The least of 5 consecutive integers

2018 Handbook

2018 Test A Math · Question 32 of 57

Also in 2026 Practice Test A Math, question 24; 2025 Practice Test A Math, question 28.

Question

The least of 5 consecutive integers is $l$ , and the greatest is $g$ . What is the value of $\frac{l+g}{2}$ in terms of $l$ ?

A $2l$
B $3l$
C $l+2$
D $l+5$

Show the answer

Correct answer: C

The first integer is $l$ , so the second is $l+1$ , the third is $l+2$ , then $l+3$ , and finally $l+4$ .

Since $g$ is the fifth and greatest of the integers,

$g=l+4$

Substitute $l+4$ for $g$ and simplify:

$\frac{l+g}{2}=\frac{l+l+4}{2}=\frac{2l+4}{2}=l+2$

Question analysis

This algebra question about five consecutive integers calls the least one l and the greatest g, then asks for (l + g)/2 written in terms of l. Students must express g using l and simplify a fraction that contains a variable. Counting how far the fifth integer is from the first is where many students slip.

NYS standards: 7.EE.1 linear expressions with rational coefficients
7.EE.2 rewriting expressions to show relationships
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students combine like terms in expressions such as 3(x + 4) − x. This item first asks them to build the expression from a description of consecutive integers, then simplify it, with only letters to work with.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers moderate
Turning words into math minor
Every choice must be checked minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 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 algebra is short once g is written as l + 4, so most errors come from the fence-post count or from testing numbers carelessly. The wrong choices in front of the answer are exactly the ones a single test value cannot rule out, which punishes students who stop at the first match. Using two test values, or simplifying (2l + 4)/2 directly, avoids both traps.

Variables instead of numbers moderate Understanding the question
Every expression is in l, so the simplification of (l + l + 4)/2 has to be done with letters and cannot be checked against a single number. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The description "the least of 5 consecutive integers is l, and the greatest is g" must become g = l + 4, four steps up, not five. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Every choice must be checked minor Choosing the answer
A single test value fits two choices: l = 1 makes 3l and l + 2 both 3, and l = 2 makes 2l and l + 2 both 4. (Embretson & Wetzel, 1987)

Strategy: Write the integers as l, l + 1, and so on up to the fifth, substitute for g and simplify; then test two different starting values, since one value can match a wrong choice by coincidence.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

A Checks only one number (Schoenfeld, 1985)
Reveals partial knowledge Expected pull: weak
Tests only l = 2, where 2 through 6 give (2 + 6)/2 = 4, and takes 2l, the first choice that also equals 4.
B Checks only one number (Schoenfeld, 1985)
Reveals partial knowledge Expected pull: moderate
Tests only l = 1, where 1 through 5 give (1 + 5)/2 = 3, and stops at 3l, which matches before l + 2 is checked.
D Counting the ends (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Writes the greatest integer as l + 5, counting five steps for five integers, then halves only the 2l in (2l + 5)/2, leaving l + 5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest of 5 consecutive integers: $g = l + 4$; l is the least (smallest) integer and g is the greatest. Consecutive integers go up by 1 each time. Five integers have only 4 jumps from the first to the last, so you add 4, not 5.
Dividing a sum: $\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$; a and b are the two parts of the top, and c is the number you divide by. To divide a sum, divide each part. You must divide every part of the top, not just one.

Sort out the words
Integers are whole numbers, like −2, 0 and 7. Consecutive integers come one right after another, like 8, 9, 10. There are 5 of them here.
The smallest is called l and the biggest is called g. You must write $\frac{l + g}{2}$ using only the letter l.
Write all five integers
Each integer is 1 more than the one before:
$l,\ l+1,\ l+2,\ l+3,\ l+4$
The fifth one is the greatest, so g = l + 4. Watch out: five integers have only four jumps between them. Writing g = l + 5 leads to choice D.
Put l + 4 in place of g
Swap g for l + 4 in the fraction:
$\frac{l + g}{2} = \frac{l + l + 4}{2} = \frac{2l + 4}{2}$
Here l + l is 2l, which means 2 times l.
Split the fraction
Divide each part of the top by 2:
$\frac{2l}{2} + \frac{4}{2} = l + 2$
So the answer is C, $l + 2$ .

Check your answer

Test two starting numbers. If l = 1, the integers are 1 to 5, and (1 + 5) ÷ 2 = 3, which is 1 + 2. If l = 10, they are 10 to 14, and (10 + 14) ÷ 2 = 12, which is 10 + 2. Use two tests, because l = 1 also makes 3l equal 3, choice B.

The original DOE question

Question 89 as printed on page 87 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 89. Open page 87 of the handbook (PDF)