Find the odd one out:
- (a)6 @ 2 = 16
- (b)7 @ 1 = 14
- (c)7 @ 3 = 20
- (d)4 @ 4 = 16
Answer
Why
Correct — B.
Rule: a @ b = 2 × (a + b).
6 @ 2: 2 × (6 + 2) = 16, as given
7 @ 3: 2 × (7 + 3) = 20, as given
4 @ 4: 2 × (4 + 4) = 16, as given
7 @ 1: 2 × (7 + 1) = 16, but it says 14
7 @ 1 = 14 breaks the rule → option (b).
Why the others are wrong
- (a)6 @ 2 = 16 — 6 @ 2 = 16 fits the rule: 6 + 2 = 8, and 8 × 2 = 16. It belongs to the group of three.
- (c)7 @ 3 = 20 — 7 @ 3 = 20 fits: 7 + 3 = 10, doubled to 20. Its sum differs from the others, but the rule still holds.
- (d)4 @ 4 = 16 — 4 @ 4 = 16 fits: 4 + 4 = 8, doubled to 16. It also equals 4 × 4, but multiplying fails 6 @ 2, so the sum rule stands.
Concept
An odd-one-out on a made-up operator asks for one rule that three of the four equations obey. Start with the options that share an answer: 6 @ 2 and 4 @ 4 both give 16, and both pairs sum to 8.
Doubling the sum fits them, and it fits 7 @ 3 = 20. It fails 7 @ 1, whose pair also sums to 8 but claims 14.
7 @ 1 = 14 would fit a rule of doubling the first number alone, but that rule fails the other three: 6 @ 2 would be 12. A rule that fits one equation and breaks three is not the group's rule.
Key facts
- Pairs with the same sum give the same result under a sum rule: 6 + 2, 7 + 1 and 4 + 4 all make 8.
- Doubling the sum gives 16, 16, 20 and 16 for the four left sides in order.
- The keyed equation claims 14 where the rule gives 16.
Study next
Common traps
- Testing multiplication on 4 @ 4 = 16 and stopping there
- Doubling only the first number, which fits 7 @ 1 but no other equation
14 Sep 2025, 12:30, Reasoning Q.15 also adds the two numbers first, then multiplies by the second: 9 @ 3 = 12 × 3 = 36, and 8 @ 2 is keyed 20.
15 Sep 2025, 12:30, Reasoning Q.21 builds its rule as a × (a + b): 5 * 2 = 35 and 4 * 3 = 28, so 6 * 4 is keyed 60.
Related PYQs
No directly related past PYQ was found.