Question 164·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

The short arrow in the figure marks only the wall thickness $x$; the outer radius is $6+x$. A cylinder’s volume is $\pi r^2h$, where $r$ is its radius and $h$ is its length. Subtract the hollow inner cylinder’s volume from the outer cylinder’s volume. Both are $10$ centimeters long, so the given wall volume means $10\pi(6+x)^2-10\pi(6)^2=520\pi$.

Type $10\pi((6+x_1)^2-6^2)\sim520\pi\{x_1>0\}$. Desmos uses $x_1$ as a value to find, and $\sim$ tells its regression to make the two sides match. The restriction keeps positive thicknesses only. Under PARAMETERS, Desmos reports $x_1\approx3.38083$ centimeters.

Type $-6+2\sqrt{22}$ on the next line. Desmos prints about $3.38083$, matching the thickness it found. That radical is an exact expression, so the pipe’s walls are $-6+2\sqrt{22}$ centimeters thick. Choice A.

Start with the wall-volume equation above. Divide by $10\pi$, since both cylinders have that factor: $(6+x)^2-6^2=52$. Evaluate the inner radius squared: $(6+x)^2-36=52$. What remains compares the two circular cross sections, without their common length.

Add $36$ to both sides to undo the subtraction: $(6+x)^2=88$. The whole outer radius, $6+x$, is squared here, not the thickness $x$ alone.

Take the square root of both sides: $6+x=\sqrt{88}$. Algebra also allows a negative square root, but $6+x$ is a radius, a distance from the center to the outside edge. It must be positive.

The radical $\sqrt{88}$ contains a factor of $4$, which is a perfect square. Rewrite it: $6+x=\sqrt{4\cdot22}$. Take $\sqrt4=2$ outside the root: $6+x=2\sqrt{22}$. Don't lose the $2$ when simplifying.

Subtract the inner radius, $6$, from both sides: $x=2\sqrt{22}-6$. This subtraction matters because $2\sqrt{22}$ is the outer radius, while $x$ is only the wall thickness. The thickness is $-6+2\sqrt{22}$ centimeters. Choice A.