Which statement must be true

2023 Handbook

2023 Test A Math · Question 26 of 57

Question

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

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

Show the answer

Correct answer: A

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

Option A:

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

Option B:

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

Option C:

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

Option D:

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

Option A is the only true inequality.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

The original DOE question

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