2016 AMC 10B Problem 2

The rule says: cube the first number, square the second number, and multiply.

1. For $2 \heartsuit 4$, the first number is $2$ and the second is $4$:
$$2 \heartsuit 4 = 2^3 \cdot 4^2 = 8 \cdot 16 = 128.$$
2. For $4 \heartsuit 2$, the first number is $4$ and the second is $2$:
$$4 \heartsuit 2 = 4^3 \cdot 2^2 = 64 \cdot 4 = 256.$$
3. Divide:
$$\frac{128}{256} = \frac12.$$

The answer is $\boxed{\textbf{(B)}\ \frac{1}{2}}$.

A made-up symbol like $\heartsuit$ is just a recipe. The only skill is putting each number into the right place in the recipe. Writing out both values in full before dividing keeps you from swapping the roles by accident.

Use letters to see a shortcut. The fraction is
$$\frac{2^3 \cdot 4^2}{4^3 \cdot 2^2}.$$
Cancel $2^2$ from $2^3$ (leaving one $2$) and cancel $4^2$ from $4^3$ (leaving one $4$ on the bottom). What is left is $\frac{2}{4} = \frac12$.

The first input is cubed and the second is squared.

$$2\heartsuit 4 = 2^3 \cdot 4^2 = 8 \cdot 16 = 128, \qquad 4\heartsuit 2 = 4^3 \cdot 2^2 = 64 \cdot 4 = 256.$$

Therefore
$$\frac{2\heartsuit 4}{4\heartsuit 2} = \frac{128}{256} = \frac12.$$

The answer is $\boxed{\textbf{(B)}\ \frac{1}{2}}$.