lerogmat
A saltwater solution completely fills a glass cylinder. If the solution is then poured into a larger rectangular aquarium, what percent more solution would be needed to completely fill the aquarium?
(1) The ratio of the diagonal of the base of the aquarium to the diameter of the cylinder is 1: sq rt 2
(2) The height and width of the aquarium are equal to the height and diameter of the cylinder.
Can someone explain me this one?
Thanks in advance
IF YOU FIND MY SOLUTION HELPFUL, PLEASE GIVE ME KUDOS
It is necessary to know the formulas for volume of a cylinder and volume of a rectangular solid. They are v = pi*r^2h and v=lwh Also, if we can get the algebraic expressions for volume of the cylinder and volume of the aquarium to contain the same variables, we could pick values for the variables to determine the relative volumes in term of a percentage.
(1) The ratio of the diagonal of the base of the aqarium to the diameter of the cylander is 1:sq rt 2 Ok so our cylinder will have diameter sq rt 2 * x and radius = (xrt2)/2. Lets call the diagonal of the base of the aquarium and the diameter of the cylinder x. Lets call the heights of the cylinder and aquarium h1 and h2. So the volume of our cylinder will be pi*(xrt2/2)^2 *h1. Now, since we have a rectangular aquarium, and we know the diagonal of the base is x, we know that the width of the base is x/2 and the length of the base is x*rt 3/2, since the diagonal of any rectangle forums a 30 - 60- 90 right triangle whose dimensions are in the ratio 1: 1rt3: 2. So, volume of the aquarium is then x*xroot3 * h2. Since the volume of the cylinder and the volume of the aquarium contain different variables, we cannot know their relative volumes. NS
(2) the height and width of the aquarium are equal to the height and diameter of the cylinder. Lets call the height H and the width and diameter of the aquarium x. Then the volume of the cylander is pi*(x/2)^2 * h, and the volume of the aquarium is h*x*L Since the volume of the aquarium contains the variable L for length, we cannot find the volume of the aquarium relative to the cylinder. NS
(1) and (2) The ratio of the diagonal of the base of the aquarium to the diameter of the cylinder is 1: root2 And the height and width of the aquarium are equal to the height and diameter of the cylinder. Ok, so the radius of the cylinder is xroot2/2, the height of the cylinder we can call H3, and the length, width and height of the aquarium are x/2, (xroot3)/2 and H3 (since the heights of the two shapes are the same).
So, our volumes of the cylinder and aquarium are pi*(xrt2/2)^2*H3 and (x/2)*(xroot3)/2*H3. Since there are now, no variables in the expressions for volume of the cylinder that are not in the expression for the volume of the aquarium ( and vice versa) we could pick values for both x and H3. Since we are only needing the volume of the aquarium in relative terms as a percentage of the cylinder, we are Sufficient.
The answer is C