A sphere has a diameter of 2,160 meters. How many meters long is “Unit X” if the surface of the sphere, measured in square Units X is equal to the volume of the sphere measured in cube Units X? Round to the nearest whole number, and enter the number only.
Are you taking a GMAT right this minute?
No man? It is for something else? Well bchadwick any anwer to this mate. Cheers
360
ohai thanks bro
Wait, so you didn’t know the answer? Did I just help you answer some sort of test?
what is the point of this?
you provide diameter - so radius is known. if the dia = 2.16 m -> rad = 1.08 m.
so surface area = 4 pi r^2 = 14.65 m^2
volume = 4 / 3 * pi * r^3 = 5.273 m^3
which are obviously not equal.
so if 4 i r^2 = 4/3 pi r ^3 by your 2nd statement
r = 3
and 4 pi r^2 = x = 113.04 <
113 seems to be the answer.
4 i r^2 = 4/3 pi r ^3
This statement is incorrect if you are measuring “r” in meters. Also, he said 2160 meters, not 2.160 meters.
why is it incorrect. 2160 or 2.160 should not be a problem.
Because (2 * x)^2 is not the same as 2 * x^2. But I haven’t actually looked at the problem and tried to solve it.
where did that ever come in to consideration?
please read before you post, would suggest that.
How many meters long is “Unit X” if the surface of the sphere, measured in square Units X is equal to the volume of the sphere measured in cube Units X?
surface of the sphere = 4 pi r^2
volume of the sphere = 4/3 * pi * r^3
Dimensional analysis is inherently under consideration – my point is that if you solve with r = 2.160 you will get an entirely different answer than if you solve with r = 2,160 because power functions are not linear. I appreciate the sentiment that I should “read before I post,” though – very classy.
A sphere has a diameter of 2,160 meters. How many meters long is “Unit X” if the surface of the sphere, measured in square Units X is equal to the volume of the sphere measured in cube Units X? Round to the nearest whole number, and enter the number only.
Diameter of sphere: 2,160 meters
Radius of sphere: 1,080 meters
The surface of the sphere (4pi r^2) in square Units X is the volume of the sphere (4/3 pi r^3) measured in cube Units X.
X = some number of meters
18,662,499 pi m^2 / x^2 = 13,436,928,000 pi m^3 / x^3
x = 720
18,662,499 pi / 720^2 = 113
13,436,928,000 pi / 720^3 = 113
X = 720 meters
what is 18,662,499?
what is 13,436,928,000
zzzzzzzzzzZZZZZZZZZZZZzzzzzzz
if r=3, then
Surface Area = 4*pi*r^2 = 36*pi
Volume = 4/3*pi*r^3 = 36*pi.
You can find this by setting 4*pi*r^2 = (4/3)*pi*r^3 and solving for r.
So now, for the problem to work, you need to have r=3x, where x is the mysterious unit you are looking for.
We know the diameter 2r = 2160m, which means 2*(3x) = 2160m.
So x= 360.
I declare ohai the winner!
Yeah, you’re right, it’s 360. My 720 is 2X because I forgot the radius/diameter conversion.
4 * 2160^2
4 / 3 * 2160^3
but it should be 1080^2 … and 1080^3
which is why you ended up with twice the answer
Here’s an interesting thing about this problem.
The number 3 is probably not an accident. It’s the number of dimensions of a sphere.
You can ask a similar question: If the figure were a circle, what would be the measure for which the area of the circle in squared units is the same as the perimiter (circumference) of the circle in ordinary units. That number is 2.
circumference = 2*pi*r == pi * r^2 = area
2 happens to be the number of dimensions of a circle.
If you had a hypersphere, presumably the surface volume in units cubed would be equal to the hypervolume in units^4 when r=4.
It is also no coincidence that an area is a sum of infinitely many circumferences of circles sizes 0 through r – the antiderivative of 2 * pi * r is in fact pi * r^2. The same applies for spheres: the volume of a sphere is the sum of the surface areas of infinitely many spheres with radius 0 through r, so the antiderivative of 4*pi*r^2 is 4/3*pi*r^3. So yeah, what you’re saying will hold for higher dimensions.