ISEE Upper Level (grades 9-12) Quantitative Reasoning

Help Questions

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

Questions 1 - 10

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.

A pair of fair dice are rolled. Which is the greater quantity?

(a) The probability that at least one die comes up 5 or 6

(b) $\frac{1}{2}$

(a) is greater

(b) is greater

(a) and (b) are equal

It is impossible to tell from the information given.

Explanation

For the roll to be unfavorable to the event that at least one of the dice is 5 or 6, both dice would have to be 1, 2, 3, or 4. There are $4 \times 4 = 16$ ways out of 36 that this can happen, so there are ways for either or both of the two dice to be 5 or 6. Since half of 36 is 18, the probability of this event is greater than $\frac{1}{2}$ .

Trapezoid

The above diagram depicts trapezoid . Which is the greater quantity?

(a) $m \angle T + m \angle P$

(b) $m \angle R + m \angle A$

(a) and (b) are equal.

(a) is greater.

(b) is greater.

It is impossible to tell from the information given.

Explanation

$\overline{TR} \parallel \overline{PA}$ ; $\angle T$ and $\angle P$ are same-side interior angles, as are $\angle R$ and $\angle A$ .

The Same-Side Interior Angles Theorem states that if two parallel lines are crossed by a transversal, then the sum of the measures of a pair of same-side interior angles is always $180^{\circ}$ .

Therefore, $m \angle T + m \angle P = 180= m \angle R + m \angle A$ , making the two quantities equal.

One side of a regular pentagon is 20% longer than one side of a regular hexagon. Which is the greater quantity?

(A) The perimeter of the pentagon

(B) The perimeter of the hexagon

(A) and (B) are equal

(B) is greater

(A) is greater

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

Explanation

Let be the length of one side of the hexagon. Then its perimeter is .

Each side of the pentagon is 20% greater than this length, or

The perimeter is five times this, or $5 \cdot 1.2s = 6s$ .

The perimeters are the same.

Parallelogram

Note: Figure NOT drawn to scale

The above figure shows Rhombus .

Which is the greater quantity?

(a)

(b)

(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

Explanation

The opposite sides of a parallelogram - a rhombus included - are congruent, so

Also, Quadrilateral form a rectangle; since $\overline{AX} || \overline{CY}$ and $\overline{AY} \perp \overline{CY}$ , it follows that $\overline{AY} \perp \overline{AX}$ , and, similarly, $\overline{XC} \perp \overline{CY}$ . Therefore, , and

Trapezoid

In the above figure, $\overline{XY}$ is the midsegment of Trapezoid . What percent of Trapezoid has been shaded in?

$37 \frac{1}{2} %$

$33 \frac{1}{3} %$

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 - the shaded trapezoid - is

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

The area of Trapezoid is

$\frac{1}{2} \cdot 2 h \cdot (TR+AP) = \frac{1}{2} \cdot 2 \cdot h \cdot (12+36) = \frac{1}{2} \cdot 2 \cdot 48 \cdot h = 48 h$

The percent of Trapezoid that is shaded in is

$\frac{18h}{48h} \cdot 100 % = \frac{18}{48} \cdot 100 % = 37 \frac{1}{2} %$

A rectangle is two feet shorter than twice its width; its perimeter is six yards. Give its area in square inches.

$2,816 \textrm{ in}^{2}$

$7,128 \textrm{ in}^{2}$

$1,386 \textrm{ in}^{2}$

$2,475 \textrm{ in}^{2}$

$2,772 \textrm{ in}^{2}$

Explanation

The length of the rectangle is two feet, or 24 inches, shorter than twice the width, so, if is the width in inches, the length in inches is

Six yards, the perimeter of the rectangle, is equal to $36 \times 6 = 216$ inches. The perimeter, in terms of length and width, is , so we can set up the equation:

$6W \div 6 = 264 \div 6$

The length and width are 64 inches and 44 inches; the area is their product, which is

$44 \times 64 = 2,816$ square inches

Given Trapezoid , where $\overline{TR} \parallel \overline{PA}$ . Also, $m \angle A > m\angle P$

Which is the greater quantity?

(a) $m\angle T$

(b) $m \angle R$

(a) is greater

(b) is greater

(a) and (b) are equal

It is impossible to tell from the information given

Explanation

$\angle T$ and $\angle P$ are same-side interior angles, as are $\angle R$ and $\angle A$ .

The Same-Side Interior Angles Theorem states that if two parallel lines are crossed by a transversal, then the sum of the measures of a pair of same-side interior angles is always $180^{\circ}$ . Therefore,

$m \angle T + m \angle P = 180$ , or $m \angle P = 180 - m \angle T$

$m \angle R + m \angle A = 180$ , or $m \angle A = 180 - m \angle R$

Substitute:

$m \angle A > m\angle P$

$180 - m \angle R > 180 - m\angle T$

$180 - m \angle R -180 > 180 - m\angle T-180$

$- m \angle R > - m\angle T$

$- (- m \angle R )<-\left ( - m\angle T \right )$

$m \angle R < m\angle T$

(a) is the greater quantity

A pair of fair dice are tossed. Which is the greater quantity?

(a) The probability that the sum will be at least eight

(b) The probability that the sum will be at most six

(a) and (b) are equal

(a) is greater

(b) is greater

It is impossible to tell from the information given

Explanation

The following table illustrates the ways that the dice can come up. The orange cells indicate a roll of 6 or less; the pink cells indicate a roll of eight or more.

Dice_table

Since each of these outcomes are equally likely, and there are fifteen outcomes that are favorable to each of the two events, the probabilities are the same.

$\frac{ \left ( 30 -2 \cdot 15\right )^{0}}{\left ( 30 -15\right )^{0}}$

The expression is undefined.

Explanation

$\frac{ \left ( 30 -2 \cdot 15\right )^{0}}{\left ( 30 -15\right )^{0}} = \frac{ \left ( 30 -30\right )^{0}}{15^{0}}=\frac{ 0^{0}}{15^{0}}$

The numerator is undefined, since 0 raised to the power of 0 is an undefined quantity. Therefore, the entire expression is undefined.

Page 1 of 100

Return to subject