ISEE Upper Level Quantitative Reasoning - ISEE Upper Level (grades 9-12) Mathematics Achievement

Which is the greater quantity?

(a)

(b)

It is impossible to tell from the information given.

(a) and (b) are equal.

(b) is greater.

(a) is greater.

Explanation

Each can be rewritten as a compound statement. Solve separately:

$45 - t =- 60 \textrm{ or }45 - t = 60$

Similarly:

$45 - u = -50 \textrm{ or } 45 - u = 50$

Therefore, it cannot be determined with certainty which of and is the greater.

In a certain quadrilateral, three of the angles are $65^{\circ}$ , $120^{\circ}$ , and $34^{\circ}$ . What is the measure of the fourth angle?

$141^{\circ}$

$139^{\circ}$

$219^{\circ}$

$41^{\circ}$

$29^{\circ}$

Explanation

A quadrilateral has four angles totalling $360^{\circ}$ . So, first add up the three angles given. The sum is $219^{\circ}$ . Then, subtract that from 360. This gives you the missing angle, which is $141^{\circ}$ .

Define . The graph of is a line with slope $\frac{1}{2}$ .

$(g \circ f) (2) = 6$ .

Which is the greater quantity?

(a)

(b)

(a) is the greater quantity

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

(b) is the greater quantity

(a) and (b) are equal

Explanation

, so $f (2) = 3 \cdot 2 = 6$ .

$(g \circ f) (2) = 6$ , so, by definition, , or .

The graph of is a line through the point with coordinates and with slope $\frac{1}{2}$ . The equation of the line can be determined by setting $g(x) = x = 6 , m = \frac{1}{2}$ in the slope-intercept form:

$6 = \frac{1}{2} \cdot 6+ b$

The equation of the line is $g(x) = \frac{1}{2} x+ 3$ , which makes this the definition of . By setting ,

$g(2) = \frac{1}{2} \cdot 2+ 3 = 1 + 3 = 4$ .

Therefore,

$x = -5\frac{1}{2} \textrm{ or } x = 5$

$x = 5\textrm{ or } x = 5\frac{1}{2}$

$x =17 \textrm{ or } x =23$

$x =-3\frac{1}{2} \textrm{ or } x =4$

$x =-4 \textrm{ or } x =3\frac{1}{2}$

Explanation

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then collecting all of the terms on the left side:

$2x \cdot x+ 2x \cdot 4 -7\cdot x -7\cdot 4 = 27$

$2x^{2}+ 8x -7 x -28 = 27$

$2x^{2}+ x -28 = 27$

$2x^{2}+ x -28-27 = 27-27$

$2x^{2}+ x- 55 = 0$

Use the method to split the middle term into two terms; we want the coefficients to have a sum of 1 and a product of . These numbers are , so we do the following:

$2x^{2}-10x+ 11x- 55 = 0$

$2x\left (x- 5 \right) + 11 \left ( x- 5 \right) = 0$

$\left (2x+ 11 \right) \left ( x- 5 \right) = 0$

Set each expression equal to 0 and solve:

$x = -5\frac{1}{2}$

The solution set is $\left { -5 \frac{1}{2}, 5\right }$ .

Which is the greater quantity?

(a) The number of odd integers such that

(b) The number of even integers such that

(a) is greater.

(b) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

Explanation

This question can be most easily answered by matching each element in the set in (a) with the next consecutive integer, which is in the set in (b):

$1 \rightarrow 2$

$3 \rightarrow 4$

$5 \rightarrow 6$

...

$197 \rightarrow 198$

Every element in the second set has a match, but there is an unmatched element in the first set. Therefore (a) is the greater quantity.

The length of the side of a cube is . Give its surface area in terms of .

$6x^{2} -60x + 150$

$6x^{2} -10x + 25$

$6x^{2} -60x - 150$

Explanation

Substitute in the formula for the surface area of a cube:

$A = 6s^{2}$

$A = 6 \left ( x - 5\right )^{2}$

$A = 6 \left ( x^{2} - 2 \cdot 5 \cdot x + 5^{2}\right )$

$A = 6 \left ( x^{2} - 10x +25\right )$

$A = 6 x^{2} - 60x +150$

Trapezoid

In the above figure, $\overline{XY}$ is the midsegment of Trapezoid .

Which is the greater quantity?

(a) Three times the area of Trapezoid

(b) Twice the area of Trapezoid

(b) is the greater quantity

(a) is the greater quantity

(a) and (b) are equal

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

Explanation

Midsegment $\overline{XY}$ divides Trapezoid into two trapezoids of the same height, which we will call ; the length of the midsegment is half sum of the lengths of the bases:

$XY = \frac{1}{2} (12 + 36) = \frac{1}{2} \cdot 48= 24$

The area of a trapezoid is one half multiplied by its height multiplied by the sum of the lengths of its bases. Therefore, the area of Trapezoid is

$\frac{1}{2} \cdot h \cdot (TR+XY ) = \frac{1}{2} \cdot h \cdot (12+24) = \frac{1}{2} \cdot 36 \cdot h = 18 h$ .

Three times this is

$3 \cdot 18h = 54 h$ .

The area of Trapezoid is, similarly,

$\frac{1}{2} \cdot h \cdot (XY +AP ) = \frac{1}{2} \cdot h \cdot (24+36 ) = \frac{1}{2} \cdot 60 \cdot h = 30 h$

Twice this is

$2 \cdot 30 h = 60 h$ .

That makes (b) the greater quantity.

If a cube has one side measuring cm, what is the surface area of the cube?

Explanation

To find the surface area of a cube, use the formula $6s^{2}$ , where represents the length of the side. Since the side of the cube measures , we can substitute in for .

$6(4)^{2}=96: cm^{2}$

A rectangle is two feet longer than it is wide; its perimeter is 11 feet. What is its area in square inches?

$945 \textrm{ in}^{2}$

$513 \textrm{ in}^{2}$

$4,212 \textrm{ in}^{2}$

$3,780 \textrm{ in}^{2}$

It is impossible to determine the area from the information given

Explanation

The length of the rectangle is 2 feet, or 24 inches, greater than the width, so, if is the width in inches, is the length in inches.

The perimeter of the rectangle is 11 feet, or $11 \times 12= 132$ inches. The perimeter, in terms of length and width, is , so we can set up the equation: