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ISEE Upper Level Quantitative : ISEE Upper Level (grades 9-12) Quantitative Reasoning

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

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Example Questions

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

Which is the greater quantity?

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

(b) 2

Possible Answers:

(b) is greater.

(a) and (b) are equal.

(a) is greater.

It is impossible to tell from the information given.

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 $\displaystyle s$, is

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

Substitute s = 1 $\displaystyle s = 1$ :

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

$\sqrt{3} \approx 1.7 < 2$ $\displaystyle \sqrt{3} \approx 1.7 < 2$, 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:

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

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

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

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

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

Correct answer:

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

Explanation:

The volume $\displaystyle V$ of a sphere with radius $\displaystyle r$ is

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

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

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

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

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

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

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

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

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

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

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Example Question #32 : Solid Geometry

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

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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 $\displaystyle t$

(b) The volume of a cube with sidelength $\displaystyle 2t$

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 sphere with radius $\displaystyle t$ has diameter $\displaystyle 2t$ and can be inscribed inside a cube of sidelength $\displaystyle 2t$. Therefore, the cube in (b) has the greater volume.

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Example Question #2 : How To Find The Volume Of A Sphere

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:

(b) is greater.

It is impossible to tell from the information given.

(a) and (b) are equal.

(a) 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) $864 \textrm{ in}^{3}$ $\displaystyle 864 \textrm{ in}^{3}$

Possible Answers:

(a) is greater.

(b) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

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 $\displaystyle r = 6$:

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

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

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

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

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

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

$\displaystyle t$ is a positive number. Which is the greater quantity?

(A) The volume of a cube with edges of length $\displaystyle 2t$

(B) The volume of a sphere with radius $\displaystyle t$

Possible Answers:

(B) is greater

(A) and (B) are equal

(A) is greater

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

Correct answer:

(A) is greater

Explanation:

No calculation is really needed here, as a sphere with radius $\displaystyle t$ - and, subsequently, diameter $\displaystyle 2t$ - can be inscribed inside a cube of sidelength $\displaystyle 2t$. 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 $36 \pi$ $\displaystyle 36 \pi$

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

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 $\displaystyle r$, is

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

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

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

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

$9 = r^{2}$ $\displaystyle 9 = r^{2}$

$r = \sqrt{9} = 3$ $\displaystyle r = \sqrt{9} = 3$

The formula for the volume of a sphere, given its radius $\displaystyle r$, is

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

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

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

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

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

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

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

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

The radius of both spheres is 3.

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

$A = 57,600 \pi$ $\displaystyle A = 57,600 \pi$

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Example Question #331 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Which is the greater quantity?

(a) The surface area of a sphere with radius 1

(b) 12

Possible Answers:

(a) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

(b) is greater.

Correct answer:

(a) is greater.

Explanation:

The surface area of a sphere can be found using the formula

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

The surface area of the given sphere can be found by substituting r = 1 $\displaystyle r = 1$:

$A = 4\pi \cdot 1^{2} = 4\pi$ $\displaystyle A = 4\pi \cdot 1^{2} = 4\pi$

$\pi > 3$ $\displaystyle \pi > 3$ so $4 \times \pi >4 \times 3$ $\displaystyle 4 \times \pi >4 \times 3$, or $4\pi > 12$ $\displaystyle 4\pi > 12$

This makes (a) greater.

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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : ISEE Upper Level (grades 9-12) Quantitative Reasoning

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

All ISEE Upper Level Quantitative Resources

Example Questions

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

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

Example Question #32 : Solid Geometry

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

Example Question #2 : How To Find The Volume Of A Sphere

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 #1 : How To Find The Surface Area Of A Sphere

Example Question #331 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

All ISEE Upper Level Quantitative Resources

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