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SAT II Math I : SAT Subject Test in Math I

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

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Example Questions

Example Question #77 : 2 Dimensional Geometry

Regular Octagon ABCDEFGH has perimeter 80. $\overline{AB }$ has as its midpoint; segment $\overline{ME}$ is drawn. To the nearest tenth, give the length of $\overline{ME}$ .

Possible Answers:

30.4

24.1

26.1

24.7

21.8

Correct answer:

24.7

Explanation:

Below is the regular Octagon ABCDEFGH , with the referenced midpoint and segment $\overline{ME}$ . Note that perpendiculars have also been constructed from and to meet $\overline{BE}$ at and , respectively.

Octagon 2

$\bigtriangleup MBE$ is a right triangle with legs $\overline{MB}$ and $\overline{ BE}$ and hypotenuse $\overline{ME}$ .

The perimeter of the regular octagon is 80, so the length of each side is one-eighth of 80, or 10. Consequently,

$MB = \frac{1}{2} AB = \frac{1}{2} \cdot 10 = 5$

To find the length of $\overline{ BE}$ , we can break it down as

BE = BX + XY + YE

Quadrilateral CDYX is a rectangle, so XY = CD = 10 .

$\bigtriangleup BXC$ is a 45-45-90 triangle with leg $\overline{BX}$ and hypotenuse $\overline{BC}$ ; by the 45-45-90 Triangle Theorem,

$BX = \frac{BC }{\sqrt{2}} \approx \frac{10}{1.4142} \approx 7.0711$

For similar reasons, $EY \approx 7.0711$ .

Therefore,

$BE = BX + XY + YE \approx 7.0711 + 10 + 7.0711 \approx 24.1421$

can now be evaluated using the Pythagorean Theorem:

$ME = \sqrt{(MB)^{2}+ ( BE ) ^{2}}$

Substituting and evaluating:

$ME = \sqrt{5 ^{2}+24.1421^{2}}$

$\approx \sqrt{25 +582.8427}$

$\approx \sqrt{ 607.8427}$

$\approx 24.7$ ,

the correct choice.

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Example Question #1 : Volume

A circular swimming pool has diameter meters and depth 2 meters throughout. Which of the following expressions give the amount of water it holds, in cubic meters?

Possible Answers:

$\frac{ \pi k^{2}} {4}$

$4\pi k$

$2\pi k^{2}$

$\frac{ \pi k^{2}} {2}$

$2\pi k$

Correct answer:

$\frac{ \pi k^{2}} {2}$

Explanation:

The pool can be seen as a cylinder with diameter - and, subsequently, radius half this, or $\frac{k}{2}$ - and depth, or height, 2. The volume of a cylinder is defined by the formula

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

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Example Question #1 : Volume

Swimming_pool

The above depicts a rectangular swimming pool for an apartment. 60% of the pool is six feet deep, and the remaining part of the pool is four feet deep. How many cubic feet of water does the pool hold?

Possible Answers:

$8,750 \textrm{ ft} ^{3}$

$9,750 \textrm{ ft} ^{3}$

$7,000 \textrm{ ft} ^{3}$

None of the other choices gives the correct answer.

$10,500\textrm{ ft} ^{3}$

Correct answer:

None of the other choices gives the correct answer.

Explanation:

The cross-section of the pool is the area of its surface, which is the product of its length and its width:

$50 \times 35 = 1,750$ square feet.

Since 60% of the pool is six feet deep, this portion of the pool holds

$0.60 \times 1,750 \times 6 = 6,300$ cubic feet of water.

Since the remainder of the pool - 40% - is four feet deep, this portion of the pool holds

$0.4 \times 1,750 \times 4 = 2,800$ cubic feet of water.

Add them together: the pool holds

6,300 + 2,800 = 9,100 cubic feet of water.

This answer is not among the choices.

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Example Question #1 : Volume

Find the volume of a cube in inches with a side of $1.5 \textup{ feet}.$

Possible Answers:

$40.5 \textup{ inches}^3$

$0.001953 \textup{ inches}^3$

$5832 \textup{ inches}^3$

$13458 \textup{ inches}^3$

$54 \textup{ inches}^3$

Correct answer:

$5832 \textup{ inches}^3$

Explanation:

Convert the side dimension to inches first before finding the volume.

$1.5 \textup{ feet} \times \left(\frac{12 \textup{ inches}}{1 \textup{ feet}}\right)= 18 \textup{ inches}$

Write the volume for a cube and substitute the new side to obtain the volume in inches.

$V=s^3 =(18 \textup{ inches})^3 = 5832 \textup{ inches}^3$

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Example Question #1 : 3 Dimensional Geometry

Example cylinder

Figure not drawn to scale.

What is the volume of the cylinder above?

Possible Answers:

52.36 in³

48.79 in³

66.13 in³

56.55 in³

94.25 in³

Correct answer:

56.55 in³

Explanation:

In order to find the volume of a cylinder, you find the area of the circular top and multiply it by the height.

Example cylinder

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

$\small \pi (3{^{2}})(2)=\pi(9)(2)=18\pi=56.55$

The volume of the cylinder is 56.55 in³

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Example Question #1 : Volume

Cone example

Figure not drawn to scale

If the volume of the cone above is 47.12 ft³, what is the radius of the base?

Possible Answers:

5 ft

2 ft

3 ft

4 ft

3.5ft

Correct answer:

3 ft

Explanation:

Because we have been given the volume of the cone and have been asked to find the radius of the base of the cone, we must work backwards using the volume formula.

Cone example

$\small Volume= {\frac{1}{3}}\pi r{^{2}}h$

$\small \small 47.12={\frac{1}{3}}\pi r{^{2}}(5)={\frac{5}{3}}\pi r{^{2}}$

$\small \left({\frac{3}{5}} \right)47.12=\left({\frac{3}{5}} \right )\left({\frac{5}{3}} \right )\pi r{^{2}}$

$\small \frac{28.27}{\pi}=r{^{2}}$

$\small 9.00=r{^{2}}$

$\small r=3$

The radius of the base of the cone is 3 ft.

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Example Question #1 : Volume

Box example

Figure not drawn to scale.

What is the volume of the above image?

Possible Answers:

64 yd³

42 yd²

42 yd³

32 yd³

84 yd²

Correct answer:

42 yd³

Explanation:

You can find the volume of a box by following the equation below:

$\small Volume=(length)(height)(depth)$

$\small (7)(3)(2)=42$

The surface area of the box is 42 yd³(remember that volume measurements are cubic units NOT square units)

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Example Question #1 : How To Find The Volume Of A Tetrahedron

What is the volume of the following tetrahedron? Assume the figure is a regular tetrahedron.

Possible Answers:

$\4\sqrt{3}m^3$

$\frac{3\sqrt{2}}{2}m^3$

$\frac{8\sqrt{2}}{3}m^3$

$\frac{\sqrt{2}}{3}m^3$

$\frac{2\sqrt{2}}{3}m^3$

Correct answer:

$\frac{2\sqrt{2}}{3}m^3$

Explanation:

A regular tetrahedron is composed of four equilateral triangles. The formula for the volume of a regular tetrahedron is:

$V=\frac{a^3\sqrt{2}}{12}$ , where represents the length of the side.

Plugging in our values we get:

$V=\frac{(2\: m)^3\sqrt{2}}{12}$

$V=\frac{8\: m^3\sqrt{2}}{12} = \frac{2\sqrt{2}}{3}m^3$

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Example Question #2 : Volume

Find the volume of a tetrahedron with an edge of $1\:cm$ .

Possible Answers:

$3\:cm^3$

$6\sqrt{2}\:cm^3$

$\frac{\sqrt2}{12}\:cm^3$

$\frac{1}{6}\:cm^3$

$\frac{\sqrt3}{4}\:cm^3$

Correct answer:

$\frac{\sqrt2}{12}\:cm^3$

Explanation:

Write the formula for the volume of a tetrahedron.

$V=\frac{s^3}{6\sqrt 2}$

Substitute in the length of the edge provided in the problem.

$V=\frac{(1\:cm)^3}{6\sqrt2}=\frac{1}{6\sqrt2}\:cm^3$

Rationalize the denominator.

$V= \frac{1}{6\sqrt2}\:cm\cdot \frac{\sqrt2}{\sqrt2} = \frac{\sqrt2}{12}\:cm^3$

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Example Question #1 : How To Find The Volume Of A Tetrahedron

Find the volume of a tetrahedron with an edge of $\sqrt2\:in$ .

Possible Answers:

$\frac{1}{6}\:in^3$

$\sqrt3\:in^3$

$\frac{1}{2}\:in^3$

$\frac{2}{3}\:in^3$

$\frac{1}{3}\:in^3$

Correct answer:

$\frac{1}{3}\:in^3$

Explanation:

Write the formula for the volume of a tetrahedron.

$V=\frac{s^3}{6\sqrt 2}$

Substitute in the length of the edge provided in the problem:

$V=\frac{(\sqrt2\:in)^3}{6\sqrt2}$

Cancel out the $\sqrt2$ in the denominator with one in the numerator:

$V= \frac{(\sqrt2)^3}{6\sqrt2}\:in^3$

A square root is being raised to the power of two in the numerator; these two operations cancel each other out. After canceling those operations, reduce the remaining fraction to arrive at the correct answer:

$V= \frac{(\sqrt2)^2}{6}\:in^3=\frac{2}{6}\:in^3=\frac{1}{3}\:in^3$

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SAT II Math I : SAT Subject Test in Math I

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

All SAT II Math I Resources

Example Questions

Example Question #77 : 2 Dimensional Geometry

Example Question #1 : Volume

Example Question #1 : Volume

Example Question #1 : Volume

Example Question #1 : 3 Dimensional Geometry

The volume of the cylinder is 56.55 in³

Example Question #1 : Volume

If the volume of the cone above is 47.12 ft³, what is the radius of the base?

The radius of the base of the cone is 3 ft.

Example Question #1 : Volume

The surface area of the box is 42 yd³(remember that volume measurements are cubic units NOT square units)

Example Question #1 : How To Find The Volume Of A Tetrahedron

Example Question #2 : Volume

Example Question #1 : How To Find The Volume Of A Tetrahedron

All SAT II Math I Resources

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SAT II Math I : SAT Subject Test in Math I

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

All SAT II Math I Resources

Example Questions

Example Question #77 : 2 Dimensional Geometry

Example Question #1 : Volume

Example Question #1 : Volume

Example Question #1 : Volume

Example Question #1 : 3 Dimensional Geometry

The volume of the cylinder is 56.55 in3

Example Question #1 : Volume

If the volume of the cone above is 47.12 ft3, what is the radius of the base?

The radius of the base of the cone is 3 ft.

Example Question #1 : Volume

The surface area of the box is 42 yd3 (remember that volume measurements are cubic units NOT square units)

Example Question #1 : How To Find The Volume Of A Tetrahedron

Example Question #2 : Volume

Example Question #1 : How To Find The Volume Of A Tetrahedron

All SAT II Math I Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

The volume of the cylinder is 56.55 in³

If the volume of the cone above is 47.12 ft³, what is the radius of the base?

The surface area of the box is 42 yd³(remember that volume measurements are cubic units NOT square units)