ISEE Upper Level (grades 9-12) Mathematics Achievement

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

Questions 1 - 10

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:

$A = 6s^{2}$

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

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

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

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

Sector TYP occupies 43% of a circle. Find the degree measure of angle TYP.

$154.8^{\circ}$

$15.5^{\circ}$

$430^{\circ}$

$366^{\circ}$

Explanation

Sector TYP occupies 43% of a circle. Find the degree measure of angle TYP.

Use the following formula and solve for x:

$\frac{x}{360^{\circ}}100=43%$

Begin by dividing over the 100

$\frac{x}{360^{\circ}}=.43$

Then multiply by 360

$x=.43*360^{\circ}=154.8^{\circ}$

A cube has a side length of , what is the volume of the cube?

Explanation

A cube has a side length of , what is the volume of the cube?

To find the volume of a cube, use the following formula:

$V_{cube}=s^3$

Plug in our known side length and solve

$V_{cube}=(9mm)^3=729mm^3$

Making our answer:

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 .

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

Icecreamcone

Refer to the above figure. The shaded region is a semicircle with area $16 \pi$ . Give the perimeter of $\bigtriangleup ABC$ .

$24 \sqrt{2}$

$24 \sqrt{3}$

$12 \sqrt{2}$

$12 \sqrt{3}$

Explanation

Given the radius of a semicircle, its area can be calculated using the formula

$A = \frac{1}{2} \pi r ^{2}$ .

Substituting $A = 16 \pi$ :

$\frac{1}{2} \pi r ^{2} = 16 \pi$

$\frac{2}{\pi} \cdot \frac{1}{2} \pi r ^{2} = \frac{2}{\pi} \cdot 16 \pi$

$r ^{2} = 32$

$r = \sqrt{32} = \sqrt{16} \cdot \sqrt{2} = 4\sqrt{2}$

The diameter of this semicircle is twice this, which is $d = 2 \cdot 4\sqrt{2} = 8 \sqrt{2}$ ; this is also the length of $\overline{BC}$ .

$\bigtriangleup ABC$ has two angles of degree measure 60; its third angle must also have measure 60, making $\bigtriangleup ABC$ an equilateral triangle with sidelength $s = 8 \sqrt{2}$ . Its perimeter is three times this, or $3 \cdot 8 \sqrt{2} = 24 \sqrt{2}$

Find the surface area of a sphere with a diameter of 20in.

$400\pi \text{ in}^2$

$200\pi \text{ in}^2$

$800\pi \text{ in}^2$

$600\pi \text{ in}^2$

$300\pi \text{ in}^2$

Explanation

To find the surface area of a sphere, we will use the following formula:

$SA = 4 \cdot \pi \cdot r^2$

where r is the radius of the sphere.

Now, we know the diameter of the sphere is 20in. We also know the diameter is two times the radius. Therefore, the radius is 10in.

Knowing this, we can substitute into the formula. We get

$SA = 4 \cdot \pi \cdot (10\text{in})^2$

$SA = 4 \cdot \pi \cdot 100\text{in}^2$

$SA = 400 \pi \text{ in}^2$

Find the circumference of a circle with a radius of 4cm.

$8 \pi \text{ cm}$

$4\pi \text{ cm}$

$12\pi \text{cm}$

$2\pi \text{ cm}$

$16\pi \text{ cm}$

Explanation

To find the circumference of a circle, we will use the following formula:

$C = 2 \cdot \pi \cdot r$

where r is the radius of the circle.

Now, we know the radius of the circle is 4cm.

Knowing this, we can substitute into the formula. We get

$C = 2 \cdot \pi \cdot 4\text{cm}$

$C = 2\pi \cdot 4\text{cm}$

$C = 8\pi \text{ cm}$

Solve the equation for :

$\frac{e^{3x}}{e^x}=3$

Explanation

First, simplify the equation as much as possible:

$\frac{e^{3x}}{e^x}=3\Rightarrow e^{(3x-x)}=3\Rightarrow e^{2x}=3$

Now we can take the of both sides:

$e^{2x}=3\Rightarrow ln\ e^{2x}=ln\ 3$

$\Rightarrow 2x=ln\ 3$

$\Rightarrow x=\frac{ln\ 3}{2}=\frac{1.098}{2}$

$\Rightarrow x=0.549$

Three of the vertices of a parallelogram on the coordinate plane are . What is the area of the parallelogram?

Insufficient information is given to answer the problem.

Explanation

As can be seen in the diagram, there are three possible locations of the fourth point of the parallelogram:

Axes_2

Regardless of the location of the fourth point, however, the triangle with the given three vertices comprises exactly half the parallelogram. Therefore, the parallelogram has double that of the triangle.

The area of the triangle can be computed by noting that the triangle is actually a part of a 12-by-12 square with three additional right triangles cut out:

Axes_1