution Below)
Conditional Matrix Question 11 Detailed Solution
Download Solution PDFThe logic follows here is:
In the word COSTARICA,
The first letter is a consonant (C) and the last letter is a vowel (A), therefore both these letters are coded as '&'
The fifth letter is a vowel and the third letter is a consonant, therefore both these letters are coded as 'K'
Now, Check each option one by one
1) # $ K @ K * + # =
Here, the First and Last letter is not '&', therefore Option 1) is incorrect.
2) & $ K @ K * # = &
Here, the first and last letter is '&' and the fifth and third letter is 'K'.
also, there are separate symbols used for the remaining letters, therefore Option 2) is correct.
3) & $ K # K = + # &
Here, the first and last letter is '&' and the fifth and third letter is 'K'.
In this option, the code for 'T' and 'C' is the same i.e. '#', therefore, Option 3) is incorrect.
4) & $ > @ = * + # &
Here, the first and last letter is '&' but the fifth and third letter is not 'K', therefore Option 4) is incorrect.
Hence, the correct answer is Option 2).
Comprehension:
Directions: Study the following information carefully and answer the questions that follow:
There is a matrix of 4*4. The rows are represented as %, $, # and @ from the first to the fourth in the same order and the columns are represented as 1 to 4 from the first to the fourth in the same order.
Following operations are to be performed in the given order to obtain the whole matrix.
1. One diagonal of matrix is IBPS and the other diagonal is 2019 in the same order. Both have their first element in the first Column and 4th column respectively.
2. The immediate neighbor of any alphabet in the same row is the alphabetical position of that number.
3. The remaining places are to be filled by taking the difference between the two numbers surrounding that place in the same row.
In the obtained matrix, the value of the sum of all digits in the second column would be less than that of the third column.
The compiler of a PC will respond to the input by calculating the value of the string as per the following conditions.
1. If there are equal number of even and odd numbers in the string, then value of the string will be the absolute difference between all the even numbers and all the odd numbers.
2. If there are equal number of alphabets and numbers in the string, then the value of string will be the absolute difference between the alphabetical position of all the alphabets and all the numbers.
3. If there is only one even number in the string, then the value of string will be the addition of all the numbers in that string.
4. If there is a vowel in the string, then value of the string will be the alphabetical position of that vowel.
5. If none of the above mentioned conditions follow, then the value of the string will be obtained by subtracting the lowest number from the highest number in that string.
Note:
1. The value of an input is obtained by adding the values of all the strings and then the digits in that value is added up to the point where it becomes a single digit number and then it is taken as an input.
2. If only a single string is given, then the digits in that value is added upto the point where it becomes a single digit number and then it is taken as input.
3. If the value of an input is a prime number, then the program will be executed.
If the value of an input is an even number, then program will not be executed.
If the value of an input is both even and prime number, then program will be executed but it will show an error.
If the value of an input is other than the above mentioned values, then PC will hang.
Let us assume a matrix:
|
|
1 |
2 |
3 |
4 |
|
% |
1 |
2 |
C |
4 |
|
$ |
J |
6 |
7 |
8 |
|
# |
9 |
10 |
T |
12 |
|
@ |
O |
14 |
15 |
16 |
AA = $3 @2 %1 #2, BB = @1 $3 %2
$3 denotes value in second row and column 3 which will be equal to 7.
Therefore AA = 7 14 1 10
So, this is satisfying condition 1. Hence the sum = (14 + 10) – (7 + 1) = 16
BB = O 7 2
So, this is satisfying condition 4. Thus, frequency = alphabetical position of O = 15
The values of AA and BB are added, and the final frequency is 16 + 15 = 31 which is a two digit number, So again adding them 3 + 1 = 4, which is a non-prime even number so the program will not be executed.
Consider the following four strings for all questions:
AA = $2 #1 @3 #3
BB = @1 %4 @3 %3
CC = $4 @2 #4 %1
DD = %4 @2 @4 #2
What is the value of input if only signal AA and BB is taken into account?
Answer (Detailed Solution Below)
Conditional Matrix Question 12 Detailed Solution
Download Solution PDF|
|
1 |
2 |
3 |
4 |
|
% |
I |
9 |
7 |
2 |
|
$ |
2 |
B |
0 |
2 |
|
# |
15 |
1 |
P |
16 |
|
@ |
9 |
10 |
19 |
S |
AA = $2 #1 @3 #3 = B 15 19 P
Since, the string has equal number of alphabet and numbers, hence it satisfies condition 2.
Final value = 34 – 18 = 16
BB = @1 %4 @3 %3 = 9 2 19 7
Since the string has one even number, hence it satisfies condition 3.
Final value = 9 + 2 + 19 + 7 = 37
The sum of AA and BB = 16 + 37 = 53 = 8 (5 + 3)
Hence, 8 is the correct answer.
In this question, a group of numbers/symbols is coded using letters as per the table given below and the conditions which follow. The correct combination of codes following the conditions is your answer.
| Number / Symbol | 6 | 8 | 4 | + | 5 | # | % | 2 | $ | 1 | ^ | 7 | 9 | * | @ |
| Code | S | U | T | W | N | P | V | D | A | Z | J | M | O | M | Y |
Conditions:
(i) If the first and last elements are numbers, then the last element will be replaced by the symbol @.
(ii) If the first element is a symbol, then the first element will be coded as ©.
(iii) If both the first and second elements are numbers, then the third element will be coded as *.
What will be the code for the following?
7 8 4 % @
Answer (Detailed Solution Below)
Conditional Matrix Question 13 Detailed Solution
Download Solution PDFAccording to the given information, the numbers and symbols are coded as:
| Number / Symbol | 6 | 8 | 4 | + | 5 | # | % | 2 | $ | 1 | ^ | 7 | 9 | * | @ |
| Code | S | U | T | W | N | P | V | D | A | Z | J | M | O | M | Y |
Now, let's do the coding for " 7 8 4 % @ ",
As per the given information, " 7 8 4 % @ " can be coded as " M U T V Y ",
Since, it meets the condition (iii) as its first and second elements are numbers, then the third element will be coded as *.
Thus, the final coding of " 7 8 4 % @ " will be " M U * V Y ".
Hence, the correct answer is " MU*VY ".
In this question, a group of digits/symbols is coded using the letters as per the table given below and the conditions which follow. The correct combination of codes following the conditions is your answer.
|
Digits / Symbols |
9 |
$ |
6 |
% |
1 |
& |
3 |
8 |
4 |
|
Code |
Y |
J |
L |
M |
T |
U |
E |
W |
Q |
Condition 1: If the first and the last elements are numbers, then both the numbers are coded as the code for the last number.
Condition 2: If the third element is a symbol, then the code for the symbol is exchanged with the code for the fifth element.
Condition 3: If the fourth element is an odd number, then it is to be coded as #.
What is the code for 94$3&3 ?
Answer (Detailed Solution Below)
Conditional Matrix Question 14 Detailed Solution
Download Solution PDF| Digits / Symbols | 9 | $ | 6 | % | 1 | & | 3 | 8 | 4 |
| Code | Y | J | L | M | T | U | E | W | Q |
| Rule No. | Conditions | Results |
| 1. | If the first and the last elements are numbers | then both the numbers are coded as the code for the last number. |
| 2. | If the third element is a symbol | then the code for the symbol is exchanged with the code for the fifth element. |
| 3. | If the fourth element is an odd number | then it is to be coded as #. |
Given Code: 94$3&3
Here Rule No. 1, If the first and the last elements are numbers, then both the numbers are coded as the code for the last number
Rule No. 2, If the third element is a symbol, then the code for the symbol is exchanged with the code for the fifth element.
Rule No. 3, If the fourth element is an odd number, then it is to be coded as #.
Now, in the given code '94$3&3' All rules apply.
So, The Given Code '94$3&3' will be decoded as → E Q U # J E.
Hence, the correct answer is 'Option 3'.
Read the following table and find the relevant code for '472938' as per the conditions given:
| Digit | 2 | 5 | 7 | 8 | 9 | 4 | 6 | 3 | 1 |
| Letter Code | M | R | T | B | W | K | D | N | J |
Conditions:
1. If both the first and last digits of the number are odd digits, then both are to be coded as 'S'.
2. If both the first and last digits of the number are even digits, then both are to be coded as 'Z'.
3. If one among the first and last digits is odd and the other one is even, then the first digit has to be coded as 'Q' and the last digit as 'W'.
Answer (Detailed Solution Below)
Conditional Matrix Question 15 Detailed Solution
Download Solution PDFGiven:
| Digit | 2 | 5 | 7 | 8 | 9 | 4 | 6 | 3 | 1 |
| Letter Code | M | R | T | B | W | K | D | N | J |
| Conditions | Coded As |
| Both the First and last digits of the number are odd digits | both are to be coded as 'S' |
| Both the First and last digits of the number are even digits | both are to be coded as 'Z' |
| One among the first and last digits is odd and the other one is even | first digit has to be coded as 'Q' and the last digit as 'W' |
Now, the group of numbers and symbols is:472938
Here, Second conditions followed in the given group,
So, codes of the First and last digit both are to be coded as 'Z'.
Then, the code for '472938' is → 'ZTMWNZ'
Hence, the correct answer is "Option 1".