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 the conditions is your answer. If none of the conditions follow, then codes for the respective Number/Symbol to be followed directly as given in the table.

NUMBER/SYMBOL

7

$

8

6

&

#

3

+

5

4

2

9

@

*

CODE

B

C

R

T

H

U

L

P

A

Q

V

N

J

M

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 second and third elements are perfect squares, the third element is to be coded as the code for the second element.

7 & * 8 4

The correct option is © H M R ©.

Step 1: Identify the given group of elements and their initial codes.
The sequence to be coded is: 7 & * 8 4.
Let's find the original letter codes for each element using the reference table:
• 7 → B
• & → H
• * → M
• 8 → R
• 4 → Q

Step 2: Check the given conditions in order.

Condition (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."
• First element is 7 (an odd number), and last element is 4 (an even number).
• Since the first element is not a symbol, Condition (i) does not apply.

Condition (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 ©."
• First element 7 is an odd number.
• Last element 4 is an even number.
• Therefore, Condition (ii) applies! The first and last elements must both be coded as ©.

Condition (iii): "If both second and third elements are perfect squares, the third element is to be coded as the code for the second element."
• The second element is & (a symbol) and the third element is * (a symbol). Neither is a perfect square.
• Therefore, Condition (iii) does not apply.

Step 3: Apply Condition (ii) to form the final code combination.
• First element (7): Replaced by ©
• Second element (&): Code remains H
• Third element (*): Code remains M
• Fourth element (8): Code remains R
• Fifth element (4): Replaced by ©

Putting all the codes together gives the final sequence: © H M R ©.