$\sqrt{12}=(\sqrt{4})(\sqrt{3})=2\sqrt{3}$
$\sqrt{72}=(\sqrt{36})(\sqrt{2})=6\sqrt{2}$

We get:
QUANTITY A: $\frac{2r \sqrt{12}}{r^2s \sqrt{72}}=\frac{(2r)(2 \sqrt{3})}{(r)(rs)(6\sqrt{2})}=\frac{(4r)(\sqrt{3})}{(r)(\sqrt{3})(6\sqrt{2})}=\frac{4}{6\sqrt{2}}=\frac{2}{3\sqrt{2}}$

QUANTITY B: $\frac{14rs^2}{42s}=\frac{14(rs)(s)}{42s}=\frac{rs}{3}=\frac{\sqrt{3}}{3}$

At this point, we can use matching operations

Multiply both quantities by 3 to get:
QUANTITY A: $\frac{2}{\sqrt{2}}$

QUANTITY B: $\sqrt{3}$



Multiply both quantities by $\sqrt{2}$ to get:
QUANTITY A: $2$

QUANTITY B: $\sqrt{6}$

Since $\sqrt{6}>2$, the correct answer is B


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