Find the odd one out:
- (a)A@1A,
- (b)B@2B,
- (c)C@4C,
- (d)D@4D
Answer
Why
Correct — C.
Rule: the number is the letter's position in the alphabet, with the letter on both sides of @.
A@1A: A = 1 ✓
B@2B: B = 2 ✓
D@4D: D = 4 ✓
C@4C: C = 3, not 4 → option (c).
Why the others are wrong
- (a)A@1A, — A@1A fits the rule: A is the 1st letter of the alphabet and carries 1.
- (b)B@2B, — B@2B fits the rule: B is the 2nd letter and carries 2.
- (d)D@4D — D@4D fits the rule: D is the 4th letter, so its 4 is right. The same 4 is wrong beside C, the 3rd letter.
Concept
An odd-one-out item gives four entries built the same way and asks which one breaks the shared rule. Look for the link between the parts of each entry.
Here each entry is a letter, @, a number and the same letter again. The link is alphabet position: A = 1, B = 2, D = 4. C is the 3rd letter, so C@4C carries the wrong number.
Options (a) to (c) are printed with a trailing comma and (d) without one. The comma sits outside the letter–number pattern, so it cannot decide the odd one out.
Key facts
- A = 1, B = 2, C = 3, D = 4 in the English alphabet.
- C@4C would fit the rule as C@3C.
Study next
Common traps
- Picking (d) because it shares the 4 with (c), when 4 is right for D
- Treating the missing comma after D@4D as the difference
Here SSC prints four letter–symbol–number codes, and the odd one breaks the link between a letter and its alphabet position.
Related PYQs
No directly related past PYQ was found.