← Explore exam analytics

GRE · Arithmetic · TOPIC LESSON

Number lines, absolute value and estimation

Read signed positions and distances, solve absolute-value conditions and use reliable bounds.

GRE General Test40 min study guide9 indexed sections

BY THE END, YOU CAN

  • Interpret absolute value as distance.
  • Solve distance equations and inequalities with endpoints.
  • Use estimates that preserve the certainty a comparison requires.

BEFORE YOU START

  • Signed arithmetic
  • Basic inequalities

Original educational lesson. Work through the examples before revealing the practice answers. This original lesson supports current GRE General Test study; it does not reproduce official questions, certify a writing score or predict test content.

LEARN THROUGH ORIGINAL PRACTICE

Original GRE-style worked examples and practice

All teaching passages, writing prompts, examples and exercises in this lesson are original. No official ETS questions, prompts or explanations are reproduced. Current structure and observed public sample metadata are separate from this original teaching; no official score or question probability is implied.

01 / UNDERSTAND & APPLY

Order describes position; distance describes separation

The real number line increases from left to right. Negative numbers farther from zero are smaller: −8<−3. Integers, fractions and irrational numbers all occupy positions on the same line. A number is not positive merely because its absolute value is large.

GRE number lines and coordinate/data graphs are drawn to scale; ordinary geometric figures need not be. Read the marked units and tick spacing rather than imposing a scale from a different figure. A point between two labeled ticks is not necessarily an integer.

02 / UNDERSTAND & APPLY

Absolute value removes direction, not information about location

|x| is the distance from x to zero, so it is never negative. It equals x for x≥0 and −x for x<0. The expression |x−a| is the distance between x and a. Distance between a and b can be written |a−b| or |b−a|.

Absolute value does not distribute over addition: |−4+7|=3, while |−4|+|7|=11. The latter sum is always at least as large as the absolute value of the sum. For a product, |ab|=|a||b|.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Distance across zero on a scaled number lineA number line from minus five to five with equal unit spacing. Marked points are minus three and two. A segment above the line spans their five-unit distance.-5-4-3-2-1012345Distance = |2 − (−3)| = 5 unitsEqual tick spacing: 1 unit per interval
The line is drawn to scale. The points −3 and 2 are five units apart; distance is nonnegative even when a coordinate is negative.

ORIGINAL WORKED EXAMPLE

Distance across zero

Find the distance between −3.5 and 2.25.

  1. Distance is |2.25−(−3.5)|.
  2. This equals |5.75|=5.75.
  3. Subtracting magnitudes would miss the fact that the points are on opposite sides of zero.

Answer 5.75 units

03 / UNDERSTAND & APPLY

An equation usually identifies two locations

For c>0, |x−a|=c means x=a−c or x=a+c. For c=0, the only solution is x=a. For c<0, there are no real solutions because a distance cannot be negative. Isolate the absolute value before splitting into cases.

Check both solutions in the original equation, especially when other expressions or denominators are present. A condition such as x>0 may eliminate one location.

ORIGINAL WORKED EXAMPLE

Solve and apply a restriction

Solve |2x−5|=7 for x>0.

  1. Split into 2x−5=7 or 2x−5=−7.
  2. The equations give x=6 and x=−1.
  3. The restriction x>0 retains only 6; |12−5|=7 checks.

Answer x=6

04 / UNDERSTAND & APPLY

Distance inequalities describe intervals or two rays

For r>0, |x−a|<r means a−r<x<a+r: the point lies within r units of a. The condition |x−a|>r means x<a−r or x>a+r: the point is outside. Replace strict symbols with inclusive ones when boundary distance r is allowed.

An open circle excludes an endpoint; a closed circle includes it. In interval notation, parentheses exclude and brackets include endpoints. Infinity is never a finite included endpoint. Distinguish “and” for an interior interval from “or” for separated exterior rays.

ORIGINAL WORKED EXAMPLE

An allowed measurement interval

A measurement x satisfies |x−12|≤0.4. State the interval.

  1. The distance from 12 is at most 0.4.
  2. Subtract and add 0.4: 11.6≤x≤12.4.
  3. Equality is allowed at both endpoints, so use brackets.

Answer [11.6, 12.4]

05 / UNDERSTAND & APPLY

When two distances appear, use the breakpoints

To remove absolute values, split the line at points where their inside expressions change sign. Within each region the sign is known, so the formula becomes linear. A formula derived for a region is valid only there; reject algebraic candidates outside it.

For |x−2|+|x−8|, any x from 2 through 8 has total distance 6. Outside that interval the sum grows. The geometric explanation is that inside the segment the two distances partition its length; outside it, the extra distance is traversed twice.

ORIGINAL WORKED EXAMPLE

A sum of distances

Solve |x−2|+|x−8|=10.

  1. For x≤2, the sum is (2−x)+(8−x)=10−2x, so x=0.
  2. For 2≤x≤8, the sum is 6, so there is no solution in that interval.
  3. For x≥8, the sum is (x−2)+(x−8)=2x−10, so x=10. Both solutions fit their regions.

Answer x=0 or x=10

06 / UNDERSTAND & APPLY

Estimation needs a direction and an error tolerance

An estimate is useful only at a precision suited to the question. To compare products of positive numbers, lower and upper bounds can establish a result without exact multiplication. If one quantity is certainly below 30 and another certainly above 30, the ordering is settled.

Rounding each factor independently does not automatically give a rigorous upper bound unless each replacement is upward and the quantities are nonnegative. When signs or reciprocals appear, revisit how inequalities change. Exact equality cannot be justified by equal rounded displays.

ORIGINAL WORKED EXAMPLE

Bound a product without multiplying

QC: Quantity A: 4.97×6.02. Quantity B: 31.

  1. Both factors are positive and 4.97<5, 6.02<6.1.
  2. Therefore A<5×6.1=30.5<31.
  3. This upper bound proves B is greater without needing the exact product.

Answer B

07 / UNDERSTAND & APPLY

Use valid cases when location is not fixed

Knowing |x| determines magnitude but often not sign. If |x|=4, x may be −4 or 4. A question involving x itself may therefore be indeterminate even when a question involving x² is fixed.

For QC, list boundary, zero, negative and fractional possibilities permitted by the prompt. A pair of different allowed relationships proves D. A single successful trial cannot prove a universal relationship.

08 / UNDERSTAND & APPLY

Choose a method and answer the requested question

All examples and practice in this lesson are original. For quantitative comparison (QC), use A if Quantity A is greater, B if Quantity B is greater, C if they are equal, and D if the relationship cannot be determined from the information given. A, B or C must hold for every permitted case. To establish D, give two allowed cases producing different relationships; an inability to calculate is not itself evidence for D.

A single-answer question requires one choice. For multiple-answer questions, obey the stated selection count or select every correct option if instructed. Test each option separately; do not stop after finding one. Multiple-answer questions receive credit only when the complete correct set is selected, with no incorrect choices. Numeric entry requires the requested value, with units handled in the calculation; round only when the question directs you to. A single numeric-entry box accepts an integer or decimal; a pair of fraction boxes accepts an integer numerator and a nonzero integer denominator. Fractions need not be reduced, but reducing makes checks easier. Do not enter a percent sign when the requested number is a percentage. Preserve exact fractions until the last step.

A calculator can check arithmetic but cannot decide the domain, interpretation or denominator for you. Estimate sign and size first, enter parentheses carefully, and compare the result with the original conditions. Cancel factors or exploit structure when doing so is faster and more reliable than decimal entry.

09 / UNDERSTAND & APPLY

Original practice: endpoints and signs

Use the QC choices defined earlier.

ORIGINAL PRACTICE · 1

Numeric entry: What is |−7|−|3−8|?

Reveal answer & reasoning

2

The two absolute values are 7 and 5, so the result is 2. Absolute values do not make the entire final expression necessarily positive; evaluate each separately.

ORIGINAL PRACTICE · 2

Select all that apply: Which satisfy |x+3|=5? A −8; B −2; C 2; D 8; E 5.

Reveal answer & reasoning

A and C

x+3=5 gives 2; x+3=−5 gives −8. The center is −3, not +3, and each answer is 5 units from that center.

ORIGINAL PRACTICE · 3

Single answer: |x−4|<3 is equivalent to A x<1; B x>7; C 1<x<7; D x<1 or x>7; E 1≤x≤7.

Reveal answer & reasoning

C

Distance less than 3 from 4 gives the interior interval with excluded endpoints. E incorrectly includes distances exactly equal to 3.

ORIGINAL PRACTICE · 4

QC: |x|=6. Quantity A: x. Quantity B: 0.

Reveal answer & reasoning

D

x=6 makes A greater, while x=−6 makes B greater. The magnitude condition leaves sign undetermined.

ORIGINAL PRACTICE · 5

Numeric entry: Find the midpoint of −7 and 5.

Reveal answer & reasoning

−1

The midpoint is their average, (−7+5)/2=−1. Half the distance is 6, but that is not the coordinate of the midpoint.

ORIGINAL PRACTICE · 6

Select all that apply: Which satisfy |x−1|≥2? A −2; B −1; C 0; D 2; E 3.

Reveal answer & reasoning

A, B and E

The solution is x≤−1 or x≥3. At −1 and 3 the distance is exactly 2 and is allowed. Values 0 and 2 are only 1 unit away.

ORIGINAL PRACTICE · 7

QC: −2<x<0. Quantity A: |x|. Quantity B: x².

Reveal answer & reasoning

D

At x=−1/2, A=1/2>B=1/4; at x=−3/2, A=3/2<B=9/4. Both are permitted. Negativity alone does not determine the comparison.

ORIGINAL PRACTICE · 8

Single answer: A positive quantity is strictly between 19.8 and 20.2. Which interval must contain its reciprocal? A (1/20.2,1/19.8); B (19.8,20.2); C (1/19.8,1/20.2); D (0,1/20.2); E (1/19.8,1).

Reveal answer & reasoning

A

Reciprocation reverses order for positive numbers, giving 1/20.2<1/x<1/19.8. C writes the larger endpoint first and does not define the required increasing interval.

KEEP THE BIG IDEAS

Quick revision

  • Absolute value is distance, so its output is nonnegative.
  • Split equations into both sign cases, then apply restrictions.
  • Inside a distance bound uses and; outside uses or.
  • An estimate must be tight and directional enough to prove the requested conclusion.
Back to the top ↑

READ WITH CONTEXT

References & scope

The teaching notes, passages, worked explanations and practice are original. Official references identify the assessment framework and skill scope; they do not endorse this lesson or license official question reproduction.

Explore this topic in context

GRE analytics distinguish current test structure from a small independently classified public sample selection. Writing instruction has no observed prompt-frequency denominator. These are not historical administered-exam frequencies.

KEEP EXPLORING

Your next idea is waiting.

Back to Quantitative Reasoning index →Browse all topic notes