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 $h=\boxed{\textbf{(B) }3"}$ .

1971 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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1971 AHSME Problems/Problem 15

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