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ISEE Upper Level Quantitative : Cylinders

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Example Questions

Example Question #342 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

What is the volume of a cylinder with a radius of 6 meters and a height of 11 meters? Use 3.14 for $\pi$ $\displaystyle \pi$.

Note: The formula for the volume of a cylinder is:

$V=\pi r^2h$ $\displaystyle V=\pi r^2h$

Possible Answers:

$\displaystyle 114.44m^3$

$\displaystyle 1243.44 m^3$

$\displaystyle 1432.23m^3$

$\displaystyle 207.24m^3$

396m^3 $\displaystyle 396m^3$

Correct answer:

$\displaystyle 1243.44 m^3$

Explanation:

To calculate the volume, you must plug into the formula given in the problem. When you plug in, it should look like this: $\displaystyle V=(3.14)(6^2)(11)$. Multiply all of these out and you get $\displaystyle 1243.44m^3$. The units are cubed because volume is always cubed.

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Example Question #1 : How To Find The Volume Of A Cylinder

The volume of a cylinder whose height is twice the diameter of its base is one cubic yard. Give its radius in inches.

Possible Answers:

$\frac{ 9 \sqrt[3]{ 2 \pi^{2} } } {4 \pi } \textrm{ in}$ $\displaystyle \frac{ 9 \sqrt[3]{ 2 \pi^{2} } } {4 \pi } \textrm{ in}$

$\frac{ 18 \sqrt[3]{ 2 \pi^{2} } } {\pi } \textrm{ in}$ $\displaystyle \frac{ 18 \sqrt[3]{ 2 \pi^{2} } } {\pi } \textrm{ in}$

$\frac{ \sqrt[3]{ 2 \pi^{2} } } {2\pi } \textrm{ in}$ $\displaystyle \frac{ \sqrt[3]{ 2 \pi^{2} } } {2\pi } \textrm{ in}$

$\frac{ 9 \sqrt[3]{ 2 \pi^{2} } } {\pi } \textrm{ in}$ $\displaystyle \frac{ 9 \sqrt[3]{ 2 \pi^{2} } } {\pi } \textrm{ in}$

$\frac{ 9 \sqrt[3]{ 2 \pi^{2} } } {2 \pi } \textrm{ in}$ $\displaystyle \frac{ 9 \sqrt[3]{ 2 \pi^{2} } } {2 \pi } \textrm{ in}$

Correct answer:

$\frac{ 18 \sqrt[3]{ 2 \pi^{2} } } {\pi } \textrm{ in}$ $\displaystyle \frac{ 18 \sqrt[3]{ 2 \pi^{2} } } {\pi } \textrm{ in}$

Explanation:

The volume of a cylinder with base radius $\displaystyle r$ and height $\displaystyle h$ is

$V = \pi r^{2}h$ $\displaystyle V = \pi r^{2}h$

The diameter of this circle is $\displaystyle 2r$; its height is twice this, or $\displaystyle 4r$. Therefore, the formula becomes

$V = \pi r^{2} \cdot 4r = 4\pi r^{3}$ $\displaystyle V = \pi r^{2} \cdot 4r = 4\pi r^{3}$

Set this volume equal to one and solve for $\displaystyle r$:

$4\pi r^{3} = 1$ $\displaystyle 4\pi r^{3} = 1$

$4\pi r^{3} \div 4\pi = 1 \div 4\pi$ $\displaystyle 4\pi r^{3} \div 4\pi = 1 \div 4\pi$

$r ^{3}= \frac{1}{4\pi }$ $\displaystyle r ^{3}= \frac{1}{4\pi }$

$r =\sqrt[3]{ \frac{1}{4\pi }} =\frac{ \sqrt[3]{ 1 }} { \sqrt[3]{ 4\pi }} =\frac{ \sqrt[3]{ 1 } \cdot \sqrt[3]{ 2 \pi^{2} } } { \sqrt[3]{ 4\pi } \cdot \sqrt[3]{ 2 \pi^{2} }}=\frac{ \sqrt[3]{ 2 \pi^{2} } } {\sqrt[3]{ 8\pi^{3} }}=\frac{ \sqrt[3]{ 2 \pi^{2} } } {2\pi }$ $\displaystyle r =\sqrt[3]{ \frac{1}{4\pi }} =\frac{ \sqrt[3]{ 1 }} { \sqrt[3]{ 4\pi }} =\frac{ \sqrt[3]{ 1 } \cdot \sqrt[3]{ 2 \pi^{2} } } { \sqrt[3]{ 4\pi } \cdot \sqrt[3]{ 2 \pi^{2} }}=\frac{ \sqrt[3]{ 2 \pi^{2} } } {\sqrt[3]{ 8\pi^{3} }}=\frac{ \sqrt[3]{ 2 \pi^{2} } } {2\pi }$

This is the radius in yards; multiply by 36 to get the radius in inches.

$\frac{ \sqrt[3]{ 2 \pi^{2} } } {2\pi } \cdot 36 = \frac{ 18 \sqrt[3]{ 2 \pi^{2} } } {\pi }$ $\displaystyle \frac{ \sqrt[3]{ 2 \pi^{2} } } {2\pi } \cdot 36 = \frac{ 18 \sqrt[3]{ 2 \pi^{2} } } {\pi }$

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Example Question #2 : How To Find The Volume Of A Cylinder

What is the height of a cylinder with a volume of $646.875\pi$ $\displaystyle 646.875\pi$ in^3 $\displaystyle in^3$ and a radius of 7.5 $\displaystyle 7.5$ $\displaystyle in$?

Possible Answers:

86.25 $\displaystyle 86.25$ $\displaystyle in$

11.5 $\displaystyle 11.5$ $\displaystyle in$

56250 $\displaystyle 56250$ $\displaystyle in$

22.5 $\displaystyle 22.5$ $\displaystyle in$

28125.5 $\displaystyle 28125.5$ $\displaystyle in$

Correct answer:

11.5 $\displaystyle 11.5$ $\displaystyle in$

Explanation:

Recall that the equation of for the volume of a cylinder is:

$V = \pi * r^2 * h$ $\displaystyle V = \pi * r^2 * h$

For our values this is:

$646.875\pi = \pi * 7.5^2 * h= \pi * 56.25 * h$ $\displaystyle 646.875\pi = \pi * 7.5^2 * h= \pi * 56.25 * h$

Solve for $\displaystyle h$:

$\displaystyle h = 11.5$

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Example Question #344 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

What is the volume of a cylinder with a height of $\displaystyle 5$ in. and a radius of $\displaystyle 12$ in?

Possible Answers:

$720\pi$ $\displaystyle 720\pi$ in^3 $\displaystyle in^3$

$300\pi$ $\displaystyle 300\pi$ in^3 $\displaystyle in^3$

$180\pi$ $\displaystyle 180\pi$ in^3 $\displaystyle in^3$

$408\pi$ $\displaystyle 408\pi$ in^3 $\displaystyle in^3$

$204\pi$ $\displaystyle 204\pi$ in^3 $\displaystyle in^3$

Correct answer:

$720\pi$ $\displaystyle 720\pi$ in^3 $\displaystyle in^3$

Explanation:

This is a rather direct question. Recall that the equation of for the volume of a cylinder is:

$V = \pi * r^2 * h$ $\displaystyle V = \pi * r^2 * h$

For our values this is:

$A = \pi * 12^2 * 5=720\pi$ $\displaystyle A = \pi * 12^2 * 5=720\pi$

This is the volume of the cylinder.

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Example Question #5 : How To Find The Volume Of A Cylinder

What is the volume of a cylinder with a height of $\displaystyle 14$ in. and a radius of 2.5 $\displaystyle 2.5$ in?

Possible Answers:

$87.5\pi$ $\displaystyle 87.5\pi$ in^3 $\displaystyle in^3$

$82.5\pi$ $\displaystyle 82.5\pi$ in^3 $\displaystyle in^3$

$42.25\pi$ $\displaystyle 42.25\pi$ in^3 $\displaystyle in^3$

$70\pi$ $\displaystyle 70\pi$ in^3 $\displaystyle in^3$

$35\pi$ $\displaystyle 35\pi$ in^3 $\displaystyle in^3$

Correct answer:

$87.5\pi$ $\displaystyle 87.5\pi$ in^3 $\displaystyle in^3$

Explanation:

This is a rather direct question. Recall that the equation of for the volume of a cylinder is:

$V = \pi * r^2 * h$ $\displaystyle V = \pi * r^2 * h$

For our values this is:

$A = \pi * 2.5^2 * 14=87.5\pi$ $\displaystyle A = \pi * 2.5^2 * 14=87.5\pi$

This is the volume of the cylinder.

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Example Question #351 : Geometry

What is the radius of a cylinder with a volume of $562.5\pi$ $\displaystyle 562.5\pi$ in^3 $\displaystyle in^3$ and a height of $\displaystyle 10$ $\displaystyle in$?

Possible Answers:

$56.25\pi$ $\displaystyle 56.25\pi$ $\displaystyle in$

56.25 $\displaystyle 56.25$ $\displaystyle in$

$\displaystyle 15$ $\displaystyle in$

$15\pi$ $\displaystyle 15\pi$ $\displaystyle in$

7.5 $\displaystyle 7.5$ $\displaystyle in$

Correct answer:

7.5 $\displaystyle 7.5$ $\displaystyle in$

Explanation:

Recall that the equation of for the volume of a cylinder is:

$V = \pi * r^2 * h$ $\displaystyle V = \pi * r^2 * h$

For our values this is:

$562.5\pi = \pi * r^2 * 10$ $\displaystyle 562.5\pi = \pi * r^2 * 10$

Solve for $\displaystyle r$:

$\displaystyle 56.25=r^2$

Using a calculator to calculate $\sqrt{56.25}$ $\displaystyle \sqrt{56.25}$, you will see that r = 7.5 $\displaystyle r = 7.5$

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Example Question #2 : Cylinders

What is the surface area of a cylinder of height $\displaystyle 15$ in., with a radius of $\displaystyle 4$ in?

Possible Answers:

$152\pi$ $\displaystyle 152\pi$ in^2 $\displaystyle in^2$

$240\pi$ $\displaystyle 240\pi$ in^2 $\displaystyle in^2$

$122\pi$ $\displaystyle 122\pi$ in^2 $\displaystyle in^2$

$136\pi$ $\displaystyle 136\pi$ in^2 $\displaystyle in^2$

$120\pi$ $\displaystyle 120\pi$ in^2 $\displaystyle in^2$

Correct answer:

$152\pi$ $\displaystyle 152\pi$ in^2 $\displaystyle in^2$

Explanation:

$A = \pi * r^2$ $\displaystyle A = \pi * r^2$

For our problem, this is:

$\pi * 4^2 = \pi * 16$ $\displaystyle \pi * 4^2 = \pi * 16$

You need to double this for the two bases:

$2 * 16 * \pi = 32 * \pi$ $\displaystyle 2 * 16 * \pi = 32 * \pi$

The area of the "outer face" is a little bit trickier, but it is not impossible. It is actually a rectangle that has the height of the cylinder and a width equal to the circumference of the base; therefore, it is:

$2 * \pi * r * h$ $\displaystyle 2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 4 * 15 = 120 * \pi$ $\displaystyle 2 * \pi * 4 * 15 = 120 * \pi$

Therefore, the total surface area is:

$32\pi + 120\pi = 152\pi$ $\displaystyle 32\pi + 120\pi = 152\pi$

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Example Question #2 : Cylinders

What is the surface area of a cylinder having a base of radius $\displaystyle 11$in and a height of $\displaystyle 3$ in?

Possible Answers:

$212\pi$ $\displaystyle 212\pi$ in^2 $\displaystyle in^2$

$308\pi$ $\displaystyle 308\pi$ in^2 $\displaystyle in^2$

$99\pi$ $\displaystyle 99\pi$ in^2 $\displaystyle in^2$

$363\pi$ $\displaystyle 363\pi$ in^2 $\displaystyle in^2$

$33\pi$ $\displaystyle 33\pi$ in^2 $\displaystyle in^2$

Correct answer:

$308\pi$ $\displaystyle 308\pi$ in^2 $\displaystyle in^2$

Explanation:

$A = \pi * r^2$ $\displaystyle A = \pi * r^2$

For our problem, this is:

$\pi * 11^2 = \pi * 121$ $\displaystyle \pi * 11^2 = \pi * 121$

You need to double this for the two bases:

$2 * 121 * \pi = 242 * \pi$ $\displaystyle 2 * 121 * \pi = 242 * \pi$

$2 * \pi * r * h$ $\displaystyle 2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 11* 3 = 66 * \pi$ $\displaystyle 2 * \pi * 11* 3 = 66 * \pi$

Therefore, the total surface area is:

$242 * \pi + 66 * \pi = 308\pi$ $\displaystyle 242 * \pi + 66 * \pi = 308\pi$

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Example Question #3 : Cylinders

What is the surface area of a cylinder with a height of $\displaystyle 7$in. and a diameter of $\displaystyle 4$in?

Possible Answers:

$22\pi$ $\displaystyle 22\pi$ in^2 $\displaystyle in^2$

$57\pi$ $\displaystyle 57\pi$ in^2 $\displaystyle in^2$

$28\pi$ $\displaystyle 28\pi$ in^2 $\displaystyle in^2$

$36\pi$ $\displaystyle 36\pi$ in^2 $\displaystyle in^2$

$14\pi$ $\displaystyle 14\pi$ in^2 $\displaystyle in^2$

Correct answer:

$36\pi$ $\displaystyle 36\pi$ in^2 $\displaystyle in^2$

Explanation:

Recall that to find the surface area of a cylinder, you need to find the surface area of its two bases and then the surface area of its "outer face." The first two are very easy since they are circles. Notice, however that the diameter is $\displaystyle 4$ inches. This means that the radius is $\displaystyle 2$. Now, the equation for one base is:

$A = \pi * r^2$ $\displaystyle A = \pi * r^2$

For our problem, this is:

$\pi * 2^2 = \pi * 4$ $\displaystyle \pi * 2^2 = \pi * 4$

You need to double this for the two bases:

$2 * 4 * \pi = 8 * \pi$ $\displaystyle 2 * 4 * \pi = 8 * \pi$

$2 * \pi * r * h$ $\displaystyle 2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 2 * 7 = 28 * \pi$ $\displaystyle 2 * \pi * 2 * 7 = 28 * \pi$

Therefore, the total surface area is:

$28\pi + 8\pi = 36\pi$ $\displaystyle 28\pi + 8\pi = 36\pi$

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Example Question #1 : How To Find The Surface Area Of A Cylinder

The volume of a cylinder with height of $\displaystyle 11$ is $275\pi$ $\displaystyle 275\pi$ in^3 $\displaystyle in^3$. What is its surface area?

Possible Answers:

$55\pi$ $\displaystyle 55\pi$ in^2 $\displaystyle in^2$

$160\pi$ $\displaystyle 160\pi$ in^2 $\displaystyle in^2$

$66\pi$ $\displaystyle 66\pi$ in^2 $\displaystyle in^2$

$82\pi$ $\displaystyle 82\pi$ in^2 $\displaystyle in^2$

$135\pi$ $\displaystyle 135\pi$ in^2 $\displaystyle in^2$

Correct answer:

$160\pi$ $\displaystyle 160\pi$ in^2 $\displaystyle in^2$

Explanation:

To begin, we must solve for the radius of this cylinder. Recall that the equation of for the volume of a cylinder is:

$V = \pi * r^2 * h$ $\displaystyle V = \pi * r^2 * h$

For our values this is:

$275\pi = \pi * r^2 * 11$ $\displaystyle 275\pi = \pi * r^2 * 11$

Solving for $\displaystyle r$, we get:

$\displaystyle r^2 = 25$

Hence, r = 5 $\displaystyle r = 5$

Now, recall that to find the surface area of a cylinder, you need to find the surface area of its two bases and then the surface area of its "outer face." The first two are very easy since they are circles. The equation for one base is:

$A = \pi * r^2$ $\displaystyle A = \pi * r^2$

For our problem, this is:

$\pi * 5^2 = \pi * 25$ $\displaystyle \pi * 5^2 = \pi * 25$

You need to double this for the two bases:

$2 * 25 * \pi = 50 * \pi$ $\displaystyle 2 * 25 * \pi = 50 * \pi$

$2 * \pi * r * h$ $\displaystyle 2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 5 * 11 = 110 * \pi$ $\displaystyle 2 * \pi * 5 * 11 = 110 * \pi$

Therefore, the total surface area is:

$50\pi + 110\pi = 160\pi$ $\displaystyle 50\pi + 110\pi = 160\pi$

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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : Cylinders

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #342 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #1 : How To Find The Volume Of A Cylinder

Example Question #2 : How To Find The Volume Of A Cylinder

Example Question #344 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #5 : How To Find The Volume Of A Cylinder

Example Question #351 : Geometry

Example Question #2 : Cylinders

Example Question #2 : Cylinders

Example Question #3 : Cylinders

Example Question #1 : How To Find The Surface Area Of A Cylinder

All ISEE Upper Level Quantitative Resources

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