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 #74 : Quadrilaterals
Give the area of the above rectangle in terms of 
The area of a rectangle is equal to the product of its length and height, which here are 5 and 
Example Question #3 : Rectangles
A rectangle is two feet shorter than twice its width; its perimeter is six yards. Give its area in square inches.
The length of the rectangle is two feet, or 24 inches, shorter than twice the width, so, if 
Six yards, the perimeter of the rectangle, is equal to 
The length and width are 64 inches and 44 inches; the area is their product, which is
Example Question #11 : How To Find The Area Of A Rectangle
The perimeter of a rectangle is 210 inches. The width of the rectangle is 40% of its length. What is the area of the rectangle?
If the width of the rectangle is 40% of the length, then
The perimeter of the rectangle is:
The perimeter is 210 inches, so we can solve for the length:
The length and width of the rectangle are 75 and 30 inches; the area is their product, or
Example Question #12 : How To Find The Area Of A Rectangle
Rectangle A has length 
(A) The area of Rectangle A
(B) The area of Rectangle B
(A) and (B) are equal
It is impossible to determine which is greater from the information given
(A) is greater
(B) is greater
(A) is greater
This might be easier to solve if you set 
Then the dimensions of Rectangle A are 
The dimensions of Rectangle B are 
Example Question #13 : How To Find The Area Of A Rectangle
In Rectangle 
The area of a rectangle is the product of its length and its width, which here are 
Example Question #1 : How To Find The Length Of The Diagonal Of A Rectangle
Which is the greater quantity?
(a) The length of a diagonal of a square with sidelength 20 inches
(b) The length of a diagonal of a rectangle with length 25 inches and width less than 10 inches
(b) is greater
It is impossible to tell which is greater from the information given
(a) is greater
(a) and (b) are equal
(a) is greater
The lengths of the diagonals of these rectangles can be computed using the Pythagorean Theorem:
(a)
(b)
Example Question #2 : How To Find The Length Of The Diagonal Of A Rectangle
In Rectangle 
Which is the greater quantity?
(a)
(b)
(a) is greater.
It is impossible to tell from the information given.
(b) is greater.
(a) and (b) are equal.
(a) and (b) are equal.
The diagonals of a rectangle are congruent and bisect each other. Therefore, 
Example Question #3 : How To Find The Length Of The Diagonal Of A Rectangle
Rectangle 
(a)
(b)
It is impossible to tell from the information given.
(b) is greater.
(a) is greater.
(a) and (b) are equal.
(a) is greater.
Two consecutive sides of a rectangle and a diagonal form a right triangle, so the length of any diagonal can be determined using the Pythagorean Theorem, substituting 
The diagonals of a rectangle bisect each other. Therefore, the distance from a vertex to the point of intersection is half this, and 
Example Question #4 : How To Find The Length Of The Diagonal Of A Rectangle
A rectangle has perimeter 140 inches and area 1,200 square inches. Which is the greater quantity?
(A) The length of a diagonal of the rectangle.
(B) 4 feet
It is impossible to determine which is greater from the information given
(A) is greater
(A) and (B) are equal
(B) is greater
(A) is greater
Let 
Since the product of the length and width is the area, we are looking for two numbers whose sum is 70 and whose product is 1,200; through trial and error, they are found to be 30 and 40. We can assign either to be 
The length of a diagonal of the rectangle 
A diagonal is 50 inches long; since 4 feet are equivalent to 48 inches, (A) is the greater quantity.
Example Question #11 : Rectangles
A rectangle has a width of 2x. If the length is five more than 150% of the width, what is the perimeter of the rectangle?
5x + 10
10(x + 1)
5x + 5
6x2 + 10x
10(x + 1)
Given that w = 2x and l = 1.5w + 5, a substitution will show that l = 1.5(2x) + 5 = 3x + 5.
P = 2w + 2l = 2(2x) + 2(3x + 5) = 4x + 6x + 10 = 10x + 10 = 10(x + 1)
All ISEE Upper Level Quantitative Resources
