Practice Test

6.NS.7 ordering and absolute value

Question 1 of 22

2026 Practice Test A Math, question 17

Question

Consider this expression:

$|(-8) - (-3) + 6| - |6 - 11|$

Create an equivalent expression.

Move the correct answer to each box. Not all answers will be used.

|| − ||

On the test you drag tiles into the boxes; here, choose each box’s tile from its menu. Each tile can be used once. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

Show the answer

Correct answer: blank 1: 1; blank 2: −5

To create an expression equivalent to $|(-8) - (-3) + 6| - |6 - 11|$ :

Step 1: Simplify the first absolute value term.

$(-8) - (-3) = -8 + 3 = -5$

$-5 + 6 = 1$

Simplify the second absolute value term.

$6 - 11 = -5$

Therefore the equivalent expression is

$|1| - |-5|$ .

Question analysis

This absolute value question shows an expression with two pairs of bars and asks students to build an equivalent one by placing number tiles in two boxes. It tests subtracting negative integers and understanding what absolute value bars do. Not every tile is used, and several tiles match the results of common sign mistakes.

NYS standards: 7.NS.1c rational number subtraction
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Simplifying signed integers inside absolute value bars, including subtracting a negative, is core grade 7 work. This asks only for the values inside the bars, not the final number, but one sign slip leads straight to a wrong tile.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 21 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every move here involves a sign, and the tiles are built from the usual misreadings of absolute value: 11 for making each number positive before combining, 17 for erasing every minus sign. A slip on − (−3) gives −5 for the first pair, but the second pair needs the only −5 tile, so that slip leaves a box with no tile to fit. Students who simplify each inside completely, and leave the bars untouched, finish quickly.

Error-prone arithmetic moderate Working it out
Inside the first bars, subtracting −3 must become adding 3, so −8 + 3 = −5, and adding 6 then crosses zero to reach 1; reading − (−3) as − 3 gives −5 instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Two pairs of bars are simplified separately, and both boxes need the right tile, kept in the original order, before the answer is right. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
Each box takes what is left inside a pair of bars, not its absolute value, so the second box needs the tile −5 rather than a 5, and the expression's value, −4, is never entered. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite subtracting −3 as adding 3, then simplify the inside of each pair of bars separately, left to right; each box replaces the inside of one pair, in the same order.

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.
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

EC A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Takes the absolute value of each number before combining, so the first bars become 8 − 3 + 6 = 11; the same rule still gives 6 − 11 = −5 in the second.
BC Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Believes absolute value bars make everything inside positive, so it drops every minus sign and adds 8 + 3 + 6 = 17, then works the short pair 6 − 11 normally.
CA Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Simplifies both insides correctly but places −5 first and 1 second, reversing the subtraction so the expression equals 5 − 1 = 4 instead of 1 − 5 = −4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a negative number: $a - (-b) = a + b$; a is any number, and b is a positive number, so −b is negative. Subtracting a negative number is the same as adding its opposite, the same number with the other sign. Think of money: taking away a debt of $3 leaves you $3 richer.
Absolute value: $|a| = \text{the distance from } a \text{ to } 0$; The bars around a number stand for its absolute value. a is any number. The absolute value is how far a is from 0 on a number line, so it is never negative. Both |5| and |−5| equal 5. Always simplify inside the bars first.

Sort out the question
The expression has two pairs of absolute value bars. The new expression keeps both pairs, with a box inside each one:
$|\square| - |\square|$
Each box gets the number you reach when you simplify inside that pair of bars. The bars stay, so you do not take the absolute value yet. Some tiles are not used.
Simplify inside the first bars
Inside the first bars is $(-8) - (-3) + 6$ . Subtracting −3 is the same as adding 3. Work from left to right:
$(-8) - (-3) = -8 + 3 = -5$
$-5 + 6 = 1$
The first box gets the tile 1.
Watch out: the bars do not make each number positive before you combine. Changing −8 and −3 to 8 and 3 first gives 8 − 3 + 6 = 11. Dropping every minus sign gives 8 + 3 + 6 = 17. Both are tiles, and both are wrong.
Simplify inside the second bars
Inside the second bars is $6 - 11$ . You take a bigger number from a smaller one, so the result is negative:
$6 - 11 = -5$
The second box gets the tile −5.
Watch out: do not change −5 to 5. The bars are still there in the new expression, so the box holds what is inside them.
Keep the boxes in order
The first bars came first, so 1 goes in the first box and −5 in the second. The new expression has the same value as the old one:
$|1| - |-5| = 1 - 5 = -4$
Swapping the tiles gives $|-5| - |1| = 5 - 1 = 4$ , which is not equal. So the boxes get 1 and −5, making |1| − |−5|.
On the test, you drag each tile into its box. On this site, you choose it from the box’s menu.

Check your answer

Use a number line for the first pair. Start at −8 and move 3 to the right to reach −5. Then move 6 more to the right to reach 1. For the second pair, start at 6 and move 11 to the left. You land on −5. Both boxes check out.

More questions like this

The original DOE question

Question 17 as shown on the DOE’s digital test, page 88 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test A Math, page 88, question 17.

Question 2 of 22

2026 Practice Test C Math, question 7

Question

$-12 > 6x$

For what values of x is the above inequality true?

A $x > -6$
B $x < -6$
C $x < -2$
D $x > -2$

Show the answer

Correct answer: C

Divide to solve for x:

$\frac{(-12)}{6} > \frac{6x}{6}$

$-2 > x$

Rewriting $-2 > x$ with the x term first requires reversing the inequality symbol in order to keep the meaning “x is less than $(-2)$ ”.

$x < -2$

Question analysis

This inequality question asks which values of x make −12 > 6x true. Students solve a one-step inequality by dividing and then state the solution with x written first. A negative number on one side and the variable on the other make the direction of the symbol the main thing to get right.

NYS standards: 7.EE.4b two-step inequalities
6.NS.7a inequality as relative position
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 inequalities usually take two steps, such as 3x + 4 > 19. This one needs a single division, and the care goes into reading the result correctly when the variable ends up on the right of the symbol.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 10, 15 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

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.

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)
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

Worked solution, step by step

Formulas used

Dividing both sides of an inequality: $\text{if } a > b \text{ and } c > 0 \text{, then } \frac{a}{c} > \frac{b}{c}$; a and b are the two sides of an inequality, and c is the number you divide by. If c is positive, the symbol stays the same. It flips only when you multiply or divide both sides by a negative number. Here c is 6, which is positive.
Reading an inequality the other way: $a > b \text{ means the same as } b < a$; a and b are any two numbers. Saying a is greater than b is the same as saying b is less than a. When you swap the two sides, turn the symbol around. Its point still aims at the smaller number.

Sort out the question
The inequality is $-12 > 6x$ . The symbol > means is greater than. So it says −12 is greater than 6 times x. You need every value of x that makes this true.
Divide both sides by 6
To get x alone, divide both sides by 6:
$\frac{-12}{6} > \frac{6x}{6}$
$-2 > x$
You divided by 6, which is positive. So the symbol does not flip.
Watch out: −12 is negative, but that does not matter. Only dividing by a negative number flips the symbol.
Put x first
$-2 > x$ says −2 is greater than x. That means x is less than −2. Swap the sides and turn the symbol around:
$x < -2$
Watch out: keeping > when you swap the sides gives choice D, $x > -2$ . That is the opposite meaning.
Choose the answer
Choices A and B use −6. That comes from adding 6 to −12 instead of dividing by 6. The right boundary is −2, and x is below it.
So the answer is C, $x < -2$ .

Check your answer

Test a number below −2, such as −3. Then $6 \times (-3) = -18$ , and $-12 > -18$ is true. Now test 0, which is above −2. Then $6 \times 0 = 0$ , and $-12 > 0$ is false. So only numbers below −2 work.

More questions like this

The original DOE question

Question 7 as shown on the DOE’s digital test, page 80 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 80, question 7.

Question 3 of 22

2026 Practice Test C Math, question 14

Question

If $x = -2$ , what is the value of $|x + x^{2} + x^{3}| - |x + 2x + 3x|$ ?

A−18
B−6
C−2
D2

Show the answer

Correct answer: B

Substitute $-2$ for each value of x in the expression:

$\left|(-2) + (-2)^{2} + (-2)^{3}\right|$
minus
$|(-2) + 2(-2) + 3(-2)|$

Evaluate the terms involving exponents:

$|(-2) + 4 + (-8)|$
minus
$|(-2) + 2(-2) + 3(-2)|$

Evaluate the terms involving multiplication:

$|(-2) + 4 + (-8)|$
minus
$|(-2) + (-4) + (-6)|$

Simplify the expression:

$|2 + (-8)| - |(-2) + (-10)|$

$\lvert -6 \rvert - \lvert -12 \rvert$

The absolute value of $(-6)$ is 6, and the absolute value of $(-12)$ is 12, so $\lvert -6 \rvert - \lvert -12 \rvert$ is the equivalent of subtracting 12 from 6:

$6 - 12 = (-6)$

Question analysis

This substitution question sets x = −2 in an expression that subtracts one absolute value from another: the first holds x, x squared and x cubed, the second x, 2x and 3x. Students evaluate the expression. It tests powers of negative numbers, absolute value and integer subtraction, and the signs of the square and the cube are where answers go wrong.

NYS standards: 7.NS.2a rational number multiplication
6.EE.2c evaluating expressions and order of operations
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Substituting a negative number and taking absolute values is typical grade 7 sign practice. This expression puts a square and a cube of a negative inside one pair of bars and then subtracts two absolute values, so there are several places to slip.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 8 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every step is routine, but the expression stacks sign decisions, and each wrong choice records a different one going wrong. The powers of −2 alternate in sign, which students who treat every power as positive, or the square as negative, miss. The last step catches students who know the rules: both absolute values are positive, yet 6 − 12 is still negative, so the answer keeps its minus sign.

Error-prone arithmetic moderate Working it out
With x = −2 the square is 4 but the cube is −8, so the first bars hold −2 + 4 − 8 = −6, and the sign changes from each term to the next. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Six terms are substituted, two pairs of bars are simplified and made positive, and only then is 12 subtracted from 6, four stages that must come in a fixed order. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out each power of −2 separately, keeping its sign, add the terms inside each pair of bars, take both absolute values, and subtract only at the end.

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

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Simplifies both insides correctly but treats the first bars like parentheses, keeping −6, so −6 − 12 = −18.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Takes (−2)³ as positive 8, as if every power of a negative were positive, so the first bars hold −2 + 4 + 8 = 10, and 10 − 12 = −2.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Takes (−2)² as −4, so the first bars hold −2 − 4 − 8 = −14, and 14 − 12 = 2; dropping the minus from x throughout also gives 2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value: $|a| = \text{the distance from } a \text{ to } 0$; a is any number. The bars mean absolute value: how far a is from 0 on the number line. A distance is never negative, so |−6| = 6 and |6| = 6. Work out what is inside the bars first, then take the absolute value.
Powers of a negative number: $(-2)^{2} = (-2) \times (-2) \quad (-2)^{3} = (-2) \times (-2) \times (-2)$; The small raised number is the exponent. It tells how many times −2 is multiplied by itself. Two negatives multiply to a positive. So an even number of negative factors gives a positive answer, and an odd number gives a negative one.

Sort out the question
You need the value of $|x + x^{2} + x^{3}| - |x + 2x + 3x|$ when x = −2. The bars | | mean absolute value. Work out each pair of bars on its own. Subtract only at the end.
Find the powers of −2
Put −2 in for x. Keep each negative sign:
$(-2)^{2} = (-2) \times (-2) = 4$
$(-2)^{3} = 4 \times (-2) = -8$
The square is positive, but the cube is negative.
Watch out: taking $(-2)^{3}$ as +8 makes the first bars hold 10. That leads to choice C, −2.
Work out the first bars
Add the three terms inside the first pair of bars:
$|(-2) + 4 + (-8)| = |-6| = 6$
−2 + 4 is 2, and 2 + (−8) is −6. The distance from −6 to 0 is 6.
Work out the second bars
Here 2x means 2 times x, so it is −4. And 3x is −6:
$|(-2) + (-4) + (-6)| = |-12| = 12$
All three terms are negative, so they add to −12. Its absolute value is 12.
Subtract the second value
Now subtract the second value from the first:
$6 - 12 = -6$
Watch out: the bars make each part positive, but the difference can still be negative. Keeping −6 inside the first bars gives −6 − 12 = −18, choice A.
So the answer is B, −6.

Check your answer

Check the second bars a faster way. $x + 2x + 3x = 6x$ , and $6 \times (-2) = -12$ . That matches. Also, the second value, 12, is bigger than the first, 6. So the difference must be negative, as −6 is.

More questions like this

The original DOE question

Question 14 as shown on the DOE’s digital test, page 87 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 87, question 14.

Question 4 of 22

2026 Practice Test C Math, question 22

Question

During a hiking trip, a group of students recorded the elevation (in feet) at several points along the trail:

$-120\text{ ft},\ -80\text{ ft},\ 0\text{ ft},\ 45\text{ ft},\ -30\text{ ft},\ 90\text{ ft}$

What is the difference between the highest and lowest elevations recorded on the trail?

Enter your answer in the space.

Fill-in question: work out the answer and type it as a number.

Show the answer

Correct answer: 210

Determine the difference between the highest and lowest elevations:

Determine the highest elevation:

Of −120, −80, 0, 45, −30, and 90, the greatest value is 90 because its location is farthest to the right on a number line.

Highest elevation $= 90\text{ ft}$

Determine the lowest elevation:

Of −120, −80, 0, 45, −30, and 90, the lowest value is −120 because its location is farthest to the left on a number line.

Lowest elevation $= -120\text{ ft}$

Determine the difference between the two elevations:

$90\text{ ft} - (-120)\text{ ft} = 90\text{ ft} + 120\text{ ft} = 210\text{ ft}$

Question analysis

A group of students on a hiking trip records six elevations in feet, some below zero and some above, and finds the difference between the highest and the lowest, typing the result. It tests ordering positive and negative numbers and subtracting integers. Spotting the lowest reading among several negatives and subtracting a negative number correctly are where errors creep in.

NYS standards: 7.NS.1c rational number subtraction
6.NS.7b order statements in context
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the gap between a positive and a negative value by subtracting is core grade 7 work with integers. Picking the extremes from six readings is quick, so the demand is the subtraction of a negative number.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Picking the highest and lowest of six readings is quick, though −120 must be recognized as lower than −80 and −30. The points are lost in the subtraction: the span runs 120 feet up to zero and 90 more above it, and students who treat the readings as plain sizes find only 30. Picturing the two readings on a vertical number line shows the whole span.

Error-prone arithmetic moderate Working it out
The difference is 90 − (−120), and subtracting a negative means adding 120, so the part of the trail below zero is easy to leave out. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The difference is typed with no choices, so 30 or −30, the results of mishandling the negative, gets no warning that the two readings lie on opposite sides of zero. (Enright, Morley & Sheehan, 2002)

Strategy: Pick the greatest and the least elevation by their places on a number line, then subtract the least from the greatest, turning the subtraction of a negative into an addition.

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).
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

30 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Subtracts the sizes, 120 − 90 = 30, as if both readings were on the same side of zero, leaving out the stretch below zero.
-30 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Combines the two extremes by adding, 90 + (−120) = −30, the net of the readings, instead of subtracting the lowest from the highest.
-210 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Subtracts in the reverse order, lowest minus highest, −120 − 90 = −210, and enters a negative difference.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Difference between highest and lowest: $\text{difference} = \text{highest} - \text{lowest}$; The difference between the highest and lowest values is how far apart they are. Subtract the lowest from the highest, so the answer is never negative.
Subtracting a negative: $a - (-b) = a + b$; a and b are numbers. Subtracting a negative is the same as adding a positive. On a number line, it means moving to the right.

Sort out the question
Elevation is height above or below a starting level, here 0 feet. Negative elevations are below 0. The six readings are −120, −80, 0, 45, −30 and 90 feet. You need the difference between the highest and the lowest.
Find the highest and the lowest
Picture the readings on a number line. The highest is farthest right: 90 feet. The lowest is farthest left: −120 feet.
For negative numbers, the bigger the number after the minus sign, the lower it is. So −120 is lower than −80.
Subtract the lowest from the highest
Use the difference formula. Subtracting a negative is the same as adding:
$90 - (-120) = 90 + 120 = 210$
So the two readings are 210 feet apart.
Watch out: $120 - 90 = 30$ treats both readings as if they were above 0. Adding them, $90 + (-120) = -30$ , is wrong too.
Enter the answer
A difference like this is a distance, so it is positive. Subtracting in the other order gives −210, which is wrong. Type only the number, with no units.
So you would enter 210.

Check your answer

Count the distance in two parts. From −120 feet up to 0 is 120 feet. From 0 up to 90 is 90 more feet. Together, $120 + 90 = 210$ feet. That matches.

More questions like this

The original DOE question

Question 22 as shown on the DOE’s digital test, page 95 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 95, question 22.

Question 5 of 22

2026 Practice Test C Math, question 37

Question

The table shows the probability of randomly picking each flavor of candy from a bowl.

PICKING CANDY FROM
A BOWL

Flavor	Probability
Butterscotch	0.57
Lemon	0.095
Maple	0.125
Strawberry	0.21

Which flavor is most likely to be randomly picked?

Abutterscotch
Blemon
Cmaple
Dstrawberry

Show the answer

Correct answer: A

The flavor most likely to be picked at random is butterscotch, because its probability is the greatest of the probabilities in the list. More than half the candies in the bowl are butterscotch.

Question analysis

A table lists the probability of picking butterscotch, lemon, maple or strawberry candy at random from a bowl, each as a decimal. Students decide which flavor is most likely to be picked. It tests the meaning of probability and comparing decimals, which have different numbers of digits here and must be compared by place value.

NYS standards: 6.SP.6 probability from 0 to 1
6.NS.7 ordering and absolute value
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Reading probabilities as likelihoods is grade 6 content, and this needs only a comparison of four decimals. The one demand is comparing decimals with different numbers of digits, a skill from earlier grades.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly understand
What makes it hard: Error-prone arithmetic minor
Solution: 1 step
Reading level: Question and choices: about grade 9, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One idea answers the question: the most likely flavor has the greatest probability. The only obstacle is comparing decimals of different lengths, where 0.125 and 0.095 look bigger than 0.57 to students who count digits. Writing each as thousandths, 570, 95, 125 and 210, or noticing that 0.57 alone is more than one half, makes the order plain.

Error-prone arithmetic minor Working it out
The probabilities have two and three decimal places, 0.57 beside 0.095 and 0.125, so comparing them by length, or by their digits as whole numbers, ranks them wrongly. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Line the four decimals up by place value, adding zeros so each has three decimal places, and compare them digit by digit starting from the tenths place.

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.

The wrong answers, classified

B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Reads lemon’s 0.095 as 0.95, overlooking the 0 in the tenths place, so it looks almost certain and larger than 0.57.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats the decimal with more digits as the larger, or compares 125 with 57 as whole numbers, so 0.125 seems bigger than 0.57.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common misreading makes 0.21 the largest, since it is neither the longest decimal nor the one with the biggest digits, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{candies of that flavor}}{\text{all candies in the bowl}}$; P is the probability of picking a flavor. It is a number from 0 to 1. The more candies of a flavor there are, the bigger its probability. So the flavor with the greatest probability is the most likely to be picked.

Sort out the question
The table gives the probability of picking each of four flavors. You need the flavor that is most likely to be picked.
The greater the probability, the more likely the pick. So you need the greatest of the four decimals.
Line up the decimals
The decimals have different numbers of digits. Add zeros at the end so each has three decimal places. This does not change their values:
Flavor Probability
Butterscotch 0.570
Lemon 0.095
Maple 0.125
Strawberry 0.210
Compare the tenths digits
Start with the first digit after the decimal point, the tenths place. The tenths digits are 5, 0, 1 and 2. The biggest is 5, so 0.570 is the greatest.
Watch out: 0.125 has more digits than 0.57, but it is smaller. Compare place by place, not by length. That mistake leads to choice C. Also, lemon is 0.095, not 0.95, so choice B is out.
Give the answer
Butterscotch has the greatest probability, 0.57. So it is the flavor most likely to be picked.
So the answer is A, butterscotch.

Flavor	Probability
Butterscotch	0.570
Lemon	0.095
Maple	0.125
Strawberry	0.210

Check your answer

0.57 is more than 0.5, which is one half. So more than half the candies are butterscotch. The other three flavors add up to $0.095 + 0.125 + 0.21 = 0.43$ , less than half. So butterscotch must be the most likely.

More questions like this

The original DOE question

Question 37 as shown on the DOE’s digital test, page 110 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 110, question 37.

Question 6 of 22

2024 Test A Math, question 12

Question

If n is a whole number, and 0.01 is between $\frac{1}{n}$ and $\frac{1}{n + 2}$ , what is the value of n?

Show the answer

Correct answer: D

$0.01 = \frac{1}{100}$ , which is between $\frac{1}{101}$ and $\frac{1}{99}$ .

Therefore, $n + 2 = 101$ and $n = 99$ .

Question analysis

A number sense question: 0.01 lies between 1 over n and 1 over n + 2, where n is a whole number, and students find n. It tests comparing a decimal with unit fractions and knowing how a fraction changes as its denominator grows. The unusual form, with n in two denominators, is what slows students down.

NYS standards: 6.NS.7a inequality as relative position
6.EE.5 solutions and substitution
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students compare fractions and decimals routinely, but not with an unknown in the denominators. The comparison is simple once 0.01 is written as a fraction; the unfamiliar setup is the main demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 18 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The setup looks unfamiliar, with n in two denominators, but one rewrite makes it routine: 0.01 is 1/100, so the two denominators must sit on either side of 100. The three small choices fail a quick check, since every fraction from 1/1 to 1/4 is far above 0.01, and 0 cannot be a denominator at all. Students who test the choices instead of rewriting the decimal spend a little longer but reach the same place.

Variables instead of numbers moderate Understanding the question
The unknown sits in two denominators, n and n + 2, so students must reason about which unit fractions lie on either side of 0.01 rather than solve an equation. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
Two facts are assumed: 0.01 equals 1/100, and a unit fraction shrinks as its denominator grows, so 1/99 is just above 0.01 and 1/101 just below. (Anderson & Krathwohl, 2001)

Strategy: Rewrite 0.01 as a unit fraction, recall that a larger denominator gives a smaller fraction, and test each choice in both 1/n and 1/(n + 2).

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.
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 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats 1/0 as 0, so 0.01 seems to lie between 0 and 1/2; but division by zero is undefined, so n cannot be 0.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the 1 from 0.01 or from 1/n as the value of n, without checking that 0.01 is far below both 1 and 1/3.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the 2 from n + 2, or the two decimal places of 0.01, without testing 1/2 and 1/4, which are both far above 0.01.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing unit fractions: $\text{if } a < b \text{, then } \frac{1}{a} > \frac{1}{b}$; A unit fraction has 1 on top, like one fifth. a and b are positive whole numbers on the bottom. A bigger bottom cuts the whole into more pieces, so each piece is smaller. So the fraction with the bigger bottom is the smaller number.

Sort out the question
n is a whole number: 0, 1, 2, 3 and so on. The number 0.01 is between $\frac{1}{n}$ and $\frac{1}{n + 2}$ . Between means one fraction is bigger than 0.01 and the other is smaller. You need the value of n.
Write 0.01 as a fraction
0.01 means one hundredth.
$0.01 = \frac{1}{100}$
Now all three numbers are fractions with 1 on top. That makes them easy to compare.
Find fractions on each side
With 1 on top, a bigger bottom makes a smaller fraction.
$\text{if } a < b \text{, then } \frac{1}{a} > \frac{1}{b}$
So $\frac{1}{99}$ is a little more than $\frac{1}{100}$ , and $\frac{1}{101}$ is a little less.
$\frac{1}{101} < \frac{1}{100} < \frac{1}{99}$
The bottoms 99 and 101 are 2 apart, just like n and n + 2.
Match the fractions to n
The smaller bottom is n, and the larger bottom is n + 2.
$n = 99 \text{ and } n + 2 = 101$
Both fit. Watch out: small choices do not work. For n = 1, the fractions are 1 and $\frac{1}{3}$ . Both are far bigger than 0.01. For n = 0, $\frac{1}{0}$ is not a number, because you can’t divide by zero. So the answer is D, 99.

Check your answer

Write the fractions as decimals. $\frac{1}{99} \approx 0.0101$ , a bit more than 0.01. $\frac{1}{101} \approx 0.0099$ , a bit less. The sign ≈ means about. So 0.01 sits right between them.

More questions like this

The original DOE question

Question 69 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 69. Open page 78 of the handbook (PDF)

Question 7 of 22

2024 Test A Math, question 35

Question

Evaluate:

$\left|(-8) - 12 + (-17) - (-31)\right| - \left|24\right|$

A−30
B−18
C18
D44

Show the answer

Correct answer: B

$\left|(-8) - 12 + (-17) - (-31)\right| - \left|24\right|$

Simplify negatives inside the first absolute value brackets.

$\left|-8 - 12 - 17 + 31\right| - \left|24\right|$

Simplify the first absolute value.

$\left|-6\right| - \left|24\right|$

Calculate the absolute values.

$6 - 24$

Perform the final subtraction.

$6 - 24 = -18$

Question analysis

This integer question asks students to evaluate an expression with two absolute values: the first holds four positive and negative numbers, and the second is the absolute value of 24. It tests subtracting negative numbers and understanding absolute value. Subtracting a negative number and handling each pair of bars separately are where most slips happen.

NYS standards: 7.NS.1c rational number subtraction
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students add and subtract integers and know absolute value as distance from zero. This expression puts four signed terms inside one pair of bars and subtracts a second absolute value, a typical but sign-heavy grade 7 exercise.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 16+, 2 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every step is routine, but the question stacks sign decisions, and one slip anywhere produces a clean wrong number. The last step is the subtle one: each pair of bars gives a positive number, but 6 minus 24 is still negative. Students who believe bars guarantee a positive result flip that subtraction, and the choices offer the flipped number beside the right one.

Error-prone arithmetic moderate Working it out
The first bars hold four signed terms, and two must be rewritten, + (−17) as − 17 and − (−31) as + 31, before they combine to −6. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
The answer, −18, and the choice 18 differ only in sign, so estimating cannot separate them; only doing the last step as 6 − 24 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Many steps minor Working it out
Each pair of bars is simplified inside and made positive before the outer subtraction, so three separate jobs must be done in the right order. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite each subtraction of a negative as addition, combine the four numbers inside the first bars, take both absolute values, and only then do the final subtraction.

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.)
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 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Simplifies the inside to −6 but treats the bars like parentheses, keeping −6, so −6 − 24 = −30.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Assumes an expression made of absolute values cannot be negative, so subtracts 6 from 24 instead of 24 from 6.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Reads − (−31) as − 31, so the inside becomes −8 − 12 − 17 − 31 = −68, and 68 − 24 = 44.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding and subtracting negatives: $a + (-b) = a - b \quad \text{and} \quad a - (-b) = a + b$; a is any number, and b is a positive number, so −b is negative. Adding a negative moves you left on the number line, just like subtracting. Subtracting a negative is like taking away a debt, so you end up with more. It is the same as adding.
Absolute value: $\left|a\right| = \text{the distance from } a \text{ to } 0$; The bars around a number mean absolute value. a is any number. Its absolute value is how far a is from 0 on the number line. A distance is never negative, so the absolute value of −6 is 6.

Sort out the expression
The expression has two absolute values. The bars mean the distance from 0, so what comes out of them is never negative. Work in this order. First simplify inside the first bars. Then take both absolute values. Last, do the subtraction between them.
Rewrite the signs inside
Two signs inside the first bars are tricky. Adding −17 is the same as subtracting 17. Subtracting −31 is the same as adding 31.
$a + (-b) = a - b \quad \text{and} \quad a - (-b) = a + b$
$(-8) - 12 + (-17) - (-31) = -8 - 12 - 17 + 31$
Watch out: minus negative 31 is plus 31. Treating it as minus 31 makes the inside −68. That leads to 44, choice D.
Combine the four numbers
Go from left to right, one number at a time.
$-8 - 12 = -20$
$-20 - 17 = -37$
$-37 + 31 = -6$
The inside of the first bars is −6.
Take the absolute values
$\left|a\right| = \text{the distance from } a \text{ to } 0$
$\left|-6\right| = 6 \quad \text{and} \quad \left|24\right| = 24$
Now the expression is $6 - 24$ . Watch out: the bars turn −6 into 6. Keeping −6 gives $-6 - 24 = -30$ , choice A.
Do the last subtraction
$6 - 24 = -18$
You take 24 away from 6. Taking a bigger number from a smaller one gives a negative answer. Watch out: absolute values are never negative, but this answer can be. Doing $24 - 6$ instead gives 18, choice C.
So the answer is B, −18.

Check your answer

Check the inside a new way. The positive part is 31. The negative parts add to $8 + 12 + 17 = 37$ . Since $31 - 37 = -6$ , the inside is right. Then, on a number line, start at 6 and move 24 to the left. Six steps reach 0, and 18 more land on −18.

More questions like this

The original DOE question

Question 92 as printed on page 85 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 85, question 92. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 8 of 22

2024 Test B Math, question 21

Question

$-\frac{1}{49},\ -49,\ -\frac{1}{7},\ -1$

Which number shown above is the greatest?

A $-\frac{1}{49}$
B $-49$
C $-\frac{1}{7}$
D $-1$

Show the answer

Correct answer: A

All the numbers are negative, so they are all less than zero. Consider the absolute value of each number and where it falls on a number line. The negative number closest to 0 will be the greatest.

The negative number with the greatest absolute value is −49, so it has the least value.

The negative number with the next greatest absolute value is −1, so it would fall next in order from least to greatest.

$-49,\ -1$

Compare the absolute values of $-\frac{1}{49}$ and $-\frac{1}{7}$ .

$\frac{1}{49}<\frac{1}{7}$

Therefore, the numbers listed from least to greatest are:

$-49,\ -1,\ -\frac{1}{7},\ -\frac{1}{49}$

So the number $-\frac{1}{49}$ is the greatest.

Question analysis

Four negative numbers are listed: negative 1/49, negative 49, negative 1/7 and negative 1. Students decide which of them is the greatest. Comparing negative fractions with negative whole numbers tests whether students order numbers by their position on the number line or by their size without the sign.

NYS standards: 6.NS.7a inequality as relative position
6.NS.7d absolute value versus order
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Ordering positive and negative rational numbers on a number line is grade 6 content in New York. This is a one-step comparison of four numbers, lighter than typical grade 7 work with signed numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 10, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

With no arithmetic to do, the question tests how firmly students picture negative numbers on a line. Comparing sizes without the sign picks the number farthest from 0, and comparing the two fractions by habit picks the wrong one of the pair. Placing all four as points left of 0, with −1 between −49 and the fractions, makes the order plain.

A very close wrong answer moderate Choosing the answer
−1/49 and −1/7 differ only in the denominator, and for negative numbers the larger denominator gives the greater value, the reverse of what their sizes suggest. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must bring the number-line fact that a negative number's value falls as its distance from 0 grows, so −49 is the least of the four, not the greatest. (Anderson & Krathwohl, 2001)

Strategy: Picture each number's place on the number line: among negative numbers, the one closest to 0 is the greatest, so compare their distances from 0.

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.)
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 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Picks the number with the largest digits, 49, judging its size without the minus sign that puts it farthest below 0.
C Negatives and direction (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Knows 1/7 is greater than 1/49 and keeps that order for the negatives, missing that the order reverses on the left of 0.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Places −1/7 and −1/49 below −1, as if dividing a negative number made it more negative, so −1 looks like the greatest.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing negative numbers: $\text{if } a \lt b \text{, then } -a \gt -b$; a and b are positive numbers. Putting a minus sign on both flips their order. The one with the smaller size is closer to 0, so it is greater. For example, 2 is less than 5, but −2 is greater than −5.

Sort out the question
All four numbers are negative, which means less than 0. You need the greatest one, the one with the highest value. On a number line, greater numbers are farther to the right.
Think about distance from 0
For negative numbers, the one closest to 0 is the greatest. The size of a number without its sign is its absolute value. It is the number’s distance from 0. So look for the smallest size.
$\text{if } a \lt b \text{, then } -a \gt -b$
Compare the sizes
The sizes are 1/49, 49, 1/7 and 1. Both 1/49 and 1/7 have 1 on top. Cutting 1 into 49 pieces gives much smaller pieces than cutting it into 7. So 1/49 is less than 1/7.
$\frac{1}{49} \lt \frac{1}{7} \lt 1 \lt 49$
So 1/49 is the smallest size. That puts $-\frac{1}{49}$ closest to 0.
Put the numbers in order
Flip the order of the sizes to order the negative numbers, from least to greatest.
$-49 \lt -1 \lt -\frac{1}{7} \lt -\frac{1}{49}$
Watch out: −49 has the biggest digits, but it is the least, so choice B is wrong. Watch out: 1/7 is more than 1/49, but −1/7 is less than −1/49. So choice C is wrong.
So the answer is A, $-\frac{1}{49}$ .

Check your answer

Change the fractions to decimals: $-\frac{1}{49} \approx -0.02$ and $-\frac{1}{7} \approx -0.14$ . Think of money you owe. Owing 2 cents is better than owing 14 cents, $1 or $49. So −1/49 is the greatest.

More questions like this

The original DOE question

Question 78 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 205, question 78. Open page 205 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 9 of 22

2024 Test B Math, question 33

Question

If $x=-4$ , what is the sum of x and 10% of $|x|$ ?

A−4.4
B−3.6
C0.4
D4.4

Show the answer

Correct answer: B

First, determine 10% of $|x|$ .

$0.10\bullet|x|=$

$0.10\bullet|-4|=$

$0.10\bullet 4=0.4$

Then determine the sum of x and 10% of $|x|$ .

$-4+0.4=-3.6$

The sum is −3.6.

Question analysis

With x equal to a negative number, students find the sum of x and 10 percent of the absolute value of x. It tests absolute value, a simple percent and adding numbers with different signs. The tricky part is keeping track of which term is negative and which is not.

NYS standards: 7.NS.1b rational number addition
6.NS.7c absolute value as distance
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students add positive and negative decimals and find simple percents. This joins those skills with an absolute value in two short steps, so it matches typical practice, with the sign of each term as the main hazard.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 11 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question is short, but it asks students to treat the same number two ways: x keeps its negative sign while |x| drops it. Each wrong choice gets the sign wrong in one place, by taking 10% of −4, by making both terms positive, or by stopping at the positive 0.4. Writing the two terms separately, −4 and +0.4, before adding prevents each slip.

Error-prone arithmetic moderate Working it out
The sum mixes a negative integer with a positive decimal, −4 + 0.4, and the result has to keep the negative sign while moving 0.4 toward zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
Students must know that |−4| = 4, so the 10% is taken of a positive number even though x itself is negative. (Anderson & Krathwohl, 2001)

Strategy: Evaluate the absolute value first, take 10 percent of that positive number, and then add the result to x itself, keeping x's negative sign.

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

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Takes 10% of x instead of |x|, getting −0.4, and adds it to −4 to reach −4.4, overlooking the absolute value bars.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Finds 10% of |x| = 0.4 correctly and stops there, without adding x as the word sum requires.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Uses 4 for x as well as for |x|, so adds 4 + 0.4 = 4.4 and loses the negative sign of x itself.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value: $|a| = \text{distance from } a \text{ to } 0$; The two bars mean absolute value. a is any number. Its absolute value is how far it is from 0 on the number line. A distance is never negative, so |−4| = 4.
Percent of a number: $\text{part} = \frac{\text{percent}}{100} \times \text{whole}$; The whole is the amount the percent is taken of. The percent tells how many out of every 100. The part is the piece you get. Use it to find a percent of any number.

Sort out the question
You know that x = −4. You need the sum of two things: x itself, and 10% of |x|. A sum means you add.
The bars around x mean absolute value. So the 10% is taken of |x|, not of x.
Find the absolute value
−4 is 4 steps from 0 on the number line. So its absolute value is 4. An absolute value is never negative.
$|x| = |-4| = 4$
Take 10 percent of it
$\text{part} = \frac{10}{100} \times 4 = 0.4$
So 10% of |x| is 0.4.
Watch out: If you take 10% of x instead, you get −0.4. Then −4 + (−0.4) = −4.4, choice A. The bars make the number positive first.
Add x and the result
Now add x, which is still −4. Its own minus sign stays.
$-4 + 0.4 = -3.6$
Start at −4 on the number line. Move 0.4 to the right, toward 0. You land on −3.6.
Watch out: Stopping at 0.4 gives choice C. Using 4 in place of x gives 4.4, choice D.
So the answer is B, −3.6.

Check your answer

Adding a small positive number should move −4 a little toward 0. The answer −3.6 is just to the right of −4, so that fits. Also, $-3.6 - (-4) = 0.4$ , the amount you added.

More questions like this

The original DOE question

Question 90 as printed on page 209 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 209, question 90. Open page 209 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 10 of 22

2024 Test B Math, question 49

Question

Suppose that w, x, y, and z are integers, none of which are zero. Under which set of conditions would the product $x^{2}yz^{3}|w|$ always be positive?

Ax is positive and w is positive.
Bx is negative and w is negative.
Cy is negative and z is negative.
Dx and y are negative, and z and w are positive.

Show the answer

Correct answer: C

First, analyze the role of each variable in the expression $x^{2}yz^{3}|w|$ .

$x^{2}$ will always be a positive number.

y can be a positive or a negative number.

$z^{3}$ can be a positive or a negative number.

$|w|$ will always be a positive number.

The expressions $x^{2}$ and $|w|$ will always be positive. Therefore, making the values of y and z negative would result in the expression $yz^{3}$ being positive, since a negative multiplied by a negative is positive. Thus the given product would always be positive.

Question analysis

Four nonzero integers appear in a product that includes a square, a cube and an absolute value. Students choose the set of conditions that makes the product positive every time. It tests the sign rules for multiplication together with exponents and absolute value, and the trap is judging the sign by counting negative variables without looking at what is done to each.

NYS standards: 7.NS.2a rational number multiplication
6.EE.1 whole-number exponent expressions
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students learn sign rules for products of signed numbers. Here they must reason about four unknown integers under an even power, an odd power and an absolute value, and decide what is always true, which is more abstract than typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 58 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One insight decides the question: x² and |w| are positive whatever x and w are, so only the signs of y and z³ matter. Students who count negative letters instead of negative factors pick conditions that look balanced but leave y as the only negative factor. Students who test a single set of values can be fooled by a choice that leaves y and z unspecified.

Variables instead of numbers moderate Understanding the question
There are four variables and no numbers, and always positive must hold for every allowed value, so one test case can rule a choice out but never confirm it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked moderate Choosing the answer
Each set of conditions has to be checked against all four factors; the first two fix only x and w and leave y and z free to make the product negative. (Embretson & Wetzel, 1987)
Something you must already know minor Working it out
Students must know that a nonzero square and an absolute value are always positive, while an odd power such as z³ keeps the sign of z. (Anderson & Krathwohl, 2001)

Strategy: Work out, factor by factor, which signs each part of the product can take, then test each set of conditions against every factor rather than counting negatives.

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.)
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 Checks only one number (Schoenfeld, 1985)
Reveals partial knowledge Expected pull: moderate
Tests the choice with every variable positive, so the product is positive and it seems to work; y = −1 with z = 1 would make it negative.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Pairs the negatives x and w as if they must cancel, not seeing that x² and |w| are positive anyway, so y and z still decide.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Counts two negative letters, x and y, as an even number of negatives, but x² is positive, so y is the only negative factor.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Signs in multiplication: $(-a) \times (-b) = a \times b$; a and b are positive numbers, so −a and −b are negative. A negative times a negative is positive. A negative times a positive is negative. So a product is positive when it has an even number of negative factors.

Sort out the question
w, x, y and z are integers, meaning whole numbers or their negatives. None of them is 0. The product $x^2 y z^3 |w|$ means $x^2 \times y \times z^3 \times |w|$ .
You need the conditions that make the product positive every time, for every number that fits them.
Look at x squared and |w|
$x^2$ means x × x. Two positives or two negatives multiply to a positive. So $x^2$ is always positive.
$|w|$ is the absolute value of w, its distance from 0. It is always positive, since w is not 0.
So x and w never change the sign of the product. Their signs do not matter.
Look at y and z cubed
y can be positive or negative. $z^3$ means z × z × z, and it has the same sign as z. For example, $(-2)^3 = -8$ .
So the sign of the whole product is the sign of $y \times z^3$ . It depends only on y and z.
Test each choice
- A and B say nothing about y and z. If y = −1 and z = 1, the product is negative. So they do not always work.
- C: y is negative, and $z^3$ is negative. A negative times a negative is positive. This always works.
- D: y is negative and z is positive. So the product is always negative.
Watch out: D has two negative letters, x and y. But $x^2$ is positive anyway, so y is the only negative factor.
So the answer is C, y is negative and z is negative.

Check your answer

Try numbers for C. Let x = −2, y = −3, z = −1 and w = −5. Then $x^2 \times y \times z^3 \times |w| = 4 \times (-3) \times (-1) \times 5 = 60$ , which is positive. Even with x and w negative, the product stays positive.

More questions like this

The original DOE question

Question 106 as printed on page 213 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 213, question 106. Open page 213 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 11 of 22

2024 Test B Math, question 53

Question

Which graph represents the solution to $|x+1|\geq 2$ ?

A $Number line from −3 to 3 with a tick at every integer. Closed dots at −3 and 1. The line is shaded from −3 to the left, ending in an arrow, and from 1 to the right, ending in an arrow.$
B $Number line from −3 to 3 with a tick at every integer. Closed dots at −1 and 1. The line is shaded from −1 to the left, ending in an arrow, and from 1 to the right, ending in an arrow.$
C $Number line from −3 to 3 with a tick at every integer. Closed dots at −3 and 1, with the segment between them shaded.$
D $Number line from −3 to 3 with a tick at every integer. Closed dots at −3 and 3. The line is shaded from −3 to the left, ending in an arrow, and from 3 to the right, ending in an arrow.$

Show the answer

Correct answer: A

Since $|x+1|\geq 2$ , either $x+1\leq -2$ or $x+1\geq 2$ .

Solving $x+1\leq -2$ for x gives $x\leq -3$ .

Solving $x+1\geq 2$ for x gives $x\geq 1$ .

A filled circle on a number line represents “or equal to.” So the correct number line representation has filled circles at −3 and 1, with the number line filled to the left of −3 for “less than” and to the right of 1 for “greater than.”

Question analysis

Students choose the number line graph that shows every solution of an inequality in which x + 1 sits inside absolute value bars. It tests absolute value as distance and graphing inequalities. The tricky part is turning one absolute value statement into two pieces and placing each boundary point and shaded ray correctly.

NYS standards: 7.NS.1c rational number subtraction
7.EE.4b two-step inequalities
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve and graph two-step inequalities and know absolute value as distance. An inequality with an absolute value has to be split into two cases, which is outside the grade 6 to 8 standards and well above typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
A very close wrong answer moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 7 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

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.

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)
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)
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

Worked solution, step by step

Formulas used

Absolute value as distance: $|x - a| = \text{distance from } x \text{ to } a$; x and a are points on the number line. The bars mean absolute value. |x − a| is how far apart the two points are, which is never negative. Since x + 1 = x − (−1), |x + 1| is the distance from x to −1.

Sort out the question
You need the graph of every x that makes $|x + 1| \geq 2$ true. The sign ≥ means greater than or equal to. The bars mean absolute value.
On a number line, a filled dot means that point counts. A thick line with an arrow means all the numbers that way count too.
Read it as a distance
Adding 1 is the same as subtracting −1, so $x + 1 = x - (-1)$ . By the formula, $|x + 1|$ is the distance from x to −1.
So the inequality says that x is at least 2 units away from −1.
Find the two boundary points
Start at −1 and go 2 units each way.
$-1 - 2 = -3$
$-1 + 2 = 1$
At −3 and at 1, the distance is exactly 2. The words or equal to let them count, so they get filled dots.
Decide which side to shade
Being at least 2 away means being farther out. Numbers between −3 and 1 are too close to −1. For example, 0 is only 1 unit away. So shade to the left of −3 and to the right of 1.
In algebra, that is $x + 1 \leq -2$ or $x + 1 \geq 2$ . These give $x \leq -3$ or $x \geq 1$ .
Match the graph
Choice A has filled dots at −3 and 1, with arrows going outward. That matches.
Watch out: Choice C shades between −3 and 1. Those are the points within 2 of −1, the answer to $|x + 1| \leq 2$ . Choices B and D are centered on 0, but the center here is −1.
So the answer is A, the graph shaded outward from −3 and 1.

Check your answer

Test a point from each part. For x = 2, $|2 + 1| = 3$ , and 3 is at least 2. For x = 0, $|0 + 1| = 1$ , which is less than 2. For x = −3, $|-3 + 1| = 2$ , which counts. Graph A agrees with all three.

More questions like this

The original DOE question

Question 110 as printed on page 214 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 214, question 110. Open page 214 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 12 of 22

2024 Test B Math, question 57

Question

If x and y are both negative integers with $x<y$ , which expression has the least value?

A $x^{2}y$
B $xy^{2}$
C $-x^{2}y$
D $-xy^{2}$

Show the answer

Correct answer: A

Since x and y are both negative integers, both $x^{2}$ and $y^{2}$ are positive.

Therefore, both $x^{2}y$ and $xy^{2}$ are negative, and both $-x^{2}y$ and $-xy^{2}$ are positive.

Because $x<y$ and both x and y are negative, $|y|<|x|$ and $y^{2}<x^{2}$ . So $|y^{2}|<|x^{2}|$ .

Since $|y|<|x|$ and x and y are integers, $|xy^{2}|<|x^{2}y|$ .

So $x^{2}y<xy^{2}$ .

$-xy^{2}=|xy^{2}|\text{ and }-x^{2}y=|x^{2}y|$

Therefore, $x^{2}y<xy^{2}<-xy^{2}<-x^{2}y$ .

Question analysis

Two negative integers x and y, with x less than y, appear in four expressions built from squares and minus signs. Students find the expression with the least value. It tests multiplying negative numbers and comparing their sizes, and the difficulty is tracking how each minus sign and each square changes both the sign and the size of the result.

NYS standards: 7.NS.2a rational number multiplication
6.NS.7d absolute value versus order
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students multiply negative numbers, but this asks them to compare four variable expressions for any pair of negative integers. Deciding how the squares compare when x is less than y goes beyond typical grade 7 sign practice.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
A very close wrong answer moderate
Every choice must be checked minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 20 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two facts settle the question: squares of negatives are positive, and the more negative number has the larger square. Students who track the minus signs carelessly think a leading minus makes an expression negative, and students who assume the larger y has the larger square choose the wrong negative expression. Substituting x = −3 and y = −2 gives −18, −12, 18 and 12, which orders all four at once.

Variables instead of numbers moderate Understanding the question
No values are given, only that x and y are negative integers with x < y, so students must reason about signs or invent a pair such as −3 and −2. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
x²y and xy² use the same letters with the exponent moved; both are negative, and only comparing x² with y² separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Every choice must be checked minor Choosing the answer
All four expressions have to be placed in order, because the minus signs in front of the last two turn those expressions positive. (Embretson & Wetzel, 1987)

Strategy: Pick two negative integers with x less than y, such as −3 and −2, evaluate all four expressions, and confirm the order with a second pair.

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.)
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: strong
Knows x²y and xy² are both negative but takes y² as the larger square because y > x, though the more negative x has the bigger square.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Reads the leading minus as making −x²y negative, missing that y is negative, so −x²y is in fact positive.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Assumes the leading minus makes −xy² negative, missing that x is negative, so −xy² is in fact positive.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Signs in multiplication: $(-a) \times (-b) = a \times b$; a and b are positive numbers, so −a and −b are negative. A negative times a negative is positive. A negative times a positive is negative. So a product is positive when it has an even number of negative factors.
Squaring a number: $x^2 = x \times x$; x is any number. x² (x squared) means x times itself. The square of a number other than 0 is always positive, because two negatives or two positives multiply to a positive.

Sort out the question
x and y are negative integers, such as −3 and −2. Since x < y, x is to the left of y on the number line. So x is more negative, farther from 0.
You need the expression with the least value. That is the most negative one.
Find the sign of each expression
Squares are positive, so $x^2$ and $y^2$ are positive.
- $x^2 y$ is a positive times a negative, so it is negative.
- $x y^2$ is a negative times a positive, so it is negative.
- $-x^2 y$ and $-x y^2$ are the opposites of these, so they are positive.
So the least value must be A or B. Watch out: The minus sign in front of C and D does not make them negative. It turns a negative into a positive.
Compare A and B with numbers
Try x = −3 and y = −2, since −3 < −2.
$x^2 y = (-3)^2 \times (-2) = 9 \times (-2) = -18$
$x y^2 = (-3) \times (-2)^2 = (-3) \times 4 = -12$
Since −18 < −12, A is smaller.
See why it always works
Both expressions contain x × y, which is positive. $x^2 y = (xy) \times x$ and $x y^2 = (xy) \times y$ . Multiplying the same positive number by the more negative x gives the more negative result.
Watch out: y is bigger than x, but $y^2$ is smaller than $x^2$ : 4 is less than 9. Thinking y² is bigger leads to choice B.
So the answer is A, $x^2 y$ .

Check your answer

Try another pair, x = −5 and y = −1. Then A is $25 \times (-1) = -25$ , B is $(-5) \times 1 = -5$ , C is 25 and D is 5. A is still the least.

More questions like this

The original DOE question

Question 114 as printed on page 215 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 215, question 114. Open page 215 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 13 of 22

2023 Test A Math, question 2

Question

$6 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2}$

What is the value of the expression shown above?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: -9

$6 - 9 \div \left\|-3\right\| + (-2)^{3} \times 1\frac{1}{2}$	Calculate the exponent and convert the mixed number to an improper fraction.
$6 - 9 \div \left\|-3\right\| + (-8) \times \frac{3}{2}$	Next, calculate the absolute value.
$6 - 9 \div 3 + (-8) \times \frac{3}{2}$	Perform the indicated multiplication and division.
$6 - 3 + (-12)$	Perform the addition and subtraction.
$3 + (-12) = -9$

Question analysis

This grid-in asks for the value of an arithmetic expression that mixes an absolute value, a negative number raised to the third power, a division and a mixed number. Students apply the order of operations with signed numbers. Cubing a negative base, and doing the division and multiplication before any adding, are where most slips happen.

NYS standards: 7.NS.2a rational number multiplication
6.EE.2c evaluating expressions and order of operations
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students routinely evaluate expressions with negative numbers in the right order. This one combines an absolute value, a cubed negative and a mixed number in a single line, so it takes more steps than usual, each with a sign to watch.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The expression packs four kinds of notation into one line: a subtraction, a division by an absolute value, a negative base raised to the third power and a mixed number. None of the operations is hard alone, but each has a rule that is easy to break under time pressure, and the slips lead to very different totals. Rewriting the power and the mixed number first, then dividing and multiplying before any addition, avoids them.

Error-prone arithmetic moderate Working it out
(−2)³ must stay negative, −8, and the mixed number 1½ must become 3/2 before multiplying, so the last product is −12; a sign or fraction slip here moves the total far from −9. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Six numbers and six operations, including a power and an absolute value, have to be worked in a planned order, and every addition waits until the division and multiplication are done. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The value is negative and the grid-in offers no choices, so the minus sign has to be gridded and nothing on the page confirms that the total falls below zero. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the absolute value and the power first, change the mixed number to an improper fraction, do the division and multiplication, and only then add and subtract from left to right.

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

15 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Takes (−2)³ as positive 8, as if every power of a negative were positive, so the last product is +12 and 6 − 3 + 12 = 15.
-13.5 Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Works strictly from left to right: 6 − 9 = −3, divided by 3 is −1, plus −8 is −9, and times 1½ gives −13.5.
-6 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Multiplies the base by the exponent, taking (−2)³ as −6, so the product with 3/2 is −9 and the total is 6 − 3 − 9 = −6.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations: $\text{grouping} \rightarrow \text{powers} \rightarrow \times \text{ and } \div \rightarrow + \text{ and } -$; This is the order for working out an expression. First work inside grouping symbols, such as parentheses and absolute value bars. Next do powers. Then multiply and divide from left to right. Last, add and subtract from left to right. Using one order means everyone gets the same answer.
A power: $a^{3} = a \times a \times a$; a is the base, the number being multiplied. The small 3 is the exponent. It says how many times the base is used as a factor, a number being multiplied. It does not mean a times 3.
Absolute value: $\left|a\right| = \text{the distance from } a \text{ to } 0$; The bars around a number mean absolute value. It is how far the number is from 0 on a number line. A distance is never negative, so the absolute value of −3 is 3.
Mixed number to fraction: $a\frac{b}{c} = \frac{a \times c + b}{c}$; a is the whole number part, and b over c is the fraction part. Each whole is c pieces, so a wholes are a × c pieces. Add the b extra pieces. The bottom number stays c.

Sort out the expression
You need the value of $6 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2}$ . It has many operations, so the order matters. It has absolute value bars, a power of a negative number and a mixed number. The answer may be negative, so keep track of every sign.
Do the absolute value and the power
Grouping symbols and powers come first.
$\left|-3\right| = 3$
$(-2)^{3} = (-2) \times (-2) \times (-2) = 4 \times (-2) = -8$
Two negatives multiply to a positive, 4. One more negative makes it negative again. Watch out: (−2)³ is −8, not +8. Using +8 leads to 15. It is also not −2 × 3. That mistake leads to a total of −6.
Change the mixed number to a fraction
A fraction is easier to multiply than a mixed number.
$1\frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2}$
Now the expression looks like this:
$6 - 9 \div 3 + (-8) \times \frac{3}{2}$
Multiply and divide next
Multiplying and dividing come before adding and subtracting.
$9 \div 3 = 3$
$(-8) \times \frac{3}{2} = \frac{-24}{2} = -12$
A negative times a positive is negative. Watch out: don’t just work from left to right. Doing 6 − 9 first, then dividing by 3, leads to −13.5.
Add and subtract last
Now only adding and subtracting are left. Work from left to right.
$6 - 3 + (-12) = 3 + (-12) = -9$
Adding −12 is the same as taking away 12. So you would grid −9, with the minus sign in its own column.

Check your answer

Look at the size of each part. The part 6 − 3 is a small positive number, 3. The last part, −8 times 1½, is a bigger negative number, −12. A big negative plus a small positive is negative. So an answer below zero, −9, makes sense.

More questions like this

The original DOE question

Question 59 as printed on page 75 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 75, question 59. Open page 75 of the handbook (PDF)

Question 14 of 22

2023 Test A Math, question 14

Question

If n is an odd integer that is less than −3.25, what is the greatest possible value of n?

A−1
B−2
C−3
D−5

Show the answer

Correct answer: D

The two greatest integers less than −3.25 are −4 and −5. Since −4 is even, the answer is −5.

Question analysis

This question asks for the greatest odd integer that is less than −3.25. Students compare negative numbers on a number line and apply two conditions at once, odd and less than. With negatives, the greater number is the one closer to zero, which makes the two conditions easy to misapply.

NYS standards: 6.NS.7a inequality as relative position
6.NS.6c positioning rational numbers and points
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Ordering negative numbers is grade 6 content, and the question adds only the condition that the integer be odd. It is lighter than typical grade 7 work, though the order of negatives still trips up students who compare their sizes.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 18 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question is short, but it asks students to reason on the negative side of 0, where greater means closer to 0 and the size of the digits misleads. The integers just below −3.25 are −4 and then −5, and the first of those is even. Sketching a number line, or reading less than as to the left of, keeps the direction straight.

Error-prone arithmetic minor Working it out
On the negative side, less means farther from 0, so the integers below −3.25 run −4, −5, −6; −3, though its digit is closer to 3.25, lies to the right. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Much to keep track of minor Working it out
Three conditions apply at once, odd, less than −3.25 and the greatest such value, and checking only two of them leads to a listed choice. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Picture −3.25 on a number line, list the integers to its left starting with the closest, and take the first one that is odd.

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Picks the greatest odd number offered, −1, without checking the condition that n be less than −3.25; −1 lies to its right on the number line.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
The choice −2 is even, so it breaks the condition that n be odd, and it is also greater than −3.25; it mostly catches guesses.
C Negatives and direction (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Looks for the odd integer just past −3.25 but steps toward 0, taking −3, though −3 is greater than −3.25, not less.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing negative numbers: $\text{if } a > b > 0 \text{, then } -a < -b$; a and b are positive numbers, and a is bigger. On a number line, numbers get smaller as you move left. A negative number farther from 0 is farther left. So it is smaller, even though its digits look bigger.

Sort out the question
n has to pass three tests. It is an integer, a whole number with no decimal part. It is odd, so it is not a multiple of 2. And it is less than −3.25. Of all the numbers that pass, you need the greatest one. That is the one farthest to the right on a number line.
Picture −3.25 on a number line
−3.25 sits between −4 and −3. Less than −3.25 means to the left of it.
$-5 < -4 < -3.25 < -3$
So −4 is less than −3.25, but −3 is greater. Watch out: −3 is not less than −3.25. It is to the right, closer to 0. For negative numbers, the one closer to 0 is greater. Picking −3 is choice C.
Find the odd integers to the left
The integers less than −3.25, starting with the closest, are −4, −5, −6 and so on. The greatest one is −4. But −4 is even, since 4 = 2 × 2. So skip it. The next one, −5, is odd. Watch out: −1 is odd, but it is greater than −3.25, so choice A breaks the rule. So the answer is D, −5.

Check your answer

Check all three tests for −5. It is an integer. It is odd. It is to the left of −3.25, so it is less. The only integer between −5 and −3.25 is −4, and it is even. So no greater number works.

More questions like this

The original DOE question

Question 71 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 79, question 71. Open page 79 of the handbook (PDF)

Question 15 of 22

2023 Test A Math, question 16

Question

Between which two consecutive integers is the fraction $\frac{29}{7}$ ?

A2 and 3
B3 and 4
C4 and 5
D5 and 6

Show the answer

Correct answer: C

First, convert $\frac{29}{7}$ to a mixed number: $\frac{29}{7} = 4\frac{1}{7}$ . The two consecutive integers on either side of $4\frac{1}{7}$ are 4 and 5.

Question analysis

This question asks between which two consecutive whole numbers the fraction 29/7 lies. Students divide to turn the improper fraction into a mixed number, or compare it with nearby multiples of 7. It is a quick check of number sense with fractions, where a remainder must be read correctly.

NYS standards: 6.NS.6c positioning rational numbers and points
6.NS.7a inequality as relative position
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Changing an improper fraction to a mixed number and placing it between whole numbers is review well before grade 7. This takes one division with a small divisor and nothing more.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 13 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Dividing 29 by 7 gives 4 with a remainder of 1, so the fraction is 4 1/7 and lies just past 4. The one decision is which side of 4 the remainder puts it on, and a positive remainder always moves the value up. Writing the mixed number before looking at the choices makes the pair obvious.

A very close wrong answer minor Choosing the answer
The pairs 3 and 4 and 4 and 5 both contain 4, so knowing that 29/7 is about 4 is not enough; only the remainder, 29 − 28 = 1, puts it above 4. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Find the largest multiple of 7 that is not more than 29, write the fraction as a mixed number, and read off the whole numbers on each side.

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).

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No division of 29 by 7 lands below 3, since 7 × 3 = 21 is far short of 29; only misreading 29/7 as 2.9 gets there, so it mostly catches guesses.
B Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Finds the whole-number part, 4, but takes it as the upper end of the pair, as if 29/7 fell just short of 4 instead of just past it.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Nothing puts 29/7 above 5, since 7 × 5 = 35 is more than 29; even rounding 4 1/7 up gives 5 as the upper end, so it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction to mixed number: $\frac{a}{b} = q\frac{r}{b}$; a over b is a fraction whose top is bigger than its bottom. q is the quotient, how many whole times b fits into a. r is the remainder, what is left over. The leftover stays over the same bottom number.

Sort out the question
Consecutive integers are whole numbers that come one right after the other, like 4 and 5. You need the two that $\frac{29}{7}$ falls between. So find the whole number just below it and the one just above it.
Divide 29 by 7
A fraction bar means divide. How many 7s fit in 29?
$7 \times 4 = 28$
$7 \times 5 = 35$
28 is the biggest multiple of 7 that is not more than 29. A multiple of 7 is 7 times a whole number. Then $29 - 28 = 1$ . So 29 ÷ 7 is 4 with 1 left over.
Write it as a mixed number
$\frac{29}{7} = 4\frac{1}{7}$
$4\frac{1}{7}$ is a little more than 4 and less than 5. Watch out: it is just past 4, not just short of it. So 4 is the lower number of the pair. Picking 3 and 4 is choice B. So the answer is C, 4 and 5.

Check your answer

Multiply each end by 7. $4 \times 7 = 28$ and $5 \times 7 = 35$ . The top of the fraction, 29, is between 28 and 35. So $\frac{29}{7}$ is between 4 and 5. As a decimal it is about 4.14.

More questions like this

The original DOE question

Question 73 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 79, question 73. Open page 79 of the handbook (PDF)

Question 16 of 22

2023 Test A Math, question 23

Question

$\{0.1, 0.01, 0.001, 0.0001, 0.00001\}$

If a person chooses a number at random from the set above, what is the probability that the number is less than 0.005?

A $\frac{1}{5}$
B $\frac{2}{5}$
C $\frac{3}{5}$
D $\frac{2}{3}$

Show the answer

Correct answer: C

Three of the five numbers listed (0.001, 0.0001, and 0.00001) are less than 0.005. So, the probability of a person selecting a number less than 0.005 is $\frac{3}{5}$ .

Question analysis

A set holds five decimals, from one tenth down to one hundred-thousandth, and one number is chosen at random. Students find the probability that the number is less than 0.005. Comparing decimals with many zeros is the real task here, along with putting the count over the right total.

NYS standards: 6.SP.8a uniform probability models
6.NS.7a inequality as relative position
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Probability with equally likely outcomes is grade 6 content in New York, and this needs one comparison per number and one fraction. The only real demand is comparing decimals with many zeros correctly.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each number in the set is a tenth of the one before, so only the three smallest fall below 0.005, which lies between 0.01 and 0.001. Students lose the point either by counting the numbers above 0.005 instead of below it or by writing the count over the wrong total. Lining the decimals up with the same number of places makes every comparison easy.

Error-prone arithmetic moderate Working it out
0.005 has to be compared with decimals from 0.1 down to 0.00001; written to five places, 0.00500 against 0.01000 and 0.00100, the order becomes a comparison of whole numbers. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
The probability is the count of numbers below 0.005 over all 5 numbers in the set; the stem leaves that part-to-whole rule to the student. (Anderson & Krathwohl, 2001)

Strategy: Write each number with five decimal places, compare it with 0.00500, count those that are smaller, and divide by the five numbers in the set.

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.
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 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Gives 1/5, the chance of picking any one particular number from the set, without counting how many of the five are below 0.005.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Counts the 2 numbers greater than 0.005, 0.1 and 0.01, instead of those less than it, finding the chance of the opposite event.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Writes a part-to-part ratio, the 2 numbers above 0.005 over the 3 below it, instead of dividing by all 5 numbers in the set.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{wanted results}}{\text{possible results}}$; P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of all the results that you want.
Adding zeros to a decimal: $0.1 = 0.10 = 0.100$; Zeros added to the right end of a decimal don’t change its value. One tenth is the same as ten hundredths. Giving every number the same number of decimal places lets you compare them like whole numbers.

Sort out the question
The set has five numbers. A person picks one at random, so each number has the same chance. You need the probability, or chance, that the number picked is less than 0.005.
Give every number five decimal places
Decimals are easier to compare with the same number of places. Add zeros to the end of each.
Number With five places
0.1 0.10000
0.01 0.01000
0.001 0.00100
0.0001 0.00010
0.00001 0.00001
0.005 0.00500
Count the numbers below 0.005
Compare each number with 0.00500 like whole numbers: 10000, 1000, 100, 10 and 1 against 500. Two numbers are bigger: 0.1 and 0.01. Three are smaller: 0.001, 0.0001 and 0.00001. Watch out: 0.01 has fewer digits than 0.005, but it is bigger. Counting the 2 bigger numbers gives $\frac{2}{5}$ , choice B.
Write the probability
$P = \frac{\text{wanted results}}{\text{possible results}} = \frac{3}{5}$
Watch out: divide by all 5 numbers in the set. Comparing the 2 bigger numbers with the 3 smaller ones gives $\frac{2}{3}$ , choice D. So the answer is C, $\frac{3}{5}$ .

Number	With five places
0.1	0.10000
0.01	0.01000
0.001	0.00100
0.0001	0.00010
0.00001	0.00001
0.005	0.00500

Check your answer

The numbers below 0.005 and the numbers above it make up the whole set: 3 + 2 = 5. So the two chances, $\frac{3}{5}$ and $\frac{2}{5}$ , add up to 1, as they should.

More questions like this

The original DOE question

Question 80 as printed on page 81 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 81, question 80. Open page 81 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 17 of 22

2023 Test A Math, question 26

Question

Which statement must be true if x is a whole number greater than or equal to 1?

A $\frac{1}{x + 1} > \frac{1}{x + 2}$
B $\frac{1}{x + 1} < \frac{1}{x + 2}$
C $\frac{1}{x + 1} - \frac{1}{x + 2} > 1$
D $\frac{1}{x + 1} - \frac{1}{x + 2} > \frac{1}{x}$

Show the answer

Correct answer: A

Let $x = 2$ , and substitute that value for x in each inequality to determine which one is correct:

Option A:

$\frac{1}{3} > \frac{1}{4}$ . This is true.

Option B:

$\frac{1}{3} < \frac{1}{4}$ . This is false.

Option C:

$\frac{1}{3} - \frac{1}{4} > 1$ , or $\frac{1}{12} > 1$ . This is false.

Option D:

$\frac{1}{3} - \frac{1}{4} > \frac{1}{2}$ , or $\frac{1}{12} > \frac{1}{2}$ . This is false.

Option A is the only true inequality.

Question analysis

This question asks which of four inequalities must be true when x is a whole number of at least 1. The inequalities compare fractions with denominators x + 1 and x + 2, or their difference. Students reason about how the size of a fraction changes as its denominator grows, and must be sure a statement holds for every allowed value.

NYS standards: 6.NS.7a inequality as relative position
6.EE.5 solutions and substitution
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Comparing fractions is familiar, but these fractions have variable denominators and the question asks what must be true for every whole number from 1 up. That abstraction makes it as demanding as typical grade 7 work, though the arithmetic is light.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 20 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Choosing a small value, such as x = 1 or 2, and testing every statement settles this quickly, because three of the four fail for any value tried. The trap is the belief that a bigger denominator makes a bigger fraction, which turns the true comparison around. Students unsure of fraction subtraction lose time on the two difference statements, though both differences are tiny, never more than 1/6.

Variables instead of numbers moderate Understanding the question
All four statements are about the unknown x, and the bold must means each has to hold for every whole number from 1 up, not just for a single test value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked moderate Choosing the answer
The statements are four separate inequalities, two comparisons and two differences, so each has to be tested; one counterexample is enough to rule a statement out. (Embretson & Wetzel, 1987)
Error-prone arithmetic minor Working it out
Testing the differences means subtracting unlike fractions, such as 1/3 − 1/4 = 1/12 when x = 2, and comparing the result with 1 and with 1/x. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Substitute a small whole number such as 1 or 2 for x in all four statements, rule out any that fail, and confirm the survivor by thinking about larger denominators.

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.
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 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Believes a bigger denominator makes a bigger fraction, so 1/(x + 2) seems larger than 1/(x + 1), the comparison turned around.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Both fractions are at most 1/2, so their difference can never exceed 1, and no common working makes it do so; this choice mostly draws guesses.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Subtracts over a common denominator but keeps only the numerator, (x + 2) − (x + 1) = 1, so the difference looks like 1, larger than 1/x once x is 2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing unit fractions: $\text{if } 0 < a < b \text{, then } \frac{1}{a} > \frac{1}{b}$; a and b are positive numbers, and b is bigger. One over a means one whole cut into a equal pieces. Cutting the same whole into more pieces makes each piece smaller. So the fraction with the bigger bottom is smaller.
Subtracting fractions: $\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a \times b}$; a and b are the bottoms of the two fractions. Rewrite both fractions over the common bottom a × b. Then subtract the tops. The bottom stays a × b.

Sort out the question
x is a whole number that is 1 or more, so x can be 1, 2, 3 and so on. You need the statement that must be true. That means it is true for every such x. One x that makes a statement false rules it out.
Test x = 1 in each statement
With x = 1, x + 1 is 2 and x + 2 is 3.
- A: $\frac{1}{2} > \frac{1}{3}$ . True.
- B: $\frac{1}{2} < \frac{1}{3}$ . False.
- C: $\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$ , which is not more than 1. False.
- D: $\frac{1}{6}$ is not more than $\frac{1}{1}$ . False.
Only A is true when x = 1. So B, C and D are out.
Make sure A is always true
Must means A has to work for every x, not just 1. In $\frac{1}{x + 1}$ and $\frac{1}{x + 2}$ , both tops are 1. The bottom x + 2 is always 1 more than x + 1. Cutting a whole into more pieces makes each piece smaller. So A is always true. Watch out: a bigger bottom makes a smaller fraction, not a bigger one. Thinking the other way gives choice B.
See why C and D always fail
Use the subtracting rule with a = x + 1 and b = x + 2.
$\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a \times b}$
$\frac{1}{x + 1} - \frac{1}{x + 2} = \frac{(x + 2) - (x + 1)}{(x + 1)(x + 2)} = \frac{1}{(x + 1)(x + 2)}$
The bottom is at least 2 × 3 = 6, so the difference is at most $\frac{1}{6}$ . That is less than 1. Its bottom is also bigger than x, so it is less than $\frac{1}{x}$ . Watch out: keep the bottom when you subtract. Keeping only the top, 1, leads to choice D. So the answer is A.

Check your answer

Try a bigger number, x = 9. A tenth of a pizza is more than an eleventh, so $\frac{1}{10} > \frac{1}{11}$ . The difference, $\frac{1}{110}$ , is tiny. It is less than 1 and less than $\frac{1}{9}$ . A still holds, and C and D still fail.

More questions like this

The original DOE question

Question 83 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 82, question 83. Open page 82 of the handbook (PDF)

Question 18 of 22

2023 Test A Math, question 56

Question

$Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. The line is shaded from the unlabeled tick halfway between 0 and 1 to the unlabeled tick halfway between 1 and 2, with closed dots at both ends.$

If all possible values of x are indicated by the shaded part of the number line above, which number line best shows all possible values of $\frac{1}{x}$ ?

A $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from the tick halfway between 1 and 2 to the tick halfway between 2 and 3, with closed dots at both ends.$
B $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from 1 to 2, with closed dots at both ends.$
C $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from the tick halfway between 0 and 1 to 2, with closed dots at both ends.$
D $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from a closed dot between the tick halfway between 0 and 1 and the tick at 1, about two-thirds of the way from 0 to 1, to a closed dot at 2.$

Show the answer

Correct answer: D

The shaded region goes from $\frac{1}{2}$ to $\frac{3}{2}$ . So, the possible values of $\frac{1}{x}$ would be between the reciprocal of $\frac{1}{2}$ and the reciprocal of $\frac{3}{2}$ , which are 2 and $\frac{2}{3}$ , respectively. Find the number line in the answer options that has a shaded region that ranges from $\frac{2}{3}$ to 2, which is option D.

Question analysis

This number line question shades all possible values of x from one half to three halves and asks which number line shows all possible values of 1/x. Students find reciprocals of fractions and work out how a whole range of values changes. Keeping track of what happens to each end of the range, and placing fractions accurately on the line, makes it tricky.

NYS standards: 6.NS.1 dividing fractions by fractions
6.NS.7a inequality as relative position
6.NS.6c positioning rational numbers and points
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Reciprocals of single fractions are grade 6 work. Applying them to every value in a shaded range at once, and working out where the new range begins and ends, is reasoning about a whole set of numbers that grade 7 classes rarely practice.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
A very close wrong answer moderate
Variables instead of numbers minor
Solution: 3 steps
Reading level: Question and choices: about grade 12, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Few students have flipped an interval before, so the question rewards testing its two ends and noticing that the order reverses. The harder part comes at the end: 2/3 falls between the tick marks, and a graph ending at the nearby 1/2 tick looks right at a glance. Students who believe a reciprocal is always at least 1, or who shift the interval instead of flipping it, reach the other graphs.

Reading a figure, table or graph moderate Understanding the question
The shaded segment's ends sit on unlabeled half-unit ticks, 1/2 and 3/2, and each choice must be read the same way, with 2/3 falling between ticks rather than on one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
A very close wrong answer moderate Choosing the answer
The graph from 1/2 to 2 matches the right graph's right end and differs at the left end by only a sixth of a unit, 1/2 against 2/3. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
The question asks about 1/x for every x in a range, so a whole interval must be transformed, and its ends swap: the smallest x gives the largest 1/x. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Take the reciprocal of each end of the shaded part, ask whether a bigger x gives a bigger or a smaller 1/x, and look for the graph whose endpoints are exactly those values.

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
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).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds 1 to each endpoint instead of taking reciprocals, sliding the shaded part from 1/2–3/2 one unit right, to 3/2–5/2.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Gets 2 from 1/2 but assumes a reciprocal cannot be less than 1, as flipping a proper fraction suggests, so it starts the shading at 1 instead of 2/3.
C Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: strong
Finds the reciprocals 2/3 and 2 but places 2/3 at the half-unit tick, 1/2, though 2/3 lies a third of the way from that tick toward 1.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Reciprocal: $\text{reciprocal of } \frac{a}{b} = \frac{b}{a}$; a and b are the top and bottom of a fraction. The reciprocal flips the fraction upside down. The number 1 over x is the reciprocal of x. A number times its reciprocal always makes 1.

Sort out the question
x can be any number in the shaded part of the first number line. You need the number line that shows every possible value of $\frac{1}{x}$ , the reciprocal of x. The reciprocal is the number flipped upside down, like $\frac{1}{2}$ for 2.
Read the shaded part
The ticks come every half unit. The shading starts at the tick halfway between 0 and 1, which is $\frac{1}{2}$ . It ends halfway between 1 and 2, which is $\frac{3}{2}$ , or $1\frac{1}{2}$ . Both dots are filled in, so both ends count.
Find the reciprocals of the ends
Flip each end of the shaded part.
$\text{reciprocal of } \frac{a}{b} = \frac{b}{a}$
$\text{reciprocal of } \frac{1}{2} = \frac{2}{1} = 2$
$\text{reciprocal of } \frac{3}{2} = \frac{2}{3}$
So the ends of the new shading are 2 and $\frac{2}{3}$ .
Notice that the ends swap
When x gets bigger, $\frac{1}{x}$ gets smaller. So the smallest x, $\frac{1}{2}$ , gives the biggest value, 2. The biggest x, $\frac{3}{2}$ , gives the smallest, $\frac{2}{3}$ . So $\frac{1}{x}$ runs from $\frac{2}{3}$ to 2, ends included.
Find the matching number line
$\frac{2}{3}$ lies between the half tick and 1, a third of the way from $\frac{1}{2}$ to 1. Line D starts there and ends at 2. Watch out: line C starts at the half tick, $\frac{1}{2}$ , not at $\frac{2}{3}$ . Line B starts at 1.
So the answer is D, the line shaded from $\frac{2}{3}$ to 2.

Check your answer

Test a value in the middle. x = 1 is shaded, and $\frac{1}{1} = 1$ . On line D, 1 is shaded too. Test an end: $\frac{3}{2} \times \frac{2}{3} = 1$ , so $\frac{2}{3}$ really is the reciprocal of $\frac{3}{2}$ .

More questions like this

The original DOE question

Question 113 as printed on page 91 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 91, question 113. Open page 91 of the handbook (PDF)

Question 19 of 22

2023 Test B Math, question 21

Question

$-mn$

$(-m)(-n)$

$|mn|$

$|-mn|$

$|(-m)(-n)|$

If m and n are positive integers, how many of the five expressions above are equivalent to $mn$ ?

Show the answer

Correct answer: D

Since m and n are positive, the value of $mn$ is also positive. Of the five options given:

$-mn$ is negative.

$(-m)(-n)$ is positive.

$|mn|$ is positive.

$|-mn|$ is positive.

$|(-m)(-n)|$ is positive.

There are 4 expressions that are positive, and each one is equal to $mn$ .

Question analysis

Five expressions are listed: −mn, (−m)(−n), |mn|, |−mn| and |(−m)(−n)|. If m and n are positive integers, students count how many are equivalent to mn. It tests the sign rules for multiplying negatives together with what absolute value does, and each expression has to be judged on its own.

NYS standards: 7.NS.2a rational number multiplication
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Multiplying signed numbers and absolute value are grade 6 and 7 content. Checking five expressions one by one takes longer than a typical single computation, but each check uses a familiar rule.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Much to keep track of minor
Something you must already know minor
Solution: 5 steps
Reading level: Question and choices: about grade 10, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each expression takes only a moment, but the answer is a count, so one misjudged expression changes it. The minus signs are designed to unsettle: four of the five expressions show at least one, and only one of those actually stays negative. Substituting small positive numbers and comparing each value with mn turns the question into arithmetic.

Variables instead of numbers moderate Understanding the question
The five expressions use only m and n, so each sign rule has to be applied in general unless the student substitutes values such as m = 2 and n = 3. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Five expressions are judged one at a time and a count kept, and two of them combine the product rule for negatives with absolute value bars. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Something you must already know minor Working it out
The rules that two negative factors give a positive product and that absolute value removes a minus sign are assumed, not stated. (Anderson & Krathwohl, 2001)

Strategy: Pick small values such as m = 2 and n = 3, evaluate each of the five expressions, and count the ones equal to 6.

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.)
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 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Treats any minus sign as making the value negative, even inside bars or in a product of two negatives, so counts only |mn|.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Accepts |mn| and (−m)(−n) but believes bars cannot cancel a minus sign written inside them, rejecting |−mn| and |(−m)(−n)|.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Thinks a negative times a negative stays negative, so rejects (−m)(−n) along with −mn and counts only the three absolute values.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying two negatives: $(-a) \times (-b) = a \times b$; a and b are positive numbers. A negative number times a negative number gives a positive number. The two minus signs cancel out. One negative and one positive give a negative.
Absolute value: $|a| = \text{the distance from } a \text{ to } 0$; The bars around a number mean absolute value. It is how far the number is from 0 on a number line. A distance is never negative, so |−6| = 6 and |6| = 6. The bars remove any minus sign.

Sort out the question
m and n are positive integers. That means whole numbers greater than 0. Writing mn means m times n, so mn is positive. You need to count how many of the five expressions always equal mn.
Pick numbers to test
Pick easy values: $m = 2$ and $n = 3$ . Then $mn = 2 \times 3 = 6$ . Any expression that equals mn must come out to 6.
Test the first two expressions
$-mn = -(2 \times 3) = -6$
This is negative, so it does not equal 6.
$(-m)(-n) = (-2)(-3) = 6$
A negative times a negative is positive, so this one equals 6. Watch out: if you think a negative times a negative stays negative, you reject this one and get 3, choice C.
Test the three absolute values
$|mn| = |6| = 6$
$|-mn| = |-6| = 6$
$|(-m)(-n)| = |6| = 6$
All three equal 6, because absolute value bars remove a minus sign. Watch out: the bars work even when the minus sign is inside them. Rejecting the last two gives 2, choice B.
Count the matches
Four expressions equal mn: $(-m)(-n)$ , $|mn|$ , $|-mn|$ and $|(-m)(-n)|$ . Only $-mn$ does not. The sign rules work the same for any positive m and n, not just 2 and 3. So the answer is D, 4.

Check your answer

Try other numbers, such as $m = 1$ and $n = 5$ . Then $mn = 5$ . The five expressions give −5, 5, 5, 5 and 5. Again, four of them match mn.

More questions like this

The original DOE question

Question 78 as printed on page 197 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 197, question 78. Open page 197 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 20 of 22

2022 Test A Math, question 3

Question

The list shows the daily low temperatures in a city during six days in the winter.

7° F, 4° F, 1° F, −2° F, −5° F, 3° F

What is the difference between the lowest and highest temperatures, in degrees Fahrenheit, for the six days?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 12

Determine the greatest temperature listed for the six days:

Of 7, 4, 1, (−2), (−5), and 3, the greatest value is 7 because its location is farthest to the right on a number line.

Greatest temperature = 7° F

Determine the lowest temperature listed for the six days:

Of 7, 4, 1, (−2), (−5), and 3, the lowest value is (−5) because its location is farthest to the left on a number line.

Lowest temperature = (−5)° F

Determine the difference between the two temperatures:

$7^{\circ}\text{ F} - (-5)^{\circ}\text{ F} = 7^{\circ}\text{ F} + 5^{\circ}\text{ F} = 12^{\circ}\text{ F}$

Question analysis

A list gives a city's daily low temperatures over six winter days, some above zero and some below. This grid-in asks for the difference between the highest and the lowest of them, in degrees Fahrenheit. Students order positive and negative numbers and subtract integers; picking the right negative reading and handling the double minus sign are the tricky steps.

NYS standards: 7.NS.1c rational number subtraction
6.NS.7b order statements in context
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Subtracting a negative number to find the distance between two temperatures is core grade 7 integer work. Picking the extremes from six one-digit readings is quick, so this is a typical, short grade 7 exercise.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Six readings, two of them below zero, must be sorted before any subtraction, and −5 has to be seen as colder than −2 even though 5 is the bigger number. The subtraction is where most points go: the gap runs 5 degrees up to zero and 7 more above it. A quick sketch of a thermometer from −5 to 7 shows the full span and catches a sign dropped along the way.

Error-prone arithmetic moderate Working it out
The span is 7 − (−5), and subtracting a negative means adding 5, so the five degrees below zero are easy to drop, leaving only 2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
With no choices to compare, nothing flags an entry of 2 or −12; the student must see alone that the gap between the extremes is a positive count of degrees. (Enright, Morley & Sheehan, 2002)

Strategy: Picture the six readings on a number line: the lowest is the one farthest to the left. Subtract the lowest from the highest, remembering that subtracting a negative number means adding.

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.)
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

2 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Subtracts the sizes, 7 − 5 = 2, as if both readings were above zero, leaving out the 5 degrees between −5° F and zero.
9 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Takes −2° F as the lowest reading because 2 is less than 5, then subtracts 7 − (−2) = 9.
-12 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Subtracts in the reverse order, lowest minus highest, −5 − 7 = −12, and enters a negative difference.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Difference between two values: $D = \text{highest} - \text{lowest}$; D is the difference. Subtract the lowest value from the highest, in that order, so D is not negative. It tells how far apart the two values are on a number line.
Subtracting a negative number: $a - (-b) = a + b$; a and b are numbers, and −b is the negative of b. Taking away a negative is the same as adding its positive. On a number line, going from −b up to a passes b steps to zero, then a more.

Sort out the question
The list has six low temperatures, in degrees Fahrenheit (° F). Some are below zero. A minus sign means below zero. You need the highest and the lowest temperatures. Then find the difference between them. The difference tells how many degrees apart they are.
Find the highest and lowest
Picture a number line, or a thermometer. Numbers farther to the right are higher. The highest temperature is 7° F. The lowest is the one farthest to the left. That is −5° F. Watch out: −5 is colder than −2. The number 5 is bigger than 2, but 5 below zero is lower than 2 below. Using −2 as the lowest leads to 9.
Subtract the lowest from the highest
$D = \text{highest} - \text{lowest}$
$D = 7 - (-5) = 7 + 5 = 12$
Subtracting a negative is the same as adding. The temperatures are 12 degrees apart. Watch out: $7 - 5 = 2$ leaves out the 5 degrees below zero. Also, a difference here is not negative, so don’t grid −12. So you would grid 12.

Check your answer

Count on a thermometer. From −5° F up to 0° F is 5 degrees. From 0° F up to 7° F is 7 more degrees. $5 + 7 = 12$ degrees. That matches.

More questions like this

The original DOE question

Question 60 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 79, question 60. Open page 79 of the handbook (PDF)

Question 21 of 22

2018 Test A Math, question 36

Question

$Number line with two points: L, labeled 4 and 5/16, and N to the right of L, with no value given.$

On the number line above, LN = $\frac{1}{8}$ . Point M (not shown) is located between point L and point N. Which value below is a possible value for M?

A4.26
B4.31
C4.35
D4.58

Show the answer

Correct answer: C

Since $LN=\frac{1}{8}$ , point $N$ is located at

$4\frac{5}{16}+\frac{1}{8}=4\frac{7}{16}$ .

So $M$ must be between point $L$ , $4\frac{5}{16}$ , and point $N$ , $4\frac{7}{16}$ . Point $L$ can also be written as 4.3125, and point $N$ can be written as 4.4375. The only option given that lies between those two points is 4.35.

Question analysis

This number line question says the distance from point L to point N is one eighth and asks which decimal could be the position of a point M between them. Students take L's position, a mixed number in sixteenths, find N, and compare with decimal choices. Converting sixteenths to decimals precisely is the tricky part.

NYS standards: 7.NS.2d fractions to decimals by long division
7.NS.1b rational number addition
6.NS.7a inequality as relative position
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students convert fractions to decimals and compare them. This item needs an exact conversion of sixteenths and a comparison to the thousandths, where rounding too early can mislead, but it stays within grade 7 content.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 28 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The interval from L to N is only an eighth wide, and three of the four choices crowd near it, so precision decides the question. Students comfortable working in sixteenths can avoid decimals until the end, since 4.35 lies between 4 5/16 and 4 7/16. Those who convert loosely, or who misread a mixed number as a decimal, are pulled to a neighboring choice.

Error-prone arithmetic moderate Working it out
Converting 4 5/16 to 4.3125 and N to 4.4375 needs four decimal places, while the choices stop at hundredths, so early rounding blurs the left boundary. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
The choices 4.31 and 4.35 both lie within 0.04 of L, so estimating cannot separate them; only the exact 4.3125 shows 4.31 just outside the interval and 4.35 inside. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Reading a figure, table or graph minor Understanding the question
L's value appears only on the number line, and the picture shows N to the right of L, which fixes which way the one-eighth gap runs. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add one eighth, or two sixteenths, to L's position to locate N, convert both endpoints to four-place decimals, and keep only a choice strictly between them.

Traps

Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
Answer in another form: The right answer is written differently from the way you worked it out.
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 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Measures the one-eighth gap to the left of L, giving 4.1875 to 4.3125, an interval that contains 4.26 but contradicts the figure.
B Rounds too early (King et al., 2004)
Reveals a slip Expected pull: strong
Rounds L's 4.3125 to 4.31 and treats 4.31 as inside the interval, missing that it lies slightly to the left of L.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Writes 4 5/16 as 4.516 by placing the fraction's digits after the decimal point, so N becomes about 4.641 and 4.58 appears to fit.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal fractions: $\frac{a}{b} = \frac{a \times k}{b \times k}$; a is the top number (numerator) and b is the bottom number (denominator). Multiplying both by the same number k gives an equal fraction. It is like cutting each piece into k smaller pieces. Use it to give fractions the same bottom before you add or subtract.
Fraction to decimal: $\frac{a}{b} = a \div b$; A fraction bar means divide. a is the top number and b is the bottom number. Dividing the top by the bottom turns the fraction into a decimal. Then you can compare it with decimal choices.

Sort out what you know
Point L is at $4\frac{5}{16}$ . Point N is to the right of L. The distance from L to N, called LN, is $\frac{1}{8}$ .
Point M is somewhere between L and N. You need a choice that fits between them. The choices are decimals, so you will change the fractions into decimals.
Find where N is
N is $\frac{1}{8}$ to the right of L, so add. First give $\frac{1}{8}$ a bottom of 16, to match L:
$\frac{1}{8} = \frac{1 \times 2}{8 \times 2} = \frac{2}{16}$
$4\frac{5}{16} + \frac{2}{16} = 4\frac{7}{16}$
Watch out: N is to the right of L in the figure. Going left instead gives a gap that holds 4.26, choice A.
Change both points to decimals
Divide the top of each fraction by the bottom:
$\frac{5}{16} = 5 \div 16 = 0.3125$
$\frac{7}{16} = 7 \div 16 = 0.4375$
So L is at 4.3125 and N is at 4.4375.
Test each choice
M must be bigger than 4.3125 and smaller than 4.4375.
- 4.26 is smaller than 4.3125. It is left of L.
- 4.31 is 4.3100, just smaller than 4.3125. It is a hair left of L.
- 4.35 is between them. It fits.
- 4.58 is bigger than 4.4375. It is right of N.
Watch out: if you round L to 4.31, choice B seems to fit. Keep all four decimal places. So the answer is C, 4.35.

Check your answer

Subtract to be sure. 4.35 − 4.3125 = 0.0375, which is more than 0, so 4.35 is right of L. 4.4375 − 4.35 = 0.0875, which is also more than 0, so 4.35 is left of N. It sits between them.

More questions like this

The original DOE question

Question 93 as printed on page 88 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 88, question 93. Open page 88 of the handbook (PDF)

Question 22 of 22

2018 Test B Math, question 3

Question

$\left | \left ( -6 \right )-\left ( -5 \right )+4.2 \right |-\left | 3-9.6 \right |$

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: -3.4

$=3.2-6.6=-3.4$

Question analysis

This grid-in asks students to evaluate an expression that mixes negative integers, decimals and two absolute values. The hard parts are subtracting a negative number inside the first pair of bars and keeping each absolute value separate until the final subtraction, with no answer choices to check against.

NYS standards: 7.NS.1c rational number subtraction
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students add and subtract signed numbers, and grade 6 students meet absolute value. This combines both in one expression with decimals and several sign decisions, a typical grade 7 exercise with no context to check against.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps moderate
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 16+, 1 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every step is familiar, but the expression packs four of them with negatives and decimals, and a single sign slip anywhere gives a clean-looking wrong number. The subtlest point comes last: absolute value bars make each part positive, yet the difference of two absolute values can still be negative. Students who believe bars always produce a positive answer reverse the final subtraction, and the grid-in format offers no choices to expose it.

Error-prone arithmetic moderate Working it out
Three sign decisions stack up: (−6) − (−5) must become −6 + 5, then −1 + 4.2 crosses zero, and 3 − 9.6 is negative before its bars make it positive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps moderate Working it out
Each pair of bars must be simplified inside and then made positive, and only after both does the outer subtraction happen; doing the bars and the subtraction in the wrong order changes the result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The result is a negative decimal to be gridded with its sign, and with no choices on screen nothing warns a student who has dropped the minus or flipped the subtraction. (Enright, Morley & Sheehan, 2002)

Strategy: Simplify inside each pair of absolute value bars first, rewriting minus a negative as plus, then take each absolute value, and only then subtract the second from the first.

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

3.4 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Believes an expression built from absolute values cannot be negative, so reports 6.6 − 3.2 instead of 3.2 − 6.6.
9.8 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Treats the bars like parentheses, leaving 3 − 9.6 as −6.6, then subtracts that negative: 3.2 + 6.6.
0.2 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Reads (−6) − (−5) as −6 − 5 = −11, so the first bars give |−6.8| = 6.8, and 6.8 − 6.6 = 0.2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value: $|a| = \text{distance from } a \text{ to } 0$; The two bars around a number mean absolute value. a is any number. Its absolute value is how far it is from 0 on the number line. A distance is never negative. So |−6.6| = 6.6 and |3.2| = 3.2.
Subtracting a negative: $a - (-b) = a + b$; a and b are numbers. Taking away a negative number is the same as adding the positive number. For example, 5 − (−2) = 7. Think of it as taking away a debt: you end up richer.

Plan the order of work
The expression has two pairs of bars. The bars stand for absolute value, a number’s distance from 0.
Work inside each pair of bars first. Then take each absolute value. Subtract the two results last.
Work inside the first bars
Start with $(-6) - (-5) + 4.2$ . Subtracting −5 is the same as adding 5.
$(-6) - (-5) + 4.2 = -6 + 5 + 4.2 = -1 + 4.2 = 3.2$
3.2 is already positive, so $|3.2| = 3.2$ .
Watch out: If you read the start as −6 − 5, you get −11. That path ends at the wrong answer 0.2.
Work inside the second bars
Next, $3 - 9.6 = -6.6$ . This result is negative. Its distance from 0 is 6.6, so the absolute value is positive.
$|-6.6| = 6.6$
Subtract the two results
Now take the second result away from the first.
$3.2 - 6.6 = -3.4$
You take a bigger number away from a smaller one, so the result is negative.
Watch out: The bars made each part positive. They do not make the final answer positive, so 3.4 is wrong. Also, bars are not parentheses. Working out 3.2 − (−6.6) = 9.8 is wrong too.
So you would grid -3.4, with the minus sign.

Check your answer

Estimate. The first part is about 3. The second part is about 7. And 3 − 7 is about −4. The answer −3.4 is close to that. It is also negative, as the estimate says it should be.

More questions like this

The original DOE question

Question 60 as printed on page 153 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 153, question 60. Open page 153 of the handbook (PDF)

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