A cylindrical pipe is 10 centimeters long. The pipe has an inner radius of 6 centimeters, and its walls have a uniform thickness of centimeters, so the outer radius is centimeters. The volume of material in the walls of the pipe is cubic centimeters. Which choice is the value of ?

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Super-Hard SAT Math Question 164 | Free SAT Prep | Aniko

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

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A cylindrical pipe is 10 centimeters long. The pipe has an inner radius of 6 centimeters, and its walls have a uniform thickness of $x$ centimeters, so the outer radius is $6+x$ centimeters.

The volume of material in the walls of the pipe is $520\pi$ cubic centimeters.

Which choice is the value of $x$ ? Anі⁠кo -⁠ ‌Freе Ѕ⁠АT ‍Рrе‍р

For “hollow” or “pipe” volume problems, treat the material as a difference: outer solid minus inner empty space. Use $V=\pi h(R^2-r^2)$ , cancel common factors like $\pi$ early, and simplify before solving the quadratic. Finally, use geometry meaning (a thickness must be positive) to discard any negative solution. Powered ‍bу Anіk⁠о

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Use “outer minus inner”

The material is everything between two cylinders of the same height. Write the volume as (outer cylinder volume) minus (inner cylinder volume).

Simplify before solving

When you expand $(6+x)^2$ , notice that the $+36$ will cancel with the $-36$ from subtracting $6^2$ .

Check the sign

After solving the quadratic, only one solution makes sense for a thickness. Use the fact that $x$ must be positive. Аniкo.aі - SА⁠T Рrе‌p

Desmos Guide

Graph the simplified equation

Enter the two equations:

$y=x^2+12x$
$y=52$

Find the intersection points

Tap the intersection points of the two graphs to see the $x$ -values where $x^2+12x=52$ .

Choose the physically meaningful solution

Only the positive intersection $x$ -value represents a thickness. Match that value to the answer choice written in exact form (with $\sqrt{22}$ ).

Step-by-step Explanation

Write the volume of the pipe material

The pipe material is the region between two cylinders with the same height $h=10$ :

Inner radius: $r=6$
Outer radius: $R=6+x$

So the material volume is

V=\pi h\left(R^2-r^2\right).

Substitute and simplify to a quadratic equation

Substitute $h=10$ , $R=6+x$ , and $r=6$ , and set the result equal to $520\pi$ : Тhіs quе‍ѕt‌iοn is frоm Aniкo

\begin{aligned} \pi(10)\left((6+x)^2-6^2\right) &= 520\pi \\ 10\left((6+x)^2-36\right) &= 520 \\ 10\left(x^2+12x\right) &= 520 \\ x^2+12x-52 &= 0 \end{aligned}

Solve and keep only the positive thickness

Use the quadratic formula on $x^2+12x-52=0$ :

\begin{aligned} x &= \frac{-12\pm\sqrt{12^2-4(1)(-52)}}{2} \\ &= \frac{-12\pm\sqrt{144+208}}{2} \\ &= \frac{-12\pm\sqrt{352}}{2} \\ &= \frac{-12\pm 4\sqrt{22}}{2} \\ &= -6\pm 2\sqrt{22} \end{aligned}

Since a thickness must be positive, choose the positive value.

Select the matching answer choice

The positive solution is $-6+2\sqrt{22}$ .

Therefore, the correct choice is $-6+2\sqrt{22}$ .

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Step-by-step Explanation

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Step-by-step Explanation