If 4 @ 2 = 12 and 6 @ 3 = 27, then what is the value of 8 @ 4?

The correct option is 48.

To solve this puzzle, let us analyze the pattern given in the operations:

First given relation: 4 @ 2 = 12

Second given relation: 6 @ 3 = 27

Let us look for a mathematical pattern for an operation $a@b$:

Notice that for 4 @ 2:

$\left(4^2\right)-\left(2^2\right)=16-4=12$

Alternatively, $(4+2)\times 2=6\times 2=12$, or $4\times (2+1)=12$.

Now let us test these patterns on the second relation, 6 @ 3 = 27:

Testing difference of squares: $\left(6^2\right)-\left(3^2\right)=36-9=27$

This pattern holds true for both given cases! The rule for the operator @ is:

$a@b=a^2-b^2$

Another equivalent rule that works for both is $a@b=(a+b)\times (a-b)$.

Now, applying this pattern to find the value of 8 @ 4:

$8@4=8^2-4^2$

$8@4=64-16=48$

Hence, the value of 8 @ 4 is 48.