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.
Determine the 10th number in the sequence, y, by making and solving:
Determine the 8th number in the sequence, x, by making and solving:
Subtract one from both sides of the equation.
Determine the 7th number in the sequence, x, by making and solving:
Subtract the 7th number in the sequence from the 10th number:
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.
- 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.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- No obvious method moderate
- Many steps moderate
- Much to keep track of minor
- 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.
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)
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)
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
A Counting the ends (Brown & Burton, 1978)
Steps back only once from the 9th number and subtracts the 8th, 127 − 31 = 96, one position short of the 7th.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Doubles 63 to 126 but forgets to add 1 on that step, then subtracts the correct 7th number: 126 − 15 = 111.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Finds the 10th number, 2 × 63 + 1 = 127, and stops without subtracting the 7th number from it.
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.
Worked solution, step by step
Formulas used
- The sequence rule
- 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
- 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.
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.
Find the 10th number
Use the rule with x = 63.
The 10th number is 127. Watch out: remember to add 1. Doubling only gives 126 and leads to 111, choice B.
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.
Find the 8th number
The 8th number is 31. Watch out: 31 is the 8th number, not the 7th. Using it gives , choice A.
Find the 7th number
Step back once more, from the 8th number. Use the same rule with y = 31.
The 7th number is 15.
Subtract the 7th from the 10th
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. , then , then . You land on 63 as the 9th number, so the list is right. And .
More questions like this
The original DOE question
