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.

Conditions:
(1) If the second element is an odd number and the last element is an even number, the codes for these two (the second and the last elements) are to be interchanged.
(ii) If the third element is an even number and the fifth element is a symbol, the codes for these two (the third and the fifth elements) are to be interchanged.
(iii) If the fifth element is a symbol and the sixth element is also a symbol, both these elements (the fifth and the sixth elements) are to be coded as R.
Question: E 75% & 4%8

Correct option is D

Given: 

Applying Conditions

Condition (i): If the second element is an odd number and the last element is an even number, the codes for these two (the second and the last elements) are to be interchanged.

Here, the second element is 7 (an odd number) and the last element is 8 (an even number).

So we need to interchange the codes for 7 and 8.

Codes for 7 and 8 are S and A, respectively.

After interchange, the sequence becomes: M A D U T L U S

Condition (ii): If the third element is an even number and the fifth element is a symbol, the codes for these two (the third and the fifth elements) are to be interchanged.

The third element is 5 (an odd number), so this condition does not apply.

Condition (iii): If the fifth element is a symbol and the sixth element is also a symbol, both these elements (the fifth and the sixth elements) are to be coded as R.

The fifth element is & and the sixth element is 4 (a number), so this condition does not apply.

Final Answer:After applying the relevant conditions, the final coded sequence is: M A D U T L U S

Correct answer: M A D U T L U S