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

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.

What will be the code for the following group?
7 2 4 @ 5

This question was previously asked in
RRB NTPC Graduate Level CBT-I Official Paper (Held On: 05 Jun, 2025 Shift 1)
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  1.  PBBFJ
  2. JBBFP
  3. PTBFJ
  4. JTBFP

Answer (Detailed Solution Below)

Option 3 : PTBFJ
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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".

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