<p>These problems are kind of related so I must have a fundamental misunderstanding somewhere.</p>
<li>A rectangular fish tank has a base 2 feet wide and 3 feet long. When the tank is partially fillled with water, a solid cube with an edge length of 1 foot is placed inthe tank. If no overflow of water from the tank is assumed, by how many inches will the level of the water in the tank rise when the cube becomes completely submerged?</li>
</ol>
<p>A. 1/6
B. 1/2
C. 2
D. 3
E. 4</p>
<li> The height of sand in a cylinder-shaped can drops 3 inches when 1 cubic foot of sand is poured out. What is the diameter, in inches, of the cylinder?</li>
</ol>
<p>A. 2/sqrt Pi
B. 4/sqrt Pi
C. 16/Pi
D. 32/sqrt Pi
E. 48/sqrt Pi</p>
<p>The answer to number one is C, and the answer to number two is E. However I do not quite understand why? Can anyone explain?</p>
<p>**1. A rectangular fish tank has a base 2 feet wide and 3 feet long. When the tank is partially fillled with water, a solid cube with an edge length of 1 foot is placed inthe tank. If no overflow of water from the tank is assumed, by how many inches will the level of the water in the tank rise when the cube becomes completely submerged?</p>
<p>A. 1/6
B. 1/2
C. 2
D. 3
E. 4**</p>
<p>We can assume the initial height of the water to be x. Thus the volume of this amount of water will be 3<em>2</em>x= 6x cubic feet.</p>
<p>The volume of the cube is 1 cubic foot. So the volume after the cube is placed in the water is 6x+1. The change in volume is therefore 6x+1(final volume) - 6x(initial volume) = 1 cubic foot(or the volume of the cube). Assuming the change in the level of the water is h, we equal the change in volume to the product of 2<em>3</em>h:</p>
<p>1= 6h
h= 1/6 feet</p>
<p>1 foot = 12 inches 1/6 feet = 2 inches</p>
<p>**2. The height of sand in a cylinder-shaped can drops 3 inches when 1 cubic foot of sand is poured out. What is the diameter, in inches, of the cylinder?</p>
<p>A. 2/sqrt Pi
B. 4/sqrt Pi
C. 16/Pi
D. 32/sqrt Pi
E. 48/sqrt Pi**</p>
<p>the initial volume of sand in the cylinder is pi.r^2.h. The final volume is pi.r^2.(h-3). The change in volume is 1 cubic foot. So the initial volume - the change in volume should equal the final volume:</p>
<p>pi.r^2.h - 1 cubic foot = pi.r^2.(h-3)</p>
<p>we can convert 1 cubic foot to cubic inches using dimensional analysis-</p>
<p>Both questions are based on the change in volume occupied within the containers. They both mix units of measure, (inches /feet) , which makes them a bit tricky;</p>
<ol>
<li><p>Volume occupied increases by 1 cu.ft = width x length x height_increase
1 = 2 * 3 * H, so H = 1/6 of a foot or 2 inches.</p></li>
<li><p>I cubic foot = 1x1x1 feet = 12 * 12 * 12 or 1720 cu. inches
cross-sectional<em>area * drop</em>in_height = reduction in volume
(pi * (D/2) ^ 2) * 3 = 1728
D^2 / 4 = 576/pi
D^2 = 2304/ pi
D = sqrt(2304) / sqrt(pi) = 48 / sqrt(pi)</p></li>
</ol>
<p>For both questions you can use the volume of 2 different shapes and make them equal to each other. </p>
<p>Ques 1: the volume of a rectangular solid is length<em>width</em>height.
Volume of a cube is (side)^3rd power. </p>
<p>So they give you the base and length of the rectangular solid, and the volume of the cube, so:
Vol rectangular solid= volume cube
l<em>w</em>height= Side^3
height= (Side^3)/l<em>w
height= (1)/2</em>3 = 1/6 foot, which equals 2 inches.
For the 2nd ques DUDE has the correct method. But you can also pretend that the volume occupied in the cylinder is the same as the 1 cubic foot of sand poured out. </p>
<p>So instead of subtracting the before and after volumes of the cylinder, just make the Vol of the cylinder = Vol of 1 cubic foot of sand. Then solve for radius, and solve for diameter. </p>
<p>Are these math II problems? Just wondering b/c I will be taking it and want to know what kind of ques are on the test.</p>