Be Negative about Negatives Negative signs cause more mischief and steal more points than anything else on the ACT Math test. We tell our students to imagine that whenever they see a minus or a negative sign, they should begin to hear creepy background music—the kind that plays in horror movies to let you know the villain is waiting on the other side of the door. (Also see error paranoia on page 45.) Any time your students encounter negative signs, they need to double- and triple-check their work. ACT test writers know that this is a weak point for most students and so they make sure to include answer choices that students who slipped up on negatives would arrive at. There is no partial credit on the ACT. If you are in the habit of giving your juniors and seniors partial credit for math answers where they have swapped up the negatives, you may want to reconsider. It’s better for students to feel the pain from one of your exams than to feel the pain of coming up a few points short on the ACT Math test because they never kicked the habit of being careless with negativity. Whenever you are reviewing a math question that involves negatives, consider asking your students, “What would happen if we messed up the negatives, here?” If the question is multiple choice, ask your students what choice represents getting the negative confused. This will help students build awareness of the fact that just because their answer showed up as a choice doesn’t mean it’s right! Refer to Math question #23 on page 26 of Preparing for the ACT 2015-16. In question #23, the whole expression simplifies down nicely to one term. But if your students drop the minus on the last 2y when they distribute y2, they would end up with two terms and erroneously select choice D. It’s a simple mistake to make, but one that all too often costs your students points. Refer to Math question #39 on page 29 of Preparing for the ACT 2015-16. Question #39 is relatively straightforward, but there is one major opportunity for students to drop a point. Check Your Understanding: How would you coach your students to avoid a costly error on question #39?
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DECODING THE ACT
Popularity Contest When students get stuck and they need to mark and move, they should always try to inform their guess using the popularity contest strategy. This is a powerful way to use the ACT test writers’ tricks against them and make two or three solid eliminations. It doesn’t always work and should only be used as a guessing strategy. But as a guessing strategy it can make the difference of a point or more on your students’ Math scores. Refer to Math question #4 on page 24 of Preparing for the ACT 2015-16. You may notice that all the answer choices look similar on question #4. They are variations on the numbers 45 and 15. There is a very good reason for that: on a typical multiple choice math test, students know they have arrived at an incorrect answer because their answer doesn’t show up as a choice. They are then free to try again! For that reason, ACT Math test writers include answer choices that match what students would arrive at if they made mistakes. They can catch the students in their errors and cost them points. For example, choice H is for students who treat the 3 as a coefficient instead of an exponent. Choices G and J are there in case students mess up resolving the exponent or get the place value wrong in their division. Choice K is there because, well, who knows how students would get there, but maybe it’s possible. The test writers’ method, which costs many students precious points, can be used to the advantage of a student in the know, because the correct answer will have numbers that are similar to it. The reverse is also true: the outlier, the answer that looks completely different, is probably wrong. To apply the popularity contest strategy, go with an answer choice that looks similar to most other answer choices. In the case of question #4, there are three answer choices with the digits 4 and 5, but only two answer choices with the digits 1 and 5. For a student blindly guessing, she would go with choice F, G, or J. Refer to Math question #3 on page 24 of Preparing for the ACT 2015-16. When your students are considering what answer choice wins the popularity contest, it does not matter whether the numbers are numerically close. It only matters if the numbers look similar or have a relationship in some way. For example, in question #3 choice C is twice the amount of choice A. Likewise, choice E is twice the amount of choice D, which is twice the amount of choice B. Choices B, D, and E win the popularity contest, so an educated guess would go to one of them. Refer to Math question #6 on page 24 of Preparing for the ACT 2015-16. With question #6, the popularity contest is rather obvious: three choices (F, G, and H) involve $12, so students should go with one of them if forced to blindly guess. Refer to Math question #11 on page 25 of Preparing for the ACT 2015-16.
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Check Your Understanding: If you had to apply the popularity contest to make a better guess on question #11, what choices would you eliminate? What would be your reasoning?
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DECODING THE ACT
Walk Through the Open Door Most ACT Math problems are multi-step, which means that students are often left with the daunting question, “Where do I begin?” It can be challenging to visualize all of the processes that are needed to get to the right answer. Fortunately, ACT Math questions are designed with minimal distractors. This is a matter of necessity: the test writers have to pack a lot of punch in a few lines of text. This means that if there is an equation your students can set up or a formula they can use, chances are it will be useful in solving the problem. Even if your students don’t see all the way to the end of a solution, if they see a step they can take, they should go ahead with it. We call this walking through the open door. Refer to Math question #9 on page 25 of Preparing for the ACT 2015-16. Question #9 has an open door. It states that the top and bottom layers are each 0.03 cm thick, and they don’t count as inner layers, so students can remove them from the total thickness of 0.32 cm. It’s not guaranteed that this step will bring them all the way to the answer, but this step has to be taken to make any headway, so they should walk through the open door. 0.32 – 0.03 – 0.03 = 0.26 Now students are faced with only one more step between them and the solution: how many 0.02 cm inner layers can fit inside of 0.26 cm? At this point, students can divide or, if they want to be extra careful, draw it out. This strategy gradually increases in importance as students near the more complicated questions in the back of the test. Refer to Math question #54 on page 30 of Preparing for the ACT 2015-16. Question #54 requires several steps and some skill in creating expressions that model real-world situations. Check Your Understanding: What is an open door that students can walk through to start solving question #54?
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MATH CONTENT MASTERY
MATH CONTENT MASTERY Content Strategies for the ACT Math Test
Red Herrings When reviewing content, it’s very important that you avoid pursuing the especially rare or tricky questions that appear on a particular Math test. We call these red herrings. We strongly recommend that you use the skill weights that provided in this section to prioritize your review. If you review red herrings instead of these essential skills, chances are that your students’ scores won’t budge. Even if your students do manage to master the irksome herring, the likelihood that the skill will be assessed on their next ACT Math test is minimal. This is because the ACT Math test has a broad set of standards with an enormous number of associated question variants. As a matter of fact, there are over 100 skills that could be tested on any given ACT Math test. Usually only 40 of those skills show up on any one assessment. This means that many of the challenging questions that appear on an ACT Math test are interesting, but reviewing them won’t increase your students’ scores. This section helps you avoid red herrings and review the content that will positively boost your students’ ACT outcomes. Refer to Math question #29 on page 27 of Preparing for the ACT 2015-16. Question #29 deals with complex numbers. This is a very interesting subject, particularly for math teachers. What’s more, the problem is very neat. The product works out cleanly to cancel out the imaginary numbers. It’s a sort of work of art as far as ACT Math problems go. If your testing data only includes one ACT Math practice test that your students have taken, this question looks important. It might be one of only 15 questions a student missed. It seems like a must-review. According to our research, however, complex numbers show up only once out of every three tests. In other words, most ACT tests don’t have a single i in them! This skill is a red herring that can eat up a massive amount of valuable prep time without giving your students any improvement on their scores. Refer to Math question #57 on page 31 of Preparing for the ACT 2015-16. Trigonometric functions on the coordinate plane such as those presented in question #57, only appear in one ACT Math test out of three.
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DECODING THE ACT
Check Your Understanding: How would you explain the red herring concept to another math teacher in the context of question #57?
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MATH CONTENT MASTERY
Content Area #1: Solving Equations The most heavily weighted standard family on the ACT Math test is Solving Equations. If only these questions were as straightforward as they sound! Many of these problems require students to convert a word problem into one or more equations, then solve to find a value. In some cases, the value requires more manipulation before students can determine the answer. These are the major test items associated with this standard family: • Evaluating expressions • Linear equations • Quadratic equations • Systems of equations
Practice Makes Perfect Refer to Math question #46 on page 29 of Preparing for the ACT 2015-16. Question #46 is one of the more difficult questions of this type. The item sounds like it wants students to set up a proportion, and this signal can cause students to miss the fact that what they really need to do is convert the word problem into an equation. Students must be very comfortable with converting word problems into equations in order to arrive at the correct answer on this question. The only way they can reach that level of skill is through a high quantity of practice. It can be difficult to find enough practice questions of this type, so one exercise we recommend is to take word problems that involve arithmetic (including ACT Math problems) and ask your students to solve the problems using algebra. Ask them, “How could we express this word problem as an equation?” Your students may have no idea how to proceed at the beginning, especially when the option to solve with arithmetic is staring them in the face. Eventually, they will get to the point where they naturally think about word problems in terms of equations. This has the added benefit of focusing your students on setting up word problems before they compute the solution, which reinforces the don’t compute word problems strategy (page 185).
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DECODING THE ACT CATEGORY: ALGEBRA & FUNCTIONS Family: Solving Equations Weight: 8-15% Standards: A 202. Solve equations in the form x + a = b, where a and b are whole numbers or decimals A 301. Substitute whole numbers for unknown quantities to evaluate expressions A 302. Solve one-step equations to get integer or decimal answers A 401. Evaluate algebraic expressions by substituting integers for unknown quantities A 403. Solve routine first-degree equations A 501. Recognize that when numerical quantities are reported in real-world contexts, the numbers are often rounded A 502. Solve real-world problems by using first-degree equations A 506. Identify solutions to simple quadratic equations A 507. Solve quadratic equations in the form (x + a)(x + b) = 0, where a and b are numbers or variables A 508. Factor simple quadratics (e.g., the difference of squares and perfect square trinomials) A 604. Solve systems of two linear equations A 605. Solve quadratic equations A 703. Apply the remainder theorem for polynomials, that P(a) is the remainder when P(x) is divided by (x – a)
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MATH CONTENT MASTERY
Content Area #2: Area, Perimeter, Volume, and Circumference This broad geometry category begins with assessing students’ knowledge of key formulas for area, perimeter, volume, and circumference, but it doesn’t end there. Students must be conversant with determining measures for parts or segments of shapes. They must also be prepared to use algebra to determine the dimensions of the shapes, and from there be able to determine the shapes’ properties. This standard family includes the following topics: • Line segments • Area of triangles and rectangles • Perimeter of polygons • Perimeter of composite shapes • Circles • Dimension relationships The ways that the ACT can assess this content area are wide and varied, so we recommend giving your students as much practice with these types of problems as possible. That way a question format can’t throw them off guard.
Subtraction Refer to Math question #34 on page 28 of Preparing for the ACT 2015-16. Question #34 provides a typical example of a test item from this standard family. Students are required to make several area calculations and select the operations that will bring them all the way to the correct answer. Many geometry problems of this type rely on the ability to subtract the area of one shape from that of another shape. If students are not familiar with this math strategy, the steps to solving questions like this might not be clear. In some cases, it is not so clear that subtraction is needed. Typically in these situations, students will have to compute the area of a larger, complete shape and then subtract what is missing in order to determine the area of a composite figure.
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DECODING THE ACT CATEGORY: GEOMETRY Family: Area, Perimeter, Volume & Circumference Weight: 8-15% Standards: G 201. Estimate the length of a line segment based on other lengths in a geometric figure G 202. Calculate the length of a line segment based on the lengths of other line segments that go in the same direction (e.g., overlapping line segments and parallel sides of polygons with only right angles) G 302. Compute the perimeter of polygons when all side lengths are given G 303. Compute the area of rectangles when whole number dimensions are given G 403. Compute the area and perimeter of triangles and rectangles in simple problems G 505. Compute the perimeter of simple composite geometric figures with unknown side lengths G 506. Compute the area of triangles and rectangles when one or more additional simple steps are required G 507. Compute the area and circumference of circles after identifying necessary information G 601. Use relationships involving area, perimeter, and volume of geometric figures to compute another measure (e.g., surface area for a cube of a given volume and simple geometric probability) G 702. Compute the area of composite geometric figures when planning and/or visualization is required
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DECODING MATH TEST 72-C
23. Algebra & Functions >> Expressions The correct answer is A. Distribute
1 2 y to the terms in parentheses. 2
1 2 y (6x + 2y + 12x – 2y) = 3xy2 + y3 + 6xy2 – y3 2 Combine like terms. 9xy2
24. Algebra & Functions >> Solve Equations The correct answer is H. Set up an equation with the given expression equal to $60,000 and solve for p. 500p – p2 = 60,000 p2 – 500p + 60,000 = 0 (p – 200)(p – 300) = 0 p = 200 or 300 Since we are looking for the fewest number of paintings, 200 is the best choice.
25. Algebra & Functions >> Arithmetic: Percent The correct answer is B. At $254, clothes are Lucie’s greatest expenditure. Divide by the total amount of expenditures, $900. $254 = 0.282 $900 Multiply by 100 to express as a percent, then round to find the answer. 0.282 = 28.2% ≈ 28%
26. Geometry >> Angles & Shapes The correct answer is G. ∠BAC plus ∠CAD must equal 90°, since these two angles make up ∠BAD. We subtract the measure of ∠BAC from 90° to find the measure of ∠CAD. 90° – (x + 20)° = 90° – x°– 20°= 70°– x°= (70 – x)°
27. Geometry >> Triangles The correct answer is E. The sides of an isosceles right triangle adhere to the following proportionality: x, x, x 2 . Since the hypotenuse has a length of 8 2 , the lengths of the other two sides must each be 8. Add the sides to find the perimeter. 8 + 8 + 8 2 = 16 + 8 2
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DECODING THE ACT
28. Algebra & Functions >> Function Graphs – Coordinate Plane The correct answer is H. There are two points on the graph where y = 0. These are the points where the line touches the x-axis. One of these points corresponds to a negative x value, the other to a positive x value. Therefore, 1 positive real solution and 1 negative real solution is the best description for the solutions for x.
29. Number & Quantity >> Complex Numbers The correct answer is C. Use the FOIL method and combine like terms. (–3i + 4)(3i + 4) = –9i2 – 12i + 12i + 16 = –9(–1) + 16 = 25
30. Geometry >> Trig Geometry The correct answer is G. is defined as the length of the opposite side divided by the length of the adjacent 7 opposite side, . The opposite length is 7, and the adjacent length is 5, so tan θ = gives the equation that, if 5 adjacent solved, provides the value of θ.
31. Statistics & Probability >> Probability The correct answer is D. To find the probability, determine the number of desired outcomes and divide by the number of possible outcomes. There are 5 desired outcomes (extra pieces) and 750 + 5 = 755 possible 5 outcomes, so the probability of selecting one of the extra pieces is . 755
32. Number & Quantity >> Fractions The correct answer is K. Convert the two fractions to share a common denominator. 2 8 3 9 = and = . The numerator between 8 and 9 is 8.5, which is not an available answer choice, so 3 12 4 12 8 16 9 18 convert the fractions to sharing denominators of 24. = and = . Now it is clear that the number 12 24 12 24 2 3 17 between the numerators is 17, so the fraction that lies exactly halfway between and is . 3 4 24
33. Algebra & Functions >> Arithmetic: Proportions The correct answer is B. The first paragraph of the text states that 0.25 inch represents 2 feet. Set up and solve a proportion to determine the length of the wall in the scale drawing. x 0.25 = 15 2 2x = 3.75 x = 1.875
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