Call Now to Set Up Tutoring:

(888) 888-0446

ISEE Upper Level Quantitative : ISEE Upper Level (grades 9-12) Quantitative Reasoning

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 #14 : Squares

The sidelength of Square A is three-sevenths that of Square B. What is the ratio of the area of Square B to that of Square A?

Possible Answers:

$7\textrm{ to }3$

$49 \textrm{ to } 9$

$14\textrm{ to }3$

$9\textrm{ to }7$

None of the other answers give the correct ratio.

Correct answer:

$49 \textrm{ to } 9$

Explanation:

Since the ratio is the same regardless of the sidelengths, then for simplicity's sake, assume the sidelength of Square B is 7. The area of Square B is therefore the square of this, or 49.

Then the sidelength of Square A is three-sevenths of 7, or 3. Its area is the square of 3, or 9.

The ratio of the area of Square B to that of Square A is therefore 49 to 9.

Report an Error

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

Five squares have sidelengths one foot, two feet, three feet, four feet, and five feet.

Which is the greater quantity?

(A) The mean of their areas

(B) The median of their areas

Possible Answers:

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

(A) is greater

(B) is greater

(A) and (B) are equal

Correct answer:

(A) is greater

Explanation:

The areas of the squares are:

$1^{2} = 1$ square foot

$2^{2} = 4$ square feet

$3^{2} = 9$ square feet

$4^{2}= 16$ square feet

$5^{2}= 25$ square feet

Therefore, we are comparing the mean and the median of the set $\left \{ 1, 4, 9, 16, 25\right \}$ .

The mean of this set is the sum divided by 5:

$(1 + 4 + 9 + 16 + 25) \div 5 = 55 \div 5 = 11$

The median is the middle element after arrangement in ascending order, which is 9.

This makes (A), the mean, greater.

Report an Error

Example Question #6 : How To Find The Area Of A Square

Four squares have sidelengths one meter, 120 centimeters, 140 centimeters, and 140 centimeters. Which is the greater quantity?

(A) The mean of their areas

(B) The median of their areas

Possible Answers:

(A) and (B) are equal

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

(A) is greater

(B) is greater

Correct answer:

(B) is greater

Explanation:

The areas of the squares are:

$100 \cdot 100 = 10,000$ square centimeters (one meter being 100 centimeters)

$120 \cdot 120 = 14,400$ square centimeters

$140 \cdot 140 = 19,600$ square centimeters

The mean of these four areas is their sum divided by four:

$(10,000+14,400+19,600+19,600) \div 4$

$= 63,600 \div 4 =15,900$ square centimeters.

The median is the mean of the two middle values, or

$( 14,400+19,600 ) \div 2 = 34,000 \div 2 = 17,000$ square centimeters.

The median, (B), is greater.

Report an Error

Example Question #9 : How To Find The Area Of A Square

The perimeter of a square is 12x+16 . Give the area of the square in terms of .

Possible Answers:

$144x^{2} + 256$

$9x^{2}+24x + 16$

$144x^{2} +384x+ 256$

None of the other responses gives a correct answer.

$9x^{2} + 16$

Correct answer:

$9x^{2}+24x + 16$

Explanation:

The length of one side of a square is one fourth its perimeter. Since the perimeter of the square is 12x+16 , the length of one side is

$(12x+16) \div 4 = 3x+ 4$

The area of the square is the square of this sidelength, or

$( 3x+ 4)^{2}= (3x)^{2}+ 2 \cdot 3x \cdot 4 + 4 ^{2} = 9x^{2}+24x + 16$

Report an Error

Example Question #5 : How To Find The Area Of A Square

The sidelength of a square is $2^{x}$ . Give its area in terms of .

Possible Answers:

$2 ^{4 x}$

$2 ^{x+1}$

$2 ^{x+2}$

$2 ^{2 x}$

$2 ^{x+4}$

Correct answer:

$2 ^{2 x}$

Explanation:

The area of a square is the square of its sidelength. Therefore, square $2^{x}$ :

$\left (2^{x} \right ) ^{2} = 2 ^{2 x}$

Report an Error

Example Question #15 : Squares

A diagonal of a square has length $2^{x}$ . Give its area.

Possible Answers:

$2^{2x- \frac{1}{2}}$

$2^{2x+ \frac{1}{2}}$

$2^{2x+1}$

$2^{2x}$

$2^{2x-1}$

Correct answer:

$2^{2x-1}$

Explanation:

A square being a rhombus, its area can be determined by taking half the product of the lengths of its (congruent) diagonals:

$A = \frac{1}{2} \cdot 2^{x} \cdot 2^{x} = 2 ^{-1} \cdot 2^{x} \cdot 2^{x} = 2^{-1 + x + x } = 2^{2x-1}$

Report an Error

Example Question #223 : Geometry

The lengths of the sides of ten squares form an arithmetic sequence. One side of the smallest square measures sixty centimeters; one side of the second-smallest square measures one meter.

Give the area of the largest square, rounded to the nearest square meter.

Possible Answers:

20 square meters

18 square meters

24 square meters

16 square meters

22 square meters

Correct answer:

18 square meters

Explanation:

Let $a_{n}$ be the lengths of the sides of the squares in meters. $a_{1} = 60 \div 100 = 0.6$ and $a_{2} = 1$ , so their common difference is

d =1 - 0.6 = 0.4

The arithmetic sequence formula is

$a_{n} = a_{1} + (n-1)d$

The length of a side of the largest square - square 10 - can be found by substituting $a_{1} = 0.6, n= 10, d = 0.4$ :

$a_{10} =0.6+ (10-1) 0.4 = 0.6 + 9 (0.4) = 0.6 + 3.6 = 4.2$

The largest square has sides of length 4.2 meters, so its area is the square of this, or $A = 4.2^{2} = 17.64$ square meters.

Of the choices, 18 square meters is closest.

Report an Error

Example Question #21 : Squares

The areas of six squares form an arithmetic sequence. The smallest square has perimeter 16; the second smallest square has perimeter 20. Give the area of the largest of the six squares.

Possible Answers:

100

Correct answer:

Explanation:

The two smallest squares have perimeters 16 and 20, so their sidelengths are one fourth of these, or, respectively, 4 and 5. Their areas are the squares of these, or, respectively, 16 and 25. Therefore, in the arithmetic sequence,

$A_{1} = 16$

$A_{2} = 25$

and the common difference is d = 25 - 16 = 9 .

The area of the th smallest square is

$A_{n}= A_{1}+ (n-1)d$

Setting $A_{1} = 16, n = 6, d=9$ , the area of the largest (or sixth-smallest) square is

$A_{n}= 16+ (6-1)9 = 16 + 5 \cdot 9 = 16 + 45 = 61$

Report an Error

Example Question #21 : Quadrilaterals

Which is the greater quantity?

(a) The area of a square with sides of length meters

(b) The area of a square with perimeter 160 X centimeters

Possible Answers:

(b) is the greater quantity

(a) is the greater quantity

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

(a) and (b) are equal

Correct answer:

(a) is the greater quantity

Explanation:

A square with perimeter 160 X centimeters has sides of length one-fourth of this, or $160 X \div 4 = 40 X$ centimeters. Since one meter is equal to 100 centimeters, divide by 100 to get the equivalent in meters - this is

$40 X \div 100 = 0.4X$

meters.

The square in (b) has sidelength less than that of the square in (a), so its area is also less than that in (a).

Report an Error

Example Question #23 : Squares

On the coordinate plane, Square A has as one side a segment with its endpoints at the origin and at the point with coordinates (A, B) . Square B has as one side a segment with its endpoints at the origin and at the point with coordinates (-B, -A) . and are both positive numbers and B > A . Which is the greater quantity?

(a) The area of Square A

(b) The area of Square B

Possible Answers:

(a) and (b) are equal

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

(b) is the greater quantity

(a) is the greater quantity

Correct answer:

(a) and (b) are equal

Explanation:

The length of a segment with endpoints (0,0) and (A, B) can be found using the distance formula with $x_{1} = y_{1} = 0$ , $x_{2} = A$ , $y_{2} = B$ :

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$d = \sqrt{(A-0)^{2}+(B-0)^{2}}$

$d = \sqrt{A ^{2}+B^{2}}$

The length of a segment with endpoints (0,0) and (-B, -A) can be found using the distance formula with $x_{1} = y_{1} = 0$ , $x_{2} = -B$ , $y_{2} = -A$ :

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$d = \sqrt{(-B-0)^{2}+(-A-0)^{2}}$

$d = \sqrt{(-B )^{2}+(-A )^{2}}$

$d = \sqrt{B^{2}+A ^{2}}$

$d = \sqrt{A ^{2}+B^{2}}$

The sides are of equal length, so the squares have equal area. Note that the fact that B > A is irrelevant to the question.

Report an Error

View Tutors

Iffy
Certified Tutor

University of Maryland-Baltimore County, Bachelor of Science, Computer Engineering, General. University of Maryland-College P...

View Tutors

Robert
Certified Tutor

University of North Carolina at Chapel Hill, Bachelor in Arts, English. North Carolina Central University, Master of Arts, En...

View Tutors

Ana
Certified Tutor

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

All ISEE Upper Level Quantitative Resources

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

Popular Subjects

Statistics Tutors in San Diego, Math Tutors in Phoenix, SAT Tutors in Atlanta, SSAT Tutors in Los Angeles, ACT Tutors in Philadelphia, Computer Science Tutors in Miami, ISEE Tutors in Denver, SAT Tutors in Denver, Statistics Tutors in Houston, ISEE Tutors in Phoenix

Popular Courses & Classes

Spanish Courses & Classes in Los Angeles, SAT Courses & Classes in Miami, Spanish Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Miami, Spanish Courses & Classes in Seattle, SSAT Courses & Classes in Seattle, GRE Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in Chicago, SSAT Courses & Classes in Atlanta

Popular Test Prep

LSAT Test Prep in Miami, SAT Test Prep in Philadelphia, ACT Test Prep in Houston, GRE Test Prep in Atlanta, ISEE Test Prep in Boston, ACT Test Prep in Denver, MCAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in Los Angeles, SSAT Test Prep in New York City, GMAT Test Prep in Seattle

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 : ISEE Upper Level (grades 9-12) Quantitative Reasoning

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

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #14 : Squares

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

Example Question #6 : How To Find The Area Of A Square

Example Question #9 : How To Find The Area Of A Square

Example Question #5 : How To Find The Area Of A Square

Example Question #15 : Squares

Example Question #223 : Geometry

Example Question #21 : Squares

Example Question #21 : Quadrilaterals

Example Question #23 : Squares

All ISEE Upper Level Quantitative Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: