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Difference between revisions of "1971 AHSME Problems/Problem 15"

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== Solution ==
 
== Solution ==
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<math>\boxed{\textbf{(B) }3"}</math>.
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Let the third dimension of the aquarium be <math>x</math>, and let the height of the water when the aquarium is level be <math>h</math>. When the aquarium is tilted, the water forms a triangular prism. The triangular faces have height <math>8</math> and, from the problem, base <math>\tfrac34x</math>. Thus, they have area <math>\tfrac12\cdot8\cdot\tfrac34x=3x</math>, so, because the prism has length <math>10</math>, the volume of the water is <math>10\cdot3x=30x</math>. When the tank is level, the water forms a rectangular prism with volume <math>h\cdot10\cdot x=10hx</math>. Because the amount of water in the tank is conserved, we can equate these two expressions for the volume of the water. Thus, <math>10hx=30x</math>, so <math>h=\boxed{\textbf{(B) }3"}</math>.
  
 
== See Also ==
 
== See Also ==
 
{{AHSME 35p box|year=1971|num-b=14|num-a=16}}
 
{{AHSME 35p box|year=1971|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:25, 9 August 2024

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 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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