GMAT Math : GMAT Quantitative Reasoning

Study concepts, example questions & explanations for GMAT Math

Create An Account

All GMAT Math Resources

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

Example Questions

Example Question #461 : Problem Solving Questions

Trapezoid

Refer to the above Trapezoid ABCD . There exists Trapezoid EFGH such that

Trapezoid $ABCD \sim$ Trapezoid EFGH , and $\overline{EF}$ has length 66.

To the nearest whole, give the area of Trapezoid EFGH .

Possible Answers:

1,089

715

613

1,271

933

Correct answer:

1,271

Explanation:

The similarity ratio of the trapezoids is the ratio of the length of one side to that of the corresponding side of the other. For these trapezoids, we take the ratio of the lengths of corresponding sides $\overline{EF}$ and $\overline{AB}$ :

$\frac{EF}{AB} = \frac{66}{12} = \frac{11}{2}$ .

The ratio of the areas is the square of this, or

$\left ( \frac{11}{2} \right )^{2} = \frac{121}{4}$ .

The area of Trapezoid ABCD is one half the product of the height and the sum of bases and :

$A _{1} = \frac{1}{2} \cdot BC \cdot (AB + CD)$

$= \frac{1}{2} \cdot 3 \cdot (12+ 16)$

$= \frac{1}{2} \cdot 3 \cdot 28$

= 42

Multiply this by the area ratio:

$A _{2} = A _{1} \cdot \frac{121}{4}$

$A _{2} = 42 \cdot \frac{121}{4} = 1,270\frac{1}{2}$ .

The correct response is 1,271.

Report an Error

Example Question #5 : Calculating Whether Quadrilaterals Are Similar

Rhombus $RHOM \sim$ Rhombus SIPN . RO = 15 ; Rhombus RHOM has area 90; Rhombus SIPN has area 360. What is the length of diagonal $\overline{IN}$ ?

Possible Answers:

Correct answer:

Explanation:

The area of a rhombus is half the product of the lengths of its diagonals, so we can take advantage of this to find $\overline{HM}$ from the known measurements of Rhombus RHOM :

$A_{1} = \frac{1}{2} (RO)(HM)$

$90= \frac{1}{2} \cdot 15 \cdot HM$

$90= \frac{15}{2} \cdot HM$

$\frac{2} {15}\cdot 90= \frac{2} {15}\cdot \frac{15}{2} \cdot HM$

HM = 12

The ratio of the area of Rhombus SIPN to that of Rhombus RHOM is 4, so their similarity ratio is the square root of this, or 2. We can calculate now:

$\frac{IN}{HM} = 2$

$\frac{IN}{12} = 2$

$\frac{IN}{12}\cdot 12 = 2 \cdot 12$

IN = 24

Report an Error

Example Question #6 : Calculating Whether Quadrilaterals Are Similar

In Quadrilateral KITE , $\overline{KI} \cong \overline{KE}$ , $\overline{TI} \cong \overline{TE}$ , $m \angle K = 60^{\circ }$ , $\angle T$ is a right angle, and diagonal $\overline{I E}$ has length 24.

There exists Quadrilateral LJUF such that Quadrilateral $KITE \sim$ Quadrilateral LJUF , and JU = 36 .

Give the length of $\overline{LJ}$ .

Possible Answers:

$18 \sqrt{2}$

$18 \sqrt{3}$

$36 \sqrt{2}$

$36 \sqrt{3}$

Correct answer:

$36 \sqrt{2}$

Explanation:

The Quadrilateral KITE with its diagonals is shown below. We call the point of intersection :

Kite

The diagonals of a quadrilateral with two pairs of adjacent congruent sides - a kite - are perpendicular. $\overline{KT}$ , the diagonal that connects the common vertices of the pairs of adjacent sides, bisects the other diagonal, making the midpoint of $\overline{IE}$ . Therefore, $IX= \frac{1}{2} \cdot IE = \frac{1}{2} \cdot 24 = 12$ .

$\overline{KT}$ also bisects the $60 ^{\circ }$ and $90^{\circ }$ angles of the kite, so the result is that two 30-60-90 triangles, $\bigtriangleup KXE$ and $\bigtriangleup KXI$ , and two 45-45-90 triangles , $\bigtriangleup TXE$ and $\bigtriangleup TXI$ , are formed; also, $\bigtriangleup TIE$ , being isosceles, is a 45-45-90 triangle.

Examine $\bigtriangleup KXI$ . Since its short leg $\overline{IX}$ has length 12, by the 30-60-90 Theorem, its hypotenuse, $\overline{KI}$ , has twice this length, or 24.

Examine $\bigtriangleup TXI$ . Since a leg $\overline{IX}$ has length 12, by the 45-45-90 Theorem, its hypotenuse, $\overline{IT}$ , has length $\sqrt{2}$ times this, or $12\sqrt{2}$ .

Since by similarity, corresponding sides are in proportion,

$\frac{LJ}{KI}= \frac{JU}{IT}$

$\frac{LJ}{24}= \frac{36}{12\sqrt{2}}$

$\frac{LJ}{24} \cdot 24 = \frac{36}{12\sqrt{2}} \cdot 24$

$LJ = \frac{72}{ \sqrt{2}} = \frac{72\cdot \sqrt{2}}{ \sqrt{2}\cdot \sqrt{2}} = \frac{72 \sqrt{2}}{2}= 36 \sqrt{2}$

Report an Error

Example Question #462 : Problem Solving Questions

Trapezoid

Refer to the above Trapezoid ABCD . There exists Trapezoid EFGH such that

Trapezoid $ABCD \sim$ Trapezoid EFGH , and $\overline{EH}$ has length 60.

Give the length of $\overline{EF}$ .

Possible Answers:

192

320

180

144

240

Correct answer:

144

Explanation:

Construct the perpendicular segment from to $\overline{CD}$ and let be its point of intersection with $\overline{CD}$ . By construction, the trapezoid is divided into Rectangle ABCZ and right triangle $\bigtriangleup AZD$ . Since opposite sides of a rectangle are congruent, AZ= BC = 3 and CZ = AB = 12 ; as a consequence of the latter, ZD = CD - CZ = 16-12 = 4 . By the Pythagrean Theorem, the length of the hypotenuse $\overline{AD}$ of right triangle $\bigtriangleup AZD$ can be calculated from the length of legs $\overline{AZ}$ and $\overline{ZD}$ :

$AD = \sqrt{(AZ)^{2}+(DZ)^{2}} = \sqrt{3^{2}+4^{2}} =\sqrt{9+16} = \sqrt{25} = 5$

The figure, with the segment and the calculated measurements, is below.

Trapezoid

Since Trapezoid $ABCD \sim$ Trapezoid EFGH , by proportionality of corresponding sides of similar figures:

$\frac{EF}{AB}=\frac{EH}{AD}$

$\frac{EF}{12}=\frac{60}{5} =12$

$\frac{EF}{12} \cdot 12 =12 \cdot 12$

EF = 144

Report an Error

Example Question #463 : Problem Solving Questions

Rhombus $RHOM \sim$ Rhombus SIPN . Rhombus RHOM has perimeter 80; Rhombus SIPN has perimeter 180; HM= 32 .

Find the length of diagonal $\overline{SP}$ .

Possible Answers:

$45 \sqrt{2}$

$45 \sqrt{3}$

Correct answer:

Explanation:

A rhombus has four sides the same length, so each side of Rhombus RHOM has length one fourth of 80, or 20; each side of Rhombus SIPN has length one fourth of 180, or 45.

The diagonals of a rhombus are each other's perpendicular bisectors, so, if we let be the point of intersection of the diagonals of Rhombus RHOM , $\overline{RO}$ and $\overline{HM}$ , we form four congruent right triangles.

We will examine $\bigtriangleup RXM$ . RM = 20 ; and, $XM = \frac{1}{2} \cdot HM = \frac{1}{2} \cdot 32 = 16$ .

By the Pythagorean Theorem,

$RX = \sqrt{(RM)^{2}-(RX)^{2}}$

$= \sqrt{20^{2}-16^{2}}$

$= \sqrt{400-256}$

$= \sqrt{144}$

= 12

and $RO = 2 \cdot 12 = 24$

Since corresponding sides of similar figures are in proportion, so are corresponding diagonals. Therefore,

$\frac{SP}{RO} = \frac{SN}{RM}$

$\frac{SP}{24} = \frac{45}{20}$

$\frac{SP}{24} \cdot 24= \frac{45}{20} \cdot 24$

SP=54

Report an Error

Example Question #9 : Calculating Whether Quadrilaterals Are Similar

Rhombus $RHOM \sim$ Rhombus SIPN . $m \angle R = 60^{\circ }$ ; HM = 12 ; SP = 72 .

Give the ratio of the area of Rhombus SIPN to that of Rhombus RHOM .

Possible Answers:

18:1

6:1

27:1

9:1

12 :1

Correct answer:

12 :1

Explanation:

The angles of the rhombus measure:

$m \angle R = 60 ^{\circ }$ ;

$m \angle O = 60 ^{\circ }$ , since opposite angles of a rhombus, as in any other parallelogram, are congruent;

$m \angle H = 120^{\circ }$ and $m \angle M = 120^{\circ }$ , since consecutive angles of a rhombus are supplementary (the sum of their degree measures is 180).

The diagonals of a rhombus are each other's perpendicular bisectors as well as the bisectors of the angles, so $\overline{HM}$ and $\overline{RO}$ , whose point of intersection we will call , divide Rhombus RHOM into four 30-60-90 triangles. If we examine $\bigtriangleup RXH$ , we see that its short leg $\overline{HX}$ has length half that of $\overline{HM}$ ; HM = 12 , so HX = 6 . By the 30-60-90 Triangle Theorem, long leg $\overline{RX}$ has length $\sqrt{3}$ this, or $6 \sqrt{3}$ , and the diagonal $\overline{RO}$ has measure twice this, or $12\sqrt{3}$ .

The ratio of the lengths of corresponding diagonals of the rhombuses is the same as the similarity ratios of the sides, so the similarity ratio of Rhombus SIPN to Rhombus RHOM is

$\frac{SP }{RO} = \frac{72}{12\sqrt{3}}= \frac{6}{\sqrt{3}}$

The ratio of the areas is the square of the similarity ratio, which is

$\left ( \frac{6}{\sqrt{3}} \right )^{2}= \frac{36}{3} = \frac{12}{1}$

That is, 12 to 1.

Report an Error

Example Question #4 : Calculating Whether Quadrilaterals Are Similar

In Quadrilateral KITE , $\overline{KI} \cong \overline{KE}$ , $\overline{TI} \cong \overline{TE}$ , $m \angle K = 60^{\circ }$ , and $\angle T$ is a right angle; KE = 16 .

There exists Quadrilateral LJUF such that Quadrilateral $KITE \sim$ Quadrilateral LJUF , and UF = 24 .

Give the length of $\overline{LF}$ .

Possible Answers:

$12\sqrt{2}$

$12\sqrt{3}$

$24 \sqrt{3}$

$24 \sqrt{2}$

Correct answer:

$24 \sqrt{2}$

Explanation:

The Quadrilateral KITE with its diagonals is shown below. We call the point of intersection :

Kite

The diagonals of a quadrilateral with two pairs of adjacent congruent sides - a kite - are perpendicular; also, $\overline{KT}$ bisects the $60 ^{\circ }$ and $90^{\circ }$ angles of the kite. Consequently, two congruent 30-60-90 triangles, $\bigtriangleup KXE$ and $\bigtriangleup KXI$ , and two congruent 45-45-90 triangles , $\bigtriangleup TXE$ and $\bigtriangleup TXI$ , are formed; also, $\bigtriangleup TIE$ , being an isosceles right triangle, is a 45-45-90 triangle. $\overline{KE}$ , the hypotenuse of $\bigtriangleup KXE$ , has length 16, so by the 30-60-90 Triangle Theorem, its shorter leg $\overline{XE}$ has length half this, or 8. Also, $\overline{XE}$ is a leg of $\bigtriangleup TXE$ , so by the 45-45-90 Theorem, the hypotenuse $\overline{TE}$ has length $\sqrt{2}$ times this, or $8\sqrt{2}$ .

Corresponding sides of similar quadrilaterals are in proportion, so

$\frac{LF}{KE} = \frac{UF}{TE}$

$\frac{LF}{16} = \frac{24}{8 \sqrt{2}}$

$\frac{LF}{16} \cdot 16 = \frac{24}{8 \sqrt{2}} \cdot 16$

$LF= \frac{48}{ \sqrt{2}} = \frac{48 \cdot \sqrt{2}}{ \sqrt{2} \cdot \sqrt{2}}= \frac{48 \sqrt{2}}{2} = 24 \sqrt{2}$

Report an Error

Example Question #11 : Calculating Whether Quadrilaterals Are Similar

In Quadrilateral KITE , $\overline{KI} \cong \overline{KE}$ , $\overline{TI} \cong \overline{TE}$ , $m \angle K = 60^{\circ }$ , $\angle T$ is a right angle.

There exists Quadrilateral LJUF such that Quadrilateral $KITE \sim$ Quadrilateral LJUF , and JU = KI .

Which of the following is true about the areas of the two quadrilaterals?

Possible Answers:

Quadrilateral KITE and Quadrilateral LJUF have the same area.

Quadrilateral KITE has area twice that of Quadrilateral LJUF .

Quadrilateral LJUF has area three times that of Quadrilateral KITE .

Quadrilateral LJUF has area twice that of Quadrilateral KITE .

Quadrilateral KITE has area three times that of Quadrilateral LJUF .

Correct answer:

Quadrilateral LJUF has area twice that of Quadrilateral KITE .

Explanation:

We will assume that $\overline{KI}$ and $\overline{JU}$ have common measure 1 for the sake of simplcity; this reasoning is independent of the actual measure of $\overline{KI}$ .

The Quadrilateral KITE with its diagonals is shown below. We call the point of intersection :

Kite

The diagonals of a quadrilateral with two pairs of adjacent congruent sides - a kite - are perpendicular; also, $\overline{KT}$ bisects the $60 ^{\circ }$ and $90^{\circ }$ angles of the kite. Consequently, $\bigtriangleup KXI$ is a 30-60-90 triangle and $\bigtriangleup TXI$ is a 45-45-90 triangle. By the 30-60-90 Theorem, since $\overline{ IX}$ and $\overline{KI}$ are the short leg and hypotenuse of $\bigtriangleup KXI$ ,

$IX = \frac{1}{2} \cdot KI = \frac{1}{2} \cdot 1 = \frac{1}{2}$ .

By the 45-45-90 Theorem, since $\overline{ IX}$ and $\overline{IT}$ are a leg and the hypotenuse of $\bigtriangleup TXI$ ,

$IT = IX \cdot \sqrt{2} =\frac{1}{2} \sqrt{2} = \frac{ \sqrt{2} }{2}$

The similarity ratio of Quadrilateral LJUF to Quadrilateral KITE can be found by finding the ratio of the length of side $\overline{JU}$ to corresponding side $\overline{IT}$ :

$\frac{JU}{IT}= \frac{1}{ \frac{\sqrt{2}}{2} } = \frac {2}{\sqrt{2}}$

The ratio of the areas is the square of the similarity ratio:

$\left ( \frac {2}{\sqrt{2}} \right )^{2}= \frac{2^{2}}{\left ( \sqrt{2} \right )^{2}} = \frac{4}{2}=2$

The correct choice is that Quadrilateral LJUF has area twice that of Quadrilateral KITE .

Report an Error

Example Question #231 : Geometry

The area of a trapezoid is 1,600 square centimeters. Its height is 24 centimeters, and one base is one decimeter longer than the other. What is the length of the longer base?

Possible Answers:

$61 \frac{2}{3}\textrm{ cm}$

$71 \frac{2}{3}\textrm{ cm}$

$716 \frac{2}{3}\textrm{ cm}$

$616 \frac{2}{3}\textrm{ cm}$

$66 \frac{2}{3}\textrm{ cm}$

Correct answer:

$71 \frac{2}{3}\textrm{ cm}$

Explanation:

Let be the longer base. Then, since this is ten centimeters (= 1 decimeter) longer than the shorter, the shorter base has length B - 10 . Substitute A = 1,600, b = B - 10, h = 24 in the formula for the area of a trapezoid, and solve for :

$\frac{1}{2} (B + b)h = A$

$\frac{1}{2} (B + B - 10) \cdot 24= 1,600$

12 (2B - 10) = 1,600

24B - 120 = 1,600

24B = 1,720

$24B \div 24 = 1,720 \div 24$

$B = 71 \frac{2}{3}$ centimeters

Report an Error

Example Question #1 : Calculating The Length Of The Side Of A Quadrilateral

The area of a trapezoid is 6,000 square inches. Its height is twice the length of its ionger base, which is three times the length of its shorter base. What is the height of the trapezoid?

Possible Answers:

$60 \sqrt{5 } \textrm{ in}$

$10 \sqrt{5 } \textrm{ in}$

$60 \textrm{ in}$

$30 \textrm{ in}$

$30 \sqrt{5 } \textrm{ in}$

Correct answer:

$60 \sqrt{5 } \textrm{ in}$

Explanation:

Let be the length of the shorter base. Then is the length of its longer base, and 2 (3b)= 6b is the height. Substitute B = 3b, h = 6b , A = 6,666 in the area formula:

$\frac{1}{2} ( B + b) h = A$

$\frac{1}{2} ( 3b + b) \cdot 6b= 6,000$

$\frac{1}{2} \cdot 4b \cdot 6b= 6,000$

$12b^{2}= 6,000$

$12b^{2} \div 12 = 6,000 \div 12$

$b^{2} = 500$

$b = \sqrt{500} = \sqrt{100}\cdot \sqrt{5 } =10 \sqrt{5 }$

The height is six times this, or $h = 6 \cdot 10 \sqrt{5 } = 60 \sqrt{5 }$ inches.

Report an Error

All GMAT Math Resources

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

Popular Subjects

Spanish Tutors in San Diego, ISEE Tutors in Atlanta, Reading Tutors in Denver, ACT Tutors in Chicago, English Tutors in Los Angeles, Spanish Tutors in Seattle, Physics Tutors in Atlanta, Math Tutors in Phoenix, Calculus Tutors in San Francisco-Bay Area, Chemistry Tutors in Los Angeles

Popular Courses & Classes

ISEE Courses & Classes in Washington DC, LSAT Courses & Classes in Houston, ISEE Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in Denver, MCAT Courses & Classes in Houston, Spanish Courses & Classes in Miami, SAT Courses & Classes in New York City, SSAT Courses & Classes in Los Angeles, Spanish Courses & Classes in Seattle, GRE Courses & Classes in Dallas Fort Worth

Popular Test Prep

ACT Test Prep in San Diego, GMAT Test Prep in San Diego, GRE Test Prep in Boston, SSAT Test Prep in Denver, MCAT Test Prep in Philadelphia, MCAT Test Prep in Phoenix, MCAT Test Prep in San Diego, ISEE Test Prep in Houston, SSAT Test Prep in Philadelphia, SSAT Test Prep in Boston

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 #461 : Problem Solving Questions

Example Question #5 : Calculating Whether Quadrilaterals Are Similar

Example Question #6 : Calculating Whether Quadrilaterals Are Similar

Example Question #462 : Problem Solving Questions

Example Question #463 : Problem Solving Questions

Example Question #9 : Calculating Whether Quadrilaterals Are Similar

Example Question #4 : Calculating Whether Quadrilaterals Are Similar

Example Question #11 : Calculating Whether Quadrilaterals Are Similar

Example Question #231 : Geometry

Example Question #1 : Calculating The Length Of The Side Of A Quadrilateral

All GMAT Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details