If 5 @ 2 = 17 and 6 @ 3 = 27, what is 7 @ 4?
- (a)35
- (b)37
- (c)39
- (d)41
Answer
Why
Correct — C.
Rule: a @ b = a × b + a + b (multiply, then add both numbers).
5 @ 2 = 10 + 5 + 2 = 17 ✓
6 @ 3 = 18 + 6 + 3 = 27 ✓
7 @ 4 = 28 + 7 + 4
= 39 → option (c)
Why the others are wrong
- (a)35 — 35 is 7 × 4 + 7, adding only the first number. That rule gives 5 × 2 + 5 = 15 for the first example, not the printed 17.
- (b)37 — 37 continues the results 17, 27 by another +10. The rule works inside each pair, not between answers: 7 × 4 + 7 + 4 = 39.
- (d)41 — 41 would need 13 added to 7 × 4 = 28. The examples add the sum of the pair, 7 for 5 @ 2 and 9 for 6 @ 3, so 7 @ 4 adds 11.
Concept
An invented operator hides a rule that uses both numbers. Try the simple combinations first: the product, the sum, the squares.
Here the product falls short of each result: 10 is 7 short of 17, and 18 is 9 short of 27. The shortfalls, 7 and 9, are the sums 5 + 2 and 6 + 3.
The same rule can be written (a + 1)(b + 1) − 1. For 7 @ 4 that is 8 × 5 − 1 = 39.
Test every candidate rule on both examples before using it. 3a + b and a² − 8 both fit 5 @ 2 = 17, and both fail 6 @ 3 = 27.
Key facts
- a × b + a + b equals (a + 1)(b + 1) − 1.
- 5 × 2 + 7 = 17 and 6 × 3 + 9 = 27.
- 7 × 4 + 11 = 39.
Study next
Common traps
- Treating 17, 27 as a series and answering 37.
- Adding only the first number to the product, which gives 35.
The same product-plus-sum rule decides 16 Sep 2025, 12:30, Reasoning Q.22, where 3 @ 4 = 19 and 5 @ 6 = 41 lead to 6 @ 7 = 55.
Reasoning Q.18 in this shift uses the same @ sign for a different rule, a² + b, so derive each item's rule from its own examples.
Related PYQs
No directly related past PYQ was found.