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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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2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Other 2 Dimensional Geometry

A circle is inscribed inside a square that touches all edges of the square. The square has a length of 3. What is the area of the region inside the square and outside the edge of the circle?

Possible Answers:

$9-\frac{9}{2}\pi$

$9-\frac{9}{4}\pi$

$9-3\pi$

$\textup{The answer is not given.}$

$9-9\pi$

Correct answer:

$9-\frac{9}{4}\pi$

Explanation:

Solve for the area of the square.

$A_{square}=s^2 = 3^2 = 9$

Solve for the area of the circle. Given the information that the circle touches all sides of the square, the diameter is equal to the side length of the square.

This means that the radius is half the length of the square: $r=\frac{3}{2}$

Substitute the radius.

$A_{circle}=\pi r^2 = \pi (\frac{3}{2})^2= \pi (\frac{3}{2}) (\frac{3}{2}) = \frac{9}{4}\pi$

Subtract the area of the square and the circle to determine the area desired.

The answer is: $9-\frac{9}{4}\pi$

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Example Question #1 : Other 2 Dimensional Geometry

Inscribed

Figure is not drawn to scale

$\overline{AC}$ is a diameter of the circle; its length is ten; furthermore we know the following:

$m \overarc {AB} = 126^{\circ }$

Give the length of $\overline{AB}$ (nearest tenth)

Possible Answers:

7.0

9.5

8.9

5.8

7.6

Correct answer:

8.9

Explanation:

Locate , the center of the circle, which is the midpoint of $\overline{AC}$ ; draw radius $\overline{OB}$ . $\bigtriangleup AOB$ is formed. The central angle that intercepts $\overarc {AB}$ is $\angle AOB$ , so $m \angle AOB = m \overarc {AB} = 126^{\circ }$ . $\overline{OA}$ and $\overline{OB}$ , being radii of the circle, have length half the diameter of ten, or five. The diagram is below.

Inscribed

By the Law of Cosines, given two sides of a triangle of length and , and their included angle of measure $\gamma$ , the length of the third side can be calculated using the formula

$c ^{2} = a^{2} + b^{2} - 2ab \cos \gamma$

Setting $a = OA = 5, b = OB = 5, \gamma = m \angle AOB = 126 ^{\circ } , c = AB$ , solve for :

$(AB) ^{2} = 5^{2} +5 ^{2} - 2 \cdot 5 \cdot 5 \cdot \cos 126^{\circ }$

$(AB) ^{2} =25 + 25 - 50 \cos 126^{\circ }$

$(AB) ^{2} \approx 50 - 50 (-0.5878)$

$(AB) ^{2} \approx 50 + 29.39$

$(AB) ^{2} \approx 79.39$

Taking the square root of both sides:

$AB \approx\sqrt{ 79.39} \approx 8.9$

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Example Question #1 : Volume

One cubic meter is equal to one thousand liters.

A circular swimming pool is meters in diameter and meters deep throughout. How many liters of water does it hold?

Possible Answers:

$500 \pi d^{2}f$

$500\pi df$

$1,000\pi df$

$1,000\pi d^{2}f$

$250\pi d^{2}f$

Correct answer:

$250\pi d^{2}f$

Explanation:

The pool can be seen as a cylinder with depth (or height) h= f , and a base with diameter - and, subsequently, radius half this, or $\frac{d}{2}$ . The volume of the pool in cubic meters is

$V = \pi r^{2}h = \pi \left ( \frac{d}{2} \right )^{2} \cdot f = \pi \cdot \frac{d^{2}}{4} \cdot f = \frac{ \pi d^{2}f}{4}$

Multiply this number of cubic meters by 1,000 liters per cubic meter:

$V = \frac{ \pi d^{2}f}{4} \cdot \1,000 = 250\pi d^{2}f$

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Example Question #2 : Volume

A water tank takes the shape of a sphere whose exterior has radius 16 feet; the tank is three inches thick throughout. To the nearest hundred, give the surface area of the interior of the tank in square feet.

Possible Answers:

$17,100 \textrm{ ft}^{2}$

$3,000 \textrm{ ft}^{2}$

$16,400 \textrm{ ft}^{2}$

$3,100 \textrm{ ft}^{2}$

$15,600 \textrm{ ft}^{2}$

Correct answer:

$3,100 \textrm{ ft}^{2}$

Explanation:

Three inches is equal to 0.25 feet, so the radius of the interior of the tank is

16 - 0.25 = 15.75 feet.

The surface area of the interior of the tank can be calculated using the formula

$SA = 4 \pi r^{2}$

$SA \approx 4 \cdot 3.14 \cdot 15.75 ^{2} \approx 3,116$ ,

which rounds to 3,100 square feet.

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Example Question #2 : Volume

A water tank takes the shape of a closed rectangular prism whose exterior has height 30 feet, length 20 feet, and width 15 feet. Its walls are one foot thick throughout. How many cubic feet of water does the tank hold?

Possible Answers:

$10,416\textup{ ft}^{3}$

$7,714\textup{ ft}^{3}$

$2,204\textup{ ft}^{3}$

$6,552\textup{ ft}^{3}$

$2,446\textup{ ft}^{3}$

Correct answer:

$6,552\textup{ ft}^{3}$

Explanation:

The height, length, and width of the interior tank are each two feet less than the corresponding dimension of the exterior of the tank, so the dimensions of the interior are 28, 18, and 13 feet. Multiply these to get the volume:

$V = 28 \cdot 18 \cdot 13 = 6,552$ cubic feet.

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Example Question #1 : 3 Dimensional Geometry

A circular swimming pool has diameter 40 meters and depth meters throughout. Which of the following expressions gives the amount of water it holds, in cubic meters?

Possible Answers:

$160 \pi t$

$40 \pi t$

$1,600 \pi t$

$400 \pi t$

$80 \pi t$

Correct answer:

$400 \pi t$

Explanation:

The pool can be seen as a cylinder with depth (or height) h= t , and a base with diameter 40 m - and radius half this, or r= 20 . The capacity of the pool is the volume of this cylinder, which is

$V = \pi r^{2}h = \pi \cdot 20^{2}\cdot t = 400 \pi t$

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Example Question #1 : Volume

One cubic meter is equal to one thousand liters.

A rectangular swimming pool is meters deep throughout and meters wide. Its length is ten meters greater than twice its width. How many liters of water does the pool hold?

Possible Answers:

$0.002 DW^{2} + 0.01 DW$

$0.002 DW^{2} - 0.01 DW$

$2,000 DW^{2} + 10,000 DW$

$2,000 DW^{2} -10,000 DW$

None of the other responses is correct.

Correct answer:

$2,000 DW^{2} + 10,000 DW$

Explanation:

Since the length of the pool is ten meters longer than twice its width , its length is 2W + 10 .

The inside of the pool can be seen as a rectangular prism, and as such, its volume in cubic feet can be calculated as the product of its length, width, and height (or depth). This product is

$V = D \cdot (2W + 10) \cdot W$

$V = 2DW^{2} + 10DW$

Multiply this by the conversion factor 1,000, and its volume in liters is

$V =1,000 \left ( 2DW^{2} + 10DW \right )$

$V = 2,000 DW^{2} + 10,000 DW$

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Example Question #1 : Volume

A circular swimming pool has diameter 80 feet and depth five feet throughout. To the nearest thousand, how many gallons of water does it hold?

Use the conversion factor: One cubic foot = 7.5 gallons.

Possible Answers:

$754,000\textrm{ gal}$

$94,000\textrm{ gal}$

$3,000\textrm{ gal}$

$188,000\textrm{ gal}$

$5,000\textrm{ gal}$

Correct answer:

$188,000\textrm{ gal}$

Explanation:

The pool can be seen as a cylinder with depth (or height) 5 feet, and a base with diameter 80 feet - and radius half this, or 40 feet. The capacity of the pool is the volume of this cylinder, which is

$V = \pi r ^{2} h \approx 3.14 \times 40^{2} \times 5 \approx 25,120$ cubic feet.

One cubic foot is equal to 7.5 gallons, so multiply:

$25,120 \times 7.5 = 188,400$ gallons

This rounds to 188,000 gallons.

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Example Question #1 : Volume

Pool

The above depicts a rectangular swimming pool for an apartment.

On the left and right edges, the pool is three feet deep; the dashed line at the very center represents the line along which it is eight feet deep. Going from the left to the center, its depth increases uniformly; going from the center to the right, its depth decreases uniformly.

In cubic feet, how much water does the pool hold?

Possible Answers:

$1,870 \textrm{ ft}^{3}$

$9,625 \textrm{ ft}^{3}$

$2,070 \textrm{ ft}^{3}$

$10,225 \textrm{ ft}^{3}$

$9,875 \textrm{ ft}^{3}$

Correct answer:

$9,625 \textrm{ ft}^{3}$

Explanation:

The pool can be looked at as a pentagonal prism with "height" 35 feet and its bases the following shape (depth exaggerated):

Pool

This is a composite of two trapezoids, each with bases 3 feet and 8 feet and height 25 feet; the area of each is

$A = \frac{1}{2} (3+8) \cdot 25 = 137.5$ square feet.

The area of the base is twice this, or

$B = 137.5 \cdot 2 = 275$ square feet.

The volume of a prism is its height times the area of its base, or

$V = 275 \cdot 35 = 9,625$ cubic feet, the capacity of the pool.

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Example Question #1 : Volume

The bottom surface of a rectangular prism has area 100; the right surface has area 200; the rear surface has area 300. Give the volume of the prism (nearest whole unit), if applicable.

Possible Answers:

1,000

6,000

4,243

3,464

2,449

Correct answer:

2,449

Explanation:

Let the dimensions of the prism be , , and .

Then, LW = 100 , LH = 200 , and WH = 300 .

From the first and last equations, dividing both sides, we get

WH = 300

LW = 100

$\frac{WH}{LW} = \frac{300}{100}$

$\frac{H}{L} = 3$

Along with the second equation, multiply both sides:

LH = 200

$\frac{H}{L} \cdot LH = 3 \cdot 200$

$H ^{2}= 600$

Taking the square root of both sides and simplifying, we get

$H = \sqrt{600} = \sqrt{100} \cdot \sqrt{6} = 10 \sqrt{6}$

Now, substituting and solving for the other two dimensions:

LH = 200

$L \cdot 10 \sqrt{6} = 200$

$L = \frac{200}{10 \sqrt{6}} = \frac{20}{\sqrt{6}}$

WH = 300

$W \cdot 10 \sqrt{6} = 300$

$W = \frac{300}{10 \sqrt{6}} = \frac{30 }{\sqrt{6}}$

Now, multiply the three dimensions to obtain the volume:

V= LWH

$= \frac{20}{\sqrt{6}} \cdot 10 \sqrt{6} \cdot \frac{30 }{\sqrt{6}}$

$= \frac{20}{\sqrt{6}} \cdot\frac{ 10 \sqrt{6}}{1} \cdot \frac{30 }{\sqrt{6}}$

$= \frac{ 6,000 \sqrt{6}} {6}$

$= 1,000 \sqrt{6}$

$\approx 1,000 \cdot 2.449$

$\approx 2,449$

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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 #2 : Other 2 Dimensional Geometry

Example Question #1 : Other 2 Dimensional Geometry

Example Question #1 : Volume

Example Question #2 : Volume

Example Question #2 : Volume

Example Question #1 : 3 Dimensional Geometry

Example Question #1 : Volume

Example Question #1 : Volume

Example Question #1 : Volume

Example Question #1 : Volume

All SAT II Math II Resources

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