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Question 1 of 32

2026 Practice Test D Math, question 12

Question

A circle with center P. Points R and T are on the circle at the ends of a horizontal diameter through P, and point S is on the circle above it. Segment SP is labeled 12. A vertical segment from S meets RT at a right angle, dividing RT into a part labeled x, from R to that point, and a part labeled 5, from that point to T.

P is the center of the circle above, and R, S, and T are points on the circle. What is the value of x?

  • A12
  • B15
  • C17
  • D19
Show the answer

Correct answer: D

SP\overline{\mathrm{SP}} represents the radius of the circle and has length 12.

Both RP\overline{\mathrm{RP}} and PT\overline{\mathrm{PT}} are radii and also have length 12.

Therefore, the length of RT\overline{\mathrm{RT}}, which is a diameter, is 24 and is also represented by x+5x + 5.

x+5=24x + 5 = 24

x=19x = 19

Question analysis

A circle with center P has points R, S and T on it, a segment SP labeled 12, and a vertical segment from S that splits RT into a piece labeled x and a piece labeled 5. Students find the value of x. It tests basic circle facts, and the challenge is seeing which lengths in the figure must be equal.

NYS standards
  • 7.G.4 circle area and circumference
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 circle work centers on area and circumference formulas. This needs only the fact that every radius has the same length, but spotting which segments are radii in a figure with extra markings is less routine than using a formula.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 3, 24 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 figure gives one measurement on a slanted radius and two labels on the diameter, and the right angle at the foot of the vertical segment is a decoy. Once RP and PT are seen as radii, RT is 24, and x is that diameter less the 5. Students who never connect the slanted 12 to the horizontal line fall back on the labeled numbers, adding 12 and 5, or take x to end at the center.

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

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

  2. No obvious method minor Working it out

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

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radius and diameter
d=2rd = 2r
r is the radius, the distance from the center to any point on the circle. d is the diameter, the distance straight across the circle through the center. A diameter is two radii laid end to end, one on each side of the center, so it is twice as long.

  1. Sort out the question

    P is the center of the circle, and R, S and T are on the circle. Segment SP is 12. Segment RT goes straight across through P. A short line from S meets RT at a right angle. It splits RT into x, from R to that point, and 5, from that point to T. You need x.

  2. See that SP is a radius

    SP runs from the center P to the point S on the circle. So SP is a radius, and the radius is 12. Every radius of a circle has the same length. RP and PT also run from the center to the circle, so RP = 12 and PT = 12.

  3. Find the diameter RT

    RT passes through the center, so it is a diameter. Use the formula with r = 12:

    RT=2r=2×12=24\mathrm{RT} = 2r = 2 \times 12 = 24

    RT is 24 long. It is the radius RP plus the radius PT.

  4. Solve for x

    RT is also split into x and 5, so:

    x+5=24x + 5 = 24

    Subtract 5 from both sides:

    x=24−5=19x = 24 - 5 = 19

    Watch out: x runs from R past the center P, so it is longer than a radius. Reading it as 12, choice A, stops at P. Adding 12 + 5 = 17, choice C, puts the 12 in the wrong place.

  5. Work from the numbers, not the picture

    The figure is not drawn to scale. In it, x looks about three times as long as 5, which suggests 15, choice B. Trust the labels instead. The right triangle under S is not needed at all.

    So the answer is D, 19.

Check your answer

The point where the short line meets RT is 19 from R. P is 12 from R, so that point is 19 − 12 = 7 past the center. Then 7 + 5 = 12, which is PT, a radius. Everything fits.

More questions like this

The original DOE question
Question 12 as shown on the DOE’s digital test, page 85 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 85, question 12.

Question 2 of 32

2026 Practice Test D Math, question 17

Question

At Midville High School, 64 students are members of either the orchestra or the choir, or both. Of these students, 38 are members of the orchestra, and 46 are members of the choir. What fraction of the total is the number of students who are members of both the orchestra and the choir?

  • A1621\frac{16}{21}
  • B1019\frac{10}{19}
  • C1023\frac{10}{23}
  • D516\frac{5}{16}
Show the answer

Correct answer: D

First, add the number of students that are members of the orchestra and the number of students that are members of the choir.

38+46=8438 + 46 = 84

Then, subtract to find the number of students that are members of both orchestra and choir.

84−64=2084 - 64 = 20

Then, find the fraction of the 64 students that are members of both orchestra and choir.

2064=516\frac{20}{64} = \frac{5}{16}

Question analysis

At Midville High School, 64 students belong to the orchestra, the choir or both; 38 are in the orchestra and 46 in the choir. Students find what fraction of the 64 belong to both groups. It is a Venn diagram counting question with a fraction, and the students in both groups are included in each membership number.

NYS standards
  • 7.SP.8b sample spaces for compound events
  • 6.RP.1 ratio concepts and language
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Overlapping groups are seldom taught directly in grade 7, though the arithmetic is whole-number addition and subtraction. Students must first find the overlap and then choose the right whole for the fraction, two decisions where errors are common.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 54 words.
Corrected from the DOE handbook
  • A misstatement in the explanation’s first step was corrected.
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 hides its key number: the students in both groups appear only as the amount by which 38 + 46 overshoots 64. A sketch of two overlapping circles, with 20 in the middle, 18 in orchestra only and 26 in choir only, shows the whole group at once. After that, the remaining risk is the denominator, since the question asks for a fraction of all 64 students, not of one group.

  1. Turning words into math moderate Understanding the question

    “Either the orchestra or the choir, or both” means the 64 counts each student once, while 38 + 46 counts the students in both groups twice; the overlap itself is never stated. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. No obvious method minor Working it out

    Nothing names the step that finds the overlap: the excess of 38 + 46 = 84 over 64 is the number in both groups, an idea students must bring themselves. (Schoenfeld, 1985)

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

    The fraction 20/64 appears only in lowest terms, so it must be simplified by 4 before it can be matched with a choice. (Enright, Morley & Sheehan, 2002)

Strategy: Add the orchestra and choir counts, subtract the 64 students to find how many were counted twice, then put that number over the total and simplify.

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.
  • Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation begins “First, add the number of students that are members of both orchestra and choir. 38 + 46 = 84.” But 38 is the number of orchestra members and 46 the number of choir members; the number of students in both is what the next step finds (84 − 64 = 20). The site now says “First, add the number of students that are members of the orchestra and the number of students that are members of the choir.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting two groups that overlap
total=O+C−both\text{total} = O + C - \text{both}
O is the number in the orchestra, and C is the number in the choir. Students in both groups are counted in O and again in C. Subtracting them once makes each student count one time. Here every student is in at least one group, so this gives everyone.
Fraction of a total
fraction=partwhole\text{fraction} = \frac{\text{part}}{\text{whole}}
The part is the group you are asked about. The whole is the full group it comes from. Here the part is the students in both groups, and the whole is all 64 students. Simplify by dividing the top and bottom by the same number.

  1. Sort out the question

    There are 64 students in all. Each one is in the orchestra, the choir or both. 38 are in the orchestra, and 46 are in the choir. Some students are in both, but the question does not say how many. You need that number as a fraction of all 64 students.

  2. Add the two groups

    Add the number in the orchestra and the number in the choir:

    38+46=8438 + 46 = 84

    That is more than the 64 students there are. Why? A student in both groups is counted once with the orchestra and once more with the choir.

  3. Find the students in both

    The extra amount over 64 is the students who were counted twice:

    84−64=2084 - 64 = 20

    So 20 students are in both groups. This is the overlap rule turned around: both = O + C − total.

    Watch out: 84 is not the number of students. Putting 64 over 84 gives 16/21, choice A.

  4. Write the fraction and simplify

    The part is 20 students, and the whole is all 64 students. Both numbers divide by 4:

    2064=20÷464÷4=516\frac{20}{64} = \frac{20 \div 4}{64 \div 4} = \frac{5}{16}

    Watch out: the whole is all 64 students, not one group. Using 38 gives 10/19, choice B. Using 46 gives 10/23, choice C.

    So the answer is D, 516\frac{5}{16}.

Check your answer

Split the 64 students into three groups that do not overlap. Only orchestra: 38 − 20 = 18. Only choir: 46 − 20 = 26. Both: 20. Add them: 18 + 26 + 20 = 64. That matches the total, so 20 is right.

More questions like this

The original DOE question
Question 17 as shown on the DOE’s digital test, page 90 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 90, question 17. This is the DOE’s version, before the corrections listed in the analysis.

Question 3 of 32

2026 Practice Test D Math, question 25

Question

A bus trip takes 9 hours if the mean speed is 50 miles per hour. How many hours would this trip take if the mean speed was 45 miles per hour?

  • A8
  • B10
  • C12
  • D14
Show the answer

Correct answer: B

First, find the length of the bus trip in miles by multiplying the time, 9 hours, by the average (mean) speed, 50 miles per hour, to get 450 miles.

Then, divide the total distance, 450 miles, by the average (mean) speed, 45 miles per hour, to get the total time:

450÷45=10 hours450 \div 45 = 10\text{ hours}

Question analysis

A bus trip takes 9 hours at a mean speed of 50 miles per hour, and students find how long the same trip takes at 45 miles per hour. It tests speed, distance and time. Working out the fixed distance first is the key, since guessing how the time should change leads many students astray.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Speed, distance and time problems are grade 6 rate work, and this one uses whole numbers in two steps. Its only twist is how the time responds when the speed changes, which makes it easier than typical grade 7 rate problems.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 3, 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

Two short calculations solve it, but only after noticing that speed and time move in opposite directions: a slower bus takes longer. Students who scale the 9 hours by 45/50, as in a direct proportion, get a little over 8 hours, a result that should feel wrong for a slower trip. Finding the distance first turns the question into one division with no direction to decide.

  1. No obvious method moderate Working it out

    The question never mentions the trip's length, yet the 450-mile distance is what stays fixed; setting up a proportion between hours and speeds, the usual reflex, points the wrong way. (Schoenfeld, 1985)

  2. Something you must already know minor Working it out

    Students must bring distance = speed × time themselves, to turn 9 hours at 50 miles per hour into miles before dividing by the new speed. (Anderson & Krathwohl, 2001)

Strategy: Find the trip's length in miles from the first speed and time, since that length stays the same, then divide it by the new mean speed.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance, speed and time
d=r×td = r \times t
d is the distance traveled. r is the rate, or speed, such as miles per hour. t is the time. It works because at 50 miles each hour, 9 hours covers 9 groups of 50 miles.
Finding the time
t=d÷rt = d \div r
This is the same rule turned around. d is the distance and r is the speed. Divide the distance by the speed to find t, the number of hours. Use it when you know how far and how fast.

  1. Sort out the question

    A bus trip takes 9 hours at a mean speed of 50 miles per hour. The mean speed is the average speed over the whole trip. You need the time for the same trip at 45 miles per hour. The length of the trip does not change, so find it first.

  2. Find the length of the trip

    Use distance equals speed times time, with a speed of 50 miles per hour and a time of 9 hours:

    d=50×9=450d = 50 \times 9 = 450

    The trip is 450 miles long.

  3. Find the new time

    Now the bus goes 45 miles per hour over the same 450 miles. Divide the distance by the speed:

    t=450÷45=10t = 450 \div 45 = 10

    The trip takes 10 hours.

  4. Check that it makes sense

    A slower bus needs more time for the same trip. 10 hours is more than 9 hours, so it fits.

    Watch out: time does not shrink with speed. Scaling 9 hours by 45/50 gives about 8 hours, choice A, which would make the slower bus arrive sooner. Adding the drop of 5 miles per hour to 9 hours gives 14, choice D, which has no reason behind it.

    So the answer is B, 10 hours.

Check your answer

Check the distance both ways. At 50 miles per hour for 9 hours, the bus goes 50 × 9 = 450 miles. At 45 miles per hour for 10 hours, it goes 45 × 10 = 450 miles. Same trip, same distance.

More questions like this

The original DOE question
Question 25 as shown on the DOE’s digital test, page 98 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 98, question 25.

Question 4 of 32

2024 Test A Math, question 5

Question

RESULTS FROM SURVEY OF 110 FAMILIES

Number of Children in the Family Number of Families
0 45
1 32
2 19
3 8
4 6

The table above shows the number of children in each of 110 families. What is the median number of children in these families?

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

Show the answer

Correct answer: 1

Since 110 families were surveyed, the median will have 55 values less than or equal to it, and 55 values greater than or equal to it.

If the values in the table were listed in order, they would look like this:

0,…,0⏞45,1,…,1⏞32,2,…,2⏞19,3,…,3⏞8,4,…,4⏞6\overbrace{0,\ldots,0}^{45},\overbrace{1,\ldots,1}^{32},\overbrace{2,\ldots,2}^{19},\overbrace{3,\ldots,3}^{8},\overbrace{4,\ldots,4}^{6}

The first row of the table indicates that 45 families had 0 children.

The second row of the table indicates that 32 families had 1 child.

So 77 families had either 0 children or 1 child, and 45 families had 0 children.

Since 55 is between 45 and 77, the median must be a value in the second row of the table.

So the median is 1.

Question analysis

This statistics grid-in gives a frequency table from a survey of 110 families, showing how many families have each number of children from none to four. Students find the median number of children per family. The table lists counts rather than the 110 data values themselves, so locating the middle value takes a running total of the families.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.SP.5a number of observations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 median problems usually list a dozen or so values. Here 110 values are hidden in a frequency table, so students must reason about positions with a running count, a less familiar and more error-prone task.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 7, 35 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

Students know a median as the middle of a list, but here the list is never written out. Treating the table's columns as the data gives quick, confident answers, 2 from the middle of the children column or 19 from the middle of the family counts, and neither can be checked against choices. The question rewards counting positions: 45 families have no children, so the 55th and 56th families both fall among the 32 with one child.

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

    The table hides 110 data values in five rows: the 0 to 4 column lists the possible values, and 45, 32, 19, 8 and 6 say how many families have each. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. No obvious method moderate Working it out

    The median of 110 values lies between the 55th and 56th; finding them takes a running count of families, 45 and then 77, a method seldom practiced with tables. (Schoenfeld, 1985)

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

    As a grid-in there are no choices to compare, so a student who takes the middle row's 2 has nothing on the page to suggest it is wrong. (Enright, Morley & Sheehan, 2002)

Strategy: With 110 values the median sits between the 55th and 56th; add the family counts row by row until the running total passes 55, and read that row's number of children.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Spot of the median
middle spot=n+12\text{middle spot} = \frac{n + 1}{2}
The median is the middle value when all the values are put in order. n is how many values there are. This gives the spot of the middle value, with the same number of values before and after it. If it ends in a half, the median is halfway between the two values on either side.

  1. Sort out the table

    The table covers 110 families. The left column is a number of children. The right column says how many families have that many children. For example, 45 families have 0 children. You need the median number of children. The median is the middle value when all the values are in order.

  2. Picture the values in order

    Think of a list with one number for each family: how many children it has. The list has 45 zeros, then 32 ones, then 19 twos, then 8 threes, then 6 fours. It is already in order, from least to most.

    45+32+19+8+6=11045 + 32 + 19 + 8 + 6 = 110

    There are 110 values. Watch out: the median is not the middle row, 2. Each row stands for a different number of families.

  3. Find the middle spot

    Use the median rule with n = 110.

    middle spot=n+12\text{middle spot} = \frac{n + 1}{2}

    110+12=55.5\frac{110 + 1}{2} = 55.5

    The middle is halfway between the 55th and 56th values. So you need those two values in the list.

  4. Count down the list

    Keep a running total of families, row by row. The zeros fill spots 1 to 45. Now add the 32 families with 1 child.

    45+32=7745 + 32 = 77

    So the ones fill spots 46 to 77. Spot 55 and spot 56 are both in that range. So both values are 1.

  5. Give the median

    The 55th and 56th values are both 1. Halfway between 1 and 1 is 1. So the median is 1 child. Watch out: the median is a number of children, not of families. Gridding 19, from the families column, is wrong. So is the mean, or average, which is about 1.07. So you would grid 1.

Check your answer

Half of 110 is 55. Families with 0 children: 45, which is less than half. So the median is not 0. Families with 0 or 1 child: 77, which is more than half. So the median is not more than 1. It must be 1.

More questions like this

The original DOE question
Question 62 as printed on page 76 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 76, question 62. Open page 76 of the handbook (PDF)

Question 5 of 32

2024 Test A Math, question 18

Question

Each number in a sequence is formed by doubling the previous number and then adding 1. If the 9th number in the sequence is 63, what is the 10th number minus the 7th number?

  • A96
  • B111
  • C112
  • D127
Show the answer

Correct answer: C

First, write an equation based on the rule described in the item. Let x be the previous number in the sequence and y be the number in the sequence. The ninth number in the sequence is 63.

2x+1=y2x + 1 = y

Determine the 10th number in the sequence, y, by making x=63x = 63 and solving:

2(63)+1=y2(63) + 1 = y

126+1=y126 + 1 = y

127=y127 = y

Determine the 8th number in the sequence, x, by making y=63y = 63 and solving:

2x+1=632x + 1 = 63

Subtract one from both sides of the equation.

2x+(1−1)=63−12x + (1 - 1) = 63 - 1

2x=622x = 62

2x2=622\frac{2x}{2} = \frac{62}{2}

x=31x = 31

Determine the 7th number in the sequence, x, by making y=31y = 31 and solving:

2x+1=312x + 1 = 31

2x+(1−1)=31−12x + (1 - 1) = 31 - 1

2x=302x = 30

2x2=302\frac{2x}{2} = \frac{30}{2}

x=15x = 15

Subtract the 7th number in the sequence from the 10th number:

127−15=112127 - 15 = 112

The difference is 112.

Question analysis

A sequence question: each number is found by doubling the one before it and adding 1, and the 9th number is 63. Students find the 10th number minus the 7th. That means running the rule forward once and backward twice, while keeping track of which position each number holds.

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve two-step equations but rarely have to run a rule backward through a sequence. This mixes forward and backward steps with position counting, so it asks for more planning than a typical grade 7 task.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 7, 32 words.
Corrected from the DOE handbook
  • A misplaced parenthesis in the explanation’s working was corrected.
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

Applying the rule forward is routine; running it backward is not, and the question needs both. Students must also count positions carefully, since the 9th number is given and the 7th is two steps back, not one. Each listed wrong answer stops or slips at a different stage, so writing the four numbers in a row, 15, 31, 63 and 127, before subtracting is the reliable way.

  1. No obvious method moderate Working it out

    The rule is stated only forward, double and add 1, but the 7th and 8th numbers come before the given 9th, so the rule must be undone: subtract 1, then halve. (Schoenfeld, 1985)

  2. Many steps moderate Working it out

    One forward step reaches the 10th number, 127, two backward steps reach the 8th and 7th, 31 and 15, and a final subtraction finishes. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Much to keep track of minor Working it out

    Four positions and their values, 7th to 10th, have to be kept straight, and the 8th number is easy to mistake for the 7th. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Apply the rule to 63 to get the 10th number; to go back a position, subtract 1 and halve, doing it twice to reach the 7th; then subtract.

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: Finding the 7th number, the handbook writes “2x + (1 − 1 = 31 − 1)”, with the closing parenthesis after the whole equation, so the line reads as an equation inside an expression (and 1 − 1 = 31 − 1 is false). The site now reads 2x + (1 − 1) = 31 − 1, the same step the handbook writes correctly for the 8th number (2x + (1 − 1) = 63 − 1). It leads to 2x = 30 and x = 15 as before.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The sequence rule
y=2x+1y = 2x + 1
A sequence is a list of numbers that follows a rule. x is a number in the list, and y is the next number. To get the next number, double x and then add 1.
Going back one number
x=(y−1)÷2x = (y - 1) \div 2
y is a number in the list, and x is the one just before it. To go back, undo the rule in reverse order. The rule doubles and then adds 1. So first subtract 1, then divide by 2.

  1. Sort out the question

    Each number is “formed by doubling the previous number and then adding 1”. Previous means the one just before. The 9th number is 63. You need the 10th number minus the 7th number. So you go forward one number and back two numbers.

  2. Find the 10th number

    Use the rule with x = 63.

    y=2x+1y = 2x + 1

    y=2×63+1=126+1=127y = 2 \times 63 + 1 = 126 + 1 = 127

    The 10th number is 127. Watch out: remember to add 1. Doubling only gives 126 and leads to 111, choice B.

  3. Plan how to go back

    The 7th number comes before the 9th, so you must undo the rule. The rule doubled, then added 1. Undo the last step first: subtract 1. Then undo the doubling: divide by 2.

    x=(y−1)÷2x = (y - 1) \div 2

  4. Find the 8th number

    x=(63−1)÷2=62÷2=31x = (63 - 1) \div 2 = 62 \div 2 = 31

    The 8th number is 31. Watch out: 31 is the 8th number, not the 7th. Using it gives 127−31=96127 - 31 = 96, choice A.

  5. Find the 7th number

    Step back once more, from the 8th number. Use the same rule with y = 31.

    x=(31−1)÷2=30÷2=15x = (31 - 1) \div 2 = 30 \div 2 = 15

    The 7th number is 15.

  6. Subtract the 7th from the 10th

    127−15=112127 - 15 = 112

    Watch out: 127 alone is choice D. You still have to subtract the 7th number. So the answer is C, 112.

Check your answer

Go forward from 15 using the rule. 2×15+1=312 \times 15 + 1 = 31, then 2×31+1=632 \times 31 + 1 = 63, then 2×63+1=1272 \times 63 + 1 = 127. You land on 63 as the 9th number, so the list is right. And 127−15=112127 - 15 = 112.

More questions like this

The original DOE question
Question 75 as printed on page 80 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 80, question 75. Open page 80 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 6 of 32

2024 Test A Math, question 21

Question

A coordinate grid with origin O, the x-axis and y-axis each marked 5, 10, 15 and 20, and grid lines every 5 units. One line segment joins the points (0, 20) and (20, 0), and another joins (0, 15) and (10, 0); all four points are marked with dots. The region between the two segments, with corners at (0, 15), (0, 20), (20, 0) and (10, 0), is shaded.

What is the area, in square units, of the shaded region shown in the figure above?

  • A75
  • B125
  • C150
  • D200
Show the answer

Correct answer: B

The area of the shaded region, in square units, is the area of the triangle created by the shaded and unshaded regions minus the area of the triangle created by the unshaded region.

The area of the triangle created by the shaded and unshaded regions is 200 square units.

12bh=12(20)(20)=12(400)=200\frac{1}{2}bh = \frac{1}{2}(20)(20) = \frac{1}{2}(400) = 200

The area of the triangle created by the unshaded region is 75 square units.

12bh=12(10)(15)=12(150)=75\frac{1}{2}bh = \frac{1}{2}(10)(15) = \frac{1}{2}(150) = 75

Subtract to find the area of the shaded region.

200−75=125200 - 75 = 125

Question analysis

A coordinate geometry question: two line segments cross a grid, one from (0, 20) to (20, 0) and one from (0, 15) to (10, 0), and the region between them is shaded. Students find the shaded area. The shape is not a standard figure, so they need to see it as one triangle with another removed.

NYS standards
  • 6.G.1 area of polygons
  • 6.G.3 polygons in the coordinate plane
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Area of triangles is grade 6 content; finding a composite area by subtraction is the kind of task grade 7 students meet regularly. Points on the axes make every length easy to read.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 6, 16 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 shaded shape has no name, so the question is about seeing it as a difference of two right triangles whose legs lie on the axes. Once that is seen, both areas are one-line calculations. The wrong choices come from stopping after either triangle or from pairing a base from one segment with a height from the other.

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

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

  2. No obvious method minor Working it out

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

  3. Something you must already know minor Working it out

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

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, the side the triangle sits on. h is the height: the straight-up distance from the base to the top corner, at a right angle to the base. You take half because a triangle is half of a rectangle with the same base and height.

  1. Sort out the figure

    The grid lines are 5 units apart. Two slanted segments run from the y-axis down to the x-axis. The outer one goes from 20 on the y-axis to 20 on the x-axis. The inner one goes from 15 on the y-axis to 10 on the x-axis. The shaded region is the strip between them. You need its area, in square units.

  2. See the shape as two triangles

    The shaded region has four sides, and it has no simple formula. But look at the big triangle under the outer segment, with its corner at O. It is the shaded strip plus the small white triangle. So the shaded area is the big triangle minus the small one.

  3. Find the area of the big triangle

    The big triangle has a right angle at O. Its base runs along the x-axis from 0 to 20, so it is 20 units. Its height runs up the y-axis from 0 to 20, so it is 20 units.

    A=12×b×hA = \frac{1}{2} \times b \times h

    A=12×20×20=200A = \frac{1}{2} \times 20 \times 20 = 200

    Watch out: this is the whole big triangle. Stopping here gives 200, choice D.

  4. Find the area of the small triangle

    The small triangle also has its right angle at O. Its base is 10 units, from 0 to 10 on the x-axis. Its height is 15 units, from 0 to 15 on the y-axis.

    A=12×10×15=75A = \frac{1}{2} \times 10 \times 15 = 75

    Watch out: 75 is the white triangle, choice A, not the shaded part. Also, pair the base and height of the same triangle. Mixing 20 and 15 gives 150, choice C.

  5. Subtract to find the shaded area

    Take the small white triangle away from the big triangle.

    200−75=125200 - 75 = 125

    The shaded area is 125 square units. So the answer is B, 125.

Check your answer

In the figure, the shaded strip looks bigger than the white triangle. So the shaded area should be more than half of the big triangle, which is 200÷2=100200 \div 2 = 100. 125 is more than 100. And 125+75=200125 + 75 = 200, the whole big triangle.

More questions like this

The original DOE question
Question 78 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 2025 Admissions, page 81, question 78. 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 7 of 32

2024 Test B Math, question 8

Question

If 92x=3y8\frac{9}{2x}=\frac{3y}{8}, where x≠0x\neq 0 and y≠0y\neq 0, what is the product of x and y?

  • A4
  • B6
  • C12
  • D16
Show the answer

Correct answer: C

92x=3y8\frac{9}{2x}=\frac{3y}{8}

9∙8=729\bullet 8=72

2x∙3y=6xy2x\bullet 3y=6xy

72=6xy72=6xy

12=xy12=xy

Question analysis

An algebra question gives a proportion with x in one denominator and y in one numerator, 9 over 2x equal to 3y over 8, and asks for the product of x and y. Neither variable can be found on its own, so students must clear the fractions and treat the product as a single quantity.

NYS standards
  • 8.EE.7 linear equations in one variable
  • 6.EE.2b parts of an expression
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students cross-multiply numeric proportions to find one unknown. Here two variables sit in the proportion and the question wants their product, a literal-equation idea from grade 8 that many grade 7 students have not met.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 14 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 looks harder than it is, because it asks for a product rather than for x or y. Cross-multiplying turns the proportion into 72 = 6xy, and dividing by 6 finishes it, so the only real hurdle is trusting that xy can be solved for as one quantity. The wrong choices come from grabbing numbers along the way, not from a nearly complete solution.

  1. No obvious method moderate Working it out

    Neither x nor y can be found alone, so students must see that cross-multiplying gives 72 = 6xy, where the product xy appears as one block to solve for. (Schoenfeld, 1985)

  2. Variables instead of numbers minor Understanding the question

    Both unknowns stay letters and only their product is asked for, so there is no value of x or y to check a result against. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Cross-multiply to clear both fractions, collect the numbers on one side and x and y together on the other, then divide so that only the product is left.

Traps

  • 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.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying
ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c
a, b, c and d are numbers or letters, and b and d are not zero. When two fractions are equal, multiply each top by the other bottom. The two products are equal. It works because it is the same as multiplying both sides by b and by d.

  1. Sort out the question

    The equation says that two fractions are equal:

    92x=3y8\frac{9}{2x} = \frac{3y}{8}

    Here 2x means 2 times x, and 3y means 3 times y. You need the product of x and y. A product is the answer to a multiplication. So you need x×yx \times y, written xy. You do not need x or y alone.

  2. Cross-multiply the fractions

    An equation without fractions is easier to work with. So cross-multiply.

    ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c

    9×8=2x×3y9 \times 8 = 2x \times 3y

    Watch out: do not just set the bottoms equal. Writing 2x = 8 gives x = 4, choice A. That is x, not the product.

  3. Multiply out both sides

    On the left, 9×8=729 \times 8 = 72. On the right, multiply the numbers and keep the letters together.

    2x×3y=2×3×x×y=6xy2x \times 3y = 2 \times 3 \times x \times y = 6xy

    So the equation is now:

    72=6xy72 = 6xy

    The xy is the product you want, all in one piece.

  4. Divide both sides by 6

    xy is being multiplied by 6. To undo that, divide both sides by 6.

    xy=72÷6=12xy = 72 \div 6 = 12

    Watch out: 6 is just the number in front of xy. Choosing it gives B.

    So the answer is C, 12.

Check your answer

Pick x = 4. Then y must be 3, so that xy = 12. Left side: 92×4=98\frac{9}{2 \times 4} = \frac{9}{8}. Right side: 3×38=98\frac{3 \times 3}{8} = \frac{9}{8}. The two sides match, so a product of 12 works.

More questions like this

The original DOE question
Question 65 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 200, question 65. Open page 200 of the handbook (PDF)

Question 8 of 32

2024 Test B Math, question 37

Question

Raoul gave two musical performances. His second performance occurred 86 days after his first performance. If Raoul’s second performance was on a Tuesday, on what day of the week was his first performance?

  • ASunday
  • BMonday
  • CWednesday
  • DThursday
Show the answer

Correct answer: A

The total number of days (86) is divided by the number of days in a week (7).

86÷7=12, with a remainder of 286\div 7=12\text{, with a remainder of }2

The 2 days are counted back from Tuesday, so the first performance was on a Sunday.

Question analysis

Raoul's second musical performance came 86 days after his first and fell on a Tuesday. Students work out the day of the week of the first performance. It tests division with remainders in a repeating weekly cycle, and the traps are counting in the wrong direction or miscounting the leftover days.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Dividing with a remainder is grade 6 arithmetic, but using the remainder to step through a seven-day cycle, and counting backward, is less routine than grade 7 classwork. It needs care with direction rather than new content.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 36 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

Counting 86 days one at a time is hopeless in the time allowed, so the question rewards seeing that whole weeks return to the same weekday. Two decisions remain after that, and each has a listed wrong answer: which way to count, and whether Tuesday itself is one of the 2 days. Saying the days aloud, Monday is 1 back and Sunday is 2 back, avoids both.

  1. No obvious method moderate Working it out

    Nothing in the stem says to divide by 7; the key is that 84 of the 86 days are exactly 12 weeks, which leaves only 2 days to count. (Schoenfeld, 1985)

  2. Turning words into math minor Understanding the question

    "His second performance occurred 86 days after his first" puts the unknown day before the Tuesday, so the leftover days are counted backward. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Remove whole weeks from the 86 days, since each full week lands on the same weekday, then count the leftover days backward from Tuesday.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Weeks and leftover days
days=7×weeks+leftover\text{days} = 7 \times \text{weeks} + \text{leftover}
days is the total number of days. weeks is the number of full weeks inside it. leftover is the days that remain, from 0 to 6. The day of the week repeats every 7 days, so only the leftover days change it.

  1. Sort out the question

    The second show was on a Tuesday. It came 86 days after the first show. So the first show was 86 days before that Tuesday.

    You need the day of the week of the first show. So you count back 86 days from Tuesday.

  2. Take out the full weeks

    Counting back 86 days one at a time is slow. Here is a shortcut. Every 7 days, the same weekday comes back. So divide 86 by 7.

    86÷7=12 remainder 286 \div 7 = 12 \text{ remainder } 2

    86=7×12+286 = 7 \times 12 + 2

    That is 12 full weeks and 2 extra days. Going back 12 weeks from a Tuesday lands on a Tuesday again.

  3. Count back the 2 extra days

    Start on Tuesday and go back one day at a time. One day back is Monday. Two days back is Sunday.

    Watch out: Do not count Tuesday itself as one of the 2 days. That lands on Monday, choice B. And count backward, not forward. Counting forward 2 days gives Thursday, choice D.

    So the answer is A, Sunday.

Check your answer

Count forward to be sure. Start on a Sunday. After 84 days, which is 12 weeks, it is Sunday again. Two more days are Monday and Tuesday. That is day 86, a Tuesday. It matches.

More questions like this

The original DOE question
Question 94 as printed on page 210 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 210, question 94. Open page 210 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 32

2024 Test B Math, question 52

Question

Two open boxes, labeled A and B. Box A holds two balls, numbered 2 and 6. Box B holds three balls, numbered 3, 4 and 5.

Five numbered balls are placed in two boxes as shown in the figure above. Davina picks one ball at random from Box B and places it in Box A. Then Yusuf picks one ball from Box A without looking. What is the probability that Yusuf will pick a ball with an odd number?

  • A29\frac{2}{9}
  • B13\frac{1}{3}
  • C25\frac{2}{5}
  • D23\frac{2}{3}
Show the answer

Correct answer: A

The probability of Davina picking an odd-numbered ball from Box B is 23\frac{2}{3}. If Davina places an odd-numbered ball in Box A, then the probability of Yusuf picking an odd-numbered ball from Box A is 13\frac{1}{3}. The probability of both of these events happening is the product of the two probabilities.

23×13=29\frac{2}{3}\times\frac{1}{3}=\frac{2}{9}

Question analysis

Two boxes hold five numbered balls, and Davina moves one ball at random from Box B into Box A before Yusuf draws one from Box A. Students find the probability that Yusuf's ball shows an odd number. It tests two-stage probability, and the tricky part is that the second draw depends on which ball was moved.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 compound probability lists outcomes for independent stages such as coins or spinners. Here the first random move changes what is in the second box, so students must follow two linked stages, a dependence typical grade 7 work does not include.
Estimated difficulty
Among the hardest on the SHSAT: a non-obvious approach, heavy reading, or a well-hidden trap.
Solution
3 steps
Reading level
Question and choices: about grade 4, 57 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 chance events are hidden in what reads like a question about a single draw. Every listed wrong answer is a reasonable single-stage answer, either one of the two factors on its own or a count of odd balls across both boxes. Listing the nine equally likely pairs, Davina's ball and then Yusuf's, shows that only two of them end with an odd ball and makes the structure visible without any formula.

  1. No obvious method major Working it out

    The stem asks one question about Yusuf's draw, but the answer depends on Davina's earlier random pick; the key is that Box A starts with only even balls, so Yusuf can draw odd only by drawing the ball she moved. (Schoenfeld, 1985)

  2. Something you must already know minor Working it out

    Multiplying the chances of the two stages, 2/3 for an odd ball moving and 1/3 for Yusuf then drawing it, is a rule students must bring themselves. (Anderson & Krathwohl, 2001)

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

    The numbers appear only in the figure, Box A holding 2 and 6 and Box B holding 3, 4 and 5, so the odd balls have to be spotted there. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: List every equally likely pair of a ball moved by Davina and a ball picked by Yusuf, and count the pairs that end with an odd ball.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number of wanted outcomestotal number of outcomesP = \frac{\text{number of wanted outcomes}}{\text{total number of outcomes}}
P is the probability, the chance that something happens. It is a number from 0 to 1. The top counts the results you want. The bottom counts all the results that can happen. It works when every result is equally likely.
Chance of events in a row
P(A then B)=P(A)×P(B after A)P(A \text{ then } B) = P(A) \times P(B \text{ after } A)
A and B are two events that happen one after the other. P(A) is the chance of the first. P(B after A) is the chance of the second, counting only what is left after the first. You multiply because the second must also happen inside the share of times the first did.

  1. Read the figure

    Box A holds balls numbered 2 and 6. Box B holds balls numbered 3, 4 and 5.

    An odd number does not divide evenly by 2. So the only odd balls are 3 and 5, and both start in Box B. Box A starts with no odd balls at all.

  2. Spot the two chance events

    Two things happen by chance, one after the other:

    1. Davina picks one of the 3 balls in Box B and moves it to Box A.
    2. Then Yusuf picks one of the 3 balls now in Box A.

    Yusuf gets an odd ball only if both go right. Davina must move an odd ball. Then Yusuf must pick that very ball.

    The question asks only about Yusuf. But his chance depends on Davina’s pick, which comes first.

  3. Find the chance for Davina’s pick

    Box B has 3 balls, and 2 of them, the 3 and the 5, are odd.

    P(odd ball moved)=23P(\text{odd ball moved}) = \frac{2}{3}

    If she moves the 4 instead, Box A holds only even balls. Then Yusuf can’t get an odd ball at all.

  4. Find the chance for Yusuf’s pick

    Say Davina moved an odd ball. Box A now holds 3 balls: the 2, the 6 and that one odd ball.

    P(Yusuf picks it)=13P(\text{Yusuf picks it}) = \frac{1}{3}

    Watch out: 13\frac{1}{3} alone is choice B. It forgets that Davina might move the 4.

  5. Multiply the two chances

    Both events must happen, so use the rule for events in a row.

    P(odd)=23×13=29P(\text{odd}) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}

    Watch out: 23\frac{2}{3} alone, choice D, stops after Davina’s pick. 25\frac{2}{5}, choice C, counts 2 odd balls out of all 5. But Yusuf picks only from Box A.

    So the answer is A, 29\frac{2}{9}.

Check your answer

List all 9 equally likely outcomes. Davina can move 3, 4 or 5. Then Yusuf picks one of 3 balls.

Ball movedBox A thenOdd picks
32, 6, 31
42, 6, 40
52, 6, 51

That is 2 odd picks out of 9, or 29\frac{2}{9}.

More questions like this

The original DOE question
Question 109 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 109. Open page 214 of the handbook (PDF)

Question 10 of 32

2023 Test A Math, question 31

Question

A circle divided into 6 equal spaces, labeled clockwise from the top right P, Q, R, S, T and U. An arrow from the center of the circle points into Space P.

The arrow starts on Space P and moves clockwise around the circle. It moves through one space each minute. What space will the arrow point to in 100 minutes?

  • AR
  • BS
  • CT
  • DU
Show the answer

Correct answer: C

There are 6 spaces, so first divide 100 by 6: 100÷6=16100 \div 6 = 16, remainder 4. So, the arrow goes around 16 full times and then 4 more spaces. Four spaces from Space P is Space T.

Question analysis

This patterns question shows a circle split into six lettered spaces, with an arrow that starts on P and moves one space clockwise each minute, and asks where it points after 100 minutes. Students use division with a remainder to skip the full trips around. Counting the leftover moves without including the starting space, and in the right direction, is the tricky part.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Division with a remainder is grade 6 arithmetic, but using that remainder to step around a six-part circle is a pattern task that grade 7 classes seldom practice. It needs careful counting rather than new content.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 1, 32 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 division is short; the points are lost in turning the remainder into a space. The arrow starts on P, so the 4 leftover moves land on Q, R, S and T, and counting P itself as a move stops one space short. Going around the wrong way, or one move too far, gives the other two choices. Checking a small case first, such as 7 minutes landing on Q, fixes the rule before it is used.

  1. No obvious method minor Working it out

    The stem never mentions division; the step that unlocks it is seeing that every 6 minutes brings the arrow back to P, so only the remainder of 100 ÷ 6 matters. (Schoenfeld, 1985)

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

    The six spaces and their clockwise order, P, Q, R, S, T, U, appear only in the circle diagram, so the cycle length and the direction of counting both come from the figure. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Divide 100 by 6: every full trip around the circle brings the arrow back to P, so only the remainder matters; count that many moves clockwise, starting with the space after P.

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).
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Division with a remainder
a=b×q+ra = b \times q + r
a is the number you divide, here the minutes. b is the number you divide by, here the spaces in one trip around. q is how many whole groups of b fit into a. r is the remainder, the amount left over. The remainder is always less than b.

  1. Sort out the question

    The circle has 6 spaces. Going clockwise, the way a clock’s hands turn, they are P, Q, R, S, T and U. The arrow starts on P and moves one space each minute. You need the space it points to after 100 minutes.

  2. Notice that the pattern repeats

    After 6 minutes, the arrow has gone all the way around and points to P again. After 12 minutes it is back on P again, and so on. So full trips of 6 minutes don’t change where it points. Only the leftover minutes matter.

  3. Divide 100 by 6

    a=b×q+ra = b \times q + r

    100=6×16+4100 = 6 \times 16 + 4

    The arrow makes 16 full trips. That takes 6×16=966 \times 16 = 96 minutes, and it is back on P. There are 4 minutes left over.

  4. Count the last 4 moves

    From P, move one space each minute: Q (minute 97), R (minute 98), S (minute 99), T (minute 100).

    Watch out: don’t count P itself as the first move. That stops on S, choice B. Also count clockwise. Counting the other way lands on R, choice A.

    So the answer is C, T.

Check your answer

Use a smaller number with the same remainder. 100 − 96 = 4, and 96 minutes is a whole number of trips. So after 100 minutes the arrow points where it does after 4 minutes. Four moves from P are Q, R, S, T. That is T again.

More questions like this

The original DOE question
Question 88 as printed on page 83 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 88. Open page 83 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 32

2023 Test A Math, question 36

Question

Josef and Mai divided some stamps between themselves. Josef got 60% of the stamps. If Josef received 500 more stamps than Mai, how many stamps did Josef receive?

  • A300
  • B800
  • C1,500
  • D2,500
Show the answer

Correct answer: C

Let j be the number of stamps Josef has. Then Mai has j−500j - 500 stamps. The total number of stamps is j+(j−500)=2j−500j + (j - 500) = 2j - 500. Since Josef has 60% of the stamps, set up an equation to solve for j:

j2j−500=60100\frac{j}{2j - 500} = \frac{60}{100} Cross multiply.
100j=60(2j−500)100j = 60(2j - 500) Apply the distributive property; multiply the 60 by 2j2j and 500.
100j=120j−30,000100j = 120j - 30{,}000 Apply the additive inverse property; subtract 120j120j from both sides of the equation.
−20j=−30,000-20j = -30{,}000 Apply the multiplicative inverse property; divide both sides of the equation by −20.
j=1,500j = 1{,}500

Question analysis

This percent word problem says Josef and Mai split some stamps, Josef got 60% of them, and he received 500 more than Mai; it asks how many stamps Josef received. Students must connect a difference in counts to a difference in percents. Stopping at the total, or taking a percent of 500, gives tempting wrong answers.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent problems usually give the whole and ask for a part. Here neither share nor the total is known; students must link the 500-stamp gap to the gap between two percents before finding the total and then Josef's part.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 4, 26 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 arithmetic is easy once the 500 is seen as the gap between two percents of one total; seeing that is the hard part. The tempting move is to apply the 60% to the only number given, the 500, which treats a difference as a whole. Students who do find the total must still take Josef's share of it, since the total waits among the choices. Checking that 1,500 and 1,000 differ by 500 confirms the answer.

  1. Turning words into math moderate Understanding the question

    The stem describes one split in two ways, Josef's 60% of the total and his 500-stamp lead over Mai, and the link, that the lead is 60% − 40% = 20% of the total, is never stated. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. No obvious method moderate Working it out

    No equation is suggested: the quick route treats the 500 as 20 percentage points and scales up to 100%, while the handbook's route, j over 2j − 500, needs cross-multiplying and a negative coefficient. (Schoenfeld, 1985)

  3. Many steps minor Working it out

    Even the short route takes three moves, the 20% gap, the 2,500 total and Josef's 60% of it, and the total is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: If Josef has 60%, Mai has the other 40%, so the 500-stamp difference is a known percent of the total; use it to find the total, then take Josef's share.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. The whole is the total amount, and the part is the piece you want. Percent means “out of 100,” so 60% is 60 out of every 100. Use it to find a part, or work backward to find the whole.

  1. Sort out the question

    Josef and Mai split some stamps. Josef got 60% of them. He also got 500 more stamps than Mai. You need how many stamps Josef got. You don’t know the total yet, so find it first.

  2. Find Mai’s percent

    All the stamps make 100%. Josef has 60%, so Mai has the rest.

    100%−60%=40%100\% - 60\% = 40\%

    Mai has 40% of the stamps.

  3. Find what the 500 stamps are

    Josef’s extra 500 stamps are the gap between his share and Mai’s share.

    60%−40%=20%60\% - 40\% = 20\%

    So 500 stamps are 20% of all the stamps.

  4. Find the total number of stamps

    part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}

    500=20100×whole500 = \frac{20}{100} \times \text{whole}

    whole=500÷20100=500×5=2,500\text{whole} = 500 \div \frac{20}{100} = 500 \times 5 = 2{,}500

    There are 2,500 stamps in all. Watch out: that is everyone’s stamps, choice D. Don’t stop here.

  5. Find Josef’s share

    part=60100×2,500=1,500\text{part} = \frac{60}{100} \times 2{,}500 = 1{,}500

    Watch out: taking 60% of 500 gives 300, choice A. The 500 is the gap between them, not the total. So the answer is C, 1,500 stamps.

Check your answer

Mai gets 40% of 2,500, which is 1,000 stamps. Josef’s 1,500 minus Mai’s 1,000 is 500, just as the question says. And 1,500 + 1,000 = 2,500, the whole amount.

More questions like this

The original DOE question
Question 93 as printed on page 84 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 93. Open page 84 of the handbook (PDF)

Question 12 of 32

2023 Test A Math, question 42

Question

{1,2,3,4,5,…,198,199,200}\{1,2,3,4,5,\ldots,198,199,200\}

How many members of the set shown above are multiples of 6 but not multiples of 9?

  • A11
  • B13
  • C20
  • D22
Show the answer

Correct answer: D

First, list the first several multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54 . . .

Every third multiple (i.e., 18, 36, 54, . . .) is also a multiple of 9. Thus, 23\frac{2}{3} of the multiples of 6 are not multiples of 9.

Next, calculate how many multiples of 6 are in the given number range. The last multiple of 6 would be 198. So, there are a total of 1986=33\frac{198}{6} = 33 multiples of 6. Since 23\frac{2}{3} of those are not multiples of 9, the answer is 23(33)=22\frac{2}{3}(33) = 22.

Question analysis

This number properties question shows the set of whole numbers from 1 to 200 and asks how many of them are multiples of 6 but not multiples of 9. Students count multiples in a range and remove the overlap, the numbers divisible by both. Finding the right common multiple, and removing only the overlap, is where it gets tricky.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Multiples and common multiples are grade 6 content, but counting the members of a set that meet one condition and fail another is rare in grade 7 work. The overlap between the two kinds of multiples must be found and removed exactly.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 6, 16 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

Counting multiples of 6 is routine; the harder decision is how to remove the ones that are also multiples of 9. Those are the multiples of 18, the least common multiple, not of 54 or 15, and students who count them correctly must still subtract them rather than report them. Listing 6, 12, 18, 24, 30, 36 shows the pattern, two kept for every one removed, which also gives the answer as two thirds of 33.

  1. No obvious method moderate Working it out

    The stem never mentions a common multiple: students must see that the multiples of 6 that are also multiples of 9 are the multiples of 18, every third multiple of 6. (Schoenfeld, 1985)

  2. Much to keep track of minor Working it out

    Three counts are in play at once, 33 multiples of 6 up to 198, 11 multiples of 18 up to 198, and their difference, and two of them are listed choices. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. Turning words into math minor Understanding the question

    The bold not in “multiples of 6 but not multiples of 9” asks for what is left after a removal, a set difference rather than a count of the numbers removed. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Count the multiples of 6 up to 200, find the smallest number that is a multiple of both 6 and 9, count its multiples in the set, and subtract.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting multiples
last multiple=n×k\text{last multiple} = n \times k
n is the number whose multiples you count, such as 6. The first multiple is n × 1, the second is n × 2, and so on. So if the largest multiple in the set is n × k, there are k multiples in all.
Least common multiple
LCM(6,9)=18\text{LCM}(6, 9) = 18
The least common multiple, or LCM, is the smallest number that two numbers both divide evenly. A number is a multiple of both 6 and 9 exactly when it is a multiple of 18.

  1. Sort out the question

    The set holds the whole numbers from 1 to 200. A multiple of 6 is a number that 6 divides evenly, like 6, 12 or 18. You need the multiples of 6 that are not multiples of 9. So count all the multiples of 6. Then take away those that are also multiples of 9.

  2. Count the multiples of 6

    200 is not a multiple of 6, but 198 is. So 198 is the last multiple of 6 in the set.

    last multiple=n×k\text{last multiple} = n \times k

    198=6×33198 = 6 \times 33

    There are 33 multiples of 6.

  3. Find the numbers to take away

    List the multiples of 6: 6, 12, 18, 24, 30, 36. The first one 9 also divides is 18, then 36. So every third multiple of 6 is also a multiple of 9.

    LCM(6,9)=18\text{LCM}(6, 9) = 18

    Watch out: the shared multiples are multiples of 18, not of 6 + 9 = 15. Using 15 leads to 13 or 20, choices B and C.

  4. Count the multiples of 18

    The last multiple of 18 in the set is also 198.

    198=18×11198 = 18 \times 11

    There are 11 of them. Watch out: these are the numbers you take away, not the answer. Stopping here gives 11, choice A.

  5. Take away the multiples of 18

    33−11=2233 - 11 = 22

    There are 22 multiples of 6 left over. So the answer is D, 22.

Check your answer

Every third multiple of 6 is a multiple of 18, so two out of every three multiples of 6 stay. Two thirds of 33 is 23×33=22\frac{2}{3} \times 33 = 22. That matches. Test one: 30 is a multiple of 6, and 30 ÷ 9 is not a whole number, so 30 counts.

More questions like this

The original DOE question
Question 99 as printed on page 86 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 86, question 99. Open page 86 of the handbook (PDF)

Question 13 of 32

2023 Test A Math, question 57

Question

Chanelle selects a marble from a container and then returns it to the container. When she does this 3 times, the probability of choosing a red marble all 3 times is 1216\frac{1}{216}. Based on this information, what is the probability of Chanelle choosing a red marble the first time she selects a marble?

  • A172\frac{1}{72}
  • B136\frac{1}{36}
  • C18\frac{1}{8}
  • D16\frac{1}{6}
Show the answer

Correct answer: D

Let 1x\frac{1}{x} be the probability of Chanelle selecting a red marble the first time. The probability of selecting a red marble three times is 1x×1x×1x\frac{1}{x} \times \frac{1}{x} \times \frac{1}{x}, which is given as 1216\frac{1}{216}. Since 216=6×6×6216 = 6 \times 6 \times 6, the value of x is 6, and the probability of Chanelle selecting a red marble the first time is 16\frac{1}{6}.

Question analysis

This probability question says Chanelle draws a marble and puts it back, three times, and the chance of red all three times is 1/216. It asks for the chance of red on her first draw. Students reverse the multiplication rule for repeated independent draws, which means finding a cube root, and dividing by 3 is the tempting shortcut.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 8.EE.2 square and cube roots
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Multiplying probabilities for repeated draws with replacement is grade 7 compound probability. Running it backward, from three draws to one, needs a cube root, a grade 8 idea, though the numbers are friendly and the work is short once set up.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 55 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

Once the three draws are seen as one chance cubed, the question is a cube root, and 216 is a perfect cube many students know. The risk lies in undoing the cube the wrong way: dividing by 3 because there are three draws, or stopping at 36 after one division by 6. A quick check, 1/6 × 1/6 × 1/6 = 1/216, confirms the answer.

  1. Something you must already know moderate Working it out

    Students must know that three draws with the marble returned each time multiply, p × p × p, and must recognize 216 as 6 × 6 × 6; the stem states neither. (Anderson & Krathwohl, 2001)

  2. No obvious method minor Working it out

    The usual direction runs from one draw to three; here the three-draw chance is given and the bold first asks for one draw, so students work backward and undo a cube. (Schoenfeld, 1985)

Strategy: Call the chance of red on one draw p, so three draws with replacement give p × p × p; then find the number whose cube equals the given probability.

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of repeated picks
P(red 3 times)=p×p×pP(\text{red 3 times}) = p \times p \times p
p is the chance of red on one pick. Chanelle puts the marble back each time, so every pick has the same chance p, and one pick doesn’t change the next. The chance of red on all three picks is the three chances multiplied.

  1. Sort out the question

    Chanelle picks a marble and puts it back, 3 times. The chance of red all 3 times is 1216\frac{1}{216}. You need the chance of red on just the first pick. Call that chance p.

  2. Write the three picks as a product

    She puts each marble back, so the container is the same every time. Each pick has the same chance, p.

    P(red 3 times)=p×p×pP(\text{red 3 times}) = p \times p \times p

    p×p×p=1216p \times p \times p = \frac{1}{216}

  3. Find the number that fits

    You need a fraction that, multiplied by itself three times, makes 1216\frac{1}{216}. Its top is 1, since 1 × 1 × 1 = 1. For the bottom, find a number that times itself three times makes 216.

    6×6×6=36×6=2166 \times 6 \times 6 = 36 \times 6 = 216

    So the bottom is 6, and p is 16\frac{1}{6}.

  4. Avoid the tempting shortcuts

    Watch out: dividing 216 by 3 gives 72, choice A. Three picks means multiplying three times, not dividing by 3. Also, 136\frac{1}{36} is the chance of red twice in a row, choice B.

    So the answer is D, 16\frac{1}{6}.

Check your answer

Multiply it out: 16×16×16=1216\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216}. That matches the question. Choice A fails fast: 172×172\frac{1}{72} \times \frac{1}{72} is already far smaller than 1216\frac{1}{216}.

More questions like this

The original DOE question
Question 114 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 114. Open page 91 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 14 of 32

2023 Test B Math, question 5

Question

Solve the equation for x:

(21)(14)x(49)(48)=0.875\frac{(21)(14)x}{(49)(48)} = 0.875

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

Show the answer

Correct answer: 7

(21)(14)x(49)(48)=0.875\frac{(21)(14)x}{(49)(48)} = 0.875

(2149)(1448)x=0.875\left(\frac{21}{49}\right)\left(\frac{14}{48}\right)x = 0.875

(37)(724)x=0.875\left(\frac{3}{7}\right)\left(\frac{7}{24}\right)x = 0.875

(18)x=0.875\left(\frac{1}{8}\right)x = 0.875

x=(0.875)(8)=7x = (0.875)(8) = 7

Question analysis

This equation puts x in a fraction with (21)(14) on top and (49)(48) below, set equal to the decimal 0.875. The grid-in asks for the value of x. Students do best to cancel common factors before multiplying anything out, and it helps to know the fraction that 0.875 equals.

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 7.NS.2c properties for multiplying and dividing
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Seventh graders solve equations like ax = b, usually with a simple coefficient. Here the coefficient is a product of four numbers that is best cancelled first, and a decimal has to be matched with a fraction.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 6 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

This is a one-unknown equation dressed in large numbers, and the work is in reducing them without slipping. Students who cancel early find that the whole coefficient is one eighth and that 0.875 is seven eighths, which makes the last step mental. Those who multiply out face a four-digit product and a decimal division, and with nothing to check against, an early slip goes unnoticed.

  1. Error-prone arithmetic moderate Working it out

    Multiplied out, the coefficient is 294 over 2,352 and the right side is the decimal 0.875; both invite slips unless 21/49 and 14/48 are cancelled first. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. No obvious method minor Working it out

    The quick route pairs 21 with 49 and 14 with 48 before multiplying, so that 3/7 × 7/24 collapses to 1/8; the layout of the fraction does not suggest that pairing. (Schoenfeld, 1985)

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

    The result is entered with no choices, so a coefficient’s denominator or a quotient taken the wrong way looks as complete as the right value. (Enright, Morley & Sheehan, 2002)

Strategy: Cancel before multiplying: reduce 21 over 49 and 14 over 48, multiply the simplified fractions, then undo that coefficient, writing 0.875 as a fraction if it helps.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying fractions
ab×cd=a×cb×d\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}
a and c are the tops, or numerators. b and d are the bottoms, or denominators. Multiply the tops together and the bottoms together. It also works backward: you can split one fraction into two smaller ones. That lets you simplify before the numbers get big.

  1. Sort out the question

    You need the value of x. On top, 21, 14 and x are multiplied. The bottom is 49 times 48. The right side is 0.875.

    Multiplying 49 by 48 makes a big number. It is easier to simplify first.

  2. Split the fraction into two

    Use the rule for multiplying fractions backward. Pair 21 with 49, and 14 with 48.

    21×1449×48=2149×1448\frac{21 \times 14}{49 \times 48} = \frac{21}{49} \times \frac{14}{48}

    Now the equation says these two fractions times x equal 0.875.

  3. Simplify each fraction

    To simplify, divide the top and bottom by the same number. Both 21 and 49 divide by 7. Both 14 and 48 divide by 2.

    2149=371448=724\frac{21}{49} = \frac{3}{7} \qquad \frac{14}{48} = \frac{7}{24}

    Watch out: 48 ÷ 2 is 24, not 12. Writing 712\frac{7}{12} leads to 3.5.

  4. Multiply the simple fractions

    37×724=3×77×24=324=18\frac{3}{7} \times \frac{7}{24} = \frac{3 \times 7}{7 \times 24} = \frac{3}{24} = \frac{1}{8}

    The 7 on top and the 7 on the bottom cancel, since 7 ÷ 7 = 1. Then divide 3 and 24 by 3 to get 18\frac{1}{8}. So the equation becomes:

    18×x=0.875\frac{1}{8} \times x = 0.875

  5. Solve for x

    One eighth of x is 0.875. To undo dividing by 8, multiply both sides by 8.

    x=0.875×8=7x = 0.875 \times 8 = 7

    Watch out: 8 is not the answer, and don’t divide 0.875 by 8. So you would grid 7.

Check your answer

0.875 is the same as 78\frac{7}{8}. One eighth of 7 is 78\frac{7}{8}, so x = 7 works. You can also multiply it all out: 21×14×7=2,05821 \times 14 \times 7 = 2{,}058 and 49×48=2,35249 \times 48 = 2{,}352. Then 2,058÷2,352=0.8752{,}058 \div 2{,}352 = 0.875.

More questions like this

The original DOE question
Question 62 as printed on page 192 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 192, question 62. Open page 192 of the handbook (PDF)

Question 15 of 32

2023 Test B Math, question 14

Question

Each number in a sequence is 3 more than twice the number that comes just before it. If 93 is a number in the sequence, what number comes just before it?

  • A92
  • B90
  • C48
  • D45
Show the answer

Correct answer: D

Let n represent a number in the sequence, and let x represent the number that comes just before n:

n=3+2xn = 3 + 2x

93=3+2x93 = 3 + 2x

90=2x90 = 2x

45=x45 = x

Question analysis

In a sequence, each number is 3 more than twice the number just before it, and 93 is one of the numbers. Students find the number that comes just before 93. That means running the rule backward, undoing the adding and the doubling in reverse order.

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students routinely solve two-step equations. The demand here is translating the wordy sequence rule into one, or running the rule backward in the right order, rather than any harder content.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 29 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 arithmetic is two short steps, so the question rests on reading a wordy rule correctly and running it in reverse. Each wrong choice marks a different place to go wrong: ignoring the rule, stopping after the subtraction, or undoing the adding with more adding. Checking the answer forward, twice the number plus 3, confirms it reaches 93.

  1. Turning words into math moderate Understanding the question

    “Each number … is 3 more than twice the number that comes just before it” must become 2x + 3 = 93, with x the earlier number, not the later one. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. No obvious method minor Working it out

    The rule is stated forward, from a number to the next, but the question runs backward, so each operation has to be undone in reverse order: subtract 3, then halve. (Schoenfeld, 1985)

Strategy: Call the earlier number x, write the rule as an expression in x set equal to 93, and solve it by undoing each operation in reverse order.

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Stops one step early: A choice is a number from partway through the solution.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The sequence rule
n=2x+3n = 2x + 3
x is any number in the sequence, a list that follows a rule. n is the number right after it. Twice means times 2, so twice x is 2x. Then 3 more means add 3. The same rule links every number to the one after it.

  1. Sort out the question

    A sequence is a list of numbers that follows a rule. Here, each number is “3 more than twice the number that comes just before it.” 93 is in the list. You need the number right before 93. Call it x.

  2. Write the rule as an equation

    Twice x is 2x. Three more than that is 2x + 3. The rule says this equals the next number, which is 93.

    n=2x+3n = 2x + 3

    93=2x+393 = 2x + 3

  3. Undo the adding of 3

    The rule doubles x first, then adds 3. To go backward, undo the steps in reverse order. First, subtract 3 from both sides.

    93−3=2x93 - 3 = 2x

    90=2x90 = 2x

    Watch out: 90 is twice the number, not the number itself. Stopping here gives choice B. And don’t add 3, which leads to 48, choice C.

  4. Undo the doubling

    Now divide both sides by 2.

    x=90÷2=45x = 90 \div 2 = 45

    Watch out: “just before” means just before in this list. It does not mean one less than 93, so 92 is wrong. So the answer is D, 45.

Check your answer

Run the rule forward from 45. Twice 45 is 90, and 3 more is 93: 2×45+3=932 \times 45 + 3 = 93. So 45 really does come just before 93.

More questions like this

The original DOE question
Question 71 as printed on page 195 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 195, question 71. Open page 195 of the handbook (PDF)

Question 16 of 32

2023 Test B Math, question 27

Question

N={6,7,…,21,22}\mathrm{N} = \{6, 7, \ldots, 21, 22\}

Set L (not shown) is a list of 25 consecutive integers. The median of the integers in set N above is equal to the least integer in set L. What is the median of the integers in set L?

  • A25
  • B26
  • C27
  • D28
Show the answer

Correct answer: B

First, find the median of set N. There are 17 integers in set N, so the 9th integer is the median. The 9th integer is 14.

The least integer in set L is equal to the median of set N, so the least integer in set L is 14.

There are 25 integers in set L, so the median is the 13th integer. That integer is 26.

Question analysis

Set N lists the consecutive integers from 6 to 22, and set L is a list of 25 consecutive integers whose least member equals the median of N. Students find the median of set L. It chains two median questions together, and counting positions in a run of consecutive integers invites off-by-one slips.

NYS standards
  • 6.SP.5c range, center, and overall pattern
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Medians of short lists are grade 6 content. This item hides both lists, one behind set notation with an ellipsis and one not shown at all, and links them, so it needs more planning and careful counting than typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 5, 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

The question chains two medians, and both depend on counting members correctly in runs of consecutive integers. The easy slip is between position and distance: the 13th member of L is 12 steps above its first, not 13. With four consecutive choices, every counting error has a home, so writing out the first and last members of L, 14 and 38, and averaging them is the safest check.

  1. A very close wrong answer moderate Choosing the answer

    The choices 25, 26, 27 and 28 are consecutive, so a count that is off by one position lands on a neighboring choice with nothing to flag it. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. No obvious method moderate Working it out

    Set L is never listed; its median has to be found from its least member, 14, and its size, 25, by working out that the middle is the 13th member. (Schoenfeld, 1985)

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

    Set N is written with an ellipsis, {6, 7, …, 21, 22}, so its 17 members must be counted, and 22 − 6 = 16 undercounts by one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find N's median by averaging its least and greatest members. L's median is its middle member, so work out that member's position and count up from L's least member.

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).
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting the integers in a list
count=last−first+1\text{count} = \text{last} - \text{first} + 1
This counts the whole numbers in a list that goes up by 1, like 6, 7, 8. Subtracting gives the number of steps from the first number to the last. You add 1 because the first number counts too.
Position of the median
position=count+12\text{position} = \frac{\text{count} + 1}{2}
The count is how many numbers are in a list that is in order. When the count is odd, the median is at this position. It has the same number of values before it and after it.

  1. Sort out the question

    Set N is 6, 7, and so on up to 22. The dots stand for every integer in between. Integers are whole numbers. Set L has 25 consecutive integers, one right after another. Its least number equals the median of N. You need the median of L.

  2. Count the numbers in set N

    count=last−first+1\text{count} = \text{last} - \text{first} + 1

    22−6+1=1722 - 6 + 1 = 17

    Set N has 17 integers. Watch out: 22−6=1622 - 6 = 16 misses one, because the first number counts too.

  3. Find the median of set N

    position=count+12\text{position} = \frac{\text{count} + 1}{2}

    position=17+12=9\text{position} = \frac{17 + 1}{2} = 9

    The median is the 9th number. 6 is the 1st, so the 9th is 8 steps up: 6 + 8 = 14. The median of N is 14.

  4. Find the middle place in set L

    So set L starts at 14 and has 25 integers. Find the position of its median.

    position=25+12=13\text{position} = \frac{25 + 1}{2} = 13

    The median of L is its 13th number.

  5. Count up to the 13th number

    14 is the 1st number. The 13th is 12 steps after it.

    14+12=2614 + 12 = 26

    Watch out: the 13th number is 12 steps up, not 13. Adding 13 gives 27, choice C. And 25 is how many numbers L has, not its median. So the answer is B, 26.

Check your answer

Set L runs from 14 to 14+24=3814 + 24 = 38. For consecutive integers, the median is halfway between the ends: (14+38)÷2=26(14 + 38) \div 2 = 26. That matches. The same trick gives N’s median: (6+22)÷2=14(6 + 22) \div 2 = 14.

More questions like this

The original DOE question
Question 84 as printed on page 199 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 84. Open page 199 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 32

2023 Test B Math, question 28

Question

The sum of the numbers x, y, and z is 40. The ratio of x to y is 1:2, and the ratio of y to z is 2:5. What is the value of y?

  • A2
  • B4
  • C8
  • D10
Show the answer

Correct answer: D

The ratio of x to y is 1:2, so xy=12\frac{x}{y} = \frac{1}{2} or x=12yx = \frac{1}{2}y. Similarly, the ratio of y to z is 2:5, so yz=25\frac{y}{z} = \frac{2}{5}. Solve for z in terms of y:

5y=2z5y = 2z

52y=z\frac{5}{2}y = z

Use the given sum of x, y, and z to find the value of y:

x+y+z=40x + y + z = 40

12y+y+52y=40\frac{1}{2}y + y + \frac{5}{2}y = 40

4y=404y = 40

y=10y = 10

Question analysis

This ratio question gives two ratios that share y: x to y is 1:2 and y to z is 2:5, and x, y and z total 40. Students find the value of y. The numbers in a ratio count parts, not amounts, so the work is to combine the ratios and find what one part is worth.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Most grade 7 ratio tasks split a total using one ratio. Here two ratios linked through y must first become a single three-part ratio; since y's term already matches, that step is short and the problem stays at grade 7.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 31 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 rewards seeing that the two ratios share y’s term, which makes one three-part ratio. From there, one part is worth 5 and y is two parts. The wrong choices are numbers that live inside the ratios, and adding all four ratio numbers, which counts y twice, also leads to the part count, so that choice draws students along two paths.

  1. No obvious method moderate Working it out

    Nothing says to merge the ratios, but y is 2 in both, so x : y : z becomes 1 : 2 : 5 at once; the explanation’s route through fractions of y is slower. (Schoenfeld, 1985)

  2. Variables instead of numbers minor Understanding the question

    x, y and z appear only in two ratios and a sum of 40, so the ratio numbers 1, 2 and 5 are easy to mistake for the values themselves. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. Many steps minor Working it out

    Adding the 8 parts, dividing 40 by 8 to get 5 per part, and taking y’s 2 parts are three steps, and the part count is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Write x : y : z as one ratio using y's matching term, add the parts, divide 40 by that sum to get one part, and multiply by y's parts.

Traps

  • 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.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Splitting a total by a ratio
one part=total÷number of parts\text{one part} = \text{total} \div \text{number of parts}
Each number in a ratio counts parts that are all the same size. Add the numbers in the ratio to find how many parts make up the total. Dividing the total by that number gives the size of one part. Each amount is its number of parts times that size.

  1. Sort out the question

    x, y and z add up to 40. A ratio compares two amounts. The ratio of x to y is 1:2, so y is twice x. The ratio of y to z is 2:5. You need the actual value of y, not its number in a ratio.

  2. Combine the two ratios

    y is 2 in both ratios. So both ratios count the same size of part, and they join into one.

    x:y:z=1:2:5x : y : z = 1 : 2 : 5

    So x is 1 part, y is 2 parts and z is 5 parts.

  3. Find the total number of parts

    1+2+5=81 + 2 + 5 = 8

    The 40 is split into 8 equal parts. Watch out: 8 counts parts. It is not y, so choice C is wrong.

  4. Find the size of one part

    one part=total÷number of parts\text{one part} = \text{total} \div \text{number of parts}

    one part=40÷8=5\text{one part} = 40 \div 8 = 5

    The 40 is shared equally among the 8 parts. So each part is worth 5.

  5. Find y

    y is 2 parts, so multiply the size of one part by 2.

    y=2×5=10y = 2 \times 5 = 10

    Watch out: the 2 in the ratio counts parts, so y is not 2, choice A. Adding the two 2s to get 4 is wrong too, choice B. So the answer is D, 10.

Check your answer

Find all three: x = 5, y = 10 and z = 25. They add to 5+10+25=405 + 10 + 25 = 40. x to y is 5 to 10, which is 1:2. y to z is 10 to 25, which is 2:5.

More questions like this

The original DOE question
Question 85 as printed on page 199 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 85. Open page 199 of the handbook (PDF)

Question 18 of 32

2023 Test B Math, question 29

Question

Which number could be the sum of 4 consecutive integers?

  • A−12
  • B−10
  • C−8
  • D−4
Show the answer

Correct answer: B

The sum of 4 consecutive integers can be expressed as:

n+(n+1)+(n+2)+(n+3)n+(n+1)+(n+2)+(n+3)

=4n+6=4n+6

Set that expression equal to each of the options and solve for n:

Option A:

4n+6=−124n+6=-12

4n=−184n=-18

n=−184n=-\frac{18}{4} is not an integer.

Option B:

4n+6=−104n+6=-10

4n=−164n=-16

n=−4n=-4 is an integer.

Option C:

4n+6=−84n+6=-8

4n=−144n=-14

n=−144n=-\frac{14}{4} is not an integer.

Option D:

4n+6=−44n+6=-4

4n=−104n=-10

n=−104n=-\frac{10}{4} is not an integer.

The only option for which n is an integer is −10.

Question analysis

A number properties question about consecutive integers: which of four negative numbers could be the sum of 4 consecutive integers? Students write the sum with a variable, or add runs of negative integers, and test each choice. Keeping signs straight with negative numbers, and remembering that the integers must be whole numbers, is the tricky part.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 consecutive-integer problems usually give the sum and ask for the integers. This one reverses the task: students test which negative total is possible at all, using the structure of the sum rather than one equation.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 9 words.
Corrected from the DOE handbook
  • A stray word in the explanation’s last sentence was removed.
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

Three of the four choices are multiples of 4, and that is the whole trap: a student who pictures the sum as four times some integer finds three choices that seem to work. The sum of four consecutive integers is 4n + 6, so it is always even but never a multiple of 4. Trying the most natural run, −4, −3, −2 and −1, lands on −10 at once and is the fastest check.

  1. Every choice must be checked moderate Choosing the answer

    The stem asks which number “could be” the sum, so no single equation gives the answer; each of −12, −10, −8 and −4 has to be tested against the form a four-integer sum must take. (Embretson & Wetzel, 1987)

  2. No obvious method minor Working it out

    Nothing points to the shortcut that n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 is always 2 more than a multiple of 4, or to simply adding a run such as −1, −2, −3 and −4. (Schoenfeld, 1985)

  3. Error-prone arithmetic minor Working it out

    Every choice is negative, so solving 4n + 6 = −12 or adding a run of integers means subtracting 6 from a negative number and dividing a negative by 4. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the integers as n, n + 1, n + 2 and n + 3, add them, and set the sum equal to each choice to see which one gives a whole-number n.

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook ends with “The only option for which n is an integer answer is −10.” The word “answer” does not belong in that sentence; the site reads “The only option for which n is an integer is −10.” The working above it is unchanged: only 4n + 6 = −10 gives an integer, n = −4, and −4 + (−3) + (−2) + (−1) = −10.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sum of 4 consecutive integers
n+(n+1)+(n+2)+(n+3)=4n+6n + (n + 1) + (n + 2) + (n + 3) = 4n + 6
n is the smallest of the four integers. Each next integer is 1 more than the one before, so they are n, n + 1, n + 2 and n + 3. Adding them gives four n’s, plus 1 + 2 + 3, which is 6. Use it to add any four integers in a row.

  1. Sort out the question

    Integers are whole numbers and their negatives, like −2, 0 and 5. They have no fractions or decimals. Consecutive integers come one right after another, like 7, 8, 9, 10 or −3, −2, −1, 0.

    You need the choice that can be the sum of four consecutive integers. The sum is what you get when you add them.

  2. Write the sum with a letter

    You don’t know the integers yet. So call the smallest one n. Each next one is 1 more: n + 1, n + 2 and n + 3. Use the formula. Collect the four n’s, then add the numbers.

    n+n+n+n+1+2+3=4n+6n + n + n + n + 1 + 2 + 3 = 4n + 6

    Any four consecutive integers add up to 4n + 6.

  3. Test each choice

    Set 4n + 6 equal to each choice and solve for n. First subtract 6 from both sides. Then divide both sides by 4.

    Choice4n + 64nn
    A−12−18−4.5
    B−10−16−4
    C−8−14−3.5
    D−4−10−2.5

    n must be an integer. Only choice B gives one: n = −4.

  4. Name the four integers

    Start at n = −4 and count up by 1. The integers are −4, −3, −2 and −1.

    Watch out: Choices A, C and D are multiples of 4. A multiple of 4 is 4 times an integer, like 4 × (−3) = −12. It is tempting to think four integers must add to one of those. But the sum is 4n + 6, and the extra 6 gets in the way. Choice D works only with n = −2.5, and that is not an integer.

    So the answer is B, −10.

Check your answer

Add the integers: −4+(−3)+(−2)+(−1)=−10-4 + (-3) + (-2) + (-1) = -10. That is choice B. Also, 4n + 6 is always 2 more than a multiple of 4. And −10 is 2 more than −12, which is 4 × (−3).

More questions like this

The original DOE question
Question 86 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 86. Open page 200 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). This is the DOE’s version, before the corrections listed in the analysis.

Question 19 of 32

2023 Test B Math, question 35

Question

In Center City, the ratio of people to dogs is 4:14:1, and the ratio of dogs to cats is 3:23:2. What is the ratio of people to cats?

  • A2:12:1
  • B7:37:3
  • C6:16:1
  • D12:112:1
Show the answer

Correct answer: C

Take each ratio and solve for dogs:

pd=41\frac{p}{d}=\frac{4}{1}

4d=p4d=p

d=14pd=\frac{1}{4}p

And

dc=32\frac{d}{c}=\frac{3}{2}

d=32cd=\frac{3}{2}c

Set the two expressions for d equal to each other and solve for pc\frac{p}{c}:

14p=32c\frac{1}{4}p=\frac{3}{2}c

p=6cp=6c

pc=61\frac{p}{c}=\frac{6}{1}

So, the ratio of people to cats is 6:16:1.

Question analysis

A ratio question about Center City, where the ratio of people to dogs and the ratio of dogs to cats are both given. Students link the two ratios through the dogs to find the ratio of people to cats. It tests combining ratios, a step that goes wrong when the given numbers are paired up as written.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.1 ratio concepts and language
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 ratio problems usually work with a single ratio. Here two ratios share the dogs but give them different numbers of parts, so one ratio must be rescaled before people and cats can be compared.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 32 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 rewards seeing that the dogs link the two ratios but are counted differently in each. Reading the people’s 4 and the cats’ 2 straight off the page gives a tidy 2 : 1, and it is the trap most students fall into. Rewriting people to dogs as 12 : 3 makes one chain, 12 : 3 : 2, from which people to cats is 12 : 2.

  1. No obvious method moderate Working it out

    Dogs are 1 part in the first ratio and 3 parts in the second, so the ratios cannot be joined until 4 : 1 is rescaled to 12 : 3; nothing in the stem says to do this. (Schoenfeld, 1985)

  2. Much to keep track of minor Working it out

    Three groups, people, dogs and cats, run through two ratios, and students must keep track of which number belongs to which group while scaling. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Rewrite both ratios with the same number for dogs, so that people to dogs to cats becomes one three-part ratio, then read off people to cats and simplify.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent ratios
a:b=(a×k):(b×k)a : b = (a \times k) : (b \times k)
a and b are the two amounts in a ratio, in order. k is any number you choose, except 0. Multiplying both amounts by the same number keeps the comparison the same. For example, 4 : 1 and 12 : 3 both mean 4 people for every dog.

  1. Sort out the question

    A ratio compares two amounts. People to dogs is 4 : 1. That means 4 people for every 1 dog. Dogs to cats is 3 : 2. That means 3 dogs for every 2 cats.

    You need people to cats. Dogs show up in both ratios, so they are the link.

  2. Make the dog numbers match

    The dogs are 1 in the first ratio and 3 in the second. To join the ratios, the dog numbers must be the same. So multiply both parts of the first ratio by 3.

    4:1=(4×3):(1×3)=12:34 : 1 = (4 \times 3) : (1 \times 3) = 12 : 3

    Now there are 12 people for every 3 dogs.

  3. Join the two ratios

    Both ratios now have 3 dogs. So you can write all three groups in one ratio.

    people:dogs:cats=12:3:2\text{people} : \text{dogs} : \text{cats} = 12 : 3 : 2

    For every 12 people, there are 3 dogs and 2 cats.

  4. Read off and simplify

    People to cats is 12 : 2. Divide both parts by 2.

    12:2=(12÷2):(2÷2)=6:112 : 2 = (12 \div 2) : (2 \div 2) = 6 : 1

    Watch out: Taking the 4 people and the 2 cats straight from the problem gives 4 : 2 = 2 : 1, choice A. That skips matching the dogs. Stopping at 12 : 1, without using the 2 cats, gives choice D.

    So the answer is C, 6 : 1.

Check your answer

Build a small town. Start with 2 cats. The second ratio gives 3 dogs. The first ratio gives 4 people for each dog, so 4 × 3 = 12 people. That is 12 people for 2 cats, or 6 people for each cat.

More questions like this

The original DOE question
Question 92 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 201, question 92. Open page 201 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 32

2023 Test B Math, question 44

Question

In a certain class, Rosa will take a total of 4 tests. She has already taken 2 of the tests and earned scores of 81 and 83. What is the least possible score Rosa can earn on the third test and still be able to finish the class with an average score of 85 on all 4 tests? (Assume that test scores can range from 0 to 100.)

  • A76
  • B82
  • C88
  • D90
Show the answer

Correct answer: A

Let x represent Rosa’s score on the third test, and let y represent her score on the fourth test.

81+83+x+y4=85\frac{81+83+x+y}{4}=85

164+x+y=340164+x+y=340

x+y=176x+y=176

If Rosa scores 100 on the fourth test, then the least score she can earn on the third test is 76.

Question analysis

Rosa has scores of 81 and 83 on the first two of four tests and wants a mean of 85 on all four. Students find the lowest score she can earn on the third test and still reach that average, with scores capped at 100. Two tests are still unknown, so the problem needs a plan, not a single equation.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.EE.8 writing simple inequalities
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 mean problems usually find one missing score. Here two tests remain, and the least possible score depends on how high the other unknown score is allowed to go, a planning step beyond typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 60 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

Most students can find that Rosa needs 176 more points, but the question is about the least third score, which only comes from pushing the fourth score to its limit. Splitting the 176 evenly gives a natural-looking 88 that ignores the word least. The parenthesis about the 0 to 100 range is the key information, so it pays to ask what each stated limit is for.

  1. No obvious method moderate Working it out

    The least third score depends on how high the fourth can go; only the parenthesis “test scores can range from 0 to 100” tells students to set the fourth test at 100. (Schoenfeld, 1985)

  2. Many steps minor Working it out

    The 340 points needed, the 164 already earned, the 176 still needed and the 100 allowed on the last test make four steps, each a chance to lose track. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply the target mean by 4 for the total points needed, subtract the two scores so far, then decide how the remaining points can be split so the third score is as low as possible.

Traps

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Average
average=sum of the scoresnumber of scores\text{average} = \frac{\text{sum of the scores}}{\text{number of scores}}
The sum is all the scores added together. The number of scores is how many there are. The average, or mean, shares the total equally among them. Turned around, the sum equals the average times the number of scores.

  1. Sort out the question

    Rosa takes 4 tests. She scored 81 and 83 on the first two. She wants an average of 85 on all 4. You need the lowest score she can get on test 3 and still reach that average. Each score is from 0 to 100.

  2. Find the total points needed

    Turn the average formula around. The sum is the average times the number of tests.

    sum of the scores=85×4=340\text{sum of the scores} = 85 \times 4 = 340

    Rosa needs 340 points in all.

  3. Find the points still needed

    Subtract the points she already has.

    81+83=16481 + 83 = 164

    340−164=176340 - 164 = 176

    Tests 3 and 4 must add up to 176 points.

  4. Make test 4 as high as possible

    To make test 3 as low as it can be, test 4 must carry as many points as it can. The highest score is 100.

    176−100=76176 - 100 = 76

    So the lowest score on test 3 is 76.

    Watch out: Splitting 176 evenly gives 88 on each test, choice C. That works, but it is not the least score. And 82 is just the average of her first two tests, choice B.

    So the answer is A, 76.

Check your answer

Add all four scores: 81 + 83 + 76 + 100 = 340. Then 340 ÷ 4 = 85, the average she wants. A score of 75 on test 3 would need 101 on test 4, which is not possible.

More questions like this

The original DOE question
Question 101 as printed on page 203 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 101. Open page 203 of the handbook (PDF)

Question 21 of 32

2023 Test B Math, question 56

Question

A coordinate grid with the origin O, the x-axis marked 1, 2, and 3, and the y-axis marked 1 through 5. Line t passes through the point (0, 1) and line s passes through the point (1, 0); they intersect at P (3, 5), and dashed lines run from P across to 5 on the y-axis and down to 3 on the x-axis. The shaded region is the four-sided figure with vertices O, (0, 1), P (3, 5), and (1, 0).

Lines s and t intersect at point P (3, 5). What is the area, in square units, of the shaded region in the graph above?

  • A2
  • B3
  • C4
  • D7
Show the answer

Correct answer: C

First, find the area of the rectangle in the graph:

3×5=153\times 5=15

Then, find the areas of the two right triangles that are not shaded:

Upper triangle:

12×4×3=6\frac{1}{2}\times 4\times 3=6

Lower triangle:

12×2×5=5\frac{1}{2}\times 2\times 5=5

The area of the shaded region, in square units, is calculated by subtracting the total area of the two right triangles from the area of the rectangle:

15−(6+5)=15−11=415-(6+5)=15-11=4

Question analysis

Two lines meet at the point (3, 5) on a coordinate grid, and the shaded region between them and the axes is a four-sided shape. Students find its area in square units. The shape is not a standard figure, so no single area formula applies to it directly.

NYS standards
  • 6.G.1 area of polygons
  • 6.G.3 polygons in the coordinate plane
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Areas of triangles and rectangles are grade 6 content, and grade 7 students often find composite areas. This region is an irregular four-sided shape with two slanted sides, so students need a plan, and the lengths come from coordinates rather than labels.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 4, 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

The question rewards a plan more than computation: the region has no name, so students must see it inside the dashed rectangle from O to P, with two right triangles cut away. Reading each triangle’s legs from the axes is the second hurdle, since the 1-unit intercepts shorten one leg of each, to 4 instead of 5 and to 2 instead of 3. Only one wrong choice, from using one coordinate as both heights, has a clear working behind it, so most errors here are guesses after a failed plan.

  1. No obvious method moderate Working it out

    The shaded shape has four sides, two of them slanted, and no formula of its own; it must be boxed in the 3-by-5 rectangle with two triangles removed, or cut along OP. (Schoenfeld, 1985)

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

    No lengths are labeled; every one comes from the coordinates of P (3, 5), the y-intercept 1 of line s and the x-intercept 1 of line t. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Many steps minor Working it out

    The rectangle, two triangle areas and a subtraction make four computations, 15, 6 and 5, then 15 − 11, each built from different coordinates. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Draw the rectangle from O to P, find the areas of the two unshaded right triangles inside it from the coordinates, and subtract both from the rectangle's area.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle
A=l×wA = l \times w
A is the area, in square units. l is the length and w is the width. Multiplying them counts the unit squares that fill the rectangle.
Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, one side. h is the height: the distance from the base straight up to the far corner. A right triangle has a square corner, and the two sides that meet there are the base and height. A triangle is half of a rectangle with the same base and height.

  1. Sort out the shaded region

    The shaded region has four corners: O at (0, 0), (0, 1) on the y-axis, P at (3, 5), and (1, 0) on the x-axis. Line t meets the y-axis at (0, 1), and line s meets the x-axis at (1, 0). There is no easy formula for this shape. But the dashed lines make a rectangle around it. So find the rectangle’s area and take away the white parts.

  2. Find the rectangle’s area

    The rectangle has corners O, (3, 0), P and (0, 5). It is 3 units wide and 5 units tall.

    A=3×5=15A = 3 \times 5 = 15

    The rectangle’s area is 15 square units.

  3. Find the upper triangle’s area

    The white triangle at the top left has corners (0, 1), (0, 5) and P. Its square corner, or right angle, is at (0, 5). One side runs up the y-axis from 1 to 5, so it is 4 units. The other runs across from 0 to 3, so it is 3 units.

    A=12×4×3=6A = \frac{1}{2} \times 4 \times 3 = 6

    Its area is 6 square units.

  4. Find the lower triangle’s area

    The white triangle at the bottom right has corners (1, 0), (3, 0) and P. Its square corner is at (3, 0). One side runs along the x-axis from 1 to 3, so it is 2 units. The other goes up from 0 to 5, so it is 5 units.

    A=12×2×5=5A = \frac{1}{2} \times 2 \times 5 = 5

    Its area is 5 square units.

    Watch out: The lines cross the axes at 1, not at O. So these sides are 4 and 2, not 5 and 3.

  5. Subtract the white triangles

    Take both triangles away from the rectangle.

    15−(6+5)=15−11=415 - (6 + 5) = 15 - 11 = 4

    The shaded area is 4 square units.

    So the answer is C, 4.

Check your answer

Cut the shaded region along a line from O to P. One piece has a base of 1 on the x-axis and a height of 5, since P is 5 units up. Its area is 2.5. The other has a base of 1 on the y-axis and a height of 3, since P is 3 units across. Its area is 1.5. Then 2.5 + 1.5 = 4. Using 3 for both heights gives 3, choice B.

More questions like this

The original DOE question
Question 113 as printed on page 208 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 113. Open page 208 of the handbook (PDF)

Question 22 of 32

2019 Test A Math, question 51

Question

The greatest common factor of 24 and x is 8. How many possible values for x are greater than 10 and less than 70?

  • A8
  • B7
  • C6
  • D5
Show the answer

Correct answer: D

First find the multiples of 8 that fall between 10 and 70:

16, 24, 32, 40, 48, 56, 64

Since the greatest common factor of 24 and x is 8, x cannot be 24 or 48 (otherwise the GCF would be 24, not 8).

There are 5 remaining values: 16, 32, 40, 56, and 64.

Question analysis

This number properties question gives the greatest common factor of 24 and an unknown number x as 8, and asks how many possible values of x are greater than 10 and less than 70. It runs the usual GCF task in reverse. Each candidate has to meet the range and the GCF condition exactly, so a quick count is risky.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Greatest common factors are grade 6 content, usually found for two given numbers. Here the factor is given and every number producing it in a range must be found and checked, a longer and less familiar task than grade 7 classwork.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 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

The question runs the familiar GCF task backward, and the method has to be built: list the multiples of 8 in range, then strike those whose GCF with 24 is larger than 8. The strike-out is where points are lost, because 24 is plainly out but 48, a multiple of 24, looks like any other multiple of 8. Each wrong count is one step of that process left undone, so a student who skips a single check still finds a matching choice.

  1. No obvious method moderate Working it out

    The usual task finds the GCF of two given numbers; here the GCF is given and x is not, so students must see that x is a multiple of 8 and then test each candidate. (Schoenfeld, 1985)

  2. Much to keep track of moderate Working it out

    A candidate must be a multiple of 8, lie strictly between 10 and 70, and share no larger factor with 24, a test that 24 and 48 fail with a GCF of 24. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. A very close wrong answer minor Choosing the answer

    The four choices are the consecutive counts 5 to 8, so keeping a single bad value, most often 48, still lands on a listed answer that looks right. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: List the multiples of 8 between 10 and 70, then test each one: its greatest common factor with 24 must be exactly 8, so cross out any that 24 divides.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor
GCF=product of the prime factors both numbers share\text{GCF} = \text{product of the prime factors both numbers share}
A factor divides a number with nothing left over. The greatest common factor, or GCF, is the biggest number that is a factor of both numbers. A prime number has only two factors, 1 and itself, like 2 or 3. Write each number as primes multiplied together, then multiply the primes they share.

  1. Sort out the question

    A factor of a number divides it with nothing left over. The greatest common factor (GCF) of two numbers is the biggest number that is a factor of both. Here the GCF of 24 and x is exactly 8. You need to count the values x can have. “Greater than 10 and less than 70” means 10 and 70 themselves don’t count.

  2. List the multiples of 8

    8 is a factor of x, so x must be a multiple of 8. A multiple of 8 is 8 times a whole number. These multiples of 8 are greater than 10 and less than 70:

    16, 24, 32, 40, 48, 56, 64

    Leave out 8 itself, since it is less than 10. Counting 8 as well makes 8 numbers, choice A.

    Watch out: stopping here gives 7, choice B. Each of these has 8 as a factor, but the GCF must be exactly 8, not more.

  3. Break 24 into prime factors

    Write 24 as primes multiplied together.

    24=2×2×2×3=8×324 = 2 \times 2 \times 2 \times 3 = 8 \times 3

    Use the GCF formula. Every multiple of 8 shares the three 2s with 24. That shared part is 8. If x also has a factor of 3, it shares the 3 too.

    GCF=product of the prime factors both numbers share\text{GCF} = \text{product of the prime factors both numbers share}

    GCF=2×2×2×3=24\text{GCF} = 2 \times 2 \times 2 \times 3 = 24

    Then the GCF is 24, not 8. 24 has no other primes to share. So the GCF is exactly 8 when x is a multiple of 8 but not a multiple of 3.

  4. Test each number for a factor of 3

    A number is a multiple of 3 when its digits add up to a multiple of 3.

    xDigit sumMultiple of 3?GCF with 24
    167no8
    246yes24
    325no8
    404no8
    4812yes24
    5611no8
    6410no8

    Watch out: 48 is easy to miss. It is 24×224 \times 2, so its GCF with 24 is 24. Keeping 48 gives 6, choice C.

  5. Count the numbers left

    Cross out 24 and 48. Five numbers are left: 16, 32, 40, 56 and 64. Each one has a GCF of exactly 8 with 24. So the answer is D, 5.

Check your answer

Test one number the long way. The factors of 32 are 1, 2, 4, 8, 16 and 32. The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. The greatest number in both lists is 8. For 48, the number 24 is in both lists, so 48 is out.

More questions like this

The original DOE question
Question 108 as printed on page 106 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 106, question 108. Open page 106 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 23 of 32

2019 Test B Math, question 26

Question

A number line with arrows at both ends and four tick marks. From left to right, the ticks are labeled 0 (below the line), x and y (above the line), and x + y (above the line) with 24 below it. A bracket above the line marks the distance from x to y as 4.

According to the number line above, what is the value of y?

  • A4
  • B12
  • C14
  • D20
Show the answer

Correct answer: C

According to the number line, y−x=4y - x = 4 and x+y=24x + y = 24. Add the two equations together and solve for y:

y−x=4+x+y=242y=28\begin{array}{l} y - x = 4 \\ + x + y = 24 \\ \hline 2y = 28 \end{array}

y=14y = 14

Question analysis

A number line marks 0, x, y and x + y, with x + y at 24 and a bracket showing that x and y are 4 units apart. Students find the value of y. The question combines reading a number line with solving for two unknowns, and the bracket and the sum have to be used together.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Two unknowns usually call for grade 8 systems, but here one unknown is 4 more than the other, so a single grade 7 equation of the form px + q = r is enough. Reading those facts from a number line adds a step.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 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

The number line holds all the information but none of it as an equation, and the drawing is not to scale, so the positions must come from the labels. Students who split 24 evenly stop at the point midway between x and y, and those who take the bracket's 4 for a position subtract it from 24. Writing the two facts down and checking the result, 10 + 14 = 24 with a gap of 4, closes off every listed slip.

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

    The two facts needed, y − x = 4 and x + y = 24, are never written; they come from the bracket over x and y and from the 24 under the tick marked x + y. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Variables instead of numbers minor Understanding the question

    Neither x nor y has a value, and the point x + y is itself built from both, so nothing can be placed until the two facts are combined. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. No obvious method minor Working it out

    Two unknowns usually call for a system of equations, which grade 7 students have not met; they must instead write y as x + 4, or see that x and y sit 2 units either side of 12. (Schoenfeld, 1985)

Strategy: Write the two facts the labels give, y − x = 4 and x + y = 24, then add the equations or replace y with x + 4; the drawing is not to scale, so don't measure.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
distance=right point−left point\text{distance} = \text{right point} - \text{left point}
On a number line, numbers get bigger to the right. The distance between two points is how many units apart they are. Subtract the left number from the right one, so the distance is never negative.
Keeping an equation balanced
if a=b, then a−c=b−c and a÷c=b÷c\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c
An equation says two sides are equal, like a balanced scale. a and b are the two sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal. Use it to get the letter by itself.

  1. Sort out the question

    The number line shows 0, x, y and x + y. You need the value of y. The drawing is not to scale, though the question doesn’t say so. That means the spacing on the page doesn’t match the numbers. If you measure, y looks like it is near 15. So don’t measure. Work only from the labels.

  2. Write the labels as equations

    An equation is a math sentence that says two amounts are equal. The bracket marked 4 runs from x to y, and y is to the right.

    distance=right point−left point\text{distance} = \text{right point} - \text{left point}

    y−x=4y - x = 4

    So y is 4 more than x, or y = x + 4. The tick marked x + y has 24 under it, so x + y = 24.

    Watch out: the 4 is the distance from x to y. It is not where x is, and it is not y. Taking x as 4 gives 24−4=2024 - 4 = 20, choice D. Taking y as 4 is choice A.

  3. Put x + 4 in place of y

    Two unknown letters are hard to solve at once. y is the same as x + 4, so swap it into x + y = 24. Now there is only one letter.

    x+(x+4)=24x + (x + 4) = 24

    2x+4=242x + 4 = 24

    Here 2x means 2 times x. Two x’s plus 4 make 24.

  4. Solve for x

    Subtract 4 from both sides. Then divide both sides by 2. Doing the same to both sides keeps the equation balanced.

    if a=b, then a−c=b−c and a÷c=b÷c\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c

    2x+4−4=24−42x + 4 - 4 = 24 - 4

    2x=202x = 20

    x=20÷2=10x = 20 \div 2 = 10

    So x is 10.

  5. Find y

    y is 4 more than x.

    y=x+4=10+4=14y = x + 4 = 10 + 4 = 14

    So y is 14.

    Watch out: half of 24 is 12, choice B. That is the point halfway between x and y. y is 2 more than that.

    So the answer is C, 14.

Check your answer

Put both numbers back. 10+14=2410 + 14 = 24, so x + y lands at 24. 14−10=414 - 10 = 4, so x and y are 4 apart. Both labels work. Notice that y is 14, not the 15 or so that the drawing suggests.

More questions like this

The original DOE question
Question 83 as printed on page 181 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 181, question 83. Open page 181 of the handbook (PDF)

Question 24 of 32

2018 Test A Math, question 17

Question

The sum of the numbers x, y, and z is 50. The ratio of x  to y is 1:4, and the ratio of y to z is 4:5. What is the value of y ?

  • A4
  • B8
  • C10
  • D20
Show the answer

Correct answer: D

Since both ratios have yy in common, solve for xx and zz in terms of yy in both equations. Using x:y=1:4x:y=1:4, solve for xx in terms of yy.

xy=14\frac{x}{y}=\frac{1}{4}

x=y4x=\frac{y}{4}

Using the ratio y:z=4:5y:z=4:5, solve for zz in terms of yy:

yz=45\frac{y}{z}=\frac{4}{5}

z=54yz=\frac{5}{4}y

The question states x+y+z=50x+y+z=50. Substitute from the two equations above and solve for yy.

14y+y+54y=50\frac{1}{4}y+y+\frac{5}{4}y=50

104y=50\frac{10}{4}y=50

10y=20010y=200

y=20y=20

Question analysis

A ratio problem with three unknowns: x, y and z add to 50, x to y is 1 to 4, and y to z is 4 to 5. Students merge the two ratios into one three-part ratio and use the sum to find y. The step that is easy to skip is turning ratio parts into actual values.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 ratio problems usually involve one ratio. Here two ratios share a term and must be merged before the total is used, though the shared term already matches, which keeps it near typical grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 31 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 rewards noticing that y has the same term in both ratios, which turns two ratios into one three-part ratio in a single move. Students who miss this often write x and z in terms of y, as the explanation does, which works but is slower. The wrong choices are numbers taken from the ratios themselves, so a student who never scales up to the total of 50 finds an answer waiting.

  1. No obvious method moderate Working it out

    The two ratios have to be merged into one, x : y : z = 1 : 4 : 5, which works only because y is 4 in both; nothing in the wording says to combine them. (Schoenfeld, 1985)

  2. Variables instead of numbers minor Understanding the question

    x, y and z are given only through two ratios and their sum of 50, so there is no actual count to anchor the work until the parts are scaled. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. Many steps minor Working it out

    Adding the parts, dividing 50 by their total and scaling y's 4 parts are three steps, and the count of parts after the first step is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Since y is 4 in both ratios, write x : y : z as one combined ratio, find the value of one part from the total of 50, and scale up.

Traps

  • 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.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sharing a total in a ratio
one part=total÷number of parts\text{one part} = \text{total} \div \text{number of parts}
In a ratio like 3:2, the amounts are made of equal-sized parts. Add the numbers in the ratio to count the parts. Divide the total by that count to get the size of one part. Then multiply by how many parts each amount has.

  1. Sort out the question

    The three numbers x, y and z add up to 50. The ratio of x to y is 1:4. The ratio of y to z is 4:5. A ratio compares sizes. You need the actual value of y.

  2. Join the two ratios

    y is 4 in both ratios. So the two ratios use the same size parts, and you can join them into one.

    x:y:z=1:4:5x : y : z = 1 : 4 : 5

    This means x is 1 part, y is 4 parts, and z is 5 parts.

  3. Count the parts

    1+4+5=101 + 4 + 5 = 10

    There are 10 parts in all. Watch out: 10 is the number of parts, not the value of y. Picking it is choice C.

  4. Find the size of one part

    Use the sharing formula. The 10 parts share the total of 50.

    one part=50÷10=5\text{one part} = 50 \div 10 = 5

    Each part is worth 5.

  5. Find y

    y is 4 parts.

    y=4×5=20y = 4 \times 5 = 20

    Watch out: the 4 in the ratio counts parts. It is not the value of y. Picking 4 is choice A. So the answer is D, 20.

Check your answer

Find all three: x = 5, y = 20, z = 25. They add up: 5+20+25=505 + 20 + 25 = 50. x to y is 5 to 20, which is 1:4. y to z is 20 to 25, which is 4:5. Everything fits.

More questions like this

The original DOE question
Question 74 as printed on page 84 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 74. Open page 84 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 25 of 32

2018 Test A Math, question 19

Question

In Centerville, 45% of the population is female, and 60% of the population commutes to work daily. Of the total Centerville population, 21% are females who commute to work daily. What percentage of the total Centerville population are males who do not commute to work daily?

  • A15%
  • B16%
  • C24%
  • D39%
Show the answer

Correct answer: B

Create a table with the information provided in the problem and use subtraction to fill in the rest of the table:

Female Male TOTAL
Commutes to work 21% 39%
(60 − 21)		 60%
Does not commute to work 24%
(45 − 21)		 16%
(40 − 24)		 40%
(100 − 60)
TOTAL 45% 100%

16% of the population is male and does not commute to work.

Question analysis

A percent problem about overlapping groups in a town's population: the percents who are female, who commute daily, and who are female commuters are given. Students find the percent who are male and do not commute. A two-way table helps, since several of its cells are easy to confuse with the one asked for.

NYS standards
  • 7.SP.8b sample spaces for compound events
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent problems usually take a percent of one quantity. This needs two overlapping categories organized in a two-way table, which is not in New York's grade 7 standards, with three subtractions and two tempting cells.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 12, 43 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 gives three overlapping percents and asks for the one group none of them describes directly. Without a table, students tend to subtract the given numbers in some order, and several orders yield a listed choice. The last trap is the word "not", since the males who do commute are one cell away. Filling every cell of a two-by-two table before answering takes under a minute and leaves nothing to guess.

  1. No obvious method moderate Working it out

    Nothing in the wording points to a two-way table, yet the four groups, female or male by commuting or not, fall into place only once one is drawn. (Schoenfeld, 1985)

  2. Much to keep track of moderate Working it out

    Three given percents, 45, 60 and 21, plus their complements, 55 and 40, have to be kept straight while the cells are filled in by subtraction. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. Turning words into math moderate Understanding the question

    "Of the total Centerville population, 21% are females who commute" is a share of everyone, not of the females, and belongs in a single cell of the table. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Make a two-by-two table of female and male against commutes and does not commute, fill in the totals, and complete each cell by subtraction.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Parts add up to the whole
a+b=Wa + b = W
In a two-way table, every row and every column splits a group into two parts. a and b are the two parts, and W is the whole group. If you know the whole and one part, subtract to find the other part.

  1. Sort out the question

    Every number is a percent of the whole town. 45% of people are female. 60% commute, which means they travel to work every day. 21% are females who commute. You need the percent who are “males who do not commute to work daily”.

  2. Make a two-way table

    A two-way table sorts people two ways at once: female or male, and commutes or not. Put in what you know. The whole town is 100%.

    FemaleMaleTotal
    Commute21%?60%
    Do not commute???
    Total45%?100%

  3. Fill in the two totals

    Everyone is female or male, so the males are what is left.

    100%−45%=55%100\% - 45\% = 55\%

    Everyone either commutes or does not.

    100%−60%=40%100\% - 60\% = 40\%

    So 55% are male, and 40% do not commute.

  4. Find the males who commute

    Use the commute row. All commuters are 60%, and the female commuters are 21%.

    60%−21%=39%60\% - 21\% = 39\%

    So 39% are males who commute. Watch out: the question says do not commute. Picking 39% is choice D.

  5. Find the males who do not commute

    Use the male column. All males are 55%, and the males who commute are 39%.

    55%−39%=16%55\% - 39\% = 16\%

    Watch out: 24% is the females who do not commute, 45%−21%45\% - 21\%. That is choice C. So the answer is B, 16%.

Check your answer

Find the same cell another way. Females who do not commute: 45%−21%=24%45\% - 21\% = 24\%. All who do not commute: 40%. So males who do not commute: 40%−24%=16%40\% - 24\% = 16\%. The answers match.

More questions like this

The original DOE question
Question 76 as printed on page 84 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 76. Open page 84 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 26 of 32

2018 Test A Math, question 22

Question

1 sind = 4 lorgs

2 plunks = 5 dalts

5 sinds = 2 harps

1 plunk = 3 harps

A nation has five types of coins: sinds, dalts, lorgs, harps, and plunks. The relationship between the coins is shown above. Which coin is most valuable?

  • Asind
  • Bdalt
  • Charp
  • Dplunk
Show the answer

Correct answer: D

1 sind = 4 lorgs,

so 1 sind > 1 lorg.

2 harps = 5 sinds,

so 1 harp > 1 sind.

1 plunk = 3 harps,

so 1 plunk > 1 harp,

meaning that 1 plunk > 1 sind and 1 lorg.

2 plunks = 5 dalts,

so 1 plunk > 1 dalt.

Therefore, the plunk is the most valuable.

Question analysis

This reasoning question invents a money system: four exchange rates link five made-up coins. Students compare the coins step by step to find which one is worth the most. The rates run in different directions, so it is easy to misjudge which side of an equation holds the more valuable coin.

NYS standards
  • 6.RP.1 ratio concepts and language
  • 6.RP.3d converting measurement units
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 6 and 7 ratio work converts between two units at a time. Chaining four invented rates and comparing values with no common unit given is less familiar, though the arithmetic stays simple.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 1, 38 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

There is little arithmetic, but the relationships have to be read the right way round and chained. Students who think a larger count means a more valuable coin rank the coins backward, and students who stop after one or two links pick a coin that beats only its neighbor. Writing each coin's worth in lorgs, 4 for a sind, 10 for a harp, 12 for a dalt and 30 for a plunk, makes the ranking plain.

  1. Turning words into math moderate Understanding the question

    In "2 plunks = 5 dalts" the coin with the larger count is worth less, the reverse of what the bigger number suggests, and the same holds in all four equations. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of moderate Working it out

    Five coins and four exchange rates have to be linked into one ranking; dalts connect only through plunks, and lorgs only through sinds. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. No obvious method minor Working it out

    Nothing says how to compare coins that never appear in the same equation; converting every coin to lorgs, or chaining inequalities, has to be chosen. (Schoenfeld, 1985)

Strategy: Rewrite every coin in terms of lorgs, which only ever appear as the smaller unit, following the rates one link at a time, then compare.

Traps

  • Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Value of a group of coins
V=n×vV = n \times v
V is what a group of coins is worth. n is the number of coins, and v is what one coin is worth. Here every value is counted in lorgs, the smallest coin. Using one unit for all the coins lets you compare them.

  1. Sort out the question

    There are five coins and four trading rules. You need the coin that is worth the most. Read each rule carefully. In 2 plunks = 5 dalts, it takes more dalts to match. So each dalt is worth less than a plunk. More coins on one side means each of those coins is worth less.

  2. Pick one coin to measure with

    Measure every coin in lorgs, the way you might measure every length in inches. The first rule says 1 sind = 4 lorgs. So one sind is worth 4 lorgs.

  3. Find a harp’s value in lorgs

    The rule 5 sinds = 2 harps links harps to sinds. First find what 5 sinds are worth.

    V=5×4=20V = 5 \times 4 = 20

    So 2 harps are worth 20 lorgs, and 1 harp is worth half of that.

    20÷2=1020 \div 2 = 10

    One harp is worth 10 lorgs.

  4. Find a plunk’s value in lorgs

    The rule 1 plunk = 3 harps says a plunk is worth 3 harps.

    V=3×10=30V = 3 \times 10 = 30

    One plunk is worth 30 lorgs.

  5. Find a dalt’s value in lorgs

    The rule 2 plunks = 5 dalts links dalts to plunks. Two plunks are worth 2×30=602 \times 30 = 60 lorgs. So 5 dalts are worth 60 lorgs too.

    60÷5=1260 \div 5 = 12

    One dalt is worth 12 lorgs.

  6. Compare all five coins

    CoinValue in lorgs
    lorg1
    sind4
    harp10
    dalt12
    plunk30

    Watch out: 5 dalts trade for only 2 plunks, so a dalt is worth less. Reading that rule backward leads to choice B. Watch out: the harp beats the sind, but the plunk beats the harp. Stopping early gives choice C. So the answer is D, plunk.

Check your answer

Test each rule with the lorg values. 2 plunks = 60 lorgs, and 5 dalts = 60 lorgs. 5 sinds = 20 lorgs, and 2 harps = 20 lorgs. 1 plunk = 30 lorgs, and 3 harps = 30 lorgs. Every rule balances.

More questions like this

The original DOE question
Question 79 as printed on page 85 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 85, question 79. Open page 85 of the handbook (PDF)

Question 27 of 32

2018 Test A Math, question 37

Question

An unmarked straight stick will be laid end over end to measure a distance of exactly 72 feet. The same stick will be used in the same way to measure a distance of exactly 30 feet. What is the length of the longest possible stick that can be used for both measurements?

  • A3 ft
  • B4 ft
  • C6 ft
  • D8 ft
Show the answer

Correct answer: C

The length of the stick must be the greatest common factor of 72 and 30.

The factors of 30 are

1, 2, 3, 5, 6, 10, 15, and 30.

Of those, only 1, 2, 3, and 6 are also factors of 72. The greatest of these is 6.

Question analysis

This word problem asks for the longest stick that can be laid end over end to measure exactly 72 feet and exactly 30 feet. Students must recognize that the stick's length has to divide both distances evenly, so the task is finding a greatest common factor. The story wording hides the factor idea.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a greatest common factor is a sixth-grade skill, and the arithmetic is light. The question is a little less routine than a bare factor problem only because students must see that the story calls for one.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 3, 54 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

Once the story is recognized as a greatest common factor problem, the arithmetic is light and the choices can simply be tested. The trap is checking only one of the two distances, because the longest choice works for 72 feet but not for 30. A student who lists the factors of 30 first, the smaller number, sees the answer quickly.

  1. No obvious method moderate Working it out

    Nothing in the stick story says "common factor"; students must see that laying the stick a whole number of times along 72 and 30 feet means it divides both. (Schoenfeld, 1985)

  2. Hard words minor Understanding the question

    The phrases "unmarked straight stick" and "laid end over end" are unusual wording for a math item; they describe whole-stick lengths, a reading hurdle rather than more math. (Perfetti, 2007; Freedle & Kostin, 1993)

Strategy: Translate 'measures both distances exactly' into 'divides both numbers evenly', list the factors of the smaller number, 30, and take the largest that also divides 72.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of sticks
n=dsn = \frac{d}{s}
d is the distance to measure. s is the length of the stick. n is how many times you lay the stick down. For an exact measure, n must be a whole number. So s must divide d evenly.

  1. Sort out the story

    You lay a stick down again and again, end to end, like taking steps. The stick has no marks, so you can only count whole stick lengths.

    It must fit exactly into 72 feet and exactly into 30 feet. You need the longest stick that does both.

  2. Turn it into factors

    Fitting exactly means no piece is left over. So the stick length must divide 72 evenly and divide 30 evenly.

    A number that divides both is a common factor. You want the greatest common factor: the biggest number that divides both.

  3. List the factors of 30

    Start with the smaller number, 30, since its list is shorter. Its factors are 1, 2, 3, 5, 6, 10, 15 and 30.

    Now test them on 72, starting from the biggest. 30, 15, 10 and 5 leave a remainder. But 6 works:

    72÷6=1230÷6=572 \div 6 = 12 \qquad 30 \div 6 = 5

    So 6 is the greatest common factor.

  4. Check the other choices

    Watch out: 8 feet fits 72 feet, 9 times. But 30 ÷ 8 = 3.75, not a whole number, so choice D fails. Choice B fails too, since 30 ÷ 4 = 7.5. A 3-foot stick works, but it is not the longest, so choice A is wrong.

    So the answer is C, 6 ft.

Check your answer

Use the stick. A 6-foot stick goes down 12 times for 72 feet, since 12 × 6 = 72. It goes down 5 times for 30 feet, since 5 × 6 = 30. Both counts are whole numbers, so 6 feet works.

More questions like this

The original DOE question
Question 94 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 94. Open page 88 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 28 of 32

2018 Test A Math, question 46

Question

SCORES ON BIOLOGY TEST

Section Lowest Score Range
I 65 28
II 62 25
III 67 22

Mr. Blake’s biology class is divided into three sections. The same test was given to each section. The table above shows both the lowest score and the range of scores on this test for each section. What is the overall range of all scores in all three sections?

  • A25
  • B27
  • C28
  • D31
Show the answer

Correct answer: D

First, calculate the highest score for each section by adding the lowest score to the range:

Section I: 65 + 28 = 93

Section II: 62 + 25 = 87

Section III: 67 + 22 = 89

The overall highest score is 93, and the overall lowest score is 62. Thus the overall range is 93 − 62 = 31.

Question analysis

This statistics question uses a table of the lowest score and the range for three sections of a biology class and asks for the range of all the scores combined. Students must rebuild each section's highest score from its lowest score and range, then compare across sections. Reversing the usual range calculation is the tricky part.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.SP.3 measures of center and variation
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students compute a range from a list of data. Here they must work backward from range to the highest score in each section and then combine three groups, one step beyond routine.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 59 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

Range is usually computed from a list, so working backward from it is the unfamiliar part. The key fact, that the overall lowest and highest scores come from different sections, is invisible until all three highest scores are written down. Every wrong choice comes from skipping that step and reasoning with the ranges as given.

  1. No obvious method moderate Working it out

    The table gives each section's lowest score and range but no highest score, so students must reverse the range formula, high = low + range, before comparing sections. (Schoenfeld, 1985)

  2. A very close wrong answer moderate Choosing the answer

    The choices 28 and 27 sit within 4 of 31, so a sense of the spread cannot separate them; only rebuilding the highest scores, which puts the top at 93, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Much to keep track of minor Working it out

    The overall lowest, 62, comes from Section II and the overall highest, 93, from Section I, so values from two rows must be combined. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Add each section's range to its lowest score to get its highest score, then subtract the lowest of all scores from the highest of all scores.

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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Range
R=H−LR = H - L
R is the range, the spread of the scores. H is the highest score and L is the lowest score. The range tells how far apart the top and bottom scores are. If you know L and R, then H = L + R.

  1. Sort out what you know

    The table gives each section’s lowest score and its range. The range is the highest score minus the lowest score.

    The bold overall means one range for all three sections together. For that you need the highest and lowest scores of all.

  2. Find each section’s highest score

    Turn the range formula around: highest = lowest + range.

    SectionLowestRangeHighest
    I652865 + 28 = 93
    II622562 + 25 = 87
    III672267 + 22 = 89

  3. Pick the overall highest and lowest

    The highest score of all is 93, from Section I. The lowest score of all is 62, from Section II. They come from different sections, and that is fine.

  4. Subtract to find the range

    R=93−62=31R = 93 - 62 = 31

    Watch out: 28 is just Section I’s range, choice C. Averaging the three ranges gives 25, choice A. Neither covers all the scores at once.

    So the answer is D, 31.

Check your answer

The overall range must be at least as big as each section’s range, because it covers all their scores. 31 is bigger than 28, 25 and 22, so it is reasonable. Also, 62 + 31 = 93, the top score.

More questions like this

The original DOE question
Question 103 as printed on page 90 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 90, question 103. Open page 90 of the handbook (PDF)

Question 29 of 32

2018 Test A Math, question 48

Question

1013=0.769230\frac{10}{13}=0.\overline{769230}

In the infinitely repeating decimal above, 7 is the first digit in the repeating pattern. What is the 391st digit?

  • A0
  • B3
  • C6
  • D7
Show the answer

Correct answer: D

There are 6 digits in the repeating decimal (769230), so 7 would be the first, seventh, thirteenth digit and so on.

To find the 391st digit, divide 391 by 6.

391 ÷ 6 = 65 R1

Since the remainder is 1, that means the 391st digit is the same as the 1st digit, which is 7.

Question analysis

This repeating decimal question shows the six-digit repeating pattern of 10/13 and asks for the 391st digit after the decimal point. Students must use the length of the repeating block and division with a remainder to find the position within the cycle. Matching the remainder to the right digit is where students slip.

NYS standards
  • 7.NS.2d fractions to decimals by long division
  • 6.NS.2 multi-digit division
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Seventh graders learn that fractions have repeating decimals. Using the length of the repeating block and a remainder to find a digit hundreds of places away is a pattern-reasoning step beyond typical seventh-grade classwork.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
2 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

Once the cycle length is used, the arithmetic is a single division, 391 ÷ 6 = 65 remainder 1. The difficulty is turning the remainder into a position: remainder 1 is the first digit of a new block, not the last digit of the previous one or the second of the new one. Checking the rule on small positions, the 1st and 7th digits are both 7, removes the doubt.

  1. No obvious method moderate Working it out

    Nothing in the stem mentions division; students must see that the six-digit block 769230 repeats, so position 391 depends only on the remainder of 391 divided by 6. (Schoenfeld, 1985)

Strategy: Divide 391 by the length of the repeating block and look only at the remainder; a remainder of r points to the r-th digit of the block.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Division with a remainder
n=d×q+rn = d \times q + r
n is the number you divide. d is what you divide by. q is the quotient, the number of whole groups. r is the remainder, what is left over. Here it tells how many full blocks of digits fit, and where you land after them.

  1. Sort out what you know

    The bar over 769230 means those six digits repeat forever: 769230769230 and so on. The 7 is digit number 1. You need digit number 391.

  2. Spot the pattern

    The block 769230 has 6 digits, so the pattern starts over every 6 digits. Digit 1 is 7, and so are digits 7, 13, 19 and so on. Each full block ends with a 0, at digits 6, 12, 18 and so on.

  3. Divide 391 by 6

    Find how many full blocks fit into 391 digits, and what is left:

    391=6×65+1391 = 6 \times 65 + 1

    So 65 full blocks use up 6 × 65 = 390 digits. The remainder is 1. That means digit 391 is the 1st digit of the next block.

  4. Read the digit

    The 1st digit of the block is 7. Watch out: digit 390 is the last digit of a full block, which is 0. That is choice A. Digit 391 is one step past it, so it starts a new block.

    So the answer is D, 7.

Check your answer

Test the idea on a small number. Digit 7: 7 = 6 × 1 + 1, so the remainder is 1 and the digit should be 7. Write it out: 7692307. The 7th digit is indeed 7. The rule works.

More questions like this

The original DOE question
Question 105 as printed on page 90 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 90, question 105. Open page 90 of the handbook (PDF)

Question 30 of 32

2018 Test B Math, question 8

Question

M=3N=P4=Q+5=R7>0M=3N=\frac{P}{4}=Q+5=\frac{R}{7}> 0

Based on the statement above, which variable has the greatest value?

  • AMM
  • BNN
  • CPP
  • DRR
Show the answer

Correct answer: D

Break the equations apart to each equal MM:

M=3NM=3N

M=P4M=\frac{P}{4}

M=Q+5M=Q+5

M=R7M=\frac{R}{7}

Pick a number to substitute into the equations, and solve the equations to find the values of MM, NN, PP, QQ, and RR.

Let M=2M=2. Since all the equations are equal to 2, substitute 2 to find each variable.

M=3NM=3N

2=3N2=3N

23=N\frac{2}{3}=N

M=P4M=\frac{P}{4}

2=P42=\frac{P}{4}

8=P8=P

M=Q+5M=Q+5

2=Q+52=Q+5

−3=Q-3=Q

M=R7M=\frac{R}{7}

2=R72=\frac{R}{7}

14=R14=R

Variable RR has the greatest value.

Question analysis

This algebra question sets five expressions equal to one another and to a positive number, then asks which variable has the greatest value. It tests whether students can rewrite each variable in terms of one shared value and compare them. What makes it tricky is that there are no numbers to compute, so each variable's size has to be reasoned out.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students rewrite expressions to see how quantities relate, usually one equation at a time. This chain of five linked expressions, with no numbers to compute and a positivity condition to use, is a reasoning task beyond routine grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 10, 16 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 rewards a strategy, not a calculation: once the shared value is named, each variable follows in one line and R is clearly largest. Without that step, students judge by how the expressions look, and the look is backwards, since the 3 in 3N makes N small and the 7 under R makes R large. The positivity condition at the end is easy to skip but is what fixes the order.

  1. Variables instead of numbers major Understanding the question

    There are no numbers to compute: five variables are tied together in the one chain M = 3N = P/4 = Q + 5 = R/7, and each must be compared through the shared value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. No obvious method moderate Working it out

    Nothing says how to compare; the efficient move is to name the common value, or pick one such as 12, and rewrite each variable, which students must think of themselves. (Schoenfeld, 1985)

  3. Much to keep track of minor Working it out

    Five relations and the condition > 0 must be held at once; the condition matters, since with a negative common value the order of M, N, P and R would reverse. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Call the shared value k, a positive number, and rewrite each variable as an expression in k; or substitute a convenient positive value such as 12 and compare the results.

Traps

  • Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing division
ab=c means a=b×c\frac{a}{b} = c \text{ means } a = b \times c
a is the top of the fraction, b is the bottom, and c is the result. Multiplying undoes dividing. So if a divided by b is c, then a is b times c.
Undoing multiplication
b×a=c means a=c÷bb \times a = c \text{ means } a = c \div b
b is the number multiplied by a, and c is the result. Dividing undoes multiplying. So to get a alone, divide c by b.

  1. Understand the chain

    The statement is a chain of equal signs. It says five expressions all have the same value. The last part says that value is greater than 0, so it is positive.

    You need to find which letter has the largest value.

  2. Pick a value to test

    All five expressions are equal, so give them one shared value. Pick a positive number that 3 divides evenly. That way N will be a whole number.

    Let the shared value be 12. Then M=12M = 12, and each of the other four expressions also equals 12.

  3. Solve for each letter

    Undo each operation to find each letter. Divide to undo times. Multiply to undo divide. Subtract to undo plus.

    3N=12, so N=12÷3=43N = 12 \text{, so } N = 12 \div 3 = 4

    P4=12, so P=12×4=48\frac{P}{4} = 12 \text{, so } P = 12 \times 4 = 48

    Q+5=12, so Q=12−5=7Q + 5 = 12 \text{, so } Q = 12 - 5 = 7

    R7=12, so R=12×7=84\frac{R}{7} = 12 \text{, so } R = 12 \times 7 = 84

  4. Compare the values

    The values are M = 12, N = 4, P = 48, Q = 7 and R = 84. R is the largest.

    Watch out: 3N does not make N big. It means 3 times N equals M, so N is only one third of M. That trap points to choice B. Also, P is big, but R is bigger. Check every letter before you pick choice C.

    So the answer is D, R.

Check your answer

Think without numbers. R divided by 7 is M, so R is 7 times M. P is only 4 times M. N is a third of M, and Q is 5 less than M. Since M is positive, 7 times M is the biggest.

More questions like this

The original DOE question
Question 65 as printed on page 154 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 65. Open page 154 of the handbook (PDF)

Question 31 of 32

2018 Test B Math, question 33

Question

A large container is partially filled with nn liters of water. Ito adds 10 liters of water to the container, making it 60% full. If Ignacio adds 6 more liters of water, the container will be 75% full. What is the value of nn ?

  • A14
  • B15
  • C26
  • D30
Show the answer

Correct answer: A

Let xx be the total number of liters the container can hold.

n+10x=60% and n+16x=75%\frac{n+10}{x}=60\%\text{ and }\frac{n+16}{x}=75\%

First, solve each equation for xx:

Equation 1:

n+10x=60100\frac{n+10}{x}=\frac{60}{100}

n+10x=35\frac{n+10}{x}=\frac{3}{5}

3x=5(n+10)3x=5(n+10)

x=5n+503x=\frac{5n+50}{3}

Equation 2:

n+16x=75100\frac{n+16}{x}=\frac{75}{100}

n+16x=34\frac{n+16}{x}=\frac{3}{4}

3x=4(n+16)3x=4(n+16)

x=4n+643x=\frac{4n+64}{3}

Now, set the two equations equal to each other and solve for nn.

5n+503=4n+643\frac{5n+50}{3}=\frac{4n+64}{3}

5n+50=4n+645n+50=4n+64

n+50=64n+50=64

n=14 litersn=14\text{ liters}

Question analysis

Water is poured into a large container in two stages: 10 liters make it 60% full, and 6 more liters bring it to 75% full. Students must work out how much water was in it to begin with. This percent word problem hides two unknowns, the starting amount and the capacity, and several in-between numbers look like answers.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent problems usually give the whole and ask for a part. Here the whole is unknown and must be found from the change between two percents before a second step reaches the answer.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 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

The container's size is the hidden key, and nothing in the stem names it. Students who write an equation for each fill level, or who notice that 6 liters equal 15 percent, get there; others grab a number from the story or stop at the 75% level. Because every wrong choice is a number that appears in the problem or arises while solving it, a careless finish is rewarded with a match.

  1. Turning words into math major Understanding the question

    Two percent-full statements about a container of unknown size have to become equations, n + 10 = 60% of the capacity and n + 16 = 75% of it, and the capacity is never given. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. No obvious method moderate Working it out

    The quick route is to see that Ignacio's 6 liters raise the level by 15 percentage points, which fixes the capacity at 40 liters; the wording does not point to it. (Schoenfeld, 1985)

  3. Many steps moderate Working it out

    After the capacity, the solution still needs the 60% amount, 24 liters, and a subtraction of Ito's 10 liters, so two unknowns are chased through four steps. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  4. A very close wrong answer minor Choosing the answer

    15 sits one unit from the right 14, so a rough estimate of the starting amount cannot choose between them; only solving exactly, 24 − 10, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Compare the two fill levels: the extra 6 liters account for the change in percent, which gives the capacity. Then use the 60% level to work back to the starting amount.

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.
  • 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.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. The whole is the full amount, here how much the container holds. The part is p percent of it. Percent means out of 100, so you divide p by 100 and then multiply.

  1. Sort out what you know

    The container starts with nn liters of water. Ito adds 10 liters, and it is 60% full. Then 6 more liters make it 75% full. You do not know how much the container holds when full. Call that its capacity. You need nn, the starting amount.

  2. See what the 6 liters did

    The extra 6 liters raised the level from 60% to 75%.

    75−60=1575 - 60 = 15

    So 6 liters is 15% of the capacity.

    Watch out: 15 is a percent, not a number of liters. Picking 15 gives choice B.

  3. Find the full capacity

    If 15% is 6 liters, then 1% is 6÷15=0.46 \div 15 = 0.4 liters. A full container is 100%.

    100×0.4=40100 \times 0.4 = 40

    The container holds 40 liters when full.

  4. Find the water at 60 percent

    Use the percent formula with 60 for p and 40 liters for the whole.

    part=60100×40=24\text{part} = \frac{60}{100} \times 40 = 24

    After Ito adds his water, there are 24 liters.

  5. Undo Ito’s 10 liters

    Ito added 10 liters to the starting amount. To go back, subtract them.

    n=24−10=14n = 24 - 10 = 14

    The container started with 14 liters.

    Watch out: 75% of 40 is 30 liters. That is the amount after both people add water, not at the start. It is choice D.

  6. Give the answer

    The question asks for nn, the amount before anyone added water. So the answer is A, 14.

Check your answer

Run the story forward. Start with 14 liters. Add 10 to get 24 liters, and 24 is 60% of 40. Add 6 more to get 30 liters, and 30 is 75% of 40. Both facts match.

More questions like this

The original DOE question
Question 90 as printed on page 159 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 90. Open page 159 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 32 of 32

2018 Test B Math, question 53

Question

At North High School, a survey asked two questions, Question A and Question B. For each question, students could answer either “yes” or “no.” Of the 800 students who responded to the survey, 720 answered “yes” to Question A, and 640 answered “yes” to Question B. What is the least possible number of these students who could have answered “yes” to both questions?

  • A80
  • B160
  • C560
  • D640
Show the answer

Correct answer: C

None of the 80 students (800 − 720) who answered “no” to Question A could have answered “yes” to both questions.

Therefore, the least possible number of students who could have answered “yes” to both questions can be found by subtracting the 80 who answered “no” to Question A from the 640 who answered “yes” to Question B or 640 − 80 = 560.

Question analysis

In a survey of 800 students at North High School, 720 said yes to the first question and 640 said yes to the second. Students must find the smallest possible number who said yes to both. It tests reasoning about overlapping groups, and asking for the least possible overlap rather than an exact count makes it unusual.

NYS standards
  • 7.SP.8b sample spaces for compound events
  • 7.EE.4 equations and inequalities from problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students work with Venn diagrams and two-way tables when all counts are given. Finding the smallest possible overlap requires reasoning about extremes rather than computing, which is rare in grade 7 and new to most students.
Estimated difficulty
Among the hardest on the SHSAT: a non-obvious approach, heavy reading, or a well-hidden trap.
Solution
2 steps
Reading level
Question and choices: about grade 8, 60 words.
Corrected from the DOE handbook
  • A repeated phrase and a stray comma in the explanation were removed.
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

Most students have never been asked for a smallest overlap, so they fall back on a picture of nested groups and give the largest one. The key move is to push the two yes groups apart as far as possible: only the 80 students who said no to Question A can absorb some of Question B's yes answers. Once that is seen, the arithmetic is one subtraction, so the question rewards reasoning far more than computation.

  1. No obvious method major Working it out

    Nothing in the wording points to the method: the smallest overlap appears only when all 80 students who said no to Question A are imagined saying yes to Question B. (Schoenfeld, 1985)

  2. Turning words into math moderate Understanding the question

    "The least possible number of these students who could have answered “yes” to both questions" asks for an extreme case, which has to be turned into an arrangement of the 800 students. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. A very close wrong answer minor Choosing the answer

    640 is only 80 above the right 560, so a rough sense of size cannot choose between them; only asking whether the overlap is the least or the greatest separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Add the two yes counts and compare the sum with the number of students; any amount above the total must come from students counted in both groups.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook prints “(800 − 720)” twice in its first sentence, after “80 students” and again after “Question A,” and puts a comma between subject and verb in “the least possible number of students who could have answered “yes” to both questions, can be found.” The site keeps the first “(800 − 720)” and drops the comma.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Least overlap of two groups
least overlap=a+b−t\text{least overlap} = a + b - t
a is the size of the first group and b is the size of the second. t is the total number of people. If a + b is more than t, some people must be in both groups. The extra amount is the smallest number that can be in both.

  1. Sort out what you know

    There were 800 students. Of them, 720 said yes to Question A and 640 said yes to Question B. You need the least number who could have said yes to both. The overlap is the group in both yes groups. You want the smallest overlap.

  2. Find who said no to A

    Students who said no to Question A cannot be in the overlap.

    800−720=80800 - 720 = 80

    So 80 students said no to Question A.

    Watch out: 80 is not the answer. It is only a step, and it is choice A.

  3. Keep the overlap small

    To make the overlap as small as you can, put as many B yes answers outside it as possible. The only students outside it are the 80 who said no to A. So pretend all 80 of them said yes to B.

  4. Count the rest of the yes group

    The other yes answers to B must come from students who also said yes to A.

    640−80=560640 - 80 = 560

    So at least 560 students said yes to both.

    Watch out: 640 is the most that could say yes to both, not the least. That is choice D.

  5. Give the answer

    The fewest students who could say yes to both questions is 560. So the answer is C, 560.

Check your answer

Use the overlap formula: 720+640−800=560720 + 640 - 800 = 560. There are 1,360 yes answers but only 800 students, so 560 students must be counted twice. You can also check the groups. 560 said yes to both, 160 to A only and 80 to B only. That makes 560+160+80=800560 + 160 + 80 = 800.

More questions like this

The original DOE question
Question 110 as printed on page 163 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 163, question 110. Open page 163 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). This is the DOE’s version, before the corrections listed in the analysis.