Practice Test

Why it is hard

The menus do much of the setting up, since each sentence already shows the shape of its equation. The second menu is the hard one: the 8 stamps leave Liam and join Kevin before the gap of 12 is measured. Solving gives Kevin's 28 first, and K + 20 then equals 48, Liam's count after the gift, which waits on the last menu beside his starting 56. Testing each starting count against the story is a quick check on the last menu.

1. Turning words into math moderate Understanding the question

   The gift changes both counts, so "Liam now has 12 more stamps than Kevin" compares Liam's L − 8 with Kevin's K + 8, and the second menu's K + 20 holds a 20 that appears nowhere in the stem. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

2. Much to keep track of moderate Working it out

   Each boy has a count before and after the gift, four numbers in all, and all three menus must be right; the last asks for Liam's start, not the 48 he has after giving 8 away. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Let K be Kevin's starting count, write each boy's count after the gift in terms of K, and set Liam's 12 above Kevin's; testing each number in the last menu is a fast check.

Traps

• 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.
• Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
• Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.

The wrong answers, classified

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

  Reveals partial knowledge Expected pull: strong

  Sets up both equations and solves for Kevin's 28, then takes the value of K + 20, 48, which is Liam's count after the gift, not at the start.

• BBC Follows a key word (Hegarty, Mayer & Monk, 1995)

  Reveals a surface strategy Expected pull: moderate

  Follows the words "12 more" to L + 12, adding 12 to Liam's own count instead of Kevin's, then finds 56 by testing the three counts against the story.

• ACC Equation written backwards (Clement, Lochhead & Monk, 1981)

  Reveals a misconception Expected pull: weak

  Writes "twice as many stamps as Kevin" backwards, as 2L = K, and picks 0.5, giving Liam half of Kevin's stamps, while the other two menus are right.

How the questions are analyzed, and the research behind it

Why it is hard

The six choices form three consistent pairs, each an expansion with its simplified total: the right pair, a pair built on a sign slip with −2 × 4 that totals 34, and a pair built on 3 × 2n read as 5n that totals 18 − n. A student who makes either slip finds two choices that agree with each other and seem to confirm the work. The question rewards one careful simplification and the confidence to accept a sum with no n left.

1. Error-prone arithmetic moderate Working it out

   The first expression puts a negative factor in front of a decimal term: −2 times 2.5n is −5n, and the −2 must reach the 4 as well, giving −8, not +8. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

2. A very close wrong answer moderate Choosing the answer

   The first two expansions are identical except for their last sign, − 8 against + 8, and the third differs from the first only in 5n for 6n, so only a term-by-term check separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

3. Many steps minor Working it out

   Two distributions come first, then the n terms and the constants are collected separately, and six choices are compared with the result; both right ones must be picked to score. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

   One right answer is the expansion before any terms are combined and the other is the plain number 18, which has to be accepted as equal to a sum in which every expression held n. (Enright, Morley & Sheehan, 2002)

Strategy: Distribute each factor across its parentheses, keeping the sign with every product, then combine the n terms and the constants separately and test each of the six choices against the result.

Traps

• Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
• 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.
• Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

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

  Reveals a misconception Expected pull: strong

  Gives the minus of −2 to 2.5n but not to 4, expanding −2(2.5n + 4) as −5n + 8 instead of −5n − 8, so this expansion totals 34.

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

  Reveals a slip Expected pull: weak

  Expands 3(2n + 7) as 5n + 21, adding 3 and 2 instead of multiplying them, so the n terms no longer cancel and leave −n.

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

  Reveals a misconception Expected pull: moderate

  Expects a sum of expressions in n to keep an n term, so it doubts the plain 18 and takes 18 − n, which the 5n expansion also simplifies to.

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

  Reveals a misconception Expected pull: strong

  Takes −2 × 4 as +8, so the constants add to 5 + 21 + 8 = 34 while −n + 6n − 5n still cancels, leaving no n term.

How the questions are analyzed, and the research behind it

Why it is hard

The layout settles the method, so the question comes down to one sign at each box. The printed plus before the first box nudges students toward the positive 4, and the pool then supplies every number that follows from it, down to −40. Students who distribute correctly can still lose the item at the last box by dividing 28 by 0.5. Substituting −56 back, −0.5 × (−48) = 24, confirms the final box.

1. Error-prone arithmetic moderate Working it out

   Two of the three boxes turn on a negative decimal: −0.5 times 8 must give −4, and dividing 28 by −0.5 means multiplying by −2, which doubles the size and makes it negative. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

2. A very close wrong answer minor Choosing the answer

   The pool pairs −4 with 4, −40 with 40 and −56 with 56, so a box filled with the right size and the wrong sign looks just as finished; estimating cannot separate −56 from 56. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

3. Many steps minor Working it out

   The three boxes form a chain and all three must be right, so a slip in the first carries through; the pool even holds the whole wrong chain 4, 20 and −40. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

   The second line prints a plus sign before its box, so the distributed −0.5x − 4 must be written as −0.5x + (−4), with the negative tile in the box. (Enright, Morley & Sheehan, 2002)

Strategy: Multiply −0.5 by both x and 8, undo the constant term by doing the opposite operation to both sides, then divide by −0.5, which is the same as multiplying by −2.

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

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

  Reveals a misconception Expected pull: strong

  Gives the minus of −0.5 to x but not to 8, writing −0.5x + 4, then correctly reaches 20 = −0.5x and x = −40 from that wrong start.

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

  Reveals a slip Expected pull: moderate

  Distributes and adds 4 correctly to reach 28 = −0.5x, then divides 28 by 0.5 instead of −0.5, dropping the negative and entering 56.

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

  Reveals a misconception Expected pull: weak

  Distributes correctly to −4 but moves it to the left side without changing its sign, so 24 − 4 = 20, and 20 ÷ (−0.5) = −40.

How the questions are analyzed, and the research behind it

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

The numbers are small and whole, so the work is in the roles. The $50 is paid once and the $20 every month, and a student who places them in the order they appear writes the fee as the monthly rate; others leave the fee out entirely. Checking the equation at x = 1 and x = 2, where Amiya has paid $70 and then $90, shows at once whether each amount is in the right place.

1. Turning words into math moderate Understanding the question

   Only the words “onetime setup fee” and “per month” tell which amount is multiplied by x and which is added once, and the $50 comes first in the sentence. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

   The whole equation, with y, an equals sign and x, is built on a keypad with no choices, so a reversed or incomplete equation looks as finished as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Work out Amiya's total for 1 month and for 2 months by hand, see which amount is added again each month, and build an equation that gives both totals.

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.
• Stops one step early: A choice is a number from partway through the solution.
• 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

• y = 20x Stops one step early (King et al., 2004; Gierl et al., 2017)

  Reveals partial knowledge Expected pull: strong

  Multiplies the $20 monthly charge by x months but never adds the one-time $50 setup fee, so the equation gives only the cost of service.

• y = 50x + 20 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)

  Reveals a misconception Expected pull: moderate

  Attaches x to the $50 fee and adds the $20 once, putting the numbers into y = mx + b in the order they appear rather than by which one repeats.

• y = 70x Numbers taken from the question (Hegarty, Mayer & Monk, 1995)

  Reveals a surface strategy Expected pull: weak

  Adds the $50 fee and the $20 charge into $70 and multiplies by the months, which would charge the setup fee again every month.

How the questions are analyzed, and the research behind it

Why it is hard

Undoing the + 4 is routine, so nearly all the risk sits in the last step, where m divides 8 rather than being divided by it. Students who read 8 ÷ m = 16 as asking how many 8s make 16 answer 2, and substituting it back, 8 ÷ 2 + 4 = 8, shows at once that it fails. The right value makes 8 ÷ 1/2 = 16, a division that gives a larger number, which feels wrong but is correct.

1. Error-prone arithmetic moderate Working it out

   The unknown is the divisor, so 8 ÷ m = 16 ends in m = 8 ÷ 16, a quotient less than 1; the familiar order, 16 ÷ 8, gives a whole number instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

2. Something you must already know minor Working it out

   Students must see that (8 ÷ m) is the fraction 8/m and that freeing m from a denominator means multiplying both sides by it, a move grade 7 equations rarely need. (Anderson & Krathwohl, 2001)

Strategy: Subtract 4 from both sides, multiply both sides by m so the unknown is no longer in a denominator, then divide; check by substituting your value back into the equation.

Traps

• 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 slip Expected pull: weak

  Adds 4 to 20 instead of subtracting it, so 8 ÷ m = 24 and m = 8/24, which simplifies to 1/3.

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

  Reveals a misconception Expected pull: strong

  Gets 8 ÷ m = 16 correctly but then divides 16 by 8, putting the divisor and the dividend the wrong way round, so m comes out as 2.

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

  Reveals a misconception Expected pull: weak

  Makes both mistakes, adding 4 to get 8 ÷ m = 24 and then dividing 24 by 8 instead of 8 by 24.

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 plan is clear from the stem: count the stamps, then take away what the six new albums hold, so the question tests careful arithmetic. Its choices punish the estimating that saves time elsewhere, since rounding either factor shifts the result by only a few tens, onto another listed number. Students who skip the album counts altogether and compare one album with one album find 198 waiting.

1. A very close wrong answer moderate Choosing the answer

   612, 620 and 632 lie within 20 of one another, so rounding 576 and 378 to friendly numbers before multiplying lands on a wrong choice; only exact products separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

2. Error-prone arithmetic minor Working it out

   Two products of three-digit numbers, 5 × 576 = 2,880 and 6 × 378 = 2,268, must be done by hand and then subtracted with regrouping. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply to find the total number of stamps, multiply again to find how many the six new albums hold, subtract, and check the last digit of each product.

Traps

• Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
• 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 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)

  Reveals a surface strategy Expected pull: moderate

  Subtracts 378 from 576, comparing one old album with one new album, and ignores that there are 5 old albums and 6 new ones.

• C Rounds too early (King et al., 2004)

  Reveals partial knowledge Expected pull: strong

  Rounds 576 to 580 and 378 to 380 before multiplying, so 2,900 − 2,280 = 620, an estimate rather than the exact number left over.

• D Rounds too early (King et al., 2004)

  Reveals partial knowledge Expected pull: weak

  Rounds only 576 up to 580, getting 2,900 stamps, then subtracts the exact 2,268, so the count comes out 20 too high.

How the questions are analyzed, and the research behind it

Why it is hard

The equation takes only two inverse operations, but their order is the trap: the 4 divides the whole sum z + 3w, so it has to be cleared first. Students who subtract 3w from 5w straight away, or who move 3w to the right without changing its sign, reach a tidy multiple of w that looks as finished as the right one. Substituting w = 1, which needs z = 17 to make (z + 3)/4 equal 5, checks any entry.

1. Variables instead of numbers moderate Understanding the question

   With w on both sides and no numbers to substitute, the result has to stay a multiple of w, and a slip such as 23w gives no sign that it is wrong. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

   The answer is typed after a printed “z =”, with no choices to compare, so the entry must be a single term in w, not an equation or a number. (Enright, Morley & Sheehan, 2002)

Strategy: Treat w like a known number: undo the division by 4 first, multiplying the whole of each side, then undo the added 3w, keeping the terms in w together.

Traps

• 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.
• Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
• 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

• 8w Order of operations (Brown & Burton, 1978)

  Reveals a misconception Expected pull: strong

  Undoes the + 3w before the ÷ 4, subtracting 3w from 5w as if only z were divided by 4, so z/4 = 2w and z = 8w.

• 23w A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)

  Reveals a misconception Expected pull: moderate

  Multiplies correctly to get z + 3w = 20w, then moves 3w to the right side without changing its sign, so z = 20w + 3w = 23w.

• -1.75w Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)

  Reveals a misconception Expected pull: weak

  Divides 5w by 4 instead of multiplying, undoing a division with another division, so z + 3w = 1.25w and z = 1.25w − 3w = −1.75w.

How the questions are analyzed, and the research behind it

Why it is hard

Every step is routine, yet there are four of them, each with a classic slip, and the typed answer catches none. Multiplying the 4 into only the a, or carrying 8b across without changing its sign, gives another whole multiple of b that looks finished. Testing b = 1, where 4(a + 2) and 28 + 2a must come out equal, confirms an entry quickly.

1. Variables instead of numbers moderate Understanding the question

   Both letters appear on both sides, and the answer must stay a multiple of b, so there is no number to check an entry against unless students invent one. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

2. Many steps moderate Working it out

   Four moves run in a chain, distributing the 4, gathering the a terms, gathering the b terms and dividing by 2, and an error at any one carries into the typed result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

   The answer is typed after a printed “a =”, with no choices, so 20b, the line before dividing by 2, can be entered without anything on screen questioning it. (Enright, Morley & Sheehan, 2002)

Strategy: Distribute the 4 to both terms in the parentheses, gather the a terms on one side and the b terms on the other, then divide by the coefficient of a.

Traps

• 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.
• Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

• 13b A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)

  Reveals a misconception Expected pull: strong

  Multiplies only the a by 4, writing 4a + 2b = 28b + 2a, so 2a = 26b and a = 13b.

• 18b A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)

  Reveals a misconception Expected pull: moderate

  Distributes correctly but moves 8b to the right without changing its sign, so 2a = 28b + 8b = 36b and a = 18b.

• 20b Stops one step early (King et al., 2004; Gierl et al., 2017)

  Reveals partial knowledge Expected pull: moderate

  Reaches 2a = 20b correctly and enters 20b, stopping before dividing both sides by 2 to get a alone.

How the questions are analyzed, and the research behind it

Why it is hard

The formula is short, but solving it for w means undoing two operations in reverse order, with letters only. The tempting shortcut multiplies p by r and leaves the f untouched, which yields a formula that looks as finished as the right one. Swapping p and w without undoing anything is the other quick trap. Substituting a small book, 5,000 words at 250 a page with 4 front pages, shows which formula gives back 5,000.

1. Variables instead of numbers moderate Understanding the question

   Four letters and no numbers mean each choice has to be judged by its structure, unless students pick values such as r = 250, w = 5,000 and f = 4 to test it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

2. Many steps minor Working it out

   Two inverse operations must come in order, subtracting f before multiplying by r; multiplying first means the f would have to be multiplied by r as well. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Treat it like a two-step equation with numbers: undo what was done to w last, then the step before, applying each operation to the whole of the other side.

Traps

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

  Reveals a misconception Expected pull: moderate

  Treats solving for w as trading the places of p and w, keeping the 1/r and the + f exactly where they were.

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

  Reveals a misconception Expected pull: strong

  Multiplies p by r to undo the 1/r but leaves the + f alone, neither subtracting it first nor multiplying it by r too.

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

  Reveals a surface strategy Expected pull: weak

  Adds the other three letters, r + p + f, keeping the formula’s letters but none of its multiplication or division.

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 equation has three fractions and three letters, and the work is three routine moves: multiply through by 10, distribute, collect the x terms. Most errors come from the middle term, whose denominator 5 differs from the other two; treating it as tenths makes z simply y. Students who test values should avoid x = y, since that pair makes y and 2y − x agree, and x = y = 1 even makes the number 1 fit.

1. Variables instead of numbers moderate Understanding the question

   The answer is z in terms of x and y, not a number, yet two choices are plain numbers, 1 and 5, which tempt students who expect the equation to give z a single value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

2. Error-prone arithmetic moderate Working it out

   Clearing the denominators turns (y − x)/5 into 2(y − x), and the −2x from distributing must then combine with the lone x to leave −x, two places where a factor or a sign slips. (Brown & Burton, 1978; Junker & Sijtsma, 2001)