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Quantitative Aptitude · Linear Inequalities

Linear Inequalities: Word Problems and Applications

Updated 1 October 2026 · Fact-checked

A linear inequality word problem describes a limit, such as a budget or capacity, in words. You define a variable, translate phrases like 'at most' into ≤ or 'at least' into ≥, solve using the standard rules, then check the answer against real-world conditions like whole numbers and non-negativity.

Understand Word Problems and Applications

A word problem on linear inequalities describes a business limit. A budget cannot be exceeded. A machine can run only so many hours. A shop must earn at least a target profit. Each limit becomes an inequality instead of an equation.

The real skill is translation. Pick a variable for the unknown quantity, such as the number of units. Write each cost or time as a quantity built from that variable. Then compare it with the limit using the right symbol.

The symbol comes from the wording. 'At most', 'not more than', 'up to' and 'maximum' mean ≤. 'At least', 'not less than', 'minimum' and 'a minimum of' mean ≥. 'More than' means > and 'less than' means <.

Solving is the same as for equations, with one extra rule: if you multiply or divide both sides by a negative number, reverse the sign. After solving, read the answer in context. Items are usually whole numbers, so you round in the direction that keeps the limit satisfied. Quantities such as units or hours cannot be negative.

Questions often ask for the maximum or minimum value that works. That is simply the largest or smallest whole number in your solution range.

Key formulas to remember

Phrase to symbol
at most / not more than / maximum → ≤ ; at least / not less than / minimum → ≥
Strict signs < and > are used for 'less than' and 'more than'.
Total cost model
Total cost = Fixed cost + (Variable cost per unit × x)
Use this when a cost limit is given with a fixed part.
Profit model
Profit = Revenue − Cost = (Selling price × x) − (Fixed cost + Variable cost × x)
For a profit target, set Profit ≥ target.
Negative multiplier rule
If a > b, then −a < −b (multiplying or dividing by a negative reverses the sign)
Most common sign error in solving.
Resource constraint
a₁x + a₂y ≤ available resource
a₁ and a₂ are resource use per unit of two products.
Average condition
(Sum of values) ÷ n ≥ required average, so Sum ≥ required average × n
Used in marks-type and sales-type problems.

How to solve Word Problems and Applications questions

Use this method for any linear inequality word problem. It keeps translation and checking separate, which prevents most lost marks.

  1. 1Read the question and mark exactly what is asked: a maximum, a minimum or a range.
  2. 2Define the variable, for example let x = number of units. Write it with its unit.
  3. 3Convert each statement into an expression in x, such as cost = fixed + variable × x.
  4. 4Pick the symbol from the wording: 'at most' is ≤, 'at least' is ≥.
  5. 5Form the inequality and solve it. Reverse the sign only if you multiply or divide by a negative.
  6. 6Apply real conditions: x ≥ 0 and x must be a whole number if items are counted.
  7. 7Round in the safe direction. For a maximum under ≤, round down. For a minimum under ≥, round up.
  8. 8Substitute the answer, and the next whole number, to check the limit holds.

Quickest way: Option testing with boundary values

When to use it: Use in the MCQ paper when the question asks for a maximum or minimum number of units and the options are whole numbers.

  1. Write the inequality quickly, or just the boundary equation (replace the sign with =).
  2. Solve the boundary equation to get a decimal value.
  3. For a maximum under ≤, take the whole number just below the boundary. For a minimum under ≥, take the one just above.
  4. If the boundary is exactly a whole number, include it when the sign is ≤ or ≥.
  5. Check the chosen option and its neighbour in the original limit if you have time.
  6. If the setup is unclear and takes more than about a minute, mark it for later. A wrong answer costs 0.25.

Common mistakes in Word Problems and Applications

  • Choosing the wrong symbol for 'at least' or 'at most'.

    Students read quickly and link 'at least' with 'less' and pick ≤.

    Fix: Remember that 'at least' means the smallest allowed value, so the quantity can be that or more: ≥. Underline the phrase before writing the sign.

  • Not reversing the sign when dividing by a negative number.

    The same steps as equations are applied automatically.

    Fix: Whenever the coefficient of x is negative, pause and flip the sign after dividing. Better, move terms so that x has a positive coefficient.

  • Ignoring the fixed cost in the cost expression.

    Students focus on the per-unit cost and forget the one-time amount.

    Fix: Always write Total = Fixed + Variable × x as the first line of any cost problem.

  • Giving a decimal answer for items that must be whole numbers.

    The algebra ends with a decimal and students stop.

    Fix: Round to a whole number in the direction that keeps the condition true. Maximum under ≤ rounds down; minimum under ≥ rounds up.

  • Rounding in the wrong direction.

    Students round to the nearest number instead of the safe one.

    Fix: Check the result by substitution. For x ≤ 12.7, the maximum is 12, not 13.

  • Forgetting x ≥ 0 in production problems.

    The condition is not written in the question.

    Fix: State non-negativity yourself. It can remove options that are negative or change the lower limit of the range.

Worked examples

Example 1

A firm has a monthly budget of ₹50,000 for making a product. The fixed cost is ₹14,000 and the variable cost is ₹300 per unit. The maximum number of units it can make is: (a) 110 (b) 120 (c) 130 (d) 140

Show the solution
  1. Let x = number of units.
  2. Total cost = 14,000 + 300x.
  3. Budget limit: 14,000 + 300x ≤ 50,000.
  4. 300x ≤ 36,000.
  5. x ≤ 120.
  6. Check: 14,000 + 300 × 120 = 14,000 + 36,000 = 50,000, which is within the budget.

Answer: (b) 120

Example 2

A trader sells an item at ₹80 per unit. The cost is ₹50 per unit and the fixed cost is ₹2,000. To earn a profit of at least ₹1,000, the minimum number of units to be sold is: (a) 90 (b) 100 (c) 110 (d) 120

Show the solution
  1. Let x = units sold.
  2. Revenue = 80x. Cost = 2,000 + 50x.
  3. Profit = 80x − 2,000 − 50x = 30x − 2,000.
  4. Condition: 30x − 2,000 ≥ 1,000.
  5. 30x ≥ 3,000, so x ≥ 100.
  6. Check: at x = 100, profit = 3,000 − 2,000 = ₹1,000, which meets the target.

Answer: (b) 100

Example 3

A machine runs at most 40 hours a week. Product A needs 3 hours per unit. 4 units of product B are made each week, using 2 hours per unit. The maximum whole number of units of A that can be made is: (a) 9 (b) 10 (c) 11 (d) 12

Show the solution
  1. Let x = units of A.
  2. Time for A = 3x. Time for B = 4 × 2 = 8 hours.
  3. Condition: 3x + 8 ≤ 40.
  4. 3x ≤ 32, so x ≤ 10.67.
  5. x must be a whole number, so round down to 10.
  6. Check: 3 × 10 + 8 = 38 ≤ 40. For 11 units: 33 + 8 = 41 > 40, which fails.

Answer: (b) 10

Exam tips

  • Underline the limit phrase ('at most', 'at least', 'not exceed') before you do anything else. Most errors start here.
  • Questions on this topic are usually short and have one variable. Aim to finish each in about a minute.
  • Test the boundary value first, then pick the option on the correct side of it. This often saves a full solution.
  • Watch for options that are the decimal answer rounded the wrong way. The examiner places them deliberately.
  • Practise with past MTP and RTP style questions on budgets, production and profit targets, since these are the usual settings.

Practice questions from Linear Inequalities

Word Problems and Applications: frequently asked questions

How do I form an inequality from a word problem?

Define a variable for the unknown. Write each cost, time or quantity as an expression in that variable. Then connect the expression to the limit using ≤ or ≥ based on the wording.

What is the difference between 'at least' and 'at most'?

'At least' sets a lower limit, so the quantity can equal that value or exceed it (≥). 'At most' sets an upper limit, so the quantity can equal that value or fall below it (≤).

Do I always round down in these problems?

No. Round down when finding a maximum under a ≤ limit. Round up when finding a minimum under a ≥ requirement. Always check by substitution.

Are word problems on linear inequalities asked often in CA Foundation?

Linear inequalities are part of the Business Mathematics section of Paper 3, and application-style MCQs on budgets or production limits are a usual way to test it. Practise questions from the ICAI MTP and RTP papers to see the pattern.