## Tuesday, March 6, 2012

### this week's MGRE Math Beast Challenge

From here:

Go to it! My answer will eventually appear in the comments, but right now I suspect the correct response is (D)-- cannot be determined. Why? Because there are four pieces of data we need to know, and we don't know any of them:

1. initial height
2. "diminished" height
3. initial radius
4. "diminished" radius

Still, that's just a conjecture at this point.* When I finally write out my attempted solution, I may disagree with what I've just written. Stay tuned-- but feel free to leave your answer in the comments if you think you've arrived at it before I have.

*If we think of the problem two-dimensionally, i.e., in terms of similar triangles, it may be that there is a solution. Again, I'm typing all this in a rush, so I really don't know yet. Will have to stare hard at the problem before I give it a go.

_

#### 1 comment:

Kevin Kim said...

I think the answer may in fact be (A): Quantity A is greater.

Because I can't use subscript and superscript fonts, I'm going to designate our four relevant variables this way:

H = the height of the full cone with 1000π units of water in it.

h = the height of the cone whose water volume has been reduced by 60%.

R = the radius of the larger cone.

r = the radius of the smaller cone.

The relationships between the heights and radii of the two cones can be expressed as:

r/R = h/H

Why? Think: similar triangles.

Imagine that we're looking at a cross-section of the two cones. It might look something like this:

__________ (H,R)
_________
________
_______
______
_____
____ (h,r)
___
__
_

--where (H,R) represents "height H, radius R," and (h,r) represents "height h, radius r." This cross-section represents two similar triangles: they share the same bottom angle, and the respective radii extend outward from the vertical edge at 90 degrees. That's enough to prove AAA (angle-angle-angle) similarity, which means the corresponding sides are proportionally related. So, once again:

r/R = h/H

The volumes of these two cones can be expressed this way:

V(sm) = (1/3)π(r^2)h

V(lg) = (1/3)π(R^2)H

We know that the volume of the smaller cone is only 40% that of the larger cone, so:

V(sm)/V(lg) = 400π/1000π = 400/1000

Let's leave that unreduced for the moment. You'll see why.

We can now set up a ratio:

V(sm)/V(lg) = 400/1000 =

[(1/3)π(r^2)h]/[(1/3)π(R^2)H]

Let's rewrite the above fraction in a more "vertical" form so you can see what's going to happen next.

(1/3)π(r^2)h
----------------
(1/3)π(R^2)H

We can immediately cancel out the [(1/3)π] on the top and bottom:

(r^2)h
----------
(R^2)H

...and now it's a good time to remember that r/R = h/H. This means that we can replace the r and R with h and H, and still preserve the ratio:

h^3/H^3 = 400/1000

So then we take the cube root of both sides (which I'll write using fractional powers since there's no cube-root symbol that I can use in the comment thread), and we get:

h/H = 400^(1/3)/10

So what's the cube root of 400? I can't say right away, but I do know it's a number greater than 4. In fact, I can reason that it's between 4 and 10: 4^3 = 64 and 10^3 = 1000. The cube root of 400 lies between these whole numbers. Since h/H is the very quantity we're comparing with 0.4, then the answer is clearly that Quantity A is greater.

(In case you're wondering, my calculator says the cube root of 400 is approximately 7.368063.)

QED.