# Coding Decoding Practice Quiz 3

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# Coding Decoding Practice Quiz 3

### Introduction

A CODE is a ‘system of signals’ . Therefore, coding is a method of transmitting a message between the sender and the receiver without a third person knowing it. Coding and Decoding test is set up to judge the candidate’s ability to decipher the rule that codes a particular word/ message and break the code to decipher the message. The article Coding Decoding Practice Quiz 3 provides information about Coding Decoding, a important topic of Reasoning Consists of different types Coding Decoding questions with solutions useful for candidates preparing for different competitive examinations like RRB, RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC CGL,SSC CHSL, IBPS, SBI PO, SBI Clerks, CAT and etc.

### Quiz

Directions (1 – 5): Study the following information carefully and answer the questions given below.
There is 3*5 matrix which can produce signals which in turn help in the illumination of some bulbs. The row of the matrix are denoted by \$, & and * from bottom to top and the columns are denoted by the alphabets P, Q, R, S and
T from left to right. \$ row contains number which are consecutive multiple of 7, staring from 28 (from left to right). & row contains numbers which are consecutive multiple of 11, starting from 11 (from left to right). * row contains number which are consecutive multiple of 13, starting from 13 (from left to right). The matrix helps in producing signals which can be either a single string of number X- or two-line string X and Y. There are 4 lights A, B, C and D. Based on the outcome of the strings mentioned above one of the light blinks.
1. If the outcome is below 85, then A will blink 2. If outcome range is 85-110, then B blinks 3. If outcome range is 111-210, then C blinks 4. If outcome is greater than 210, then D blinks
For outcome of the string:
1. If the string has all even numbers, then outcome of the string is obtained by adding all the numbers. 2. If an odd number is followed by an even number then the one’s places of all the two-digit numbers are deleted and, tenth place are multiplied to get the outcome 3. If the string contains 2 prime number, then the tenth’s place is deleted from each of the two-digit number and remaining number are multiplied. 4. If no above logic is followed, then simple outcome is addition of the numbers.
1. If X = *R &S *P *T, then which bulb blink?
A. D B. B C. A D. C E. Either C or D

2. If X = \$P \$R *S &Q, then which bulb blink?
A. D B. B C. A D. C E. Either A or B

3. If X = *P &T &P \$Q, then which bulb blink?
A. D B. B C. A D. C E. Either A or B

4. If X = \$P *Q \$S &R, then which bulb blink?
A. D B. B C. A D. C E. Either C or D

5. If X = \$T *S \$S *T, then which bulb blink?
A. D B. B C. A D. C E. Either C or D

Explanation -
For matrix formation:
There is 3 × 5 matrix.
The row of the matrix are denoted by \$, & and * from bottom to top and the columns are denoted by the alphabets P, Q, R, S and T from left to right.

\$ row contains number which are consecutive multiple of 7, starting from 28 (from left to right).
& row contains number which are consecutive multiple of 11, starting from 11 (from left to right).
* row contains number which are consecutive multiple of 13, starting from 13 (from left to right).

X = *R &S *P *T
X = 39 44 13 65
Condition 2 is applicable here: If an odd number is followed by an even number then the one’s places of all the two-digit numbers are deleted and, tenth place are multiplied to get the outcome
X = 3 × 4 × 1 × 6 = 72
As the outcome is below 85, hence bulb A will blink.
Explanation -
X = \$P \$R *S &Q
X = 28 42 52 22
Condition 1 is applicable here: If the string has all even numbers, then outcome of the string is obtained by adding all the numbers.
X = 28 + 42 + 52 + 22 = 144
As the outcome of the string is between 111 – 210, hence bulb C will blink.
Explanation -
X = *P &T &P \$Q
X = 13 55 11 35
Condition 3 is applicable here: If the string contains 2 prime number, then the tenth’s place is deleted from each of the two-digit number and remaining number are multiplied.
X = 3 × 5 × 1 × 5 = 75
As the outcome is below 85, hence bulb A will blink.
Explanation -
X = \$P *Q \$S &R
X = 28 26 49 33
Condition 4 is applicable here: If no above logic is followed, then simple outcome is addition of the numbers.
X = 28 + 26 + 49 + 33 = 136
As the outcome of the string is between 111 – 210, hence bulb C will blink.
Explanation -
X = \$T *S \$S *T
X = 56 52 49 65
Condition 4 is applicable here: If no above logic is followed, then simple outcome is addition of the numbers.
X = 56 + 52 + 49 + 65 = 222
As the outcome of the string is greater than 210, hence bulb D will blink.
Directions (1 – 5): Study the following information carefully and answer the questions given below.
In a certain code language, “Regular fitness sound health” is coded as “J&10 X&15 X#4 P&1” “Prove gear what times” is coded as “P@7 D#4 Z&4 J&4” “Health problems increasing daily” is coded as “J&5 R&11 X&15 R*0” “Wireless telephone physical group” is coded as “D&3 J#10 D*9 F&7”
1. What is the code for word ‘solution’?
A. R@2 B. R#1 C. R*0 D. J!17 E. None of these

2. Which of the following is the code for ‘phone call’?
A. R&5 B@12 B. R@6 B!12 C. D*6 B!11 D. D&6 B@11 E. None of these

3. ‘B#1’ is the code for which of the following?
A. Spherical B. Physical C. Education D. Important E. Cannot be determined

4. What is the code for ‘delete data’?
A. K#12 B#14 B. K@15 B*14 C. J#15 B@5 D. J#15 B&19 E. None of these

5. What is the code for ‘parenting’?
A. R!12 B. R#13 C. E%12 D. R*13 E. J@12

Explanation [1-5]-
Logic:
Number – Subtract the position of 2nd alphabet from right end with 2nd alphabet from left end.
Symbol – Symbol is coded according to number of vowels in the word:
# – 3 @ – 1 * – 4 ! – 5 & – 2
Alphabet – Position of 3rd alphabet from right end. Add same number to it to get the alphabet
For example – Wireless
Number – Position of ‘I’ in English alphabet = 9
Position of ‘S’ in English alphabet = 19
Subtract 19 – 9 = 10
Symbol – Number of vowels in ‘wireless’ = 3
Symbol – #
Alphabet – 3rd alphabet from right end = O
Position of ‘E’ = 5
5 + 5 = 10
Alphabet – J
Hence, ‘wireless’ is coded as ‘J#10’.

Hence, the code for word ‘solution’ is ‘R*0’.
Explanation -
Hence, code for ‘phone call’ is D&6 B@11.
Explanation -
Hence, ‘B#1’ is the code for ‘important’.
Explanation -
Hence, code for ‘delete data’ is ‘J#15 B&19’.
Explanation -
Hence, code for ‘parenting’ is ‘R#13’.
Directions (1 – 5): Study the following information carefully and answer the questions given below.

Condition:
i. If the first number is odd number and the last number is even number the codes are to be interchanged. ii. If the first number is even number and the last number is a odd number,both are to be coded as the code for the last letter. iii. If both the first and the third numbers are odd numbers, both are to be coded as ‘+’. iv. If even number is immediately proceeded and followed by odd number, then the even number is coded as %
1. 45239
A. XF%@X B. XF%@# C. #F%@# D. XFG@# E. None of these

2. 72163
A. +%+%@ B. !%+Z@ C. !G+%@ D. +G+Z@ E. None of these

3. 61283
A. BJG*@ B. +JG*+ C. &JG*& D. @JG*@ E. None of these

4. 85642
A. *FBX* B. *F+XG C. *F+X% D. *FBXG E. None of these

5. 962738
A. #BG!@# B. *BG!@# C. *BG!@* D. &B+!@& E. None of these

Explanation -
45239 [latex]\Rightarrow[/latex] XFG@# [latex]\Rightarrow[/latex] #F%@#
Explanation -
72163 [latex]\Rightarrow[/latex] !GJB@ [latex]\Rightarrow[/latex] +%+%@
Explanation -
61283 [latex]\Rightarrow[/latex] BJG*@ [latex]\Rightarrow[/latex] @JG*@
Explanation -
85642 [latex]\Rightarrow[/latex] *FBXG
Explanation -
962738 [latex]\Rightarrow[/latex] #BG!@* [latex]\Rightarrow[/latex] *BG!@#

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