A lift has two maximum capacity constraints:
1. Constraint 1: It can carry 20 bundles each weighing $x$ lbs.
2. Constraint 2: It can carry 30 bundles each weighing $y$ lbs.

A person has already placed $\(\mathbf{1 2}\)$ bundles each weighing $\(y \mathbf{~ l b s}\)$ in the lift. We need to determine the maximum number of additional bundles weighing $x$ Ibs that can be added without exceeding the lift's capacity.

Step 1: Determine the Lift's Total Weight Capacity
The lift's maximum capacity can be expressed in two ways based on the given constraints:
1. From Constraint 1:

Total capacity $=20 x$ lbs
2. From Constraint 2:

Total capacity $=30 y$ lbs
Since both expressions represent the same lift's capacity, we equate them:

$$
\(20 x=30 y \Longrightarrow x=\frac{30 y}{20}=1.5 y\)
$$


So, the weight of one $x$-bundle is $\(\mathbf{1 . 5}\)$ times the weight of one $y$-bundle.
Step 2: Calculate the Weight Already in the Lift
The person has already placed 12 bundles of $y$-weight:
Weight already in lift $=12 y$ lbs
Step 3: Determine the Remaining Capacity
Using the total capacity from Constraint 2 (since it's in terms of $y$ ):
Remaining capacity $\(=30 y-12 y=18 y\)$ lbs

Step 4: Convert Remaining Capacity to $x$-Bundles
Since $x=1.5 y$, the number of additional $x$-bundles that can be added is:

$$
\(\text { Number of } x \text {-bundles }=\frac{\text { Remaining capacity }}{x}=\frac{18 y}{1.5 y}=12\)
$$


However, we must also ensure that this does not violate Constraint 1 (maximum $\(20 x\)$-bundles).
Since we're adding bundles to an initially empty lift (with respect to $x$-bundles), the total $x$-bundles after adding would be 12, which is within the limit of 20 .

But wait! Let's cross-validate using Constraint 1:
Total weight from $\(x\)$-bundles $\(\leq 20 x\)$

$$
\(\begin{aligned}
& \text { Weight from } y \text {-bundles }=12 y=12\left(\frac{2 x}{3}\right)=8 x \\
& \text { Total weight }=\text { Weight of } x \text {-bundles }+8 x \leq 20 x
\end{aligned}\)
$$


Weight of $x$-bundles $\leq 12 x$

$$
\(\text { Number of } x \text {-bundles } \leq \frac{12 x}{x}=12\)
$$


Both methods confirm that the maximum number of additional $x$-bundles is $\mathbf{1 2}$.
Step 5: Verify Against Answer Choices
The options are:
- (A) 4
- (B) 8
- (C) 10
- (D) 12
- (E) 18

Our calculation shows the correct answer is 12, which corresponds to Option D.
Final Answer
\(\boxed\{D\}\)