The hash marks on the number line

2018 Handbook

2018 Test B Math · Question 44 of 57

Also in 2026 Practice Test B Math, question 37; 2025 Practice Test B Math, question 44.

Question

$A number line with eight evenly spaced hash marks. Point M is on the second mark, labeled −1/4, and point T is on the seventh mark, labeled 5/8. Point R, with no label, is on the fifth mark, three marks to the right of M.$

The hash marks on the number line above are evenly spaced. What is the coordinate of point $R$ ?

A $\frac{7}{40}$
B $\frac{9}{40}$
C $\frac{11}{40}$
D $\frac{21}{40}$

Show the answer

Correct answer: C

There are 5 sections between M and T. To find the length of one of these sections, find the distance between M and T and divide by 5:

$\left(\frac{5}{8}-\left(-\frac{1}{4}\right)\right)\div\frac{5}{1}=$

$\left(\frac{5}{8}+\frac{2}{8}\right)\div\frac{5}{1}=$

$\frac{7}{8}\div\frac{5}{1}=\frac{7}{8}\times\frac{1}{5}=\frac{7}{40}$

R is 3 sections away from M, so add:

$-\frac{1}{4}+3(\frac{7}{40})=-\frac{10}{40}+\frac{21}{40}=\frac{11}{40}$

R is located at $\frac{11}{40}$ .

Question analysis

Students find the fraction at point R on a number line with evenly spaced hash marks, where M is at −1/4 and T is at 5/8. It tests fraction arithmetic with negative numbers. Working out the size of one gap and counting gaps from M to R takes several careful steps.

NYS standards: 7.NS.1c rational number subtraction
6.NS.6c positioning rational numbers and points
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students place fractions and negatives on number lines, but here they must subtract a negative fraction, divide by five, and then add back. Four fraction steps with unlike denominators make it harder than typical classwork.
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: Error-prone arithmetic major
Reading a figure, table or graph moderate
Solution: 4 steps
Reading level: Question and choices: about grade 6, 22 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

Fractions, a negative starting point and a count of spaces all have to come out right together. Students lose points by counting tick marks instead of spaces, by subtracting −1/4 as if it were positive, or by measuring R from 0 instead of from M. Converting both labels to fortieths first, −10/40 and 25/40, makes each gap a whole number of fortieths.

Error-prone arithmetic major Working it out
The gap length needs 5/8 − (−1/4) = 7/8, then 7/8 ÷ 5 = 7/40, then −10/40 + 21/40: negatives and three different denominators meet in one problem. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Reading a figure, table or graph moderate Understanding the question
The hash marks have to be counted as spaces: five from M to T, with R three spaces past M, while the unlabeled marks outside M and T play no part. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the distance from M to T by subtracting their coordinates, divide it by the number of equal gaps, then step from M toward T by the right number of gaps.

Traps

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

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the length of one gap, 7/8 ÷ 5 = 7/40, and gives it as R's coordinate without stepping from M.
B Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Subtracts as 5/8 − 1/4 = 3/8 instead of adding, gets gaps of 3/40, and counts three gaps from 0.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Multiplies the 7/40 gap by 3 to get the distance from M to R, then forgets to add it to M's −1/4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line: $\text{distance} = \text{right number} - \text{left number}$; The right number is the point farther to the right on the number line. The left number is the other point. Subtracting tells how many units apart they are. Taking the left from the right keeps the answer positive.
Dividing a fraction by a whole number: $\frac{a}{b} \div c = \frac{a}{b \times c}$; a over b is the fraction, and c is the whole number you divide by. Splitting a fraction into c equal parts makes each part c times smaller. So you multiply the bottom number, b, by c.

Read the number line
M is at $-\frac{1}{4}$ and T is at $\frac{5}{8}$ . The hash marks are evenly spaced, so every space is the same length. Count the spaces. From M to T there are 5 spaces. From M to R there are 3 spaces. You need the number at point R.
Find the distance from M to T
Use the distance formula. First write $-\frac{1}{4}$ in eighths, as $-\frac{2}{8}$ .
$\frac{5}{8} - \left(-\frac{2}{8}\right) = \frac{5}{8} + \frac{2}{8} = \frac{7}{8}$
Taking away a negative is the same as adding. M and T are $\frac{7}{8}$ apart.
Find the length of one space
The 5 spaces are equal, so divide by 5.
$\frac{7}{8} \div 5 = \frac{7}{8 \times 5} = \frac{7}{40}$
Each space is $\frac{7}{40}$ long.
Watch out: this is the length of one space, not where R is. Stopping here gives choice A.
Step from M to R
R is 3 spaces to the right of M.
$3 \times \frac{7}{40} = \frac{21}{40}$
Now add that to M. Write $-\frac{1}{4}$ as $-\frac{10}{40}$ .
$-\frac{10}{40} + \frac{21}{40} = \frac{11}{40}$
Watch out: $\frac{21}{40}$ is how far R is from M, not where R is. Choosing it gives choice D.
Give the answer
Point R is at $\frac{11}{40}$ , a little to the right of 0. So the answer is C, $\frac{11}{40}$ .

Check your answer

Go backward from T instead. T is 2 spaces past R, and $\frac{5}{8} = \frac{25}{40}$ . So R is at $\frac{25}{40} - \frac{14}{40} = \frac{11}{40}$ . Both ways give the same point.

The original DOE question

Question 101 as printed on page 161 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 161, question 101. Open page 161 of the handbook (PDF)