Containers X and Y are rectangular containers. The base area of Container X is 30 cm^2 and the base area of Container Y is 100 cm^2. Container X contained water to a height of 120 cm and Container Y is empty.

1. Calculate the initial volume of water in Container X

$\begin{aligned} \text{Volume of water} &= \text{Base area of Container X} \times \text{Height of water} \\ &= 30\text{ cm}^2 \times 120\text{ cm} \\ &= 3600\text{ cm}^3 \end{aligned}$

1. Calculate the total volume of water initially in Container X.

The volume of water initially in Container X is given by:
$\text{Volume} = \text{Base Area} \times \text{Height} = 30 \times 120 = 3600\text{ cm}^3$

2. Set up an equation expressing the total volume of water after pouring in terms of the new height in Container Y.

Let $h$ be the height of the water level in Container Y in cm.
Then, the height of the water level in Container X is $2h$ cm.
The sum of the volumes in both containers remains equal to the initial total volume of $3600\text{ cm}^3$.
$\text{Volume in X} + \text{Volume in Y} = 3600$
$(30 \times 2h) + (100 \times h) = 3600$
$60h + 100h = 3600$
$160h = 3600$

3. Solve for the height of water in Container Y and calculate the volume of water poured into Container Y.

Solve for $h$:
$h = \frac{3600}{160} = 22.5\text{ cm}$

The amount of water poured into Container Y is the volume of water now in Container Y:
$\text{Volume in Y} = 100 \times h = 100 \times 22.5 = 2250\text{ cm}^3$