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GMAT Inequalities: Why You’re Losing Points (And How to Stop)

Inequalities typically account for 3 to 5 questions on the GMAT, appearing in Quant and resurfacing inside Data Sufficiency in the Data Insights section. Master four things and most of...

Devmitra Sen
Devmitra Sen · Head of Academics
Published Aug 2020 · Updated Jul 2026
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TL;DR

Inequalities typically account for 3 to 5 questions on the GMAT, appearing in Quant and resurfacing inside Data Sufficiency in the Data Insights section. Master four things and most of them fall quickly: the flip rule, the number line, the wavy curve method, and knowing when you cannot multiply, divide, or square. Concepts first, number line second, plugging in last of the last resorts.

GMAT inequalities punch above their weight. The topic claims its fair share of the exam, roughly 3 to 5 questions, yet it sinks more Quant scores than that count suggests, and it sits squarely in the algebra portion of the GMAT syllabus.

The reason is simple. Test-takers treat inequalities like equations. They multiply both sides by a variable without knowing its sign, and the trap snaps shut. This guide fixes that habit for good.

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01 — Strategy first

How to Approach Inequalities on the GMAT

One strategic rule governs this entire topic, so it comes before the maths.

Key Rule

Plugging in is the last of the last of the last resorts. Inequality questions are dense with must-be-true scenarios, and those carry implied conditions the question never spells out. Under the timer, it is dangerously easy to plug in a value that quietly violates one of them, and one bad plug-in “proves” a wrong answer. Random plug-ins also burn time you do not have. Work with basic concepts and the number line instead. The one situation where plugging in is legitimate: when the answer choices themselves are concrete values that can be tested directly against the conditions. That is elimination, not guessing.


02 — Foundations

What Are GMAT Inequalities?

An inequality relates two expressions, just like an equation. The difference is the sign in the middle.

SignExpressionMeaning
>x > yx is greater than y
<x < yx is less than y
x ≥ yx is greater than or equal to y
x ≤ yx is less than or equal to y

The single best habit you can build is representing every inequality on a number line. An inequality is not a value. It is a range, and seeing the range is what lets you solve hard questions in under two minutes.

Take x ≤ 2. A closed (shaded) circle at 2 shows the endpoint is included, because the sign is ≤. Everything to its left is part of the solution.

-4-2-1 2 46 x ≤ 2

Closed circle: endpoint included

Now take x > 5. An open (unshaded) circle at 5 shows the endpoint is excluded, because the sign is a strict >. The solution starts just after 5.

134 57 x > 5

Open circle: endpoint excluded

Compound ranges combine both ideas. Here is −3 ≤ x ≤ 4: a shaded segment between two closed circles. Both endpoints are in, and so is everything between them.

-4 -3 -102 4 5 -3 ≤ x ≤ 4

Both endpoints included


03 — Basic rules

The Two Basic Rules of GMAT Inequalities

Two rules govern every linear inequality. Get these right and the rest follows.

Rule 1: Adding or subtracting the same quantity on both sides changes nothing. The sign holds. Multiplying or dividing both sides by the same positive number also changes nothing.

Think of it this way. If A has more money than B and both receive 300 rupees, A still has more. Take 300 from each, A still has more. The relationship survives.

Rule 2: Multiplying or dividing both sides by a negative number reverses the sign. This is the flip rule, and it is the single most tested inequality concept on the GMAT.

Watch it with a true inequality, 4 < 8:

+

Multiply by +2: 8 < 16. Still true, sign unchanged.

Multiply by −2: −8 > −16. Sign flips.

÷

Divide by +2: 2 < 4. Still true, sign unchanged.

Divide by −2: −2 > −4. Sign flips.

Can you multiply or divide both sides by a variable?

Not unless you know its sign. This is where most wrong answers are born.

If xy > 1, the tempting move is to multiply both sides by y and conclude x > y. But nobody told you y is positive. If x = 3 and y = 2, the inequality holds and x > y. If x = −3 and y = −2, the inequality still holds, but x is less than y.

The only safe deduction from xy > 1 is that x and y share the same sign.

GMAT tip

We cannot divide by an unknown variable. Now, can we divide by z²? It feels safe, since z² is never negative. The answer is still no. You do not know whether z equals zero, and division by zero is undefined. Dividing by z² is legitimate only when the question tells you z ≠ 0.

How to solve a linear inequality

Three moves, always in this order: isolate the variable and keep it positive where possible, apply the properties above without cancelling or cross-multiplying blind, then represent the answer on a number line.

Worked Example

Solve: −6x + 4 ≤ −2

Show solution

Subtract 4 from both sides: −6x ≤ −6. Divide both sides by −6 and flip the sign, because the divisor is negative: x ≥ 1.

Practice Question

If a, b, c are non-zero integers and a > bc, which of the following must be true?

  • I. ab > c
  • II. ac > b
  • III. abc > 1
  • A. I only
  • B. II only
  • C. III only
  • D. I, II, and III
  • E. None of these
Show answer & solution
Answer: E

The trap answer is D. Dividing a > bc by b, or c, or bc looks harmless, but the signs of b and c are unknown. If either is negative, the sign flips and the statement fails. Since this is a must-be-true question, none of the three survives.


04 — Advanced rules

Advanced Rules for GMAT Inequalities

Beyond linear expressions, the GMAT keeps returning to six patterns. Each has one governing idea.

1. Signs of products and quotients

When the right-hand side of an inequality is zero, the inequality is really a statement about signs. If a > 0 and b > 0, multiplying gives ab > 0. Reading that logic backwards is where the marks are:

ab > 0:

a and b have the same sign. Both positive or both negative, and neither is zero.

ab < 0:

a and b have opposite signs, and neither is zero.

ab ≥ 0:

Same signs, or at least one of a and b equals zero. The ≥ opens the door to zero.

ab > 0 : same side of zero 0 a b (or both here) ab < 0 : opposite sides of zero 0 a b

A positive product keeps a and b together. A negative product separates them across zero

Quotients behave identically, with one extra guardrail:

ab > 0:

Same signs, both non-zero. Identical to ab > 0.

ab ≥ 0:

Same signs, or a = 0. But b can never be zero, because it sits in the denominator.

ab < 0:

Opposite signs, both non-zero.

2. Fractions between 0 and 1

For any x where 0 < x < 1, this chain always holds: √x > x > x².

Check it with x = 14. The square root is 12, the square is 116, and 12 > 14 > 116. Squaring a positive fraction shrinks it. Rooting it grows it.

Practice Question

If x = 0.888, y = √0.888 and z = (0.888)², which of the following is true?

  • A. x < y < z
  • B. x < z < y
  • C. y < x < z
  • D. z < y < x
  • E. z < x < y
Show answer & solution
Answer: E

0.888 sits between 0 and 1, so √x > x > x², which means y > x > z. Reading left to right: z < x < y.

3. Adding and subtracting inequalities

The core rule: two inequalities can be added when their signs point the same way. If a > b and c > d, then a + c > b + d. Bigger plus bigger beats smaller plus smaller.

What about subtraction? Subtraction is nothing but addition after aligning the signs. Say a > b and c < d. The signs point in opposite directions, so multiply the second inequality by −1 and flip it: −c > −d. Now both signs match, so add: a − c > b − d. That is a subtraction, done safely, through addition.

Practice Question

If 4a + 2b < n and 4b + 2a > m, then b − a must be:

  • A. < m − n2
  • B. ≤ m − n2
  • C. > m − n2
  • D. ≥ m − n2
  • E. ≤ m + n2
Show answer & solution
Answer: C

Multiply the second inequality by −1 to align the signs: −4b − 2a < −m. Add it to the first: 2a − 2b < n − m, so a − b < n − m2. Multiply by −1 and flip: b − a > m − n2.

GMAT Tip

When both right-hand sides are zero, you can also multiply. If mn > 0 and pq > 0, then their product mnpq > 0. If one is positive and the other negative, say mn > 0 and pq < 0, the product is negative: mnpq < 0. This shortcut collapses some hard-looking questions into one line.

Practice Question

If mn > 0 and np < 0, which of the following must be negative?

  • A. mnp
  • B. mnp²
  • C. mn²p
  • D. mn²p²
  • E. m²n²p²
Show answer & solution
Answer: C

Approach 1: the sign table. mn > 0 means m and n share a sign, neither zero. np < 0 means n and p have opposite signs, neither zero. Only two sign patterns exist:

mnp
Case 1++
Case 2+

C. mn²p: Case 1 gives (+)(+)(−), negative. Case 2 gives (−)(+)(+), negative. Negative in both. This is the must.

Approach 2: multiply the inequalities. Both right-hand sides are zero, so multiplication is allowed:

mn > 0
np < 0
mn²p < 0

One positive quantity times one negative quantity gives a negative product. Option C in three lines.

Approach 3: read the structure. Option C is simply the split (mn)(np), the two quantities whose signs you already know.

4. Reciprocal inequalities

Start from x < y and ask: what happens to 1x and 1y? Three cases:

Case 1: both positive. Reciprocal can be taken, and the sign flips.

Take x = 2 and y = 3. Then 1x = 0.5 and 1y ≈ 0.33. So 1x > 1y.

Case 2: both negative. Reciprocal can be taken, and the sign flips.

Take x = −3 and y = −2. Then 1x ≈ −0.33 and 1y = −0.5. So 1x > 1y.

Case 3: one positive, one negative. Reciprocal can be taken, no flip.

Take x = −2 and y = 3. Then 1x = −0.5 and 1y ≈ 0.33. Still 1x < 1y.

1

Same sign: reciprocal can be taken, and the sign of the inequality flips.

2

Opposite signs: reciprocal can be taken, and the sign does not flip.

3

Signs unknown: the reciprocal cannot be taken at all.

!

Why it matters: knowing the sign is what enables the move. The GMAT hides the sign in the structure of the expression.

Practice Question

What is the largest integer x such that 12x > 0.01?

  • A. 5
  • B. 6
  • C. 7
  • D. 10
  • E. 51
Show answer & solution
Answer: B

Both sides positive, same sign, so the reciprocal can be taken with a flip: 2x < 100. The largest power of 2 below 100: 2⁶ = 64 works, 2⁷ = 128 does not. So x = 6.

5. Square roots and squaring

Two templates cover nearly every square root question:

1

If x² < a²: then −a < x < a. Example: x² < 100 gives −10 < x < 10.

2

If x² > a²: then x > a or x < −a. Example: x² > 100 gives x > 10 or x < −10.

Worked Example

If (y − 5)² < 36, find the range of y.

Show solution

Template 1 with a = 6: −6 < y − 5 < 6. Add 5 throughout: −1 < y < 11.

And the reverse direction: when can you square an inequality? Only when you know the signs of both sides. Four cases:

Both sides positive: square, no flip.

If a > 4, both sides are positive, so a² > 16.

Both sides negative: square and flip.

If a < −4, both sides are negative, so a² > 16.

One positive, one negative: squaring tells you nothing.

No conclusion possible.

Signs unknown: do not square.

If x > −3, x could be negative, zero or positive, and x² ranges from 0 upward with no usable bound.

This is exactly the kind of pattern a private GMAT tutor catches in your practice sets long before the exam does.

6. Max-Min inequalities

When two ranges are given and you need the extremes of x + y, x − y, or xy, write the ranges one under the other, then combine the extreme values to get the candidates. The largest is the max, the smallest is the min.

Worked Example

If −13 < 7x + 1 < 29 and 19 < 2 − y < 23, what is the maximum possible integer value of x + y?

  • A. −23
  • B. −18
  • C. −14
  • D. −13
  • E. −12
Show answer & solution
Answer: C

Clean up the first range: −2 < x < 4. Clean up the second: −21 < y < −17.

−2 < x < 4
−21 < y < −17
−23 < x + y < −13

The maximum integer strictly below −13 is −14. Only option C sits inside the range.

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05 — Wavy curve method

Quadratic Inequalities and the Wavy Curve Method

When the expression has an x² term, you cannot isolate x the way you would in a linear inequality.

Suppose the GMAT hands you this:

Solve: x² − 5x + 6 < 0

The x² term changes the game: the expression can dip below zero in one stretch of the number line and climb above it elsewhere. The answer is not a single cut-off. It is an interval, sometimes two.

The wavy curve method finds those intervals in six steps. Below, each step shows a small teaching example, and alongside it we check the same step on our main question.

1

Make the coefficient of the highest power positive

In ax² + bx + c, check whether a > 0. If not, multiply the whole inequality by −1 and flip the sign.

Example: −x² + x + 2 > 0. Multiply by −1 and flip: x² − x − 2 < 0.
Our question: x² − 5x + 6 < 0. Coefficient of x² is 1, already positive. Nothing to do.
2

Make the right-hand side zero

Move everything to the left so the inequality reads: expression > 0, < 0, ≥ 0, or ≤ 0.

Example: x² − x > 2. Bring the 2 across: x² − x − 2 > 0.
Our question: the right-hand side is already zero. Nothing to do.
3

Find the critical points

Factorise the expression. The critical points are the values where the entire expression becomes zero.

Example: x² − x − 2 = (x − 2)(x + 1). Critical points: 2 and −1.
Our question: x² − 5x + 6 = (x − 2)(x − 3). Critical points: 2 and 3.
4

Plot the zero points on the number line

With n distinct critical points, the line splits into n + 1 regions.

Our question: plotting 2 and 3 splits the line into three regions.
23 critical points plotted
5

The rightmost region is positive

Anything to the right of the largest critical point makes the overall expression positive. Mark it +.

Our question: take x = 4, to the right of 3. Then (4 − 2)(4 − 3) = 2, positive.
23 +
6

Alternate signs moving left, then read off your regions

Each region to the left takes the opposite sign of its neighbour: +, −, +, − and so on.

Our question: regions read +, −, + from the right. We need < 0, so the answer is: 2 < x < 3.
23 + + alternate from the right, moving left

Here is the finished picture for x² − 5x + 6 < 0. The dotted wave shows the alternation. We need the negative stretch, so the solution is 2 < x < 3, with open circles because the inequality is strict.

+ + 23 solution: 2 < x < 3

(x − 2)(x − 3) < 0. Wave above the line means +, below means −

Same expression, different signs

The picture stays identical. Only the regions you read off change:

<

< 0: the middle, endpoints out. 2 < x < 3.

>

> 0: the outside, endpoints out. x < 2 or x > 3.

≤ 0: the middle, endpoints in. 2 ≤ x ≤ 3.

≥ 0: the outside, endpoints in. x ≤ 2 or x ≥ 3.

Your turn: solve 3x² − 7x + 4 ≤ 0

Try It Yourself

Solve: 3x² − 7x + 4 ≤ 0

Show solution

Factorise: (3x − 4)(x − 1) ≤ 0. Critical points: 1 and 43.

Rightmost region +, alternate leftward: +, −, +. We need ≤ 0, so take the negative region. Closed circles.

Answer: 1 ≤ x ≤ 43

+ + 14/3 solution: 1 ≤ x ≤ 4/3

(3x − 4)(x − 1) ≤ 0. Closed circles: endpoints included

Fractions inside the method

On the GMAT, do not work with a fractional inequality directly. Convert it into a non-fractional one first. Two steps, every time.

Example 1: x + 1x − 3 < 0

1

Multiply the numerator and the denominator by the denominator term

The overall sign of the fraction does not change:

(x + 1)(x − 3)(x − 3)² < 0

(x − 3)² is never negative and cannot be zero. Mental note: x ≠ 3.

2

Transfer the sign to the numerator, then run the wavy curve

The denominator is positive. So the negative sign comes from the numerator: (x + 1)(x − 3) < 0. The fraction is gone.

From here it is the usual wavy curve: critical points at −1 and 3. Solution: −1 < x < 3.

+ + -13 solution: -1 < x < 3

06 — Cheat sheet

Seven Rules to Remember on Test Day

Pin these. Every inequality question traces back to one of them.

1

Add or subtract freely. Any quantity, both sides. The sign never changes.

x > 5 → x + 3 > 8, and x − 3 > 2
2

Positive multipliers are safe. Multiply or divide by a positive value. The sign never changes.

x > 4 → 3x > 12
3

Negative multipliers flip. Multiplying or dividing by a negative number always reverses the sign.

x > 4 → −2x < −8
4

Square only when signs are known. Both positive: square, keep the sign. Both negative: square and flip. Mixed or unknown: no squaring.

x > 4 → x² > 16
5

Unknown sign, no operation. Never multiply or divide by a variable whose sign is unknown.

xy > 1 does not give x > y
6

Even powers are not automatically safe. z² is never negative, but it can be zero. Divide by z² only when z ≠ 0 is given.

z²x > z²y gives x > y only if z ≠ 0
7

Combine inequalities through addition. Align the signs first; subtraction is just multiplying one by −1 and adding.

a > b, c < d → a − c > b − d

07 — Practice

GMAT Inequalities Practice Questions

Seven questions in a graded ladder: two easy, two medium, three hard. Concepts and the number line first; use answer choices only to eliminate.

Question 1Easy

How many integers x satisfy 1 < 5x + 5 < 25?

  • A. 1
  • B. 2
  • C. 3
  • D. 4
  • E. 5
Show answer & solution
Answer: D

Subtract 5: −4 < 5x < 20. Divide by 5: −45 < x < 4. The integers in range: 0, 1, 2 and 3. Four of them.

Question 2Easy

If 1 < a < b < c, which of the following has the greatest value?

  • A. c(a + 1)
  • B. c(b + 1)
  • C. a(b + c)
  • D. b(a + c)
  • E. c(a + b)
Show answer & solution
Answer: E

Expand and compare in pairs. E versus B: cancel cb, get ca versus c. Since a > 1, E wins. E versus D: cancel bc, get ca versus ab, i.e. c versus b. E wins. E beats every option.

Two for two? Well played. The easy ones test whether you know the rules. The next two test whether you can spot which rule applies before the clock notices.
Question 3Medium

If m < 12, then it must be true that:

  • A. −m < −12
  • B. −m − 2 < 14
  • C. −m + 2 < −10
  • D. m + 2 < 10
  • E. m − 2 < 11
Show answer & solution
Answer: E

E: m − 2 < 11 rearranges to m < 13. Given m < 12, m is certainly less than 13. Must be true.

Question 4Medium

If 6a(a + 1) > 1, which of the following could be the value of a?

  • A. −3.5
  • B. −2.5
  • C. 2.5
  • D. 3.5
  • E. 4.5
Show answer & solution
Answer: B

The real condition is: 0 < a(a + 1) < 6. Test: B gives (−2.5)(−1.5) = 3.75. Positive and below 6. Works.

Medium, cleared. Now the real filter. Three hard questions where the trap is baked into the reading itself.
Question 5Hard

If a > −2 and a < 7, which of the following must be true?

  • A. a > 2
  • B. a > −7
  • C. a < 2
  • D. −7 < a < 2
  • E. None of the above
Show answer & solution
Answer: B

The given range is −2 < a < 7. Place −7 on the line. The entire given range sits to the right of −7. Every possible value of a is greater than −7. B is the must. Since B holds, E is out automatically.

Question 6Hard

If y > 0 and 0 < 1 − xy < 1, which of the following must be true?

  • I. x > 0
  • II. xy < 1
  • III. x² + y² > 1
  • A. I only
  • B. II only
  • C. I and II only
  • D. II and III only
  • E. I, II and III
Show answer & solution
Answer: C

Break the given down. Subtract 1: −1 < −xy < 0. Multiply by −1 and flip: 0 < xy < 1.

I: xy > 0, same sign as y > 0, so x > 0. True.

II: xy < 1 is directly in the breakdown. True.

III: x = 0.0002 and y = 0.0003 satisfy the given but x² + y² is nowhere near 1. Not always true.

Question 7Hard

If x, y, z and w are positive integers and xy < zw, which of the following must be true?

  • I. x + zy + w < zw
  • II. x + zy + w < xy
  • III. x + zy + w = xy + zw
  • A. None
  • B. I only
  • C. II only
  • D. I and II
  • E. I and III
Show answer & solution
Answer: B

Cross-multiply the given: xw < yz. Statement I cross-multiplies to xw < yz, exactly what we were given. Must be true. Statement II yields yz < xw, the opposite. Not true. Statement III has no chance. Only I holds.

Seven for seven? Inequalities are officially off your weak-spot list. The same discipline carries straight into the rest of your Quant.
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08 — Common questions

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Typically 3 to 5 questions draw on inequalities. They appear as Problem Solving questions in the 45-minute Quant section, and inequality-based reasoning also surfaces inside Data Sufficiency questions in the Data Insights section.
Not realistically. With only 21 questions in the Quant section, every topic carries weight, and inequalities interlock with number properties, absolute values, and algebra. Skipping it caps your score; the topic is compact enough to master in a focused week.
Yes. Some Data Sufficiency questions are built directly on inequality reasoning, and a strong grip on ranges pays off in Two-Part Analysis and Multi-Source Reasoning questions where values must be compared against thresholds.
Usually one of three habits: multiplying or dividing by a variable of unknown sign, forgetting the zero possibility hidden inside signs like greater-than-or-equal-to, or skipping the number line and misreading where a range starts and stops. The fix is procedural: check signs before every operation and draw the line every time.
If you are targeting a competitive Quant score, yes. The wavy curve method reaches the answer in seconds once the critical points are plotted, and speed is the real constraint in a 45-minute section.

What to do next

The Rules Fit on an Index Card. The Score Difference Comes from Applying Them.

Inequalities reward discipline more than brilliance. Work the number line, respect the flip rule, and never operate on a variable of unknown sign. If the broader section is the struggle, start with how to improve your GMAT Quant score systematically, then anchor it all inside a sensible GMAT study schedule.

The bottom line

Your score earns you a read. Your discipline on these rules earns you the marks. Sign checks, number lines, no blind operations. Take it into the rest of your Quant.

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Devmitra Sen
Written by
Devmitra Sen
Head of Academics · Crackverbal

Devmitra Sen is Head of Academics at Crackverbal and has trained over 4,000 students. Her scorers tell the story: GMAT 745, 725, 715, 705 alongside turnarounds like 575→715 and 375→675. She has produced multiple Q90 scores, including a perfect 100th percentile on GMAT Quant — a benchmark very few coaches can claim consistently. On Data Insights, her superpower is changing how students see, observe, and comprehend data: breaking it down, reasoning through it, and zeroing in on exactly what the question asks. The results follow: multiple 90+ percentile DI scores, including a 1st to 99th percentile turnaround in under two and a half months. She carries a quiet interest in the history of mathematical thought — particularly ideas rooted in India long before they were formalised elsewhere — a perspective that gives her an unusually grounded sense of why the subject matters.

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