Bunuel wrote:
jax91 wrote:
(1) The top surface of the water in the cone is currently 9pi square centimeters in area. -- sufficient on its own (we can get D)
(2) The top surface of the water in the cone currently is exactly 4 centimeters below the cone's top, measuring vertically. -- sufficient on its own ( we have H)
As the ratio of diameter:hieght of any sub-cone formed in this cone will be 1:2
So we need either D or H to get the volume.
IMO D.
Disagree.
A right circular cone, twice as tall as it is wide at its greatest width, is pointing straight down. The cone is partially filled with water, which is dripping out of a tiny hole in the cone's tip at a rate of 2 cubic centimeters per hour. If the water were to continue to drip out at this rate, how much longer would it take for the cone to empty, assuming that no water is added to the cone and that there is no loss of water from the cone by any other means?Attachment:
Untitled.png
H - height of cone, R radius of cone.
h - height of water in cone, r radius of top surface of water in cone.
(1) The top surface of the water in the cone is currently \(9\pi\) square centimeters in area. Top surface area of water = \(9\pi=\pi*r^2\) --> \(r=3\) --> \(\frac{R}{H}=\frac{r}{h}=\frac{1}{4}\) --> \(h=12\) --> \(V=\frac{1}{3}*\pi*r^2*h=36*\pi\) cubic centimeters.
Leak rate 2 cubic centimeters per hour --> \(time=\frac{36\pi}{2}\) hours.
Sufficient.
(2) The top surface of the water in the cone currently is exactly 4 centimeters below the cone's top, measuring vertically. Not sufficient we know that H=h+4, but h can be any value and thus the Volume can be any.
Answer: A.
The top surface of the water in the cone is currently \(9\pi\) square centimeters in area. Top surface area of water = \(9\pi=\pi*r^2\) --> \(r=3\) --> \(\frac{R}{H}=\frac{r}{h}=\frac{1}{4}\)
after you solved for r=3, how did you get
\(\frac{R}{H}=\frac{r}{h}=\frac{1}{4}\)
Aren't R, H, and h, all unknown?