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ISEE Upper Level Quantitative : Solid Geometry

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Example Questions

Example Question #322 : Geometry

What is the length of one side of a cube that has a surface area of 1944 in^2 ?

Possible Answers:

44.1

12.48

17.31

Correct answer:

Explanation:

Recall that the formula for the surface area of a cube is:

SA = 6 * s^2 , where is the length of a side of the cube. This equation is easy to memorize because it is merely a multiplication of a single side ( s^2 ) by because a cube has equal sides.

Now, we know that is 1944 ; therefore, we can write:

1944 = 6*s^2

Solve for :

s^2 = 324

Take the square root of both sides:

s = 18

This is the length of one of your sides.

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Example Question #1 : Tetrahedrons

Which is the greater quantity?

(a) The surface area of a regular tetrahedron with edges of length 1

(b) 2

Possible Answers:

It is impossible to tell from the information given.

(a) is greater.

(b) is greater.

(a) and (b) are equal.

Correct answer:

(b) is greater.

Explanation:

A regular tetrahedron has four faces, each of which is an equilateral triangle. Therefore, its surface area, given sidelength , is

$A = 4 \cdot \frac{s^{2} \sqrt{3}}{4} = s^{2} \sqrt{3}$ .

Substitute s = 1 :

$A = s^{2} \sqrt{3} =1^{2} \cdot \sqrt{3} = \sqrt{3}$

$\sqrt{3} \approx 1.7 < 2$ , so (b) is greater.

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Example Question #1 : How To Find The Radius Of A Sphere

The volume of a sphere is one cubic yard. Give its radius in inches.

Possible Answers:

$\frac{ 18\sqrt[3]{6 \pi^{2}} } {\pi} \textrm{ in}$

$\frac{ 6\sqrt[3]{6 \pi^{2}} } {\pi} \textrm{ in}$

$\frac{ \sqrt{3 \pi } } {2 \pi} \textrm{ in}$

$\frac{ \sqrt[3]{6 \pi^{2}} } {2\pi}\textrm{ in}$

$\frac{ 18\sqrt{3 \pi } } {\pi} \textrm{ in}$

Correct answer:

$\frac{ 18\sqrt[3]{6 \pi^{2}} } {\pi} \textrm{ in}$

Explanation:

The volume of a sphere with radius is

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

To find the radius in yards, we set V = 1 and solve for .

$1 = \frac{4}{3} \pi r^{3}$

$\frac{3}{4} \cdot 1 =\frac{3}{4} \cdot \frac{4}{3} \pi r^{3}$

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

$\frac{1}{\pi} \cdot \frac{3}{4} =\frac{1}{\pi} \cdot \pi r^{3}$

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

$r = \sqrt[3]{ \frac{3}{4\pi} } = \sqrt[3]{ \frac{3\cdot 2 \pi^{2}}{4\pi \cdot 2 \pi^{2}} }= \sqrt[3]{ \frac{6 \pi^{2}}{8 \pi^{3}} } = \frac{ \sqrt[3]{6 \pi^{2}} } {2\pi}$ yards.

Since the problem requests the radius in inches, multiply by 36:

$36 \times \frac{ \sqrt[3]{6 \pi^{2}} } {2\pi} = \frac{ 18\sqrt[3]{6 \pi^{2}} } {\pi}$

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Example Question #31 : Solid Geometry

In terms of $\pi$ , give the volume, in cubic feet, of a spherical tank with diameter 36 inches.

Possible Answers:

$\frac{9}{2} \pi \textrm{ ft}^{3}$

$9 \pi \textrm{ ft}^{3}$

$3 \pi \textrm{ ft}^{3}$

$36 \pi \textrm{ ft}^{3}$

$18 \pi \textrm{ ft}^{3}$

Correct answer:

$\frac{9}{2} \pi \textrm{ ft}^{3}$

Explanation:

36 inches = $36 \div 12 = 3$ feet, the diameter of the tank. Half of this, or $\frac{3}{2}$ feet, is the radius. Set $r = \frac{3}{2}$ , substitute in the volume formula, and solve for :

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

$V = \frac{4}{3} \pi\left \cdot \left( \frac{3}{2} \right ) ^{3}$

$V = \frac{4}{3}\cdot \frac{3}{2} \cdot \frac{3}{2} \cdot \frac{3}{2} \cdot \pi\left$

$V = \frac{1}{1}\cdot \frac{1}{1} \cdot \frac{3}{1} \cdot \frac{3}{2} \cdot \pi\left$

$V = \frac{9}{2} \pi$

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

Which is the greater quantity?

(a) The volume of a sphere with radius

(b) The volume of a cube with sidelength

Possible Answers:

It is impossible to tell from the information given

(b) is greater

(a) is greater

(a) and (b) are equal

Correct answer:

(b) is greater

Explanation:

A sphere with radius has diameter and can be inscribed inside a cube of sidelength . Therefore, the cube in (b) has the greater volume.

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Example Question #32 : Solid Geometry

Which is the greater quantity?

(a) The volume of a cube with sidelength inches.

(b) The volume of a sphere with radius inches.

Possible Answers:

It is impossible to tell from the information given.

(a) and (b) are equal.

(a) is greater.

(b) is greater.

Correct answer:

(a) is greater.

Explanation:

You do not need to calculate the volumes of the figures. All you need to do is observe that a sphere with radius inches has diameter inches, and can therefore be inscribed inside the cube with sidelength inches. This give the cube larger volume, making (a) the greater quantity.

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Example Question #3 : How To Find The Volume Of A Sphere

Which is the greater quantity?

(a) The volume of a sphere with diameter one foot

(b) $864 \textrm{ in}^{3}$

Possible Answers:

(a) is greater.

It is impossible to tell from the information given.

(b) is greater.

(a) and (b) are equal.

Correct answer:

(a) is greater.

Explanation:

The radius of the sphere is one half of its diameter of one foot, which is six inches, so substitute r = 6 :

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

$V = \frac{4}{3} \pi \cdot 6 ^{3}$

$V = \frac{4}{3} \pi \cdot 216$

$V = 288 \pi \approx 288 \cdot 3.14 = 904.32$ cubic inches,

which is greater than $864 \textrm{ in}^{3}$ .

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Example Question #4 : How To Find The Volume Of A Sphere

is a positive number. Which is the greater quantity?

(A) The volume of a cube with edges of length

(B) The volume of a sphere with radius

Possible Answers:

(B) is greater

(A) is greater

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

(A) and (B) are equal

Correct answer:

(A) is greater

Explanation:

No calculation is really needed here, as a sphere with radius - and, subsequently, diameter - can be inscribed inside a cube of sidelength . This makes (A), the volume of the cube, the greater.

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Example Question #6 : Spheres

Which is the greater quantity?

(a) The radius of a sphere with surface area $36 \pi$

(b) The radius of a sphere with volume $36 \pi$

Possible Answers:

It cannot be determined which of (a) and (b) is greater

(a) is the greater quantity

(a) and (b) are equal

(b) is the greater quantity

Correct answer:

(a) and (b) are equal

Explanation:

The formula for the surface area of a sphere, given its radius , is

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

The sphere in (a) has surface area $36 \pi$ , so

$36 \pi = 4 \pi r^{2}$

$36 \pi\div 4 \pi = 4 \pi r^{2} \div 4 \pi$

$9 = r^{2}$

$r = \sqrt{9} = 3$

The formula for the volume of a sphere, given its radius , is

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

The sphere in (b) has volume $36 \pi$ , so

$36 \pi = \frac{4}{3} \pi r^{3}$

$\frac{3} {4} \cdot 36 \pi = \frac{3} {4} \cdot \frac{4}{3} \pi r^{3}$

$27 \pi = \pi r^{3}$

$27 \pi\div \pi = \pi r^{3} \div \pi$

$27 = r^{3}$

$r = \sqrt[3]{27} = 3$

The radius of both spheres is 3.

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Example Question #6 : Spheres

In terms of $\pi$ , give the surface area, in square inches, of a spherical water tank with a diameter of 20 feet.

Possible Answers:

$18,432,000\pi\textrm{ in}^{2}$

$57,600 \pi\textrm{ in}^{2}$

$2,304,000 \pi \textrm{ in}^{2}$

$230,400 \pi\textrm{ in}^{2}$

$307,200\pi\textrm{ in}^{2}$

Correct answer:

$57,600 \pi\textrm{ in}^{2}$

Explanation:

feet = $20 \times 12 = 240$ inches, the diameter of the tank; half of this, or 120 inches, is the radius. Set r = 120 , substitute in the surface area formula, and solve for :

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

$A =4\pi \cdot 120^{2}$

$A =4\pi \cdot 14,400$

$A = 57,600 \pi$

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ISEE Upper Level Quantitative : Solid Geometry

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #322 : Geometry

Example Question #1 : Tetrahedrons

Example Question #1 : How To Find The Radius Of A Sphere

Example Question #31 : Solid Geometry

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

Example Question #32 : Solid Geometry

Example Question #3 : How To Find The Volume Of A Sphere

Example Question #4 : How To Find The Volume Of A Sphere

Example Question #6 : Spheres

Example Question #6 : Spheres

All ISEE Upper Level Quantitative Resources

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