GMAT Math : GMAT Quantitative Reasoning

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

22 Diagnostic Tests 693 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #5 : Prisms

A box with dimensions 8 inches, 10 inches, and 5 inches needs to be gift wrapped. Gift wrapping is priced at $0.10 per square inch of surface of a box. How much will it cost to wrap the gift?

Possible Answers:

$\$17$

$\$20$

$\$40$

$\$34$

Correct answer:

$\$34$

Explanation:

Find the surface area of the box by summing the area of all six faces: two 8 by 10, two 8 by 5, two 10 by 5.

$2(80)+2(40)+2(50)=340\ in^{2}$

Since the price is $.10 for each square inch,

$Total\ cost=340(.10)=\$34$

Report an Error

Example Question #1 : Calculating The Surface Area Of A Prism

The sum of the length, the width, and the height of a rectangular prism is one meter. The length of the prism is sixteen centimeters greater than its width, which is three times its height. What is the surface area of this prism?

Possible Answers:

$25,872 \textrm{ cm}^{2}$

$7,672 \textrm{ cm}^{2}$

$22,464 \textrm{ cm}^{2}$

$6,104 \textrm{ cm}^{2}$

$5,856 \textrm{ cm}^{2}$

Correct answer:

$5,856 \textrm{ cm}^{2}$

Explanation:

Let be the height of the prism. Then the width is , and the length is 3h+ 16 . Since the sum of the three dimensions is one meter, or 100 centimeters, we solve for in this equation:

h + 3h + (3h+16) = 100

7h+16 = 100

7h+16-16 = 100-16

7h = 84

$7h \div 7 = 84 \div 7$

h = 12

The height of the prism is 12 cm; the width is three times this, or 36 cm; the length is sixteen centimeters greater than the width, which is 52 cm.

Set l= 52,w=36,h=12 in the formula for the surface area of a rectangular prism:

A = 2lw + 2wh + 2lh

$= 2 \cdot 52 \cdot 36 + 2 \cdot 36 \cdot 12 + 2 \cdot 52 \cdot 12$

= 3,744 + 864 + 1,248

= 5,856 square centimeters

Report an Error

Example Question #771 : Gmat Quantitative Reasoning

The volume of the rectangular solid above is 120. The area of ABFE is 20 and the area of ABCD is 30. What is the area of BFGC?

Possible Answers:

Correct answer:

Explanation:

We can set the lengths of three sides to be , , , respectively. The volume is 120 means that xyz=120 . Also we know the areas of two sides, so we can use to represent the area of 20 and to represent the area of 30. Now the question is to figure out . Then we can use

xy * xz * yz = x^2 * y^2 * z^2 =(xyz)^2 to solve .

20 * 30 * yz = 120^2

yz=24

Report an Error

Example Question #5 : Calculating The Surface Area Of A Prism

A right prism has as its bases two right triangles, each of whose legs have lengths 12 and 16. The height of the prism is half the perimeter of a base. Give the surface area of the prism.

Possible Answers:

1,152

2,304

1,344

768

584

Correct answer:

1,344

Explanation:

The area of a right triangle is equal to half the product of its legs, so each base has area

$\frac{1}{2} \cdot 12 \cdot 16 = 96$

The measure of the hypotenuse of each base is determined using the Pythagorean Theorem:

$c = \sqrt{12^{2}+16^{2}}= \sqrt{144+256} = \sqrt{400}= 20$

Therefore, the perimeter of each base is

P=12+16+20 = 48 ,

and the height of the prism is half this, or $h = \frac{1}{2} \cdot 48 = 24$ .

The lateral area of the prism is the product of its height and the perimeter of a base; this is

$L= Ph= 24 \cdot 48 = 1,152$

The surface area is the sum of the lateral area and the two base areas, or

S= L+B+B =1,152+96+96 =1,344 .

Report an Error

Example Question #11 : Prisms

A right prism has as its bases two isosceles right triangles, each of whose hypotenuse has length 10. The height of the prism is the length of one leg of a base. Give the surface area of the prism.

Possible Answers:

$150+ 50 \sqrt{2}$

250

$150\sqrt{2}$

$200\sqrt{2}$

$200+ 50 \sqrt{2}$

Correct answer:

$150+ 50 \sqrt{2}$

Explanation:

By the 45-45-90 Theorem, dividing the length of the hypotenuse of an isosceles right triangle by $\sqrt{2}$ yields the length of one leg; therefore, the length of one leg of each base is

$\frac{10}{\sqrt{2}} = \frac{10 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{10 \sqrt{2}}{2} = 5\sqrt{2}$ .

The area of a right triangle is half the product of its legs, so the area of each base is

$B = \frac{1}{2} \cdot 5 \sqrt{2} \cdot 5 \sqrt{2} = \frac{1}{2} \cdot 25 \cdot 2 = 25$

The perimeter of each base is the sum of its sides, which here is

$P = 5 \sqrt{2}+ 5 \sqrt{2}+ 10 = 10 + 10 \sqrt{2}$

The height of the prism is the length of one leg of a base, which is $h = 5\sqrt{2}$ .

The lateral area of the prism is equal to the product of the height of the prism and the perimeter of a base, so

L = Ph

$= \left (10+10 \sqrt{2} \right ) \cdot 5 \sqrt{2}$

$= 10 \cdot 5 \sqrt{2} +10 \sqrt{2} \cdot 5 \sqrt{2}$

$= 50 \sqrt{2} +10\cdot 5 \cdot 2$

$= 100+ 50 \sqrt{2}$

The surface area is the sum of the lateral area and the areas of the bases:

$S=L+B+B= 100+ 50 \sqrt{2}+25+25 = 150+ 50 \sqrt{2}$

Report an Error

Example Question #7 : Calculating The Surface Area Of A Prism

A right prism has as its bases two equilateral triangles, each of whose sides has length 6. The height of the prism is three times the perimeter of a base. Give the surface area of the prism.

Possible Answers:

$972 + 9 \sqrt{3}$

$486 \sqrt{3}$

972

$972 + 18 \sqrt{3}$

$486\sqrt{2}$

Correct answer:

$972 + 18 \sqrt{3}$

Explanation:

The area of each equilateral triangle base can be determined by setting s = 6 in the formula

$B= \frac{s^{2}\sqrt{3}}{4} = \frac{6^{2}\sqrt{3}}{4} = \frac{36\sqrt{3}}{4} = 9\sqrt{3}$

The perimeter of each base is $P=6 \cdot 3 = 18$ , and the height of the prism is three times this, or $h = 18 \cdot 3 = 54$ . The lateral area of a prism is equal to the perimeter of a base multiplied by the height, so

$L =Ph= 18 \cdot 54 = 972$

Add this to the areas of two bases - the surface area is

$S= L+ B + B= 972 + 9\sqrt{3}+ 9\sqrt{3}= 972 + 18 \sqrt{3}$ .

Report an Error

Example Question #4 : Calculating The Surface Area Of A Prism

A right prism has as its bases two isosceles right triangles, each of whose legs has length 16. The height of the prism is the length of the hypotenuse of a base. Give the surface area of the prism.

Possible Answers:

$768 + 512 \sqrt{2}$

$512+ 512 \sqrt{2}$

$384+ 512 \sqrt{2}$

$640+ 512 \sqrt{2}$

$256+ 512 \sqrt{2}$

Correct answer:

$768 + 512 \sqrt{2}$

Explanation:

By the 45-45-90 Theorem, multiplying the length of a leg of an isosceles right triangle by $\sqrt{2}$ yields the length of its hypotenuse; therefore, the length of the hypotenuse of each base is $16 \sqrt{2}$ .

The area of a right triangle is half the product of its legs, so the area of each base is

$B = \frac{1}{2} \cdot 16 \cdot 16 =128$

The perimeter of each base is the sum of its sides, which here is

$P =16+16+16 \sqrt{2} = 32 + 16 \sqrt{2}$

The height of the prism is the length of the hypotenuse of the base, which is $h = 16\sqrt{2}$ .

The lateral area of the prism is equal to the product of the height of the prism and the perimeter of a base, so

L = Ph

$= \left ( 32 + 16 \sqrt{2}\right ) \cdot 16 \sqrt{2}$

$= 32 \cdot16 \sqrt{2} +16 \sqrt{2} \cdot 16 \sqrt{2}$

$= 32 \cdot16 \sqrt{2} +16 \cdot 16 \cdot \sqrt{2} \cdot \sqrt{2}$

$= 512 \sqrt{2} +16 \cdot 16 \cdot 2$

$= 512+ 512 \sqrt{2}$

The surface area is the sum of the lateral area and the areas of the bases:

$S=L+B+B= 512+ 512 \sqrt{2} + 128 +128 = 768 + 512 \sqrt{2}$

Report an Error

Example Question #774 : Problem Solving Questions

A right prism has as its bases two right triangles, each of which has a hypotenuse of length 20 and a leg of length 10. The height of the prism is equal to the length of the longer leg of a base. Give the surface area of the prism.

Possible Answers:

500

$500+300 \sqrt{3}$

600

$300+400 \sqrt{3}$

$400+300 \sqrt{3}$

Correct answer:

$300+400 \sqrt{3}$

Explanation:

A right triangle with one leg half the length of the hypotenuse is a 30-60-90 triangle. Its other leg has measure $\sqrt{3}$ times the length of the first leg, which in the case of each base is $10 \sqrt{3}$ .

The area of each base is half the product of the legs, or

$B = \frac{1}{2} \cdot 10 \cdot 10 \sqrt{3} = 50 \sqrt{3}$ .

The perimeter of each base is

$P = 20+10+10 \sqrt {3} = 30 + 10 \sqrt {3}$

and the height of the prism is equal to the length of the longer leg, or

$h= 10 \sqrt{3}$ .

The lateral area of the prism is equal to the product of the height of the prism and the perimeter of a base, so

L = Ph

$=\left ( 30 + 10 \sqrt {3} \right ) \cdot 10 \sqrt {3}$

$= 30 \cdot 10 \sqrt {3}+ 10 \sqrt {3}\cdot 10 \sqrt {3}$

$= 30 \cdot 10 \sqrt {3}+ 10 \cdot 10\cdot \sqrt {3}\cdot \sqrt {3}$

$= 30 \cdot 10 \sqrt {3}+ 10 \cdot 10\cdot 3$

$= 300+ 30 0 \sqrt {3}$

The surface area is the sum of the lateral area and the areas of the bases:

$S=L+B+B= 300+300 \sqrt{3} + 50 \sqrt{3} + 50 \sqrt{3}= 300+400 \sqrt{3}$

Report an Error

Example Question #10 : Calculating The Surface Area Of A Prism

The length and width of a rectangular solid are 18 and 20; its volume is 5,400. Calculate its surface area.

Possible Answers:

1,620

2,160

1,800

1,898

1,860

Correct answer:

1,860

Explanation:

The volume of a rectangular solid is the product of its length, height, and width. Since L = 18 , W = 20 , and H = 5,400 ,

LWH = V

$18 \cdot 20 \cdot H =5,400$

360H = 5,400

H = 15

The surface area of a rectangular solid is

S = 2LH + 2LW + 2WH

$= 2 \cdot 18 \cdot 15 + 2 \cdot 18 \cdot 20 + 2 \cdot 20\cdot 15$

= 540+720 + 600

= 1,860

Report an Error

Example Question #775 : Problem Solving Questions

The length of a rectangular prism is twice its width and five times its height. If is the width, give the surface area in terms of .

Possible Answers:

$\frac{13}{5}W^{2}$

$\frac{32}{5}W^{2}$

$\frac{13}{10}W^{2}$

$16W^{2}$

$\frac{16}{5}W^{2}$

Correct answer:

$\frac{32}{5}W^{2}$

Explanation:

If we let be the width, then, since the length is twice this,

L = 2W .

Since the length is five times the height, the height is one fifth the length, so $H = \frac{1}{5}L = \frac{1}{5} \cdot 2W = \frac{2}{5} W$ .

The surface area of a rectangular solid is

S = 2LH + 2LW + 2WH .

which can be rewritten as

$S = 2 \cdot 2W \cdot \frac{2}{5} W + 2 \cdot 2W \cdot W + 2W \cdot \frac{2}{5} W$

$S = \frac{8}{5} W^{2} + 4W^{2} + \frac{4}{5} W^{2} = \frac{32}{5}W^{2}$

Report an Error

All GMAT Math Resources

22 Diagnostic Tests 693 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Computer Science Tutors in New York City, SSAT Tutors in Los Angeles, Statistics Tutors in Miami, Reading Tutors in Denver, GMAT Tutors in Houston, MCAT Tutors in Denver, Statistics Tutors in Phoenix, English Tutors in Washington DC, GRE Tutors in Los Angeles, GMAT Tutors in Denver

Popular Courses & Classes

ISEE Courses & Classes in Chicago, SSAT Courses & Classes in Miami, LSAT Courses & Classes in Los Angeles, SAT Courses & Classes in Denver, GRE Courses & Classes in Philadelphia, ISEE Courses & Classes in Boston, ACT Courses & Classes in Boston, LSAT Courses & Classes in Atlanta, MCAT Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in Miami

Popular Test Prep

GMAT Test Prep in Phoenix, SSAT Test Prep in Philadelphia, SAT Test Prep in Boston, GMAT Test Prep in San Francisco-Bay Area, ISEE Test Prep in Boston, ACT Test Prep in Boston, ISEE Test Prep in Atlanta, MCAT Test Prep in Los Angeles, LSAT Test Prep in Seattle, LSAT 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)

Email address:
Your name:
Feedback:

GMAT Math : GMAT Quantitative Reasoning

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #5 : Prisms

Example Question #1 : Calculating The Surface Area Of A Prism

Example Question #771 : Gmat Quantitative Reasoning

Example Question #5 : Calculating The Surface Area Of A Prism

Example Question #11 : Prisms

Example Question #7 : Calculating The Surface Area Of A Prism

Example Question #4 : Calculating The Surface Area Of A Prism

Example Question #774 : Problem Solving Questions

Example Question #10 : Calculating The Surface Area Of A Prism

Example Question #775 : Problem Solving Questions

All GMAT Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details