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SAT II Math II : SAT Subject Test in Math II

Study concepts, example questions & explanations for SAT II Math II

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All SAT II Math II Resources

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Example Questions

Example Question #1 : 3 Dimensional Geometry

The width of a box is two-thirds its height and three-fifths its length. The volume of the box is 6 cubic meters. To the nearest centimeter, give the width of the box.

Possible Answers:

$223 \textrm{ cm}$

$164 \textrm{ cm}$

$134 \textrm{ cm}$

$201 \textrm{ cm}$

$246 \textrm{ cm}$

Correct answer:

$134 \textrm{ cm}$

Explanation:

Call , , and the length, width, and height of the crate.

The width is two-thirds the height, so

$W = \frac{2}{3} H$ .

Equivalently,

$\frac{3} {2} \cdot \frac{2}{3} H = \frac{3} {2} W$

$H = \frac{3} {2} W$

The width is three-fifths the length, so

$W = \frac{3}{5} L$ .

Equivalently,

$\frac{5} {3} \cdot \frac{3}{5} L = \frac{5} {3} \cdot W$

$L = \frac{5} {3 } W$

The dimensions of the crate in terms of are , $\frac{3} {2} W$ , and $\frac{5} {3 } W$ . The volume is their product:

V= LWH ,

Substitute:

LWH = V

$\frac{5} {3 } W \cdot W \cdot \frac{3} {2} W = V$

$\frac{5} {3 } \cdot \frac{3} {2} \cdot W \cdot W \cdot W = 6$

$\frac{5} {2} \cdot W ^{3}= 6$

$\frac{2} {5} \cdot \frac{5} {2} \cdot W ^{3}= \frac{2} {5} \cdot 6$

$W ^{3}= \frac{12} {5}$

Taking the cube root of both sides:

$W =\sqrt[3]{ \frac{12} {5}}$

$W \approx 1.339$ meters.

Since one meter comprises 100 centimeters, multiply by 100 to convert to centimeters:

$1.339 \cdot 100 = 133.9$ centimeters,

which rounds to 134 centimeters.

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Example Question #1 : Volume

Box 2

The shaded face of the rectangular prism in the above diagram is a square. The volume of the prism is ; give the value of in terms of .

Possible Answers:

$x =\frac{ \sqrt{V}}{25}$

$x =\frac{ \sqrt{V}}{5}$

$x =\frac{ \sqrt{V}}{625}$

$x =\frac{ V}{25}$

$x =\frac{ V}{625}$

Correct answer:

$x =\frac{ \sqrt{V}}{5}$

Explanation:

The volume of a rectangular prism is the product of its length, its width, and its height; that is,

LWH = V

Since the shaded face of the prism is a square, we can set L = H = x , and W = 25 ; substituting and solving for :

$x \cdot 25 \cdot x = V$

$25x^{2} = V$

$\frac{25x^{2}}{25} = \frac{V}{25}$

$x^{2} = \frac{V}{25}$

Taking the positive square root of both sides, and simplifying the expression on the right using the Quotient of Radicals Rule:

$x =\sqrt{ \frac{V}{25}} = \frac{\sqrt{V}}{\sqrt{25}} = \frac{\sqrt{V}}{5}$

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Example Question #11 : 3 Dimensional Geometry

Find the volume of a sphere with a diameter of 10.

Possible Answers:

$100\pi$

$25\pi$

$1333.33\pi$

$166.67\pi$

$33.33\pi$

Correct answer:

$166.67\pi$

Explanation:

The surface area of a sphere is found using the formula $A=\frac{4}{3}\pi r^3$ . We are given the diameter of the circle and so we have to use it to find the radius (r).

$Radius=\frac{Diameter}{2}=\frac{10}{2}=5$

Plug r into the formula to find the surface area

$A=\frac{4}{3}\pi 5^3=\frac{4}{3}\pi 125=166.67\pi$

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Example Question #11 : Volume

Determine the volume of the cube with a side length of 3x^2 .

Possible Answers:

3x^8

27x^6

3x^6

9x^6

27x^8

Correct answer:

27x^6

Explanation:

Write the formula for the volume of a cube.

V = s^3

Substitute the length into the formula.

V = (3x^2)^3=(3x^2)(3x^2)(3x^2) = 27x^6

The volume is: 27x^6

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Example Question #11 : Volume

Billy has a ice cream cone that consists of a cone and hemisphere. Suppose the cone has a height of 4 inches, and the radius of the hemisphere is 2 inches. Assuming that the combined shape is not irregular, what is the total volume?

Possible Answers:

$\frac{64}{9}\pi$

$\frac{16}{3}\pi$

$\frac{64}{3}\pi$

$\frac{8}{3}\pi$

$\frac{32}{3}\pi$

Correct answer:

$\frac{32}{3}\pi$

Explanation:

Write the volume for a cone.

$V_{cone } = \frac{1}{3}\pi r^2 h$

Substitute the radius and height. The radius is 2.

$V_{cone } = \frac{1}{3}\pi (2)^2 (4) = \frac{16}{3}\pi$

Write the volume for a hemisphere. This should be half the volume of the full sphere.

$V_{hemisphere} = \frac{1}{2}[ \frac{4}{3}\pi r^3] = \frac{2}{3}\pi r^3$

Substitute the radius.

$V_{hemisphere}= \frac{2}{3}\pi (2)^3 = \frac{16}{3}\pi$

Add the volumes of the cone and hemisphere to determine the total volume.

$\frac{16}{3}\pi+\frac{16}{3}\pi$

The answer is: $\frac{32}{3}\pi$

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Example Question #391 : Sat Subject Test In Math Ii

Find the volume of a sphere with a diameter of $\frac{\sqrt[3]{2}}{2}$ .

Possible Answers:

$\frac{1}{24}\pi$

$\frac{1}{9}\pi$

$\frac{1}{3}\pi$

$\frac{1}{6}\pi$

$\frac{1}{12}\pi$

Correct answer:

$\frac{1}{24}\pi$

Explanation:

Divide the diameter by two to get the radius. This is also the same as multiplying the diameter by one-half.

$r =\frac{\sqrt[3]{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt[3]{2}}{4}$

Write the formula for the volume of the sphere.

$V=\frac{4}{3}\pi r^3$

Substitute the radius.

$V=\frac{4}{3}\pi ( \frac{\sqrt[3]{2}}{4})^3$

Simplify the terms.

$\frac{4}{3}\pi ( \frac{\sqrt[3]{2}}{4})^3 = \frac{4}{3}\pi ( \frac{\sqrt[3]{2}}{4})( \frac{\sqrt[3]{2}}{4})( \frac{\sqrt[3]{2}}{4}) = \frac{2}{3(4)(4)}\pi$

The answer is: $\frac{1}{24}\pi$

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Example Question #13 : 3 Dimensional Geometry

If the side of a cube is 3x^3 , what must be the volume?

Possible Answers:

27x^9

$27x^{27}$

9x^9

27x^3

27x^6

Correct answer:

27x^9

Explanation:

Write the formula for the volume of a cube.

V =s^3

Substitute the side length. When we are multiplying common bases with exponents, we are adding the exponents instead.

V =(3x^3)^3 = (3x^3)(3x^3)(3x^3) = 27x^9

The answer is: 27x^9

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Example Question #14 : 3 Dimensional Geometry

Determine the volume of a cube if the side length is $2\sqrt6$ .

Possible Answers:

$8\sqrt6$

$24\sqrt3$

$48\sqrt6$

$72\sqrt3$

Correct answer:

$48\sqrt6$

Explanation:

Write the formula for the volume of a cube.

V= s^3

Substitute the side length into the equation.

$V= (2\sqrt6)^3 = (2\sqrt6)(2\sqrt6)(2\sqrt6) = 8\cdot 6\cdot \sqrt6$

The answer is: $48\sqrt6$

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Example Question #15 : 3 Dimensional Geometry

The radius and the height of a cylinder are equal. If the volume of the cylinder is , what is the diameter of the cylinder?

Possible Answers:

3.21

1.77

2.95

1.61

Correct answer:

3.21

Explanation:

Recall how to find the volume of a cylinder:

$\text{Volume}=\pi r^2 h$

Since we know that the radius and the height are equal, we can rewrite the equation:

$\text{Volume}=\pi r^2 (r)=\pi r^3$

Using the given volume, find the length of the radius.

$13=\pi r^3$

r^3=4.138

r=1.605

Since the question asks you to find the diameter, multiply the radius by two.

d=3.21

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Example Question #11 : Volume

Determine the volume of the cube if the side lengths are $\frac{2}{5}$ .

Possible Answers:

$\frac{8}{15}$

$\frac{8}{125}$

$\frac{6}{5}$

$\frac{6}{125}$

Correct answer:

$\frac{8}{125}$

Explanation:

The volume of a cube is: V =LWH

Substitute the dimensions.

$V =(\frac{2}{5})(\frac{2}{5})(\frac{2}{5}) = \frac{8}{125}$

The answer is: $\frac{8}{125}$

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SAT II Math II : SAT Subject Test in Math II

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #1 : 3 Dimensional Geometry

Example Question #1 : Volume

Example Question #11 : 3 Dimensional Geometry

Example Question #11 : Volume

Example Question #11 : Volume

Example Question #391 : Sat Subject Test In Math Ii

Example Question #13 : 3 Dimensional Geometry

Example Question #14 : 3 Dimensional Geometry

Example Question #15 : 3 Dimensional Geometry

Example Question #11 : Volume

All SAT II Math II Resources

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