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

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Example Questions

Example Question #5 : How To Find The Volume Of A Cylinder

What is the volume of a cylinder with a height of $\displaystyle 14$ in. and a radius of 2.5 $\displaystyle 2.5$ in?

Possible Answers:

$87.5\pi$ $\displaystyle 87.5\pi$ in^3 $\displaystyle in^3$

$82.5\pi$ $\displaystyle 82.5\pi$ in^3 $\displaystyle in^3$

$42.25\pi$ $\displaystyle 42.25\pi$ in^3 $\displaystyle in^3$

$70\pi$ $\displaystyle 70\pi$ in^3 $\displaystyle in^3$

$35\pi$ $\displaystyle 35\pi$ in^3 $\displaystyle in^3$

Correct answer:

$87.5\pi$ $\displaystyle 87.5\pi$ in^3 $\displaystyle in^3$

Explanation:

This is a rather direct question. Recall that the equation of for the volume of a cylinder is:

$V = \pi * r^2 * h$ $\displaystyle V = \pi * r^2 * h$

For our values this is:

$A = \pi * 2.5^2 * 14=87.5\pi$ $\displaystyle A = \pi * 2.5^2 * 14=87.5\pi$

This is the volume of the cylinder.

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Example Question #351 : Geometry

What is the radius of a cylinder with a volume of $562.5\pi$ $\displaystyle 562.5\pi$ in^3 $\displaystyle in^3$ and a height of $\displaystyle 10$ $\displaystyle in$?

Possible Answers:

$56.25\pi$ $\displaystyle 56.25\pi$ $\displaystyle in$

56.25 $\displaystyle 56.25$ $\displaystyle in$

$\displaystyle 15$ $\displaystyle in$

$15\pi$ $\displaystyle 15\pi$ $\displaystyle in$

7.5 $\displaystyle 7.5$ $\displaystyle in$

Correct answer:

7.5 $\displaystyle 7.5$ $\displaystyle in$

Explanation:

Recall that the equation of for the volume of a cylinder is:

$V = \pi * r^2 * h$ $\displaystyle V = \pi * r^2 * h$

For our values this is:

$562.5\pi = \pi * r^2 * 10$ $\displaystyle 562.5\pi = \pi * r^2 * 10$

Solve for $\displaystyle r$:

$\displaystyle 56.25=r^2$

Using a calculator to calculate $\sqrt{56.25}$ $\displaystyle \sqrt{56.25}$, you will see that r = 7.5 $\displaystyle r = 7.5$

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Example Question #351 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

What is the surface area of a cylinder of height $\displaystyle 15$ in., with a radius of $\displaystyle 4$ in?

Possible Answers:

$122\pi$ $\displaystyle 122\pi$ in^2 $\displaystyle in^2$

$152\pi$ $\displaystyle 152\pi$ in^2 $\displaystyle in^2$

$120\pi$ $\displaystyle 120\pi$ in^2 $\displaystyle in^2$

$136\pi$ $\displaystyle 136\pi$ in^2 $\displaystyle in^2$

$240\pi$ $\displaystyle 240\pi$ in^2 $\displaystyle in^2$

Correct answer:

$152\pi$ $\displaystyle 152\pi$ in^2 $\displaystyle in^2$

Explanation:

$A = \pi * r^2$ $\displaystyle A = \pi * r^2$

For our problem, this is:

$\pi * 4^2 = \pi * 16$ $\displaystyle \pi * 4^2 = \pi * 16$

You need to double this for the two bases:

$2 * 16 * \pi = 32 * \pi$ $\displaystyle 2 * 16 * \pi = 32 * \pi$

The area of the "outer face" is a little bit trickier, but it is not impossible. It is actually a rectangle that has the height of the cylinder and a width equal to the circumference of the base; therefore, it is:

$2 * \pi * r * h$ $\displaystyle 2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 4 * 15 = 120 * \pi$ $\displaystyle 2 * \pi * 4 * 15 = 120 * \pi$

Therefore, the total surface area is:

$32\pi + 120\pi = 152\pi$ $\displaystyle 32\pi + 120\pi = 152\pi$

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Example Question #2 : Cylinders

What is the surface area of a cylinder having a base of radius $\displaystyle 11$in and a height of $\displaystyle 3$ in?

Possible Answers:

$212\pi$ $\displaystyle 212\pi$ in^2 $\displaystyle in^2$

$308\pi$ $\displaystyle 308\pi$ in^2 $\displaystyle in^2$

$99\pi$ $\displaystyle 99\pi$ in^2 $\displaystyle in^2$

$363\pi$ $\displaystyle 363\pi$ in^2 $\displaystyle in^2$

$33\pi$ $\displaystyle 33\pi$ in^2 $\displaystyle in^2$

Correct answer:

$308\pi$ $\displaystyle 308\pi$ in^2 $\displaystyle in^2$

Explanation:

$A = \pi * r^2$ $\displaystyle A = \pi * r^2$

For our problem, this is:

$\pi * 11^2 = \pi * 121$ $\displaystyle \pi * 11^2 = \pi * 121$

You need to double this for the two bases:

$2 * 121 * \pi = 242 * \pi$ $\displaystyle 2 * 121 * \pi = 242 * \pi$

$2 * \pi * r * h$ $\displaystyle 2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 11* 3 = 66 * \pi$ $\displaystyle 2 * \pi * 11* 3 = 66 * \pi$

Therefore, the total surface area is:

$242 * \pi + 66 * \pi = 308\pi$ $\displaystyle 242 * \pi + 66 * \pi = 308\pi$

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Example Question #3 : Cylinders

What is the surface area of a cylinder with a height of $\displaystyle 7$in. and a diameter of $\displaystyle 4$in?

Possible Answers:

$22\pi$ $\displaystyle 22\pi$ in^2 $\displaystyle in^2$

$57\pi$ $\displaystyle 57\pi$ in^2 $\displaystyle in^2$

$28\pi$ $\displaystyle 28\pi$ in^2 $\displaystyle in^2$

$36\pi$ $\displaystyle 36\pi$ in^2 $\displaystyle in^2$

$14\pi$ $\displaystyle 14\pi$ in^2 $\displaystyle in^2$

Correct answer:

$36\pi$ $\displaystyle 36\pi$ in^2 $\displaystyle in^2$

Explanation:

Recall that to find the surface area of a cylinder, you need to find the surface area of its two bases and then the surface area of its "outer face." The first two are very easy since they are circles. Notice, however that the diameter is $\displaystyle 4$ inches. This means that the radius is $\displaystyle 2$. Now, the equation for one base is:

$A = \pi * r^2$ $\displaystyle A = \pi * r^2$

For our problem, this is:

$\pi * 2^2 = \pi * 4$ $\displaystyle \pi * 2^2 = \pi * 4$

You need to double this for the two bases:

$2 * 4 * \pi = 8 * \pi$ $\displaystyle 2 * 4 * \pi = 8 * \pi$

$2 * \pi * r * h$ $\displaystyle 2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 2 * 7 = 28 * \pi$ $\displaystyle 2 * \pi * 2 * 7 = 28 * \pi$

Therefore, the total surface area is:

$28\pi + 8\pi = 36\pi$ $\displaystyle 28\pi + 8\pi = 36\pi$

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Example Question #51 : Solid Geometry

The volume of a cylinder with height of $\displaystyle 11$ is $275\pi$ $\displaystyle 275\pi$ in^3 $\displaystyle in^3$. What is its surface area?

Possible Answers:

$55\pi$ $\displaystyle 55\pi$ in^2 $\displaystyle in^2$

$135\pi$ $\displaystyle 135\pi$ in^2 $\displaystyle in^2$

$82\pi$ $\displaystyle 82\pi$ in^2 $\displaystyle in^2$

$160\pi$ $\displaystyle 160\pi$ in^2 $\displaystyle in^2$

$66\pi$ $\displaystyle 66\pi$ in^2 $\displaystyle in^2$

Correct answer:

$160\pi$ $\displaystyle 160\pi$ in^2 $\displaystyle in^2$

Explanation:

To begin, we must solve for the radius of this cylinder. Recall that the equation of for the volume of a cylinder is:

$V = \pi * r^2 * h$ $\displaystyle V = \pi * r^2 * h$

For our values this is:

$275\pi = \pi * r^2 * 11$ $\displaystyle 275\pi = \pi * r^2 * 11$

Solving for $\displaystyle r$, we get:

$\displaystyle r^2 = 25$

Hence, r = 5 $\displaystyle r = 5$

Now, recall that to find the surface area of a cylinder, you need to find the surface area of its two bases and then the surface area of its "outer face." The first two are very easy since they are circles. The equation for one base is:

$A = \pi * r^2$ $\displaystyle A = \pi * r^2$

For our problem, this is:

$\pi * 5^2 = \pi * 25$ $\displaystyle \pi * 5^2 = \pi * 25$

You need to double this for the two bases:

$2 * 25 * \pi = 50 * \pi$ $\displaystyle 2 * 25 * \pi = 50 * \pi$

$2 * \pi * r * h$ $\displaystyle 2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 5 * 11 = 110 * \pi$ $\displaystyle 2 * \pi * 5 * 11 = 110 * \pi$

Therefore, the total surface area is:

$50\pi + 110\pi = 160\pi$ $\displaystyle 50\pi + 110\pi = 160\pi$

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Example Question #52 : Solid Geometry

What is the surface area of a cylinder of height $\displaystyle 15$ in, with a radius of $\displaystyle 4$ in?

Possible Answers:

$120\pi$ $\displaystyle 120\pi$ in^2 $\displaystyle in^2$

$122\pi$ $\displaystyle 122\pi$ in^2 $\displaystyle in^2$

$152\pi$ $\displaystyle 152\pi$ in^2 $\displaystyle in^2$

$240\pi$ $\displaystyle 240\pi$ in^2 $\displaystyle in^2$

$136\pi$ $\displaystyle 136\pi$ in^2 $\displaystyle in^2$

Correct answer:

$152\pi$ $\displaystyle 152\pi$ in^2 $\displaystyle in^2$

Explanation:

$A = \pi * r^2$ $\displaystyle A = \pi * r^2$

For our problem, this is:

$\pi * 4^2 = \pi * 16$ $\displaystyle \pi * 4^2 = \pi * 16$

You need to double this for the two bases:

$2 * 16 * \pi = 32 * \pi$ $\displaystyle 2 * 16 * \pi = 32 * \pi$

$2 * \pi * r * h$ $\displaystyle 2 * \pi * r * h$

For our problem, this is:

$2 * \pi * 4 * 15 = 120 * \pi$ $\displaystyle 2 * \pi * 4 * 15 = 120 * \pi$

Therefore, the total surface area is:

$32\pi + 120\pi = 152\pi$ $\displaystyle 32\pi + 120\pi = 152\pi$

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Example Question #1 : Greatest Common Factor

3/5 + 4/7 – 1/3 =

Possible Answers:

3/37

72/89

7/9

88/105

Correct answer:

88/105

Explanation:

We need to find a common denominator to add and subtract these fractions. Let's do the addition first. The lowest common denominator of 5 and 7 is 5 * 7 = 35, so 3/5 + 4/7 = 21/35 + 20/35 = 41/35.

Now to the subtraction. The lowest common denominator of 35 and 3 is 35 * 3 = 105, so altogether, 3/5 + 4/7 – 1/3 = 41/35 – 1/3 = 123/105 – 35/105 = 88/105. This does not simplify and is therefore the correct answer.

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Example Question #1 : How To Find The Greatest Common Factor

25 is the greatest common factor of 175 and which of these numbers?

Possible Answers:

150 $\displaystyle 150$

$\displaystyle 77$

$\displaystyle 90$

$\displaystyle 70$

None of the answers are correct.

Correct answer:

150 $\displaystyle 150$

Explanation:

Of the four numbers given, 25 is only a factor of 150, since all multiples of 25 end in the digits 25, 50, 75, or 00. To determine whether 150 is correct, we inspect the factors of 150 and 175:

Factors of 150: $\left \{ 1, 2, 3,5,6,10,15,\underline{25},30, 50, 75, 150\right \}$ $\displaystyle \left \{ 1, 2, 3,5,6,10,15,\underline{25},30, 50, 75, 150\right \}$

Factors of 175: $\left \{ 1,5,7,\underline{25}, 35,175\right \}$ $\displaystyle \left \{ 1,5,7,\underline{25}, 35,175\right \}$

Since 25 is the greatest number in both lists, $\displaystyle GCF (150,175) = 25$.

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Example Question #3 : How To Find The Greatest Common Factor

$\displaystyle N$ is an odd prime.

Which is the greater quantity?

(a) $\displaystyle GCF (N,16)$

(b) $GCF (N^{4},2)$ $\displaystyle GCF (N^{4},2)$

Possible Answers:

It is impossible to tell from the information given.

(b) is equal.

(a) is greater.

(a) and (b) are equal.

Correct answer:

(a) and (b) are equal.

Explanation:

The greatest common factor of two numbers is the product of the prime factors they share; if they share no prime factors, it is $\displaystyle 1$.

(a) $16 = 2 \times 2\times 2\times2$ $\displaystyle 16 = 2 \times 2\times 2\times2$. Since $\displaystyle N$ is an odd prime, $\displaystyle N$ and $\displaystyle 16$ share no prime factors, and $\displaystyle GCF (N,16) = 1$.

(b) $N^{4} = N \times N \times N \times N$ $\displaystyle N^{4} = N \times N \times N \times N$, since $\displaystyle N$ is prime. Since $\displaystyle 2$ is an even prime, $N^{4}$ $\displaystyle N^{4}$ and $\displaystyle 2$ share no prime factors, and $GCF (N^{4},2) = 1$ $\displaystyle GCF (N^{4},2) = 1$.

The quantities are equal since each is equal to $\displaystyle 1$.

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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 #5 : How To Find The Volume Of A Cylinder

Example Question #351 : Geometry

Example Question #351 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #2 : Cylinders

Example Question #3 : Cylinders

Example Question #51 : Solid Geometry

Example Question #52 : Solid Geometry

Example Question #1 : Greatest Common Factor

Example Question #1 : How To Find The Greatest Common Factor

Example Question #3 : How To Find The Greatest Common Factor

All ISEE Upper Level Quantitative Resources

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