Tips for Using the Math Benchmark Assessments

© 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 SheetGeometry

Area, Volume, and Surface Area of Polygons and Solids

Triangle A = bh

Parallelogram A = bh

Trapezoid A = (b1 + b2)h

Regular Polygon A = ap

Regular Prism V = BhSA = 2B + Ph

Right Circular Cylinder V = π r 2hSA = 2π r 2 + 2π rh

Pyramid V = Bh

SA = B + Pl

Right Circular Cone V = π r 2h

SA = π r 2 + π rl

Sphere V = π r 3

SA = 4π r 2

Angles of Polygons

Sum of Degree Measures of the 180(n − 2) n = number of sidesInterior Angles of a Polygon

Degree Measure of an Interior Angleof a Regular Polygon

Right Triangles

Right Triangle Trigonometry sin A =

cos A =

tan A =

Pythagorean Theorem a2 + b2 = c2

a__b

b__c

a__c

180(n − 2)_________n

4__3

1__3

1__2

1__3

1__2

1__2

1__2

A = areab = baseh = heighta = apothemp = perimeterV = volumeB = area of baseSA = surface areaP = perimeter of baser = radiusl = slant heightπ ≈ 3.14

continued

b

aC

A

c

B

Given:

Equations of a Line

Standard Form Ax + By = C

Slope-Intercept Form y = mx + b

Point-Slope Form y − y1 = m(x − x1)

Coordinate Geometry

Slope m =

Midpoint M = � , �Distance d =

Circles

Equation of a Circle (x − h)2 + (y − k)2 = r 2

Area Formula A = π r 2

Circumference Formula C = πd = 2π r

Area of a Sector with Central Angle θ A = π r 2θ____360

��(x2 − x1)2 + (y2 − y1)

2

y1 + y2_______2

x1 + x2_______2

y2 − y1_______x2 − x1

Key to Symbols

∆ABC . . . . . . . . . . . . . . . triangle ABC

∠ABC . . . . . . . . . . . . . . . angle ABC

m∠ABC . . . . . . . . . . . . . the degree measure of angle ABC

AB↔

. . . . . . . . . . . . . . . . . line AB

AB___

. . . . . . . . . . . . . . . . . line segment AB

AB . . . . . . . . . . . . . . . . . the length of line segment AB

Circle O . . . . . . . . . . . . . circle with center point O

AB� . . . . . . . . . . . . . . . . . arc AB

⊥ . . . . . . . . . . . . . . . . . . . . is perpendicular to

� . . . . . . . . . . . . . . . . . . . . is parallel to

≅ . . . . . . . . . . . . . . . . . . . . is congruent to

∼ . . . . . . . . . . . . . . . . . . . . is similar to

≈ . . . . . . . . . . . . . . . . . . . . is approximately equal to

A, B, and C are constants withA and B not both equal to zero.(x1,y1) is a point.m = slopeb = y-intercept

(x1,y1) and (x2,y2) are 2 points.m = slopeM = midpointd = distance

center (h,k)r = radiusA = areaC = circumferenced = diameterπ ≈ 3.14

© 2010 by ACT, Inc. All rights reserved.14194 *019246080* Rev 2

QualityCore® Benchmark Assessment Geometry – Benchmark 1 Logic and Proof

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 00326 C L1 C.1.a 00352 D L1 C.1.g 00469 B L1 C.1.h 00321 A L2 C.1.b 00322 B L2 C.1.c 00360 D L2 C.1.d 00327 A L2 C.1.e 00374 D L2 C.1.f 00404 C L2 C.1.i 00325 D L3 C.1.a 00391 A L3 C.1.e 00460 C L3 C.1.b 00819 D L3 C.1.c 00445 B L3 C.1.e 00446 A L3 C.1.h 00932 - L3 C.1.b

C.1.e


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:

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Scoring Criteria: A 4-point response may include, but is not limited to, the following points: A. To use indirect proof, you assume that the conclusion is false; in this case you assume

none of the 3 integers is odd. Then, to finish the proof, you arrive at a contradiction of the given information; that is, you contradict that when 3 consecutive integers are added together, the sum is odd.

B. First, I assume that none of the 3 integers is odd. Therefore, they must all be even. I will

let 2n represent the smallest even number, since 2 times any number gives an even number. Then, I will add 2 to 2n + 2 twice more to obtain 2 even numbers (2n + 2 and 2n + 4). The sum of the 3 integers is 2n + (2n + 2) + (2n + 4) = 6n + 6. However, 6n + 6 must be even because 6 is an even number, and an even number multiplied by any number is an even number (6n). Adding 6 to 6n still produces an even number because the sum of any 2 even numbers is even. This is a contradiction of the given information, which stated that the sum is odd. Therefore, at least 1 of the integers must be odd.

Note: Other forms of indirect proof are also acceptable.

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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 Geometry – Benchmark 2 Properties of Geometric Figures


ID Key Cognitive

Level Standard 00771 D L1 D.1.d 00764 A L1 D.1.e 00371 D L1 D.2.a 00847 A L1 D.3.a 00436 B L1 D.4.b 00331 A L2 D.1.b 00361 C L2 D.1.f 00576 B L2 D.2.c 00450 C L2 D.2.f 00384 D L2 D.2.g 00397 B L2 D.2.h 00350 B L2 D.2.j 00603 A L2 D.3.d 00340 A L3 D.1.a 00461 D L3 D.2.d 00465 C L3 D.2.e 00355 B L3 D.2.i 00605 A L3 D.3.b 00593 C L3 D.3.c 00820 B L3 D.4.a 00947 - L3 D.2.i

D.2.j D.3.a D.3.b D.3.c





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Scoring Criteria:

A 4-point response may include, but is not limited to, the following points: Correct angle measure: °24 Appropriate work leading to the answer: ∠ = ° − ∠ − ∠ = ° − ° − ° = °180 180 25 25 130m TXV m TVX m VTX ∠ = ° − ∠ = ° − ° = °180 180 130 50m VXW m TXV

= ∠ = ° = °2 2(72 ) 144mTZY m TWY

= ° − = ° − ° = °144 144 106 38mTZ mYZ = ° − − − = ° − ° − ° − ° = °360 360 50 130 144 36mWY mWV mTV mTZY = + = ° + ° = °50 36 86mVWY mWV mWY

( )m R mVWY mTZ∠ = − = ° − ° = °1 1 12 2 2

(86 38 ) (48 )

Explanation of how the answer was found: Since the center of the circle is on TW , the line segment must be a diameter of the circle. Since T and V are on the circle, then TX and VX are radii of the circle, and ΔTVX is an isosceles triangle. Therefore, ∠ = ∠ = °25m TVX m VTX . I used the fact that the sum of the measures of the interior angles of a triangle is °180 to find ∠m TXV . Then, I used the fact that ∠TXV and ∠VXW are supplementary to find m∠VXW . I

used the fact that the measure of an inscribed angle is 12

its intercepted arc to find mTZY . I

subtracted mYZ from mTZY to find mTZ , which is one of the intercepted arcs of the secants. Since the measure of a central angle is equal to its intercepted arc, = °50mVW and

= °130mTV . Then, I subtracted all the arc measures from °360 to find mWY . Then, I added mVW and mWY to find mVWY , the other intercepted arc of the secants. I used the fact that the measure of an angle formed by 2 secants that meet at a point outside the circle is 1

2 the

difference of the intercepted arcs to find ∠m R .

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QualityCore® Benchmark Assessment Geometry – Benchmark 3 Comparing Geometric Figures


ID Key Cognitive

Level Standard 00765 B L1 E.1.a 00865 D L1 E.1.b 00767 A L1 E.1.e 00421 D L1 E.1.f 00486 D L1 E.1.h 00774 A L2 E.1.a 00351 A L2 E.1.b 00455 C L2 E.1.c 00481 D L2 E.1.d 00905 D L2 E.1.e 00437 A L2 E.1.f 00577 C L2 E.1.f 00478 B L2 E.1.g 00462 B L2 E.1.h 00487 D L2 E.1.h 00780 A L3 E.1.a 00482 C L3 E.1.d 00777 B L3 E.1.e 00457 C L3 E.1.g 00497 C L3 E.1.h 00943 L3 E.1.c

E.1.d E.1.f





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Scoring Criteria:

A 4-point response may include, but is not limited to, the following points: A. Correct proportion: 8

481 756x=

Correct length: 12.6 inches

Appropriate work leading to the answer: 8

481 756x=

=481 8(756)x 6048481

12.6x = ≈

Explanation of why the proportion was set up the way it was: I made two fractions from the dimensions of the model over the corresponding dimensions of the actual pyramid. Note: Students may use and describe different correct proportions.

B. Correct volume: 91,722,887.7 ft3

Appropriate work leading to the answer: 4818

60.125=

= 3422(60.125)V Explanation of how the answer was found: I found the scale factor by dividing the height of the actual pyramid by the height of the model. Then, I multiplied the volume of the model by the cube of the scale factor since volume is the product of 3 dimensions.

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QualityCore® Benchmark Assessment Geometry – Benchmark 4 Coordinate Geometry


ID Key Cognitive

Level Standard 00438 B L1 F.1.a 00422 B L1 F.1.d 00200 C L1 H.1.a 00599 D L1 H.1.b 00845 A L1 H.1.c 00415 D L2 F.1.a 00416 D L2 F.1.b 00429 A L2 F.1.c 00439 A L2 F.1.e 00578 B L2 F.2.c 00431 D L3 F.1.e 00846 C L3 F.2.a 00419 C L3 F.2.c 00464 C L3 H.1.a 00585 B L3 H.1.b 00939 - L3 F.1.a

F.1.c F.1.e





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Scoring Criteria: A 4-point response may include, but is not limited to, the following points: A. Correct shape: Sector of the circle

Correct probabilities: Sector of the circle ≈ 16.0%, Rectangle on the board: ≈ 14.6%, Triangle: ≈ 15.6%

Appropriate work leading to the answer:

Sector of the circle: ≈ ≈ ≈2110

360(4) 15.36 0.160 16.0%

8(12) 96

π

Rectangle on the board: = ≈ ≈2(7) 14 0.146 14.6%8(12) 96

Triangle: = ≈ ≈12(5)(6) 15 0.156 15.6%8(12) 96

Explanation of how the answer was found: I found the area of each shape and divided the result by the area of the bigger rectangle. Then, I converted my decimal to a percent. Since the sector of the circle has the greatest area, choosing the sector of the circle gives Keisha the greatest probability of winning.

B. Correct probability: 53.8%

Note: There might be a slightly different answer if the student uses exact values for the areas throughout the problem. Appropriate work leading to the answer:

− − − =100% 16.0% 14.6% 15.6% 53.8%

Explanation of how the answer was found: I took the 3 probabilities that I found in Part A and subtracted them from 100%. Since the entire board represents 100% of the area, subtracting the 3 probabilities gave me the probability of the remaining area, which is where no one could win.

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QualityCore® Benchmark Assessment Geometry – Benchmark 5 Measurement and Trigonometry


ID Key Cognitive

Level Standard 00318 C L1 G.1.a 00343 D L1 G.1.a 00319 C L1 G.1.b 00320 B L1 G.1.c 00587 C L1 G.1.d 00324 A L2 G.1.b 00406 D L2 G.1.c 00601 A L2 G.1.c 00588 D L2 G.1.d 00770 B L2 G.1.e 00344 C L3 G.1.b 00386 C L3 G.1.c 00873 A L3 G.1.d 00783 B L3 G.1.e 00789 D L3 G.1.e 00951 - L3 G.1.e





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Scoring Criteria:

A 4-point response may include, but is not limited to, the following points: Correct image points: ′′X (−1,3), ′′Y (2,−2), and ′′Z (−3,−4) Appropriate work leading to the answer: ′X (−3,−1), ′Y (2,2), and ′Z (4,−3)

Explanation of how the answer was found: First, I reflected ΔXYZ across the y-axis to get

′ ′ ′ΔX Y Z by taking the opposite of each x-coordinate. Then, I rotated ′ ′ ′ΔX Y Z clockwise about the origin to get ′′ ′′ ′′ΔX Y Z by switching the x- and y-coordinates and taking the opposite of the new y-coordinate.

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Tips for Using the Math Benchmark Assessments

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