The same operation(s) are followed in all the given number pairs except one. Find that odd number pair. (NOTE : Operations should be performed on the whole numbers, without breaking down the numbers into its constituent digits. E.g. 13 – Operations on 13 such as adding /deleting /multiplying etc., to 13 can be performed. Breaking down 13 into 1 and 3 and then performing mathematical operations on 1 and 3 is not allowed.)bxs

Correct — D. Test one relation across all four pairs rather than trusting whichever pair you read first.

Rule: second = first × 3 + 3 — it fits three of the four.

15 × 3 + 3 = 48 ✓
12 × 3 + 3 = 39 ✓
13 × 3 + 3 = 42 ✓
11 × 3 + 3 = 36, but the pair on offer is 11 : 35 ✗

11 : 35 runs on × 3 + 2, so it is the pair that does not belong.

Why the others are wrong

• (a)
  15 : 48 — 15 × 3 + 3 = 48 exactly. It obeys the rule the majority share, so it is inside the group and cannot be the odd pair.
• (b)
  12 : 39 — 12 × 3 + 3 = 39 exactly. Same rule as the majority, so it belongs.
• (c)
  13 : 42 — 13 × 3 + 3 = 42 exactly. It follows the group rule, so it is not the outlier.

In an odd-one-out you want the rule that fits the most pairs, not the rule suggested by the first pair you happened to read.

So compute one candidate relation on every pair before deciding anything. When each second number is about three times its partner, write out the remainder after multiplying by 3.

48 − 45 = 3, 39 − 36 = 3, 42 − 39 = 3, 35 − 33 = 2. The remainder column makes the outlier obvious and takes seconds.

Key facts

• 15 : 48, 12 : 39 and 13 : 42 all follow second = first × 3 + 3.
• 11 × 3 + 3 = 36, not 35, so 11 : 35 breaks the rule.
• 11 : 35 follows first × 3 + 2 instead.

Common traps

• Deriving the rule from 11 : 35 first and then declaring one of the other three odd.
• Testing plain × 3 with no offset and concluding that every pair is broken.
• A slip on 13 × 3 + 3, which is 42 exactly and not 41.

No directly related past PYQ was found.