A solid steel object is formed by drilling a cylindrical hole through the center of a larger right circular cylinder. - The larger cylinder has radius 4 centimeters and height 15 centimeters. - The drilled hole is a right circular cylinder with radius 2 centimeters and the same height. - The density of steel is 7,800 kilograms per cubic meter. To the nearest whole number, what is the mass of the object, in kilograms?

Solution available with step-by-step explanation

Super-Hard SAT Math Question 142 | Free SAT Prep | Aniko

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Question 142·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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A solid steel object is formed by drilling a cylindrical hole through the center of a larger right circular cylinder.

The larger cylinder has radius 4 centimeters and height 15 centimeters.
The drilled hole is a right circular cylinder with radius 2 centimeters and the same height.
The density of steel is 7,800 kilograms per cubic meter.

To the nearest whole number, what is the mass of the object, in kilograms?

For “density and mass” solids problems, write mass as (density)×(volume). Compute volume using a known formula (here $\pi r^2h$ ) and subtract any removed parts. Before multiplying by density, confirm the volume units match the density’s units—unit conversion (cm $^3$ to m $^3$ ) is often the main trap. Finish by rounding only at the end.

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Subtract volumes

Find the volume of the larger cylinder and subtract the volume of the cylindrical hole.

Watch the units

Your volume will come out in cm $^3$ , but the density uses m $^3$ . Convert cm $^3$ to m $^3$ .

Use the density formula

Mass equals density times volume. After multiplying, round to the nearest whole kilogram.

Desmos Guide

Enter the net volume in cm³

In Desmos, type V = pi*(4^2)*15 - pi*(2^2)*15 to represent the net volume in cm^3.

Convert to m³

Type Vm = V/1000000 to convert cm^3 to m^3 (since $1\text{ m}^3=1{,}000{,}000\text{ cm}^3$ ).

Compute mass and round

Type M = 7800*Vm.

Use the value of M (about 4.41) and round it to the nearest whole number to get the correct choice.

Step-by-step Explanation

Compute the net volume in cubic centimeters

Volume of a cylinder is $V=\pi r^2h$ .

Larger cylinder: $\pi(4^2)(15)=\pi(16)(15)=240\pi$ cm $^3$
Hole: $\pi(2^2)(15)=\pi(4)(15)=60\pi$ cm $^3$

Net volume:

240\pi-60\pi=180\pi\text{ cm}^3

Convert the volume to cubic meters

Since $1\text{ m}=100\text{ cm}$ , then

1\text{ m}^3=(100\text{ cm})^3=1{,}000{,}000\text{ cm}^3

180\pi\text{ cm}^3=\frac{180\pi}{1{,}000{,}000}\text{ m}^3

Multiply by density and round

Mass $=$ density $\times$ volume:

\text{mass}=7800\left(\frac{180\pi}{1{,}000{,}000}\right) =1.404\pi\approx 4.41

Rounded to the nearest whole number, the mass is 4 kilograms.

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Step-by-step Explanation

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Step-by-step Explanation