Using the property that the exterior angle of a triangle is equal to the sum of the interior opposite angles, we get angle $\(\mathrm{ADB}=\mathrm{y}+\mathrm{z}\)$ (Angle ADB is an exterior angle of triangle ADC )

We are given that $\(\mathrm{AB}<\mathrm{BD}\)$ which implies angle opposite to AB is less than that opposite to BD i.e. angle $\(\mathrm{BDA}<\)$ angle BAD which is same as $\(\mathrm{y}+\mathrm{z}<\mathrm{x}\)$

Now, $\(x>y+z \Rightarrow \frac{x}{2}>\frac{y+z}{2}$, so $\frac{x}{2}+\frac{y}{2}=\frac{x+y}{2}>\frac{y+z}{2}\)$

Hence column $A$ has higher quantity when compared with column $B$, so the answer is $(A)$.