If 9 @ 1 = 82 and 6 @ 2 = 38, what is the value of 7 @ 3?

Correct — C.
Rule: a @ b = a² + b (square the first number, add the second).

9 @ 1 = 81 + 1 = 82 ✓
6 @ 2 = 36 + 2 = 38 ✓

7 @ 3 = 7² + 3
= 49 + 3 = 52 → option (c)

Why the others are wrong

• (a)
  50 — 50 is 7² + 1, adding the 1 from the first example. The amount added is the second number of each pair, 1 and then 2, so 7 @ 3 adds 3.
• (b)
  60 — 60 would need 11 added to 7² = 49. The examples add only the second number, 81 + 1 and 36 + 2, so 7 @ 3 is 49 + 3.
• (d)
  66 — 66 would need 17 added to 7² = 49. Nothing in the examples adds more than the second number: 81 + 1 = 82 and 36 + 2 = 38.

When results sit just above perfect squares, start there. 82 is 1 more than 81 = 9², and 38 is 2 more than 36 = 6². The amount over the square is the second number each time, which gives a² + b.

The first example alone cannot separate a² + b from a² + b², because 1² = 1. The second settles it: 6² + 2² = 40, not the printed 38.

Key facts

• 9² + 1 = 82 and 6² + 2 = 38.
• 7² + 3 = 52.
• When b = 1, a² + b and a² + b² give the same value, so test the rule on the other example.

Common traps

• Squaring both numbers because 9 @ 1 fits a² + b². That rule gives 40 for 6 @ 2, not 38.
• Adding the first example's 1 to every square, which gives 50.

No directly related past PYQ was found.