End-of-Unit Assessment

1.

Select all the numbers that are solutions to the equation $x^3 = 27$ .

A.

$\sqrt{27}$

B.

3

C.

$\sqrt[3]{27}$

D.

$27^3$

E.

9

Answer: B, C

Teaching Notes

Students who select A may be confused between the square root and cube root symbols. Students who select D may need more work understanding what it means for a number to be a solution to an equation. Students who select E may think $x^3$ and $3x$ have the same meaning.

2.

Each of the following gives the lengths, in inches, of the sides of a triangle. Which one is a right triangle?

A.

\sqrt{3}

,

\sqrt{4}

,

\sqrt{5}

B.

3,

\sqrt{7}

, 4

C.

\sqrt{5}

,

\sqrt{12}

, 13

D.

4, 5, 9

Answer: 3, $\sqrt{7}$ , 4

Teaching Notes

Students who select A or C may have ignored the square root symbols. Students who select D may think that

a^2

is the same as

2a

.

3.

Which of these is equal to $0.2\overline{5}$ ?

A.

$\dfrac{25}{99}$

B.

$\dfrac {23}{90}$

C.

$\dfrac14$

D.

$2\dfrac59$

Answer:

$\dfrac {23}{90}$

Teaching Notes

Students who select A may have found the fraction form of $0.\overline {25}$ . Students who select C ignored the repeating decimal bar. Students who select D are giving the fraction form of $\frac{23}{9}$ instead of $\frac{23}{90}$ .

4.

For each number, write the letter of the point that shows its location on the number line.

$\sqrt{2}$ $\hspace{.1in}\underline{\hspace{.5in}}$
$\sqrt[3]{25}$ $\hspace{.1in}\underline{\hspace{.5in}}$
$\sqrt{15}$ $\hspace{.1in}\underline{\hspace{.5in}}$
$\sqrt{25}$ $\hspace{.1in}\underline{\hspace{.5in}}$
$\sqrt[3]{8}$ $\hspace{.1in}\underline{\hspace{.5in}}$
$\sqrt{5}$ $\hspace{.1in}\underline{\hspace{.5in}}$

Answer:

$\sqrt{2}$ : A
$\sqrt[3]{25}$ : D
$\sqrt{15}$ : E
$\sqrt{25}$ : F
$\sqrt[3]{8}$ : B
$\sqrt{5}$ : C

Teaching Notes

Sample reasoning students may use to determine the approximate location of each number:

\sqrt{2}

is between 1 and 2 because 2 is greater than

1^2=1

but less than

2^2=4

.

\sqrt[3]{25}

is a little less than 3 because 25 is a little less than

3^3=27

.

\sqrt{15}

is a little less than 4 because 15 is a little less than

4^2=16

.

\sqrt{25}

is equal to 5, though students may confuse

\sqrt{25}

and

\sqrt[3]{25}

.

\sqrt[3]{8}

is equal to 2, though students may confuse it with

\sqrt{5}

, which is close to the location of

\sqrt[3]{8}

.

\sqrt{5}

is a little greater than 2 since 5 is a little greater than

2^2=4

.

5.

Find the exact length of the segment that joins the points $(\text-5, 4)$ and $(6, \text-3)$ .

A coordinate plane with the origin labeled "O." The x-axis has the numbers negative 7 through 7 indicated. The y-axis has the numbers negative 5 through 5 indicated.

Answer:

$\sqrt{170}$ units

Teaching Notes

Students may graph the points and sketch the segment on the coordinate axes provided, though this is not required. Look out for students making sign errors. Students might use either a horizontal or vertical length of 1, thinking that $4-3$ or $6-5 = 1$ is a correct calculation. Drawing the diagram will make this error much less likely. Students who use signed distances like -11 or -7 and square them incorrectly might wind up with an incorrect answer like $\sqrt{72}$ using $\sqrt{11^2 - 7^2}$ .

6.

Mai’s younger brother tells her that $\frac{10} 7$ is equal to $\sqrt 2$ . Mai knows this can’t be right, because $\frac{10} 7$ is rational and $\sqrt 2$ is irrational. Write an explanation that Mai could use to convince her brother that $\frac{10} 7$ cannot be the square root of 2.

Answer:

Sample responses:

$\frac{10} 7$ squared is $\frac{100}{49}$ , which does not equal 2. Therefore, $\frac{10} 7$ does not equal $\sqrt 2$ .
(With accompanying long division) $\frac{10}7$ is about 1.43, while $\sqrt 2$ is about 1.41, so they cannot be the same number.

Minimal Tier 1 response:

Work is complete and correct.
Sample: $(\frac{10}{7})^2 = \frac{100}{49}$ , which is not 2.

Tier 2 response:

Work shows general conceptual understanding and mastery, with some errors.
Sample errors: Response does not explicitly compare $\frac{100}{49}$ to 2; Response compares $\frac{10}7$ to $\sqrt 2 \approx 1.4$ , which is not an accurate enough approximation of $\sqrt 2$ to show that the two numbers are different; arithmetic error in squaring $\frac{10}7$ ; response simply states that $\frac{10}7$ is rational, but $\sqrt 2$ is irrational.

Tier 3 response:

Significant errors in work demonstrate lack of conceptual understanding or mastery.
Sample errors: Explanation does not appeal to the work of the unit; no explanation.

Teaching Notes

The explanation that most closely follows the development in this unit is the first sample explanation: $(\frac{10}{7})^2 \neq 2$ . Students may also use their knowledge that $\sqrt 2 \approx 1.41$ to argue that the decimal expansion of $\frac{10}{7}$ is not the same.

7.

Elena wonders how much water it would take to fill her cup. She drops her pencil in her cup and notices that it just fits diagonally. (See the diagram.) The pencil is 17 cm long and the cup is 15 cm tall. How much water can the cup hold? Explain or show your reasoning.

(The surface area of a cylinder is $2 \pi r^2 + 2 \pi rh$ . The volume of a cylinder is $\pi r^2 h$ .)

An image of a pencil resting at a diagonal in a cylinder. The top of the pencil touches the top at one side of the cylinder, and the bottom of the pencil touches the opposite side of the cylinder.

Answer:

$240\pi$ cm³, or approximately 754 cm³. The diameter of the cylindrical cup is 8 cm, because $\sqrt{17^2 - 15^2} = 8$ . That means the radius is 4 cm. The amount of water the cup can hold is the volume of the cylinder: $V = \pi \boldcdot 4^2 \boldcdot 15 \approx 754$ cm³.

Minimal Tier 1 response:

Work is complete and correct, with complete explanation or justification.
Sample: $\sqrt{17^2 - 15^2} = 8$ . $V = \pi \boldcdot 4^2 \boldcdot 15 \approx 754$ . The cup can hold about 754 cubic centimeters of water.

Tier 2 response:

Work shows good conceptual understanding and mastery, with either minor errors or correct work with insufficient explanation or justification.
Sample errors: Omission of units in the final answer; correct mathematical calculations but failure to state how much water the cup can hold; use of 8 cm as the radius of the cup.

Tier 3 response:

Work shows a developing but incomplete conceptual understanding, with significant errors.
Sample errors: Correct use of the Pythagorean Theorem to calculate the diameter of the cup, but no significant further progress; calculation of the surface area of the cup rather than the volume; work involves visual estimation of the radius of the cylinder rather than calculation; calculation for the diameter of the cup treats this length as the hypotenuse of the triangle.

Tier 4 response:

Work includes major errors or omissions that demonstrate a lack of conceptual understanding and mastery.
Sample errors: Work uses neither the Pythagorean Theorem nor the volume formula for a cylinder; two or more error types under Tier 3 response.

Teaching Notes

In order to get started, students will need to realize that the pencil, diameter of the cup, and height of the cup form a right triangle. This may be tricky for some students because of the round base of the cup. Students will also need to recognize that this question is asking for a volume.