A closed rectangular prism with height $8 \mathrm{~cm}$ is standing on a face with dimensions $2 \mathrm{~cm}$ by $5 \mathrm{~cm}$. The prism contains water with a depth of $6 \mathrm{~cm}$, as shown. When the prism is tipped so that it stands on a face with the greatest area, the depth of the water is
(A) $0.75 \mathrm{~cm}$
(B) $1 \mathrm{~cm}$
(C) $1.25 \mathrm{~cm}$
(D) $1.5 \mathrm{~cm}$
(E) $1.75 \mathrm{~cm}$

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-004.jpg?height=323&width=404&top_left_y=188&top_left_x=1313)
The water is in the shape of a rectangular prism with a $2 \mathrm{~cm}$ by $5 \mathrm{~cm}$ base and depth $6 \mathrm{~cm}$.

Therefore, the volume of water is $2 \mathrm{~cm} \times 5 \mathrm{~cm} \times 6 \mathrm{~cm}=60 \mathrm{~cm}^{3}$.

A face of the prism having the greatest area has dimensions $5 \mathrm{~cm}$ by $8 \mathrm{~cm}$.

When the prism is tipped so that it stands on a $5 \mathrm{~cm}$ by $8 \mathrm{~cm}$ face, the water is once again in the shape of a rectangular prism with a $5 \mathrm{~cm}$ by $8 \mathrm{~cm}$ base and unknown depth.

Suppose that after the prism is tipped, the water's depth is $d \mathrm{~cm}$.

Since the volume of water is still $60 \mathrm{~cm}^{3}$ when the prism is tipped, then $5 \mathrm{~cm} \times 8 \mathrm{~cm} \times d \mathrm{~cm}=60 \mathrm{~cm}^{3}$ or $40 d \mathrm{~cm}^{3}=60 \mathrm{~cm}^{3}$, and so $d=\frac{60}{40}=\frac{3}{2}$.

When the prism is tipped so that it stands on a face with the greatest area, the depth of the water is $\frac{3}{2} \mathrm{~cm}=1.5 \mathrm{~cm}$.

ANSWER: (D)

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