1971 AHSME Problems/Problem 15

Problem

An aquarium on a level table has rectangular faces and is $10$ inches wide and $8$ inches high. When it was tilted, the water in it covered an $8\times 10$ end but only $\tfrac{3}{4}$ of the rectangular bottom. The depth of the water when the bottom was again made level, was

$\textbf{(A) }2\textstyle{\frac{1}{2}}"\qquad \textbf{(B) }3"\qquad \textbf{(C) }3\textstyle{\frac{1}{4}}"\qquad \textbf{(D) }3\textstyle{\frac{1}{2}}"\qquad \textbf{(E) }4"$

Solution

Let the third dimension of the aquarium be $x$, and let the height of the water when the aquarium is level be $h$. When the aquarium is tilted, the water forms a triangular prism. The triangular faces have height $8$ and, from the problem, base $\tfrac34x$. Thus, they have area $\tfrac12\cdot8\cdot\tfrac34x=3x$, so, because the prism has length $10$, the volume of the water is $10\cdot3x=30x$. When the tank is level, the water forms a rectangular prism with volume $h\cdot10\cdot x=10hx$. Because the amount of water in the tank is conserved, we can equate these two expressions for the volume of the water. Thus, $10hx=30x$, so $h=\boxed{\textbf{(B) }3"}$.

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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