Practice Test

Why it is hard

Solving takes two routine moves; the points are lost in turning 1 ≤ x into a picture. Because x sits on the right, the symbol seems to point left, and a student who shades the way it points draws the values below 1, which all fail. The dot must also be solid, since x = 1 makes both sides −3. Testing x = 2, where 2x − 5 = −1 and −3 ≤ −1 holds, confirms the direction.

1. Reading a figure, table or graph moderate Understanding the question

   The solution comes out as 1 ≤ x, with x on the right, so the ≤ sign points away from the shaded side; it has to be read as x ≥ 1 before the ray is drawn. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

2. Many steps minor Working it out

   Two solving moves, adding 5 and then halving, come before three graph choices, the endpoint, a solid dot and a direction, and a mistake in any of the five loses the item. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

3. An answer in an unexpected form minor Choosing the answer

   The ray is built, by picking one of four ray shapes and dragging its endpoint, rather than chosen from four finished graphs, so nothing on screen flags a ray that starts at the right number but points the wrong way. (Enright, Morley & Sheehan, 2002)

Strategy: Undo the subtraction and then the multiplication on both sides, rewrite the result with x first, decide whether the endpoint is included, and test one value on your ray in the original inequality.

Traps

• Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
• Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
• A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

• 1|closed|left Negatives and direction (Brown & Burton, 1978)

  Reveals a misconception Expected pull: strong

  Solves to 1 ≤ x but shades the way the ≤ symbol points, a rule that holds only with x on the left, so the ray covers x ≤ 1.

• 1|open|right Counting the ends (Brown & Burton, 1978)

  Reveals partial knowledge Expected pull: moderate

  Finds the right endpoint and direction but draws an open dot, leaving out 1 itself, though the "or equal to" in ≤ includes it.

• -4|closed|right A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)

  Reveals a misconception Expected pull: weak

  Moves the −5 to the left side without changing its sign, writing −3 − 5 = −8 ≤ 2x, so the ray starts at −4.

How the questions are analyzed, and the research behind it

Why it is hard

One subtraction solves the inequality, so the points are lost in the count that follows. The boundary 16 is the last number the solving produces and is easy to type in, or to count up to; checking that 16 + 7 is 23, which is not less than 23, rules it out. Listing 1 through 15, the positive integers below the boundary, settles the count.

1. An answer in an unexpected form minor Choosing the answer

   Solving gives x < 16, but the box asks how many positive integers make the inequality true, a count that still has to be made after the solving is done. (Enright, Morley & Sheehan, 2002)

2. Much to keep track of minor Working it out

   Two conditions set the ends of the count at once: the word positive starts it at 1 rather than 0, and the strict < stops it below 16. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Undo the addition to find the boundary for x, then count the positive whole numbers below it, checking the largest one and the boundary in the original inequality.

Traps

• Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
• A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

• 16 Counting the ends (Brown & Burton, 1978)

  Reveals partial knowledge Expected pull: strong

  Solves to x < 16 and then enters 16, either the boundary itself or a count of 1 through 16, though x = 16 makes x + 7 equal 23, not less than it.

• 29 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)

  Reveals a misconception Expected pull: weak

  Moves the 7 to the right side without changing its sign, 23 + 7 = 30, and counts the 29 positive integers below 30.

How the questions are analyzed, and the research behind it

Why it is hard

One division finds the boundary, so the question is really about direction. Two things pull students the wrong way: the −12 on the left makes the inequality look like one that must be flipped, and the result −2 > x has x on the right, where it is easy to read backward. Testing x = −3, which gives −12 > −18, shows that values below −2 work.

1. A very close wrong answer moderate Choosing the answer

   x < −2 and x > −2 share the boundary −2 and differ only in direction, so finding the right number settles nothing until the direction is read correctly. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

2. Something you must already know minor Working it out

   Students must know that the symbol flips only when both sides are multiplied or divided by a negative number, and that −2 > x says the same as x < −2. (Anderson & Krathwohl, 2001)

Strategy: Divide both sides by 6, asking whether you divided by a negative number, then rewrite the result with x on the left so that the symbol still opens toward the larger side.

Traps

• Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
• Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

• A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)

  Reveals a surface strategy Expected pull: weak

  Adds 6 to −12 instead of dividing by 6, getting −6, and also reverses the direction, so both the boundary and the symbol are wrong.

• B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)

  Reveals a misconception Expected pull: weak

  Undoes the 6 by adding it to −12 rather than dividing, as if 6x meant x − 6, getting −6 > x and so x < −6.

• D Negatives and direction (Brown & Burton, 1978)

  Reveals a misconception Expected pull: strong

  Finds −2 > x but flips the symbol because −12 is negative, or reads −2 > x as x > −2, though dividing by a positive 6 keeps the direction.

How the questions are analyzed, and the research behind it

Why it is hard

The inequality is handed over, so the solving is two routine moves; the judgment comes after. Dividing 17 by 2 leaves a half, and since a basket cannot be split, the boundary has to move up to 9: eight baskets give 2(8) + 8 = 24 points, one short of 25. Students who flip the symbol while dividing, or who move the 8 without changing its sign, still find a choice to match.

1. An answer in an unexpected form minor Choosing the answer

   Solving gives b ≥ 8.5, which no choice shows; it has to be recognized as b ≥ 9, because Marcus cannot make half a basket. (Enright, Morley & Sheehan, 2002)

2. Complex sentences minor Understanding the question

   The 33-word question restates the given inequality as “at least his minimum total points” and “make his mean number of free throws”, wording harder than the math it describes. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

Strategy: Undo the added 8 and then the 2 to solve for b, then ask which whole numbers of baskets actually reach 25 points, testing the boundary value in the original inequality.

Traps

• Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
• A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
• Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

• A Negatives and direction (Brown & Burton, 1978)

  Reveals a misconception Expected pull: moderate

  Reverses the inequality when dividing by 2, as if dividing always flipped it, so b ≤ 8.5, and whole baskets give b ≤ 8.

• C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)

  Reveals a misconception Expected pull: weak

  Moves the 8 to the right without changing its sign, so 2b ≥ 33 and b ≥ 16.5, then drops the half to match b ≥ 16.

• D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)

  Reveals a surface strategy Expected pull: weak

  Takes the 25-point total from the inequality as a limit on baskets and writes at most, a number copied from the question rather than solved for.

How the questions are analyzed, and the research behind it

Why it is hard

The prices are given in the order they are used, so the expression for the total is the easier half. The points are lost on the symbol: “at most” has to be heard as no more than, with equality allowed, and the keypad offers all four inequality signs side by side. Pricing one and two notebooks with the bottle, $17.25 and $22.00, shows which amount repeats and confirms the total.

1. Turning words into math moderate Understanding the question

   The $12.50 bottle is bought once and the $4.75 is paid for each of n notebooks, and “at most $35.00” has to become ≤, which allows a total of exactly $35.00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

2. An answer in an unexpected form minor Choosing the answer

   The whole inequality, with n, its symbol and both prices, is built on a keypad offering ≤, <, ≥ and >, so a wrong symbol looks as complete as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Price a few cases by hand, such as 1 and 2 notebooks with the water bottle, to see which amount repeats, then decide whether a total of exactly $35.00 is allowed.

Traps

• Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
• Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
• Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

• 12.50 + 4.75n ≥ 35.00 Misreads the question (Hegarty, Mayer & Monk, 1995)

  Reveals a misconception Expected pull: strong

  Builds the total correctly but reads “at most” as at least, writing ≥, which would mean Elijah must spend $35.00 or more.

• 12.50 + 4.75n < 35.00 Counting the ends (Brown & Burton, 1978)

  Reveals partial knowledge Expected pull: moderate

  Builds the total correctly but uses a strict less-than sign, leaving out a total of exactly $35.00, which “at most” allows.

• 17.25n ≤ 35.00 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)

  Reveals a surface strategy Expected pull: weak

  Adds the bottle’s price to each notebook’s, $12.50 + $4.75 = $17.25, and multiplies by n, as if a bottle came with every notebook.

How the questions are analyzed, and the research behind it

Why it is hard

The boundary 2 comes out of two routine moves; the direction is where points are lost. Whichever side the x terms are collected on, the last step either divides by a negative or leaves x on the right, and both moments invite a reversed answer. A second risk is moving the 9 without changing its sign, which yields 20. Substituting x = 0, which makes 11 > 9 true, confirms the direction.

1. Error-prone arithmetic moderate Working it out

   Subtracting 3x from both sides leaves −x > −2, and dividing by −1 must reverse the sign; subtracting 2x instead leaves 2 > x, which still has to be read from the x side. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

2. A very close wrong answer moderate Choosing the answer

   x < 2 and x > 2 share the boundary and differ only in direction, so finding the 2 settles nothing; only the sign rule or a test value such as x = 0 decides. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Subtract 2x from both sides so the x term stays positive, undo the + 9, then rewrite the result with x first; testing a value in the original inequality checks the direction.

Traps

• Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
• A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

• B Negatives and direction (Brown & Burton, 1978)

  Reveals a slip Expected pull: strong

  Reaches −x > −2 but divides by −1 without reversing the inequality, or reads 2 > x as x > 2, so the direction comes out backwards.

• C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)

  Reveals a misconception Expected pull: moderate

  Moves the 9 across the inequality without changing its sign, 11 + 9 = 20 instead of 11 − 9 = 2, and keeps the direction right.

• D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)

  Reveals a misconception Expected pull: weak

  Moves the 9 without changing its sign to get 20 > x, then also rewrites that statement backwards as x > 20, two errors at once.

What was corrected from the DOE handbook, and why

• Explanation: The DOE’s explanation says “Another way to solve is to keep the coefficient of x positive and then recognizing that 2 > x is the same as x < 2.” The two verbs joined by “and then” should have the same form, so the site now says “and then recognize that”.

How the questions are analyzed, and the research behind it

Why it is hard

The arithmetic is small, so the points ride on two decisions the keypad leaves open: which symbol, and whether to stop before solving. A budget means Jordan may spend all $120 but no more, so the total cost must stay at or under it, and a strict < leaves out the five-hour rental that costs exactly $120. Testing x = 5 and x = 6 in the typed inequality confirms both the symbol and the setup.

1. Turning words into math moderate Understanding the question

   The stem never says “at most”: the limit comes only from “a budget of $120”, while “$18 per hour” must multiply x and the “one-time cleaning fee of $30” is added just once. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

2. An answer in an unexpected form minor Choosing the answer

   The whole inequality is typed on a keypad that offers <, ≤, > and ≥ side by side, with no choices to compare, and the solved form x ≤ 5 is not accepted. (Enright, Morley & Sheehan, 2002)

Strategy: Work out what 1 hour and 2 hours would cost Jordan in all, see which amount repeats each hour, build the total for x hours, then compare it with the budget and test a few values.

Traps

• Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
• Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

• 30 + 18x < 120 Counting the ends (Brown & Burton, 1978)

  Reveals partial knowledge Expected pull: strong

  Builds the cost 30 + 18x correctly but uses a strict less-than sign, which shuts out a five-hour rental costing exactly the $120 Jordan has.

• 30 + 18x ≥ 120 Misreads the question (Hegarty, Mayer & Monk, 1995)

  Reveals partial knowledge Expected pull: moderate

  Builds the cost correctly but points the symbol the wrong way, treating the $120 budget as an amount Jordan must spend at least instead of a limit.

• x ≤ 5 Misreads the question (Hegarty, Mayer & Monk, 1995)

  Reveals a slip Expected pull: weak

  Writes and solves the inequality correctly, then enters its solution, x ≤ 5, though the item asks for the inequality showing the fee, the hourly charge and the budget.

What was corrected from the DOE handbook, and why

• Explanation: The DOE’s explanation says “The onetime cleaning fee of $30 is added to the hourly cost …”. “Onetime” means “former”; a fee charged once is a “one-time” fee, as the question itself spells it. The site now says “one-time”.

How the questions are analyzed, and the research behind it

Why it is hard

The choices form a two-by-two grid: which number goes with x, and which way the symbol points. Each decision can be made separately, so a student who settles both gets the item, while one mistake lands on a listed choice. Testing two numbers of cans, 5, which leaves the team at 78 points, and 10, which lifts it to 103, rules out every wrong inequality.

1. Turning words into math moderate Understanding the question

   The 53 points the team “has” are a fixed start, the “5 points for each can” go with x, and “at least 100” calls for ≥, though “least” can sound like less. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

2. A very close wrong answer minor Choosing the answer

   The four choices are two expressions, 5 + 53x and 53 + 5x, each with ≤ and with ≥, so the right one and the strongest wrong one differ only in the symbol. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work out the team's total points after 2 cans and then after x cans, translate "at least 100" into a symbol, and compare the result with each choice.

Traps

• Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
• Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

• A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)

  Reveals a misconception Expected pull: weak

  Attaches x to the 53 points the team already has and also reverses “at least”, combining the mistakes behind the other two wrong choices.

• B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)

  Reveals a misconception Expected pull: moderate

  Keeps the right symbol but multiplies the number of cans by 53 and adds 5, putting the starting points where the points per can belong.

• C Misreads the question (Hegarty, Mayer & Monk, 1995)

  Reveals a misconception Expected pull: strong

  Writes the points correctly as 53 + 5x but reads “at least 100” as at most 100, reversing the inequality.

How the questions are analyzed, and the research behind it

Why it is hard

Halving 4 1/2 feet is the only computation; the rest is deciding whether each end of the range can actually happen. A shorter piece of 2 1/4 feet would make the two pieces equal, and a piece of 0 feet is no piece at all, so both ends are excluded. Students who find the right bounds but never test the ends are drawn to the version with the inclusive symbols.

1. Turning words into math moderate Understanding the question

   Two phrases set the two ends: "cut into 2 pieces" means x is more than 0, and "of different lengths" with "the shorter piece" means x is less than half of 4 1/2. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

2. A very close wrong answer moderate Choosing the answer

   The inequalities 0 < x < 2 1/4 and 0 ≤ x ≤ 2 1/4 share both bounds and differ only in their symbols; only checking whether each end can happen separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

3. Variables instead of numbers minor Understanding the question

   The question asks for every possible length of the shorter piece, a range rather than one length, so students must picture the cut moving along the board. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Picture cutting the board in different places, find the largest and smallest values the shorter piece could take, and check whether each boundary value is actually possible.

Traps

• Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
• Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
• Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

• B Counting the ends (Brown & Burton, 1978)

  Reveals partial knowledge Expected pull: strong

  Finds the right bounds but includes both ends, allowing a piece of 0 feet or two equal halves, though the stem says 2 pieces of different lengths.

• C Misreads the question (Hegarty, Mayer & Monk, 1995)

  Reveals partial knowledge Expected pull: moderate

  Ignores that x is the shorter piece and allows any length up to the whole 4 1/2-foot board.

• D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)

  Reveals a slip Expected pull: weak

  Describes the longer piece, whose length lies between half the board and the whole board, instead of the shorter piece the stem calls x.

How the questions are analyzed, and the research behind it

Why it is hard

Solving takes a single subtraction, and the graphs differ only in where the solid dot sits, so the question comes down to 3 − 4. Students who lose the negative sign land on the graph that starts at 1, while those who never isolate x pick a graph that starts at a number printed in the inequality. Testing 0, which satisfies 0 + 4 ≥ 3, shows that the shading must include 0.

1. Reading a figure, table or graph minor Understanding the question

   All four graphs share a closed dot and a rightward arrow, so the only thing to read is where the dot sits, against tick labels printed with raised minus signs. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

2. Error-prone arithmetic minor Working it out

   Undoing the + 4 means computing 3 − 4, the one step that crosses zero; a dropped minus sign moves the dot from −1 to 1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract 4 from both sides to get x alone, then find the graph whose solid dot sits at that number; test one shaded value in the original inequality to confirm.

Traps

• Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
• Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

• B Negatives and direction (Brown & Burton, 1978)

  Reveals a slip Expected pull: strong

  Computes 4 − 3 instead of 3 − 4, or drops the minus sign of −1, and starts the shaded ray at positive 1.

• C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)

  Reveals a surface strategy Expected pull: moderate

  Starts the ray at 3, the number on the right side of the inequality, as if x ≥ 3, without subtracting the 4.

• D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)

  Reveals a surface strategy Expected pull: weak

  Starts the ray at 4, the number added to x, reading it off the inequality instead of solving for x.

How the questions are analyzed, and the research behind it

Why it is hard

The question stacks two classic traps: reversing the inequality signs when dividing by −3, and matching filled or open circles to ≤. All four graphs use the endpoints −1 and 2, so finding those numbers proves little; the choice turns on the circles and the direction of the shading. Testing x = 0, x = 3 and the endpoint x = 2 in the original inequality checks both decisions in under a minute.

1. Error-prone arithmetic moderate Working it out

   Dividing all three parts of −6 ≤ −3x ≤ 3 by −3 reverses both signs and the order, giving 2 ≥ x ≥ −1, which must be rewritten as −1 ≤ x ≤ 2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

2. A very close wrong answer moderate Choosing the answer

   Two graphs share the endpoints −1 and 2 and the shading between them; only the circles, filled for ≤ and open for <, separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

3. Reading a figure, table or graph minor Understanding the question

   Each graph must be read for three things at once: the endpoints, the fill of each circle, and whether the shading runs between them or outward. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Subtract 1 from all three parts, divide all three by −3 while reversing both inequality signs, then rewrite the result from smallest to largest before matching circles and shading.

Traps

• Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
• Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

• A Counting the ends (Brown & Burton, 1978)

  Reveals partial knowledge Expected pull: strong

  Solves for the right endpoints and shading but draws open circles, as if ≤ excluded −1 and 2 instead of including them.

• C Negatives and direction (Brown & Burton, 1978)

  Reveals a misconception Expected pull: moderate

  Divides by −3 without reversing the signs, so the solutions seem to lie outside −1 and 2; with open circles, this graph shows exactly the values that fail.

• D Negatives and direction (Brown & Burton, 1978)

  Reveals a misconception Expected pull: weak

  Shades outward as if the signs had not reversed, but mixes the circles, filled at −1 and open at 2, which no single reading of ≤ gives.

How the questions are analyzed, and the research behind it

Why it is hard

Every move is standard algebra, but the inequality is built so that small slips change the answer completely. The −2 has to reach both terms in the parentheses, and once the boundary −2 is found its direction still has to be read correctly, since the variable ends up on the right as −2 < x. Rewriting that as x > −2, or substituting x = 0 into the original, catches the trap.

1. Error-prone arithmetic moderate Working it out

   Distributing −2 over (5 − 3x) gives −10 + 6x, two sign decisions in one move, and the x terms then sit on both sides of the inequality. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

2. A very close wrong answer moderate Choosing the answer

   x < −2 and x > −2 share the boundary and differ only in direction, which depends on reading −2 < x correctly or flipping the sign when dividing by a negative. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

3. Many steps minor Working it out

   Four moves follow one another, distributing, collecting 5x + 6x, moving the x terms and dividing by 5, and a slip at any of them changes the result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Distribute the −2 to both terms in the parentheses, combine the x terms on the right, then gather the x terms on the side that keeps their coefficient positive.

Traps

• Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
• Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
• A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

• A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)

  Reveals a misconception Expected pull: weak

  Treats 5 − 3x as 2, combining a number with an x term, so the right side becomes 5x − 4 and the solution x < 16.

• B Order of operations (Brown & Burton, 1978)

  Reveals a misconception Expected pull: moderate

  Multiplies the −2 by the 5 but not by the −3x, writing 5x − 10 − 3x, which leads to 4x < 10 and x < 5/2.

• C Negatives and direction (Brown & Burton, 1978)

  Reveals a slip Expected pull: strong

  Finds the boundary −2 but points the inequality the wrong way, by misreading −2 < x or by not flipping the sign after dividing by −5.

How the questions are analyzed, and the research behind it

Why it is hard

Reading |x + 1| as the distance from x to −1 makes this quick: the points 2 units away are −3 and 1, and greater than or equal means outside them. Students who solve only one case and mirror it across 0 land on graphs centered on 0, and students who find both endpoints may still shade between them. The graphs differ in just one dot or one shading choice, so each must be read carefully.

1. Something you must already know moderate Working it out

   Solving |x + 1| ≥ 2 needs two cases, x + 1 ≤ −2 or x + 1 ≥ 2, joined by or; absolute value inequalities go beyond grade 7, and the stem gives no hint. (Anderson & Krathwohl, 2001)

2. A very close wrong answer moderate Choosing the answer

   The right graph and the segment from −3 to 1 share both filled endpoints, so only deciding between outside and between the boundaries separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

3. Reading a figure, table or graph minor Understanding the question

   Four number lines must be read for filled dots and arrow directions, and two of them differ from the right one only in where a single dot sits. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Read the absolute value as distance from −1 on the number line, find the points exactly 2 units away, and decide whether the solutions lie between or outside them.

Traps

• Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
• Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

• B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)

  Reveals a misconception Expected pull: moderate

  Solves x + 1 ≥ 2 to get x ≥ 1, then mirrors it across 0 to x ≤ −1, as if the solution were centered on 0 rather than on −1.

• C Negatives and direction (Brown & Burton, 1978)

  Reveals a misconception Expected pull: strong

  Finds −3 and 1 but shades between them, the solution of |x + 1| ≤ 2, reversing which side of the boundaries lies at least 2 units from −1.

• D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)

  Reveals a misconception Expected pull: weak

  Solves x + 1 ≤ −2 to get x ≤ −3, then mirrors it across 0 to x ≥ 3, assuming the boundaries are symmetric about 0.

How the questions are analyzed, and the research behind it

Why it is hard

The solving takes two moves, but the second carries the rule the question tests: dividing an inequality by a negative number flips its direction. The four graphs cross two features, boundary at −1 or 1 and shading left or right, so forgetting the flip, dropping a sign, or both each land on a graph. Testing a shaded value such as x = −2 in the original inequality, where 6 − 7 = −1 is greater than −4, confirms the choice.

1. Something you must already know moderate Working it out

   The step that decides the question, reversing the symbol when dividing both sides by −3, is a rule students must bring themselves; −3x > 3 becomes x < −1, not x > −1. (Anderson & Krathwohl, 2001)

2. A very close wrong answer moderate Choosing the answer

   The graph shaded to the right from an open circle at −1 matches the right graph's boundary exactly and differs only in direction, the result of skipping the reversal. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

3. Reading a figure, table or graph minor Understanding the question

   Each graph must be read for two things, where its open circle sits, at −1 or 1, and which way the arrow runs, so a boundary alone does not identify the answer. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add 7 to both sides, divide by −3 and reverse the inequality symbol, then match both the boundary point and the shading direction; test one shaded value in the original.

Traps

• Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

• B Negatives and direction (Brown & Burton, 1978)

  Reveals a slip Expected pull: moderate

  Reverses the symbol correctly but drops the negative in 3 ÷ (−3), so the boundary lands at 1 with the shading to its left.

• C Negatives and direction (Brown & Burton, 1978)

  Reveals a misconception Expected pull: strong

  Divides both sides by −3 without reversing the symbol, solving the inequality as if it were an equation, so x > −1 and the shading runs right.

• D Negatives and direction (Brown & Burton, 1978)

  Reveals a slip Expected pull: weak

  Treats −3x as 3x, so it solves 3x > 3 to get x > 1 with no reversal, which puts both the boundary and the direction wrong.

What was corrected from the DOE handbook, and why

• Explanation: The handbook’s steps say “add 7 to both sides of the equation” and “divide both sides of the equation by −3” (p. 142, question 95). But −3x − 7 > −4 is an inequality, not an equation, and the step’s own note is the point of the question: dividing an inequality by a negative number reverses the sign. The site now says “both sides of the inequality” in both steps, as the DOE’s later guides word this step (“Subtract 4 from both sides of the inequality” in the 2024–2025 guide). The working and the answer are unchanged.

How the questions are analyzed, and the research behind it

Why it is hard

The math is one inequality between fractions, but the long final sentence hides it, so students who skim fall back on comparing slice counts. Reasoning that the second pizza has 3 fewer slices, so the Ramirez family should eat 3 fewer, is the strongest pull. Comparing 11/18 with x/15, or turning 11/18 into about 61% and taking 61% of 15, keeps the comparison about shares of a pizza.

1. Complex sentences moderate Understanding the question

   The question is one 42-word sentence ending in a nested condition, “without eating a greater percentage of a pizza than the Barnes family ate”, a reading hurdle rather than more math. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

2. Turning words into math moderate Understanding the question

   The limit of no “greater percentage of a pizza” must become the inequality x/15 ≤ 11/18, comparing fractions of two pizzas cut into different numbers of slices, not slice counts. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

3. Error-prone arithmetic minor Working it out

   Solving gives x ≤ 11 × 15 ÷ 18 = 165/18, a division by 18 that leaves 9 1/6, which must then be cut back to whole slices. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the Barnes family's share as a fraction of their pizza, then find the largest whole number of fifteenths that does not exceed it, by cross-multiplying or by converting to decimals.

Traps

• Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
• Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

• A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)

  Reveals a slip Expected pull: weak

  Finds that 9 slices is the limit but gives the slices left on the second pizza, 15 − 9 = 6, answering a different question.

• B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)

  Reveals a surface strategy Expected pull: weak

  Subtracts the stem’s numbers, 18 − 11 = 7, the slices left on the Barnes family’s pizza, without comparing shares of the two pizzas.

• C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)

  Reveals a misconception Expected pull: strong

  Reasons that the second pizza has 3 fewer slices, so the Ramirez family should eat 3 fewer, 11 − 3 = 8, comparing by subtraction instead of by fractions.

How the questions are analyzed, and the research behind it

Why it is hard

Solving takes one step, multiplying through by 2, so most of the work is matching the result to a picture. The choices pair each common slip with a graph: the bounds left as written, the bounds halved, and the right bounds shaded on the wrong side. Reading the open circles against the labeled ticks at −5 and 5 settles it.

1. Reading a figure, table or graph moderate Understanding the question

   The four number lines are labeled only at −10, −5, 0, 5 and 10, so open circles at −8 and 4, or at −4 and 2, have to be located by counting ticks. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

2. Error-prone arithmetic minor Working it out

   All three parts of −4 < x/2 < 2 must be multiplied by 2, and the negative bound −4 becomes −8; halving instead gives −2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)