Tips for Using the Math Benchmark Assessments · PDF fileTips for Using the QualityCore®...

© 2008 by ACT, Inc. Permission granted to reproduce this page for QualityCore® educational purposes only.

Tips for Using the QualityCore® Mathematics Benchmark Assessments Each QualityCore® course has its own set of Benchmark Assessments based on the QualityCore Formative Item Pool. Algebra I has four Benchmark Assessments and Algebra II, Geometry, and Precalculus each have five Benchmark Assessments. Each assessment consists of 15 to 25 multiple-choice items and one constructed-response item. The assessments are presented as a PDF file to maintain the visual consistency of graphics, special characters, and symbols. Each assessment is “bookmarked” for easy navigation through the PDF file. The PDF file also contains the corresponding QualityCore Reference Sheet. Each Benchmark Assessment is introduced by a cover sheet displaying the item Identification Number (ID), the correct answer (Key), the cognitive level, and the alphanumeric code for each ACT Course Standard covered by that item. (See the applicable ACT Course Standards document.) The scoring criteria and a scoring rubric follow the constructed-response item.

QualityCore® Reference SheetPrecalculus

Triangles

Law of Sines = =

Law of Cosines a 2 = b 2 + c 2 − 2bc cos A

Area of a Triangle Area = bc sin A

Conic Sections

Circle (x − h)2 + (y − k)2 = r 2

Parabola, y = a(x − h)2 + k

opening vertically

Parabola,x = a(y − k)2 + h

opening horizontally

Ellipse,major axis horizontal

Ellipse,major axis vertical

Area of an ellipse A = πab

Hyperbola,transverse axis horizontal

Hyperbola,transverse axis vertical

Sequences and Series

Arithmetic Sequence an = a1 + (n − 1)d

Arithmetic Series sn = (a1 + an)

Geometric Sequence an = a1�r n − 1�

Finite Geometric Series sn = where r ≠ 1

Infinite Geometric Series s = where ⎪r⎪ < 1

Exponential Functions

Discretely Compounded Interest

ContinuouslyCompounded Interest

Discrete, ContinuousExponential Growth

a1_____1 − r

a1 − a1rn

________1 − r

n__2

1__2

c_____sin C

b_____sin B

a_____sin A

(h,k) = center, r = radius

axis of symmetry x = h

focus �h, k + �, directrix y = k −

axis of symmetry y = k

focus �h + , k�, directrix x = h −

foci (h ± c, k) where c 2 = a 2 − b 2

foci (h, k ± c) where c 2 = a 2 − b 2

foci (h ± c, k) where c 2 = a 2 + b 2

foci (h, k ± c) where c 2 = a 2 + b 2

an = nth terma1 = first termn = number of the termd = common differencer = common ratiosn = sum of the first n termss = sum of all the terms

1___4a

1___4a

1___4a

1___4a

continued

s = semi-perimeter = (a + b + c)_________2

b

aC

A

c

B

�Area = ��s(s − a)(s − b)(s − c)

A = amount of money after t yearsp = starting principal r = interest rate

n = compound periods per yeart = number of years e ≈ 2.718

Nt = value after t time periodsr = rate of growth t = time periods

+ = 1, a > b

+ = 1, a > b

− = 1

− = 1

A = p �1 + �nt

A = pert

Nt = N0(1 + r)t, Nt = N0ert

r__n

(x − h)2________

b 2

(y − k)2________

a 2

(y − k)2________

b 2

(x − h)2________

a 2

(x − h)2________

b 2

(y − k)2________

a 2

(y − k)2________

b 2

(x − h)2________

a 2

Polar Coordinates and Vectors

De Moivre’s Theorem [r(cos θ + i sin θ)]n = r n(cos nθ + i sin nθ)

Conversion: Polar to x = r cos θRectangular Coordinates y = r sin θ

Conversion: Rectangularto Polar Coordinates

Product of ComplexNumbers in Polar Form

Inner Product of Vectors

Matrices

Determinant of a 2×2 Matrix det� � = ad − bc

Determinant of a 3×3 Matrix det� � = a•det� � − b•det� � + c•det� �

Inverse of a 2×2 Matrix M −1 = � � where M = � �

Trigonometry

Sum and Difference Identities sin(α ± β ) = sin α cos β ± cos α sin βcos(α ± β ) = cos α cos β sin α sin β

tan(α ± β ) =

Double-Angle Identities sin 2θ = 2 sin θ cos θcos 2θ = cos2 θ − sin2 θ

tan 2θ =

Half-Angle Identities sin = ±

cos = ±

tan = ± , where cos α ≠ −11 − cos α_________1 + cos α

α__2

1 + cos α_________2

α__2

1 − cos α_________2

α__2

2 tan θ_________1 − tan2 θ

tan α ± tan β_____________1 ± tan α tan β

±bd

ac

−ca

d−b

1______det M

eh

dg

fj

dg

fj

eh

cfj

beh

adg

bd

ac

r = radius, distance from originθ = angle in standard positionn = exponent

a = ⟨a1,a2⟩ vector in the planea = ⟨a1,a2,a3⟩ vector in space

α, β, θ = angles, frompositive x-axis

r = , θ = arctan , when x > 0

θ = arctan + π, when x < 0

r1(cos θ1 + i sin θ1)•r2(cos θ2 + i sin θ2) =r1r2[cos(θ1 + θ2) + i sin(θ1 + θ2)]

a•b = a1b1 + a2b2 + a3b3

y__x

y__x

��x 2 + y 2

±

© 2010 by ACT, Inc. All rights reserved.14195 *0190D1080* Rev 2

QualityCore® Benchmark Assessment Precalculus – Benchmark 1 Polynomial Expressions, Equations, and

Functions; Sequences and Series The following pages contain one of the Benchmark Assessments for this course. The table below gives the ID number for each item, the correct answer (Key), the cognitive level, andthe alphanumeric code for each ACT Course Standard measured by the item. (The language associated with each code appears in the ACT Course Standards document for this course.) The items in this PDF file appear in the order presented in the table. Multiple-choice (MC)directions follow the table and are followed by a name sheet and the MC items. Following the MC items, you will find a constructed-response (CR) item followed by itsscoring criteria and/or scoring rubric. DO NOT DISTRIBUTE SCORING CRITERIA TOSTUDENTS. The scoring rubric can be included or excluded at your discretion.

ID Key Cognitive

Level Standard 00469 D L1 E.1.b 00475 D L1 E.2.a 00487 C L1 E.2.c 00461 D L2 E.1.a 00472 C L2 E.1.b 00478 B L2 E.2.a 00483 C L2 E.2.b 00484 D L2 E.2.e 00525 C L2 E.2.f 00528 B L2 E.2.g 00713 B L2 G.2.a 00716 A L2 G.2.b 00463 A L3 E.1.a 00488 D L3 E.2.c 00499 A L3 E.2.d 00502 D L3 E.2.e 00524 A L3 E.2.e 00711 C L3 G.2.a 00715 D L3 G.2.b 00719 A L3 G.2.c 00726 - L3 B.1.c

C.1.b E.1.a E.1.b


Directions: Solve each problem, choose the best answer, and then circle the corresponding letter. Do not linger over problems that take too much time. Solve as many as you can; then return to the others in the time you have left for this test. You are permitted to use a calculator on this test. You may use your calculator for any problems you choose, but some of the problems may best be solved without using a calculator. Note: Unless otherwise stated, all of the following assumptions apply to these problems. 1. Illustrative figures are NOT necessarily drawn to scale. 2. Geometric figures lie in a plane. 3. The word line indicates a straight line. 4. The word average indicates the arithmetic mean.


Name: Date:

Teacher: Class/Period:

1)

2)

3)

4)

5)

6)

7)

8)

9)

10)

11)

12)

13)

14)

15)

16)

17)

18)

19)

20)

21)

Scoring Criteria:

A 4-point response may include, but is not limited to, the following points:

A. Correct sketch:

Correct polynomial equation: 32 = πx3 + 5πx2 or πx3 + 5πx2 – 32 = 0

Appropriate work needed to find the equation: V = πr 2h 32 = πx2(x + 5)

B. Solutions of the equation: x = –4.50, –1.79, and 1.27

Explanation of how the answers were found: First, I rewrote the equation in the form πx3 + 5πx2 – 32 = 0 and then set the expression on the left side equal to y. This yielded y = πx3 + 5πx2 – 32. Then, I graphed y = πx3 + 5πx2 − 32 on my calculator. This produced a graph that crossed the x-axis 3 times. Finally, I used the CALC zero function of the calculator to find each zero.

Note: Depending on the type of calculator the students use, the solutions and explanation will vary.

C. Radius and height of the can: Radius = 1.27 inches, height = 6.27 inches

Explanation of how the answers were found: The solutions –4.5 and –1.79 must be excluded because they are negative lengths for the radius, which is not possible. Since x = 1.27, this produces a can with a radius of 1.27 inches. I know the height is 5 inches more than the radius, so I added 5 to x to get the height.

21)

Rubric: 4 A response at this level provides evidence of thorough knowledge and

understanding of the subject matter. • The response addresses all parts of the question or problem correctly. • The response demonstrates efficient and accurate use of appropriate procedures. • The explanation of strategies used in the response shows evidence of a good

understanding of mathematical concepts and principles, and it does not contain any

misconceptions. • The explanation in the response is clear and coherent.

3 A response at this level provides evidence of competent knowledge and understanding of the subject matter. • The response addresses most parts of the question or problem correctly.

• The response includes some minor errors but generally uses appropriate procedures

accurately.

• The explanation of strategies used in the response shows some evidence of a good

understanding of mathematical concepts and principles, and it contains few, if any,

misconceptions.

• The explanation in the response is mostly clear and coherent.

2 A response at this level provides evidence of a basic knowledge and understanding of the subject matter. • The response addresses some parts of the question or problem correctly. • The response includes a number of errors but demonstrates some use of

appropriate procedures. • The explanation of strategies used in the response shows a little evidence of

understanding of mathematical concepts and principles, but it may contain some

evidence of misconceptions. • The explanation in the response is partially clear, but some parts may be difficult to

understand.

1 A response at this level provides evidence of minimal knowledge and understanding of the subject matter. • The response addresses a few parts of the problem correctly, but the response is

mostly incorrect.

• The response includes inappropriate procedures or simple manipulations that show

little or no understanding of correct procedures.

• The explanation of strategies used in the response shows little or no evidence of

understanding of mathematical concepts and principles, and it may contain evidence

of significant misconceptions.

• Many parts of the explanation are difficult to understand.

0 A response at this level is not scorable. The response is off-topic, blank, hostile, or

otherwise not scorable.

QualityCore® Benchmark Assessment Precalculus – Benchmark 2 Conic Sections

The following pages contain one of the Benchmark Assessments for this course. The table below gives the ID number for each item, the correct answer (Key), the cognitive level, andthe alphanumeric code for each ACT Course Standard measured by the item. (The language associated with each code appears in the ACT Course Standards document for this course.) The items in this PDF file appear in the order presented in the table. Multiple-choice (MC)directions follow the table and are followed by a name sheet and the MC items. Following the MC items, you will find a constructed-response (CR) item followed by itsscoring criteria and/or scoring rubric. DO NOT DISTRIBUTE SCORING CRITERIA TOSTUDENTS. The scoring rubric can be included or excluded at your discretion.

ID Key Cognitive

Level Standard 00426 D L1 C.1.b 00450 C L1 D.1.d 00423 C L2 C.1.a 00425 D L2 C.1.a 00429 D L2 C.1.b 00442 B L2 D.1.b 00451 B L2 D.1.d 00454 D L2 D.1.e 00430 B L3 C.1.b 00433 C L3 C.1.b 00432 C L3 C.1.b 00439 A L3 D.1.a 00446 A L3 D.1.b 00445 A L3 D.1.b 00456 B L3 D.1.e 00740 - L3 B.1.c

D.1.e


Name: Date:


1)

5)

6)

7)

8)

10)

11)

12)

13)

14)

15)

Scoring Criteria:


A. Correct equation of the ellipse: ( )2236

x − + ( )2416

y − = 1

Appropriate work needed to find the answer:

Center of ellipse falls at 4 8 4 4,2 2− + +⎛ ⎞⎜ ⎟

⎝ ⎠ = (2,4) midpoint of major axis

Length of major axis = 12 ft a = 122 = 6 a2 = 36

Length of minor axis = 8 ft b = 82 = 4 b2 = 16

B. Coordinates of the foci of the ellipse: ( ).2 47,4− and ( ).6 47,4


c2 = 36 – 16 = 20

c = 20

c ≈ 4.47

Explanation of how the answer was found: The foci are ( ),h c k± where c2 = a2 – b2 and (h,k) is the center of the ellipse. To solve for c2, I substituted 36 and 16 for a2 and b2 that I found in Part A. Then, I wrote the coordinates of the foci based on the coordinates of the center, which I found in Part A as well.

C. Sum of the lengths of the two line segments: 12 ft


Distance from F1 to point on major axis (8,4) = 8 – ( )2 2 5− = 6 + 2 5

Distance from F2 to point on major axis (8,4) = 8 – ( )2 2 5+ = 6 – 2 5

( )6 2 5+ + ( )6 2 5−

16)

Explanation of how the answer was found: Since, by definition, the sum of the distances from the foci to any point on the ellipse is constant, I used the sum of the distances from the foci to one of the endpoints of the major axis, (8,4) to find the sum of the lengths. I found the lengths from both foci to (8,4), and then added the two lengths together.

Note: Students may use any point on the ellipse.

Rubric:

4 A response at this level provides evidence of thorough knowledge and


understanding of mathematical concepts and principles, and it does not contain any misconceptions.

• The explanation in the response is clear and coherent. 3 A response at this level provides evidence of competent knowledge and

understanding of the subject matter. • The response addresses most parts of the question or problem correctly. • The response includes some minor errors but generally uses appropriate procedures

accurately. • The explanation of strategies used in the response shows some evidence of a good

understanding of mathematical concepts and principles, and it contains few, if any, misconceptions.

• The explanation in the response is mostly clear and coherent. 2 A response at this level provides evidence of a basic knowledge and

understanding of the subject matter. • The response addresses some parts of the question or problem correctly. • The response includes a number of errors but demonstrates some use of


understanding of mathematical concepts and principles, but it may contain some evidence of misconceptions.

• The explanation in the response is partially clear, but some parts may be difficult to understand.

1 A response at this level provides evidence of minimal knowledge and

understanding of the subject matter. • The response addresses a few parts of the problem correctly, but the response is

mostly incorrect. • The response includes inappropriate procedures or simple manipulations that show

little or no understanding of correct procedures. • The explanation of strategies used in the response shows little or no evidence of

understanding of mathematical concepts and principles, and it may contain evidence of significant misconceptions.

• Many parts of the explanation are difficult to understand. 0 A response at this level is not scorable. The response is off-topic, blank, hostile, or



QualityCore® Benchmark Assessment Precalculus – Benchmark 3 Advanced Functions


ID Key Cognitive

Level Standard 00529 D L1 F.2.a 00536 A L1 F.2.b 00507 A L2 F.1.a 00516 B L2 F.1.b 00534 C L2 F.2.a 00542 D L2 F.2.b 00549 A L2 F.2.c 00562 C L2 F.2.e 00673 B L2 F.2.f 00513 A L3 F.1.a 00517 B L3 F.1.b 00548 D L3 F.2.c 00558 B L3 F.2.d 00564 C L3 F.2.e 00677 C L3 F.2.f 00728 - L3 F.2.f


Name: Date:


1)

2)

3)

4)

5)

6)

7)

8)

9)

10)

11)

12)

13)

14)

15)

16)

Scoring Criteria:


A. Number of years she should leave the money in the account: 7.5 yr

Appropriate work needed to find the answer: 2500 = 2000(1 + 0.03

4)4t

2500

2000 = (1 + 0.03

4)4t

log ( )5

4 = log(1 + 0.03

4)4t

log ( )5

4 = 4t log(1 + 0.03

4)

4t =

( )5

4

0.03

4

log

log 1 +

4t = 29.864

t = 7.466

B. Correct answer: No

Explanation needed to find the answer: When Janie first invested the money, her

starting principal was only $2,000. It took 7.5 yr to earn $500. If she left the money

in the account to earn an additional $500, her starting principal would now be

$2,500. As a result, it would take less than 7.5 yr to earn an additional $500. The

longer money stays in the account, the more quickly the principal increases.

Therefore, more interest is earned each time it is compounded.

Note: Students may find the actual time (13.6 yr) it would take to earn $1,000 to

show that it is not twice as long.

C. Amount she would have to invest: $3,979.31


( ) ( ).. 4 7 503

41000 1P P+ = +

P + 1000 = 1.2513P

1000 = .2513P

P = 3979.3076

16)







accurately.



misconceptions.






understand.


mostly incorrect.









QualityCore® Benchmark Assessment Precalculus – Benchmark 4 Trigonometric and Periodic Functions


ID Key Cognitive

Level Standard 00613 C L1 F.3.e 00699 D L1 F.3.l 00598 C L2 F.3.a 00602 C L2 F.3.b 00609 B L2 F.3.d 00619 D L2 F.3.f 00572 B L2 F.3.g 00682 A L2 F.3.h 00690 A L2 F.3.j 00695 B L2 F.3.k 00701 D L2 F.3.l 00599 B L3 F.3.a 00575 C L3 F.3.c 00607 A L3 F.3.c 00611 D L3 F.3.d 00616 C L3 F.3.e 00621 C L3 F.3.f 00683 B L3 F.3.h 00687 D L3 F.3.i 00693 A L3 F.3.j 00734 - L3 F.3.i

F.3.j


Name: Date:


1)

2)

3)

4)

6)

7)

8)

9)

10)

11)

12)

13)

14)

15)

16)

17)

18)

19)

20)

21)

Scoring Criteria:


Correct answer: x = 0, π; 32π , 2π


sin2 x + cos2 x = 1

cos2 x = 1 – sin2 x

sin x – cos2 x + 1 = 0 sin x – (1 – sin2 x) + 1 = 0 sin x – 1 + sin2 x + 1 = 0 sin2 x + sin x = 0 sin x(sin x + 1) = 0 sin x = 0 or sin x = –1

Explanation of how the answer was found: I used the Pythagorean identity sin2 x + cos2 x = 1. First, I solved the identity for cos2 x. Then, I substituted 1 – sin2 x for cos2 x. This altered the equation so all the trigonometric functions were in terms of sin x. The equation could then be factored and solved.

Note: Students can also solve the identity for sin2 x, substitute sin2 x for 1 − cos2 x, and continue from there.

21)







accurately.



misconceptions.






understand.


mostly incorrect.









QualityCore® Benchmark Assessment Precalculus – Benchmark 5 Polar Coordinates and Vectors


ID Key Cognitive

Level Standard 00628 C L1 I.1.a 00631 A L2 I.1.a 00634 D L2 I.1.b 00639 C L2 I.1.c 00646 C L2 I.1.e 00657 C L2 I.1.g 00660 A L2 I.1.h 00664 D L2 I.1.i 00669 D L2 I.1.j 00637 B L3 I.1.b 00644 B L3 I.1.d 00652 A L3 I.1.f 00656 B L3 I.1.g 00663 C L3 I.1.h 00668 B L3 I.1.i 00739 - L3 I.1.f


Name: Date:


1)

2)

3)

4)

5)

6)

7)

8)

9)

10)

11)

12)

13)

14)

Scoring Criteria:


A. Correct answers: u = 4,3 and v = −7, 2


u = – – 5 1,5 2

v = − − −13 6, 1 1

Explanation of how answer was found: To get u and v in component form, I had

to find the change in both the x and y directions. Thus, I subtracted the coordinates

of the initial points from the terminal points for both u and v .

B. Correct answers: u + v = 11,1 , u – v = −3,5


u + v = 4,3 + –7, 2

= (–+ +4 7,3 2)

u – v = 4,3 – –7, 2

= ( ) –− −4 7,3 2

C. Correct graph:

16)







accurately.



misconceptions.






understand.


mostly incorrect.









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