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Inequalities

Inequalities questions — isolate x, then count the integers

Inequalities questions typically give a compound form such as 4 < 3x < 14 and ask how many integers satisfy it. Divide every part by the coefficient, keeping the direction unchanged because the divisor is positive, then count the integers strictly inside the resulting bounds. Whether each endpoint is included is the whole difficulty.

The algebra in these is trivial and the counting is where marks go. A bound of 1.33 to 4.67 contains 2, 3 and 4 — three integers — but a bound of 2 to 6.67 contains 3, 4, 5 and 6, because the strict inequality excludes the 2 itself. Candidates who divide correctly and count carelessly still lose the mark.

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How to answer inequalities questions by counting integers in the range

  1. Divide all three parts at once

    For 4 < 3x < 14, divide throughout by 3 to get roughly 1.33 < x < 4.67. Operating on every part in the same step is what keeps a compound inequality intact.

  2. Keep the direction unless you divide by a negative

    A positive divisor leaves both signs as they are. Dividing or multiplying by a negative flips both, and that is the only situation in which the direction changes.

  3. List the integers strictly inside the bounds

    Between 1.33 and 4.67 the integers are 2, 3 and 4, giving a count of three. Fractional bounds are easy here because neither endpoint can be an integer.

  4. Treat whole-number bounds with extra care

    For 2 < x < 6.67 the integers are 3, 4, 5 and 6 — the 2 is excluded by the strict sign. This is where nearly every avoidable error in the topic occurs.

Which endpoints count as solutions

The bound isThe endpoint
x > 2 (strict)2 is excluded
x ≥ 2 (inclusive)2 is counted
x < 4.67 (fractional)Highest integer is 4
x < 5 (whole number)Highest integer is 4
x ≤ 5 (inclusive whole)Highest integer is 5

3 sample inequalities questions, with worked solutions

Generated by the same engine that mints the Quant Daily daily. Each answer is computed from the numbers printed in the question, and the walkthrough below each one is the engine's own working — not a solution written afterwards. Reload this page's live drill and you get different numbers.

Question 1

2 < 2x < 13

How many integer values of x satisfy the inequality?

  1. 5
  2. 11
  3. 7
  4. 8
  5. 6
Show the worked solution
  1. Divide through by 2 (positive, so the inequality direction stays): 1 < x < 6.5.
  2. Integers: 2, 3, 4, 5, 6.
  3. Answer: 5.

Question 2

3 < 4x < 12

How many integer values of x satisfy the inequality?

  1. 3
  2. 1
  3. 2
  4. 9
  5. 5
Show the worked solution
  1. Divide through by 4 (positive, so the inequality direction stays): 0.75 < x < 3.
  2. Integers: 1, 2.
  3. Answer: 2.

Question 3

4 < 3x < 17

How many integer values of x satisfy the inequality?

  1. 5
  2. 2
  3. 4
  4. 3
  5. 13
Show the worked solution
  1. Divide through by 3 (positive, so the inequality direction stays): 1.3333333333333333 < x < 5.666666666666667.
  2. Integers: 2, 3, 4, 5.
  3. Answer: 4.

Common mistakes this drill is built from

The wrong options are not random numbers. Each one is the result of a specific careless error for this topic, so picking one tells you which habit is costing you marks.

Including a strict whole-number endpoint

With 2 < x, the value 2 does not satisfy the inequality. Counting it adds one to the total and the inflated count is a printed option.

Dividing only one part of a compound inequality

All three parts move together. Dividing the middle alone changes the relationship rather than simplifying it.

Forgetting to flip on a negative divisor

Dividing by a negative reverses both inequality signs. Missing this inverts the entire solution range.

Why your answer came out wrong

The specific wrong numbers this topic produces, and what each one tells you about the step you took. If you have just got a question wrong and want to know which habit did it, start here.

For 6 < 3x < 20, is the answer 5 integers?

It is 4. Dividing gives 2 < x < 6.67, and the integers are 3, 4, 5 and 6. Counting 2 as well gives 5, but the sign is strict, so 2 is excluded. Whole-number lower bounds are where this topic does most of its damage.

I divided 1 < 2x < 8 and counted 1, 2, 3, 4 — why is the answer 3?

Dividing gives 0.5 < x < 4, and the upper bound is a whole number with a strict sign, so 4 is excluded. The integers are 1, 2 and 3. A whole-number bound at either end always needs the sign checked before you count it.

If −2x > 6, is x > −3?

No — dividing by −2 flips the sign, so x < −3. The unflipped version is always offered as an option and looks entirely reasonable on the page. Any time a negative coefficient is involved, reverse the direction as a reflex before continuing.

FAQ

How many integers satisfy 4 < 3x < 14?

Three. Dividing throughout by 3 gives approximately 1.33 < x < 4.67, and the integers strictly inside that range are 2, 3 and 4. Because both bounds are fractional, neither endpoint is in question, which makes this the easier version of the question.

When does the inequality sign flip?

Only when you multiply or divide by a negative number. Adding or subtracting anything, and multiplying or dividing by a positive, all leave the direction alone. Since exam coefficients are usually positive, the flip is rare — which is exactly why it gets missed when it does appear.

Does a strict inequality ever include its endpoint?

No. A strict sign excludes the endpoint by definition, so for x > 2 the value 2 is not a solution. Only the inclusive forms, written with a bar underneath, count their endpoints, and questions mix the two deliberately.

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