Answer: D

Video solution:

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Written solution:

Let $a$, $b$, and $c$ be the numbers of $\$1$, $\$3$, and $\$4$ pairs, respectively. Then $$a+b+c=12,$$ $$a+3b+4c=24.$$

Subtracting gives $2b+3c=12$. Since at least one pair of each type was bought, $b>0$ and $c>0$. Also $3c<12$, so $c<4$. Modulo $2$, the equation gives $c$ even, so $c=2$.

Then $2b+6=12$, so $b=3$, and $a=12-3-2=7$.

Thus, D is the correct answer.