Aunt and Uncle's fuel oil tank dip stick problem This problem first came to me in high school, and a couple times since, and I even assigned it for extra credit in one of my calculus classes after I became a teacher.  So I know the solution.  What I am looking for is other WAYS to obtain the solution.  I've been told there exists a solution using only arithmetic, but have never figured it out.  Other solutions using ordinary calculus, trigonometry, algebra of conic sections, and so on are also possible.
The problem is usually stated in the form of a letter from an Aunt and Uncle:

Dear niece/nephew, How are things
  going for you and your folks?  We hear
  you are doing quite well it school. 
  Keep it up!  Given this success, we
  were hoping you could help us figure
  out a little dilemma.  As you know,
  our home is heated by fuel oil, and we
  have a big tank buried in the side
  yard.  The tank is a cylinder, 20 feet
  long and 10 feet in diameter, lying on
  its side five feet deep, with a narrow
  tube coming to a fill cap at ground
  level.  Your uncle has a 15 foot
  length of old pipe that we'd like to
  utilize as a dip stick in order to
  know when we are getting close to
  needing a fill-up.  We know that 0
  feet is empty, 5 feet is half full,
  and 10 feet is completely full. 
  Trouble is, we don't know how to mark
  any other points. We are pretty sure
  they will not be uniformly spaced. 
  What we really want is to know, within
  the nearest 0.01 foot, where to mark
  the dip stick for every multiple of
  10% from 0% to 100%.  Can you figure
  this out for us?  Of course, we will
  want to see details of your solution
  and check it ourselves, and it would
  especially help if you could draw us a
  scale model of the dip stick. Love,
  Auntie Flo and Uncle Jim

That last sentence shows the teacher influence on the problem.
So, my challenge to this community is not to find any old solution, but to find the solution at the lowest possible grade level, so to speak.
Thanks.
UPDATE:  To those who are focusing in on the .01 feet accuracy, I apologize.  The intent was merely to state, it is acceptable to estimate.  If the exact answer is sqrt(2)*pi/2 or some other silly thing, go ahead and just write 2.22 feet, for example.
 A: It is sufficient to consider the circular cross-section of the tank and volumes below 50% (marks for those above 50% are the reflection image of those below 50% across the 50% mark).  Consider the radius of the tank to be 1 unit.  For some amount of oil in the tank, consider the central angle formed by the points on the circular cross-section at the top of the oil and the center of the circle--call this $\alpha$, measured in radians.

The area of the cross-section of the oil is the area of the sector determined by $\alpha$ ($\frac{\alpha}{2\pi}\pi r^2=\frac{\alpha}{2}$) minus the area of the triangle that is part of the sector but not part of the cross-section of oil ($\frac{1}{2}ab\sin C=\frac{\sin\alpha}{2}$), $\frac{1}{2}(\alpha-\sin\alpha)$.  The portion of the circular cross-section (and hence the portion of the volume) corresponding to this angle is $\frac{\alpha-\sin\alpha}{2\pi}$.  Setting this expression equal to 0.1, 0.2, 0.3, and 0.4 and solving for $\alpha$ will give the values of $\alpha$ corresponding to 10%, 20%, 30%, and 40% full--solving here is done numerically/graphically, as there is no algebraic method to solve these equations.  For each value of $\alpha$, the distance from the center of the circle to the oil level is $\cos\frac{\alpha}{2}$, so the depth of the oil is $1-\cos\frac{\alpha}{2}$.  (Note that these are for a radius of 1 unit and need to be rescaled for the original problem's specific numbers.)
A: I will give the calculus-based solution myself just for the sake of argument, taking note that I am still hoping to obtain a wide variety of other solutions, if possible.
For this, I am going to mentally rotate the tank (or the oil within it) $90^\circ$ clockwise, cut it in half, and center it at $(0,0)$, so that the upper half of the tank is represented by $y=\sqrt{25-x^2}$ and the volume of the oil by $\int_{-5}^h \sqrt{25-x^2}\mathrm{d}x$.
I know from simple $A=\pi r^2$ that the total cross-section area of the tank is $25\pi$ and thus for the upper half from -5 to +5 is $12.5\pi$.  I will set the result of the integral to the various 10% proportions of this value, knowing that 10% of the upper half will occur at the same position as 10% of the entire circle etc.
The integral is $\frac12\left(x\sqrt{25-x^2} + 25\arcsin\left(\frac{x}{5}\right)\right)$ evaluated from -5 to h, so the equation we need to solve is:
 $\frac12\left(h\sqrt{25-h^2}+25\arcsin(\frac{h}{5})\right)+\frac{25\pi}{4}=12.5\pi P$
Substituting values of .1, .2, .3, .4, and .5 successively for $P$ and using various tools to estimate $h$, my results are -3.43424, -2.45931, -1.59846, -0,788681, 0.  Adding 5 to account for the central displacement, rounding off, then reflecting these values for the right (upper) values, confirms the values given earlier by Americo.
A: The direct approach, expressing the (relative) volume as a function of the (relative) height isn't so practical as it requires to invert the relation $v=v(h)$, which can only be done numerically for all desired volumes.
It can be more attractive to express the height as a function of the volume by a differential equation,
$$\frac{dh}{dv}=\frac1{\sqrt{h(1-h)}},$$
that one can solve with a fixed step method such as Runge-Kutta.
To avoid the singularity, we can start from $(h,v)=(50\%,50\%)$, in steps of $10\%$ or smaller.
