We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

ISEE Upper Level Quantitative : Geometry

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

Create An Account

All ISEE Upper Level Quantitative Resources

8 Diagnostic Tests 234 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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

Sphere A has volume $972 \pi \textrm{ in}^{3}$ . Sphere B has surface area $400 \pi \textrm{ in}^{2}$ . Which is the greater quantity?

(a) The radius of Sphere A

(b) The radius of Sphere B

Possible Answers:

(a) is greater

(a) and (b) are equal

It is impossible to tell from the information given

(b) is greater

Correct answer:

(b) is greater

Explanation:

(a) Substitute $V = 972 \pi$ in the formula for the volume of a sphere:

$\frac{4}{3}\pi r^{3}= V$

$\frac{4}{3}\pi r^{3}= 972 \pi$

$\frac{4}{3}\pi r^{3} \div \frac{4}{3}\pi = 972 \pi \div \frac{4}{3}\pi$

$\frac{4}{3}\pi r^{3} \cdot \frac{3}{4\pi} = 972 \pi\cdot \frac{3}{4\pi}$

$r^{3} = 729$

$r =\sqrt[3]{ 729} = 9$ inches

(b) Substitute $A = 400 \pi$ in the formula for the surface area of a sphere:

$4\pi r^{2}= A$

$4\pi r^{2}= 400 \pi$

$4\pi r^{2} \div 4\pi = 400 \pi \div 4\pi$

$r^{2} = 100$

$r = \sqrt{100} = 10$ inches

(b) is greater.

Report an Error

Example Question #43 : Solid Geometry

is a positive number. Which is the greater quantity?

(A) The surface area of a sphere with radius

(B) The surface area of a cube with edges of length

Possible Answers:

(A) is greater

(A) and (B) are equal

It is impossible to determine which is greater from the information given

(B) is greater

Correct answer:

(B) is greater

Explanation:

The surface area of a sphere is $4 \pi$ times the square of its radius, which here is ; the surface area of the sphere in (A) is $4 \pi t^{2}$ .

The area of one face of a cube is the square of the length of an edge, which here is , so the area of one face of the cube in (B) is $(2t) ^{2} = 4t^{2}$ . The cube has six faces so the total surface area is $6 \cdot 4t^{2} = 24t^{2}$ .

$\pi < 4$ , so $4 \pi t^{2} < 16 t^{2} < 24 t^{2}$ , giving the sphere less surface area. (B) is greater.

Report an Error

Example Question #1 : Cones

The height of Cone B is three times that of Cone A. The radius of the base of Cone B is one-half the radius of the base of Cone A.

Which is the greater quantity?

(a) The volume of Cone A

(b) The volume of Cone B

Possible Answers:

(a) is greater.

(b) is greater.

It is impossible to tell from the information given.

(a) and (b) are equal.

Correct answer:

(a) is greater.

Explanation:

Let r,h be the radius and height of Cone A, respectively. Then the radius and height of Cone B are $\frac{1}{2}r$ and , respectively.

(a) The volume of Cone A is $\frac{1}{3} \pi r^{2} h$ .

(b) The volume of Cone B is

$\frac{1}{3} \pi \left ( \frac{1}{2} r\right ) ^{2} \cdot 3 h = \frac{1}{3} \cdot \frac{1}{4} \cdot 3\cdot \pi r ^{2} h= \frac{1}{4} r ^{2} h$ .

Since $\frac{1}{4} r ^{2} h < \frac{1}{3} r ^{2} h$ , the cone in (a) has the greater volume.

Report an Error

Example Question #1 : Volume Of A Cone

The volume of a cone whose height is three times the radius of its base is one cubic yard. Give its radius in inches.

Possible Answers:

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

$\frac{ 36 \sqrt[3]{ \pi^{2} }}{ \pi}$

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

$\frac{ 12 \sqrt[3]{ \pi^{2} }}{ \pi}$

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

Correct answer:

$\frac{ 36 \sqrt[3]{ \pi^{2} }}{ \pi}$

Explanation:

The volume of a cone with base radius and height is

$V =\frac{1}{3} \pi r^{2}h$

The height is three times this, or . Therefore, the formula becomes

$V =\frac{1}{3} \pi r^{2} \cdot 3r$

$V = \pi r^{3}$

Set this volume equal to one and solve for :

$\pi r^{3} = 1$

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

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

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

This is the radius in yards; since the radius in inches is requested, multiply by 36.

$\frac{ \sqrt[3]{ \pi^{2} }}{ \pi} \times 36 = \frac{ 36 \sqrt[3]{ \pi^{2} }}{ \pi}$

Report an Error

Example Question #44 : Solid Geometry

The height of a given cylinder is one half the height of a given cone. The radii of their bases are equal.

Which of the following is the greater quantity?

(a) The volume of the cone

(b) The volume of the cylinder

Possible Answers:

It cannot be determined which of (a) and (b) is greater

(a) is the greater quantity

(a) and (b) are equal

(b) is the greater quantity

Correct answer:

(b) is the greater quantity

Explanation:

Call the radius of the base of the cone and the height of the cone. The cylinder will have bases of radius and height $\frac{1}{2} H$ .

In the formula for the volume of a cylinder, set r = R and $h = \frac{1}{2} H$ :

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

$V = \pi R ^{2} \cdot \frac{1}{2}H = \frac{1}{2} \pi R ^{2}H$

In the formula for the volume of a cone, set r = R and h = H :

$V = \frac{1}{3} \pi r^{2}h$

$V = \frac{1}{3} \pi R^{2}H$

$\frac{1}{2} > \frac{1}{3}$ , so

$\frac{1}{2} \pi R ^{2}H > \frac{1}{3} \pi R ^{2}H$ ,

meaning that the cylinder has the greater volume.

Report an Error

Example Question #1 : Cones

The radius of the base of a given cone is three times that of each base of a given cylinder. The heights of the cone and the cylinder are equal.

Which of the following is the greater quantity?

(a) The volume of the cone

(b) The volume of the cylinder

Possible Answers:

(a) is the greater quantity

(b) is the greater quantity

(a) and (b) are equal

It cannot be determined which of (a) and (b) is greater

Correct answer:

(a) is the greater quantity

Explanation:

If we let be the radius of each base of the cylinder, then is the radius of the base of the cone. We can let be their common height.

In the formula for the volume of a cylinder, set r = R and h = H :

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

$V = \pi R ^{2}H$

In the formula for the volume of a cone, set r = 3 R and h = H :

$V = \frac{1}{3} \pi r^{2}h$

$V = \frac{1}{3} \pi (3R)^{2}H = \frac{1}{3} \pi (9R^{2})H = 3\pi R ^{2}H$

3 > 1 , so $3\pi R ^{2}H > \pi R ^{2}H$ . The cone has the greater volume.

Report an Error

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 $\pi$ .

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

$V=\pi r^2h$

Possible Answers:

114.44m^3

1243.44 m^3

1432.23m^3

207.24m^3

396m^3

Correct answer:

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: V=(3.14)(6^2)(11) . Multiply all of these out and you get 1243.44m^3 . The units are cubed because volume is always cubed.

Report an Error

Example Question #343 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

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:

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

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

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

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

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

Correct answer:

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

Explanation:

The volume of a cylinder with base radius and height is

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

The diameter of this circle is ; its height is twice this, or . Therefore, the formula becomes

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

Set this volume equal to one and solve for :

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

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

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

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

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

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

Report an Error

Example Question #343 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

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

Possible Answers:

56250

86.25

22.5

28125.5

11.5

Correct answer:

11.5

Explanation:

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

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

For our values this is:

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

Solve for :

h = 11.5

Report an Error

Example Question #51 : Solid Geometry

What is the volume of a cylinder with a height of in. and a radius of in?

Possible Answers:

$300\pi$ in^3

$204\pi$ in^3

$180\pi$ in^3

$408\pi$ in^3

$720\pi$ in^3

Correct answer:

$720\pi$ in^3

Explanation:

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

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

For our values this is:

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

This is the volume of the cylinder.

Report an Error

View Tutors

Gabby
Certified Tutor

University of Oregon, Bachelor in Arts, Journalism. University of Oregon, Current Grad Student, Communication, General.

View Tutors

Ana
Certified Tutor

Grand Canyon University, Bachelors, Education. Grand Canyon University, Masters, Education.

View Tutors

Crystal
Certified Tutor

SUNY at Binghamton, Bachelor of Science, Mechanical Engineering.

All ISEE Upper Level Quantitative Resources

8 Diagnostic Tests 234 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

English Tutors in Los Angeles, Math Tutors in Atlanta, ACT Tutors in Phoenix, ISEE Tutors in San Francisco-Bay Area, Reading Tutors in Houston, French Tutors in Boston, Algebra Tutors in Phoenix, Spanish Tutors in Philadelphia, Biology Tutors in San Francisco-Bay Area, Algebra Tutors in San Francisco-Bay Area

Popular Courses & Classes

MCAT Courses & Classes in Chicago, Spanish Courses & Classes in San Diego, ACT Courses & Classes in Washington DC, LSAT Courses & Classes in Miami, SSAT Courses & Classes in Philadelphia, GMAT Courses & Classes in Atlanta, Spanish Courses & Classes in Philadelphia, SAT Courses & Classes in Philadelphia, LSAT Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Houston

Popular Test Prep

MCAT Test Prep in Denver, ACT Test Prep in Washington DC, SSAT Test Prep in San Francisco-Bay Area, SSAT Test Prep in Philadelphia, SSAT Test Prep in New York City, GMAT Test Prep in Atlanta, GMAT Test Prep in Phoenix, ISEE Test Prep in Phoenix, ACT Test Prep in San Francisco-Bay Area, MCAT Test Prep in Dallas Fort Worth

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

ISEE Upper Level Quantitative : Geometry

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

All ISEE Upper Level Quantitative Resources

Example Questions

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

Example Question #43 : Solid Geometry

Example Question #1 : Cones

Example Question #1 : Volume Of A Cone

Example Question #44 : Solid Geometry

Example Question #1 : Cones

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

Example Question #343 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #343 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #51 : Solid Geometry

All ISEE Upper Level Quantitative Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: