Question

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 just covered an $8$ -inch by $10$ -inch end but only three-fourths of the rectangular bottom. The depth of the water when the bottom was again made level was:

$2\tfrac12$ inches

$3$ inches

$3\tfrac14$ inches

$3\tfrac12$ inches

$4$ inches

Answer: B

Concepts:volume triangle area ratio and proportion

Difficulty rating: 1760

Small Hint:

Compare the water volume in the tilted and level positions

Big Hint:

The tilted side view is a triangle with height $8$ and base three-fourths of the aquarium length

Solution:

Let the aquarium length be $L$ and the level-water depth be $h.$ In the tilted position the longitudinal cross-section of the water is a triangle with base $\frac{3L}{4}$ and height $8.$ Therefore $10\left(\frac12\cdot\frac{3L}{4}\cdot8\right)=10Lh,$ so $h=3$ inches.

Therefore, the correct answer is B.

1971 AMC 12 Problem 15

Answer: B

Difficulty rating: 1760

Small Hint:

Big Hint:

Solution:

Problem 15 in Other Years