Practice Test

Why it is hard

This is one formula and one square, so the question comes down to recall and to the form of the answer. The area and circumference formulas both pair π with the radius, and 2πr gives a tidy value with π in it that looks just as right. The other losses are about form: replacing π with 3.14, which the instructions rule out twice, or dropping π after squaring 5.

1. Something you must already know moderate Working it out

   The formula A = πr² is not printed with the question, and the circumference formula 2πr uses the same r and π, so recalling the wrong one gives a value that looks just as finished. (Anderson & Krathwohl, 2001)

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

   The answer is built on a keypad with a π key and must keep π as a symbol; a decimal worked out with 3.14 is not accepted, and no choices show the expected form. (Enright, Morley & Sheehan, 2002)

Strategy: Recall the area formula rather than the circumference formula, square the radius before multiplying, and leave π as a symbol instead of replacing it with 3.14.

Traps

• Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
• A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
• Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

• 10π Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: strong

  Uses the circumference formula 2πr, or doubles the radius instead of squaring it, so 2 × 5 gives 10π, the distance around the circle.

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

  Reveals partial knowledge Expected pull: moderate

  Finds the area correctly but replaces π with 3.14, 25 × 3.14 = 78.5, though the question asks for the area in terms of π.

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

  Reveals partial knowledge Expected pull: weak

  Squares the radius, 5 × 5 = 25, and stops before multiplying by π, so the entry leaves out the π the instructions ask for.

How the questions are analyzed, and the research behind it

Why it is hard

Once the angle fact is in place, 5x = 180 is a one-line equation, so the question tests recall more than computation. Students who use 90° for the pair land on half the right value, and the two smallest choices come only from combining the labels' 3 and 2. A sense check helps: x = 36 makes the angles 108° and 72°, one obtuse and one acute, just as the drawing shows.

1. Something you must already know moderate Working it out

   The figure gives no number of degrees, only 3x° and 2x°, so everything rests on recalling that the two angles at the ends of one side of a parallelogram add to 180°. (Anderson & Krathwohl, 2001)

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

   The marked angles sit at the upper and lower ends of the left side, consecutive rather than opposite, which is what makes them supplementary instead of equal. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Decide what two angles that share a side of a parallelogram add up to, set 3x + 2x equal to that total, solve for x, and check that the two angles fit the figure.

Traps

• Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
• 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

• B Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: strong

  Uses 90° as the total of the two angles, the complementary total, so 5x = 90 gives 18, though consecutive angles of a parallelogram add to 180°.

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

  Reveals a surface strategy Expected pull: weak

  Multiplies the two coefficients from the labels, 3 × 2 = 6, without using any angle fact or writing an equation.

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

  Reveals partial knowledge Expected pull: weak

  Combines 3x + 2x into 5x and gives the coefficient 5, stopping before dividing the 180° total by it.

How the questions are analyzed, and the research behind it

Why it is hard

Each fact needs only a subtraction, but the student has to plan a route from the 73° angle at line p over to r at line n. The first number the route gives, the triangle's 17° angle, is listed, and it is the acute partner of r rather than r itself. The figure helps here: r is drawn plainly obtuse, and 163 is also 90 + 73, the triangle's exterior angle, which confirms it.

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

   The 73° sits between line p and the vertical segment, not between p and m, and r is the obtuse angle above m on the left of line n, far from every marked angle. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

2. Combines ideas moderate Working it out

   Three facts chain together: the triangle's angles sum to 180°, the angles along straight line m sum to 180°, and corresponding angles at the parallel lines n and p are equal. (Embretson, 1998; Daniel & Embretson, 2010)

3. Something you must already know minor Working it out

   The angle pairs formed by a line crossing two parallel lines are a grade 8 topic, and the question uses them without naming them. (Anderson & Krathwohl, 2001)

Strategy: Find the third angle of the right triangle, use the straight line m to get the angle beside it, then carry that angle over to line n using the parallel lines.

Traps

• Stops one step early: A choice is a number from partway through the solution.
• An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

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

  Reveals partial knowledge Expected pull: strong

  Finds the triangle's third angle, 180 − 90 − 73 = 17, and stops, though r is the obtuse angle next to the 17° angle, not that angle itself.

• B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)

  Reveals a surface strategy Expected pull: weak

  No angle fact in the figure gives 117; even 180 − 73 is 107, so this choice mostly catches guesses at an obtuse angle.

• D An arithmetic slip (Junker & Sijtsma, 2001)

  Reveals a slip Expected pull: weak

  Finds the 17° angle and subtracts it from 180 but borrows without reducing the 8, so 180 − 17 comes out as 173.

How the questions are analyzed, and the research behind it

Why it is hard

Circle A is direct: its labeled radius of 5 gives an area of 25π. Circle C is where points are lost, since its size comes from a sentence rather than a label, and the segment through its center is a diameter, so its circumference is 5π while its area is only 6.25π. The row labels favor guessers, because the A and C in A = 25π and C = 5π match the circles' names, though they stand for area and circumference.

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

   Circle C's size is shown only by segment DE, drawn through center C with no label, and only the stem's sentence that DE is congruent to AB makes it a diameter of 5. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

2. Something you must already know moderate Working it out

   Neither A = πr² nor C = πd is given, and one circle is measured by a radius and the other by a diameter, so each formula needs the right length. (Anderson & Krathwohl, 2001)

3. Much to keep track of minor Working it out

   Four values are in play, areas of 25π and 6.25π and circumferences of 10π and 5π, and both rows must be matched correctly for the item to count. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Write down the radius and the diameter of each circle, compute both circles' areas and circumferences in terms of π, then match each value in the table.

Traps

• Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
• Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
• A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

• 1:B Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: weak

  Squares the diameter instead of the radius, so Circle C's area seems to be π × 5² = 25π while Circle A's becomes 100π.

• 1:C Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: moderate

  Uses 2πr for the area, so Circle A seems to have an area of 10π and Circle C one of 5π, and no circle seems to have 25π.

• 2:A Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: moderate

  Uses π times the radius for the circumference, so Circle A's radius of 5 seems to give a circumference of 5π.

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

  Reveals partial knowledge Expected pull: strong

  Overlooks that DE is congruent to AB, so Circle C has no known size, and with Circle A's circumference at 10π picks Neither.

How the questions are analyzed, and the research behind it

Why it is hard

One multiplication answers the question once the right pair of lengths is chosen. Each wrong choice pairs or combines the labeled lengths in another way: the two sides as for a rectangle, the height with the slanted side, or all four sides as a perimeter. Following the right-angle mark at T from QT down to PS identifies the base and height that belong together.

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

   Three lengths are labeled, 34 cm, 18 cm and 22 cm, but only the base PS and the height QT meet at the right-angle mark; the slanted side RS is not needed. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

2. Something you must already know minor Working it out

   The formula, base times height, is not given, and the height must be the perpendicular distance between two sides, not the length of a slanted side. (Anderson & Krathwohl, 2001)

Strategy: Multiply the base by the height measured at a right angle to it, and check that the length you use as the height meets that base at a right-angle mark.

Traps

• Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
• Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

• A Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: moderate

  Adds all four sides, 2 × (34 + 22) = 112, finding the perimeter of the parallelogram instead of its area.

• B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  Reveals a slip Expected pull: weak

  Multiplies the 18 cm height by the 22 cm slanted side, a pair that does not meet at a right angle, getting 396.

• D Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: strong

  Multiplies the two side lengths, 34 × 22 = 748, as if a parallelogram's area were side times side like a rectangle's.

How the questions are analyzed, and the research behind it

Why it is hard

This is a direct composite-volume problem with whole numbers: one cube, one hole, one subtraction. What it checks is that both volumes are real volumes, the hole as tall as the cube, and that the subtraction is done at the end. Students who stop at the hole’s volume, or who subtract only the square face, give amounts that are easy to type and hard to doubt without choices.

1. Something you must already know minor Working it out

   The volume of a rectangular prism, length times width times height, is not given, and it is needed twice, once for the 20 cm cube and once for the hole. (Anderson & Krathwohl, 2001)

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

   The figure labels only the 5 cm opening and the cube’s 20 cm edge; the hole’s 20 cm height comes from the words, while the drawing shows it reaching from top to bottom. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

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

   The volume is typed with no choices to compare, so the hole’s 500 cubic centimeters or the cube’s 8,000 can be entered without any warning. (Enright, Morley & Sheehan, 2002)

Strategy: Find the volume of the full cube, then the volume of the square hole that runs from the top face to the bottom, and subtract the hole from the cube.

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.
• Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
• Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

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

  Reveals partial knowledge Expected pull: moderate

  Finds the volume of the hole, 5 × 5 × 20 = 500 cubic centimeters, and gives the clay removed instead of the clay that is left.

• 7,875 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  Reveals a slip Expected pull: weak

  Treats the hole as a 5 cm cube, 5 × 5 × 5 = 125, though it runs the full 20 cm height, so 8,000 − 125 = 7,875.

• 7,975 Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: weak

  Subtracts only the area of the square opening, 5 × 5 = 25, from the cube’s 8,000, mixing a face’s area with the hole’s volume.

How the questions are analyzed, and the research behind it

Why it is hard

One formula and one multiplication finish the problem once the right formula is chosen. The cement in the story pulls toward area, and the area, 78.5, is listed, as is half the circumference, from multiplying 3.14 by the radius alone. Remembering that the circumference is π times the diameter, here 10 feet, and that the answer is in feet, not square feet, settles it.

1. Something you must already know moderate Working it out

   The circumference formula is not printed, and the area formula uses the same radius and π, so recalling the wrong one gives a number that looks just as finished. (Anderson & Krathwohl, 2001)

2. Turning words into math minor Understanding the question

   Pouring cement suggests covering a surface, but the question asks for “the distance, in feet, around the edge”, which is the circumference, not the area. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Decide which circle measurement the distance around the edge is, recall its formula, and substitute the radius and 3.14, doubling the radius wherever the formula calls for the diameter.

Traps

• Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
• A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
• 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 Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: moderate

  Multiplies π by the radius, 3.14 × 5 = 15.7, using the radius where C = πd needs the diameter, so it finds half the distance around.

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

  Reveals a misconception Expected pull: weak

  Starts from the area formula and squares π instead of the radius, 3.14 × 3.14 × 5 ≈ 49.3, mixing the wrong formula with a slip in squaring.

• D Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: strong

  Finds the patio’s area, 3.14 × 5² = 78.5 square feet, using the area formula instead of the circumference for the distance around.

How the questions are analyzed, and the research behind it

Why it is hard

Every number is small, so the work is choosing the right pieces: the area of the square, the area of the hole, and the difference. A radius of 1 makes the circle's area simply π, which is easy to confuse with the circumference, 2π, when the formulas are half remembered. The region asked for is the gray metal, not the hole, and the final rounding goes up, since the third decimal place is 8.

1. Something you must already know moderate Working it out

   Neither area formula is given, and the hole is labeled by its 2-inch diameter, so students must recall A = πr² and halve the diameter to get the radius 1 first. (Anderson & Krathwohl, 2001)

2. Combines ideas minor Working it out

   The shaded metal is a square with a circle removed, so two area formulas are combined by subtraction rather than either one used alone. (Embretson, 1998; Daniel & Embretson, 2010)

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

   The result is typed to the nearest hundredth with no choices to compare, so the long decimal 5.858… must be rounded up, not cut off after two places. (Enright, Morley & Sheehan, 2002)

Strategy: Halve the diameter to get the radius, find the area of the square and of the circle, subtract the circle from the square, and round only at the end.

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.
• Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
• A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

• 2.72 Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: strong

  Uses the circumference, 2π × 1 ≈ 6.28, in place of the circle's area and subtracts it from the square's 9 square inches, getting about 2.72.

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

  Reveals partial knowledge Expected pull: moderate

  Finds the area of the circular hole, π × 1² ≈ 3.14 square inches, and enters it instead of the area of the shaded metal left around it.

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

  Reveals a slip Expected pull: weak

  Works the area out correctly as about 5.858 but cuts off the extra digits instead of rounding up to the nearest hundredth.

What was corrected from the DOE handbook, and why

• Explanation: The DOE’s explanation gives the area of the hole as “π(1 in.)² ≅ 3.14 sq in.” The symbol ≅ means “is congruent to”, which an area and a number cannot be; the line means “is approximately equal to”, written ≈. The site now says “π(1 in.)² ≈ 3.14 sq in.”

How the questions are analyzed, and the research behind it

Why it is hard

A scale drawing multiplies every length by one factor, and the perimeter shares that factor, so the work is short once the two perimeters are paired. The common loss is reasoning with differences: the perimeter grows by 5, so a side seems to grow by 5 as well. Converting to miles per inch first, then checking that the three real sides add back to 25 miles, guards against that error and against scaling the wrong side.

1. Turning words into math minor Understanding the question

   The 25 miles is the park's perimeter, not a side, so it must be matched with the drawing's perimeter, 5 + 7 + 8 = 20 inches, before any side can be scaled. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

2. Many steps minor Working it out

   Three operations in a row, adding the sides, dividing 25 by 20 for 1.25 miles per inch, and multiplying the 5-inch side, each produce a number the next step needs. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the three drawn sides to get the drawing's perimeter, divide the real perimeter by it to find the miles per inch, then multiply the shortest drawn side by that scale.

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.
• Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

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

  Reveals a slip Expected pull: weak

  Finds the scale of 1.25 miles per inch but applies it to the 7-inch side, the middle one, giving 8.75 miles instead of the shortest side.

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

  Reveals a misconception Expected pull: strong

  Keeps a constant difference instead of a constant ratio: the perimeter grows by 5, from 20 to 25, so it adds 5 to the 5-inch side.

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

  Reveals a slip Expected pull: weak

  Pairs the 25 miles with the longest side, 8 inches, instead of the 20-inch perimeter, so each inch seems to be 3.125 miles and 5 inches 15.625.

How the questions are analyzed, and the research behind it

Why it is hard

The figure gives one measurement on a slanted radius and two labels on the diameter, and the right angle at the foot of the vertical segment is a decoy. Once RP and PT are seen as radii, RT is 24, and x is that diameter less the 5. Students who never connect the slanted 12 to the horizontal line fall back on the labeled numbers, adding 12 and 5, or take x to end at the center.

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

   The 12 labels the slanted segment SP, not the horizontal one, and the arrow for x runs from R past the center P to the foot of the vertical segment, so x is neither a radius nor a whole diameter. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

2. No obvious method minor Working it out

   The right-angle mark and the triangle under S suggest the Pythagorean theorem, which is no help; the way in is seeing that RP and PT are radii as well. (Schoenfeld, 1985)

Strategy: Mark every segment from P to the circle as 12, notice that R and T are the ends of a diameter, then write an equation for that diameter using x and the 5.

Traps

• Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
• Information you don't need: The question gives numbers or facts that play no part in the answer.
• 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 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  Reveals a slip Expected pull: moderate

  Reads x as the radius RP, as if its arrow ended at the center P, though it runs past P to the foot of the vertical segment.

• B Trusting the picture (Lesh, Post & Behr, 1987)

  Reveals a surface strategy Expected pull: weak

  Judges x from the drawing, where it looks about three times as long as the piece labeled 5, instead of working from the radius of 12.

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

  Reveals a surface strategy Expected pull: strong

  Adds the two labeled numbers, 12 + 5 = 17, without placing the 12 on the diameter, as if x were a radius plus the 5.

How the questions are analyzed, and the research behind it

Why it is hard

Setting the sum of the two labeled angles to 90° is the decision; the algebra after it is routine. The trap waits at the end: x = 19 is not an angle, and the expression at the top vertex, 2x + 9, gives the larger acute angle, 47°. Evaluating both expressions and checking that they add to 90° takes seconds and shows which one is smaller.

1. Combines ideas moderate Working it out

   The angle facts and the algebra meet in one equation: the right angle means the two labeled angles add to 90°, and only that sum makes (2x + 9) + (3x − 14) solvable. (Embretson, 1998; Daniel & Embretson, 2010)

2. Many steps minor Working it out

   After solving 5x − 5 = 90, both angle expressions must be evaluated and compared, because the question wants an angle, not x, and not the first angle found. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

3. Something you must already know minor Working it out

   The 180° angle sum of a triangle is not printed, and only the small square at the bottom left shows that one angle is 90°. (Anderson & Krathwohl, 2001)

Strategy: Use the right angle to see that the other two angles add to 90°, solve that equation for x, then substitute into both expressions and compare them.

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.
• A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

• B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)

  Reveals a surface strategy Expected pull: weak

  No common working gives 44°; it sits one degree from the right measure and fits no consistent value of x, so it mostly draws estimates and guesses.

• C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)

  Reveals a surface strategy Expected pull: weak

  No common working gives 46°, which would need x = 20 and two angles adding to 95°, so it mostly catches guesses near 45°.

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

  Reveals partial knowledge Expected pull: strong

  Solves x = 19 correctly but answers with 2x + 9 = 47°, the larger acute angle, instead of evaluating both expressions to find the smaller.

How the questions are analyzed, and the research behind it

Why it is hard

A horizontal cut runs parallel to the ground, and this prism stands on a square base, so the slice copies the base. The main danger is picturing the cut along the tall front face, which gives a 2-by-6 rectangle, the shape a vertical slice makes. The two triangles need a slanted cut through a corner and catch only students guessing.

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

   The prism is drawn standing on end, 6 inches tall on a 2-by-2 base, so the cut has to be pictured in a flat drawing where the tall front face is the most visible shape. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

2. Something you must already know minor Working it out

   Students must know that a slice parallel to the base of a prism has the same shape and size as the base, which here is the square at the bottom. (Anderson & Krathwohl, 2001)

Strategy: Picture a flat cut across the prism parallel to the ground, ask which face of the prism that cut runs parallel to, and read that face's edge lengths from the labels.

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

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

  Reveals a misconception Expected pull: weak

  Expects a slice to cut off a corner and make a triangle, as a cut through a pyramid might, not a cut parallel to this square base.

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

  Reveals partial knowledge Expected pull: moderate

  Describes a vertical slice, which runs down the 6-inch height and makes a 2-by-6 rectangle, mixing up horizontal and vertical.

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

  Reveals a surface strategy Expected pull: weak

  Builds a triangle from the labels in the figure, one 2-inch and two 6-inch sides, a shape no horizontal cut of this prism can make.

How the questions are analyzed, and the research behind it

Why it is hard

The question turns a statement about perimeters into a ratio of sides, and the natural guess, that the sides keep the perimeters’ 2 to 1, ignores that the square shares its perimeter among 4 sides and the triangle among 3. Choosing a triangle perimeter such as 12 makes the comparison concrete. The order of the ratio still has to be checked at the end, because its reverse is listed too.

1. Turning words into math moderate Understanding the question

   “The perimeter of square W is twice the perimeter of equilateral triangle X” must become 4s = 2 × 3t, with the 2 on the triangle’s side; written backward, it gives the triangle the larger perimeter. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

2. Variables instead of numbers moderate Understanding the question

   No side length is given, so students must either work with letters for the two sides or choose a convenient perimeter themselves, such as 12 for the triangle. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

3. A very close wrong answer minor Choosing the answer

   The ratios 3:2 and 2:3 use the same two numbers, so only keeping the square’s side first, as the question orders it, separates the answer from its reverse. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Pick an easy side length for the triangle, find its perimeter, double that for the square, divide by the square's number of sides, and write square to triangle.

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.
• 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).
• 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

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

  Reveals a misconception Expected pull: weak

  Writes the comparison backward, 2 × 4s = 3t, giving the triangle the larger perimeter, so the ratio of the sides comes out 3:8.

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

  Reveals partial knowledge Expected pull: moderate

  Finds the right relationship but turns the ratio upside down, giving the triangle’s side to the square’s instead of the square’s side first.

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

  Reveals a misconception Expected pull: strong

  Assumes the sides are in the same ratio as the perimeters, 2:1, ignoring that the square has 4 sides and the triangle only 3.

How the questions are analyzed, and the research behind it

Why it is hard

Once supplementary is read as a 180° total, the problem is a one-variable equation with whole numbers. Students lose the point in two ways: by stopping at the smaller angle, which the equation produces first, or by using 90° for the total. Checking that the gridded angle is the larger of the pair, and that the two angles add to 180°, catches both.

1. Something you must already know moderate Working it out

   The stem never says what supplementary means; the 180° total has to be recalled, and a student who confuses it with complementary's 90° halves every angle. (Anderson & Krathwohl, 2001)

2. Turning words into math minor Understanding the question

   The sentence "The measure of angle R is 5 times the measure of angle M" has to become R = 5M, so the 180° splits into six equal parts, not five. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

   As a grid-in there is nothing to compare against, so gridding angle M's 30°, the first value the equation gives, looks just as finished as angle R. (Enright, Morley & Sheehan, 2002)

Strategy: Call angle M one part and angle R five parts; together the six parts make the supplementary total, so find one part and then multiply for angle R.

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.
• Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

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

  Reveals partial knowledge Expected pull: strong

  Solves 6m = 180 correctly for angle M, 30°, and grids it instead of multiplying by 5 to get angle R, the angle the question asks for.

• 75 Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: moderate

  Uses 90° for supplementary, the complementary total, so 6m = 90 gives m = 15 and angle R = 75, half its real size.

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

  Reveals a surface strategy Expected pull: weak

  Divides 180 by the 5 in the stem, combining the two numbers without an equation that gives angle M its own share of the 180°.

How the questions are analyzed, and the research behind it

Why it is hard

The question chains three short facts: from the square's area to its side, from the side to the radius, and from the whole circle to a quarter of it. None is hard, but each wrong choice marks a broken link, most often the whole circle's area, which looks like a finished answer. Pointing to S as the center before computing anything settles the radius.

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

   The radius is never labeled: students must see that S is both a corner of the square and the circle's center, so sides PS and RS are radii. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

2. Something you must already know minor Working it out

   Neither the circle area formula, π times the radius squared, nor the step from a square's area to its side length is given. (Anderson & Krathwohl, 2001)

3. Combines ideas minor Working it out

   Square and circle are linked in one chain: the area 4 gives a side of 2, the side is the radius, and the shaded part is a quarter of π × 2². (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Take the square root of the square's area to get its side, which is also the radius, then take one quarter of pi times the radius squared.

Traps

• Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
• Stops one step early: A choice is a number from partway through the solution.
• A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

• A Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: moderate

  Halves the square's side as if it were a diameter, so it uses a radius of 1, and one quarter of π × 1² gives π/4.

• C Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: weak

  Misremembers the circle's area as 2πr², blending it with the circumference 2πr, so one quarter of 2π × 2² gives 2π.

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

  Reveals partial knowledge Expected pull: strong

  Finds the whole circle's area, π × 2² = 4π, and stops before taking the quarter that is shaded.

How the questions are analyzed, and the research behind it

Why it is hard

Scale questions usually convert one length; this one needs a round trip, from the drawing to the real room and back. The real width of 24 feet appears partway through and is a choice, so students who lose track of units stop there. The last conversion, 24/20, then has to be written as a mixed number, and its fraction part is where a hurried student slips.

1. Combines ideas moderate Working it out

   The 960 square feet is a real-world area, so the scale cannot be applied to it directly; the length must go to feet, the width come from the area, and then return to inches. (Embretson, 1998; Daniel & Embretson, 2010)

2. Much to keep track of minor Working it out

   Four lengths in two units are in play, 2 inches, 40 feet, 24 feet and the width on the drawing, and each must stay labeled with its unit. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Convert the 2-inch length to real feet, divide the area by it to get the real width, then use the scale to change that width back to inches.

Traps

• Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
• 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 An arithmetic slip (Junker & Sijtsma, 2001)

  Reveals a slip Expected pull: weak

  Gets the real width of 24 feet and divides by 20 correctly, but writes the remainder 4/20 as 1/4 instead of 1/5.

• C Units or scale (King et al., 2004)

  Reveals partial knowledge Expected pull: strong

  Finds the room's real width, 960 ÷ 40 = 24 feet, and gives it as the drawing's width without converting back at 1 inch per 20 feet.

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

  Reveals partial knowledge Expected pull: weak

  Converts the 2-inch length to the real 40 feet and gives that number, answering for the length instead of the width and in the wrong units.

How the questions are analyzed, and the research behind it

Why it is hard

The shaded shape has no name, so the question is about seeing it as a difference of two right triangles whose legs lie on the axes. Once that is seen, both areas are one-line calculations. The wrong choices come from stopping after either triangle or from pairing a base from one segment with a height from the other.

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

   All four lengths come from where the two segments meet the axes, at 20 and 15 on the y-axis and at 10 and 20 on the x-axis, read against grid lines 5 units apart. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

2. No obvious method minor Working it out

   The shaded region is a four-sided shape with no formula of its own; it has to be seen as the large triangle with the small one removed. (Schoenfeld, 1985)

3. Something you must already know minor Working it out

   The area of a triangle, one half of base times height, is assumed and not given. (Anderson & Krathwohl, 2001)

Strategy: Find the area of the large right triangle under the outer segment and subtract the area of the smaller triangle under the inner segment, reading the legs from the axes.

Traps

• Stops one step early: A choice is a number from partway through the solution.
• 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.
• Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

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

  Reveals partial knowledge Expected pull: moderate

  Finds the area of the small unshaded triangle, 1/2 × 10 × 15 = 75, and gives that region instead of the shaded one.

• C Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  Reveals a slip Expected pull: weak

  Pairs the outer segment's x-intercept, 20, with the inner segment's y-intercept, 15, as if they bounded one triangle: 1/2 × 20 × 15 = 150.

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

  Reveals partial knowledge Expected pull: strong

  Finds the area of the large triangle, 1/2 × 20 × 20 = 200, and stops before subtracting the unshaded triangle inside it.

How the questions are analyzed, and the research behind it

Why it is hard

This is a one-step formula question with whole numbers, made slightly harder by one length too many. A student who multiplies the two side lengths, as for a rectangle, or brings in a triangle's one half, finds a listed choice. Matching the base to the height marked with a right angle is all it takes.

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

   The figure labels three lengths, but only the 30-foot side and the 20-foot height drawn at a right angle to it belong together; the slanted 25-foot side is extra. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

2. Something you must already know minor Working it out

   The parallelogram area formula, base times height, is not given, and its height must be the perpendicular distance, not the slanted side. (Anderson & Krathwohl, 2001)

Strategy: Use base times height, where the height is the perpendicular distance between the two parallel sides chosen as bases.

Traps

• Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
• Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

• A Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: strong

  Multiplies the two side lengths, 30 × 25 = 750, as if a parallelogram's area were side times side like a rectangle's.

• C Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  Reveals a slip Expected pull: weak

  Pairs the 20-foot height with the slanted 25-foot side, 25 × 20 = 500, though the height is drawn at a right angle to the 30-foot sides.

• D Wrong formula (King et al., 2004)

  Reveals a misconception Expected pull: moderate

  Uses the triangle formula, 1/2 × 30 × 20 = 300, halving the base times the height.

How the questions are analyzed, and the research behind it

Why it is hard

The method is short, but the question invites a shortcut: the sides differ by 2, so it feels as if the areas should differ by 2 squared. Two choices reward working with the side lengths and one rewards perimeters. Squaring each side before subtracting avoids all three, and a sketch of the 13-inch square inside the 15-inch one shows the gap is an L-shaped strip, far more than 4 square inches.

1. Something you must already know minor Working it out

   The area of a square, side times side, is not given; students must square 13 and 15 themselves rather than work with the side lengths. (Anderson & Krathwohl, 2001)

2. Error-prone arithmetic minor Working it out

   Squaring 13 and 15 by hand gives 169 and 225, and the subtraction 225 − 169 needs a borrow, where an arithmetic slip creeps in. (Brown & Burton, 1978; Junker & Sijtsma, 2001)