Algebra Practice

Linear Inequality Word Problems Practice Test

Model maximums, minimums, capacities, goals, and whole-number decisions with linear inequalities.

Linear Inequality Word Problems Practice Test

20 practical inequality problems with worked solutions and context-based rounding.

Instant feedback · Worked explanations
Constraint planner · linear inequality word problems

After the test

Linear inequality word problems model limits: how much is allowed, required, affordable, safe, or possible

An equation asks for an exact balance. An inequality asks for a range of acceptable values. The key modeling skill is to identify the boundary described by the story and decide whether equality at that boundary is allowed. After solving, the final value must still be interpreted in context — especially when x counts whole people, tickets, boxes, or days.

At leastAt mostNo more thanFewer than BudgetCapacityMinimum requirementMaximum allowed
A correct inequality answer is a condition, not just a number. “x ≤ 7” means every allowed value up to 7 must be considered, then restricted further by the real-world context if x must be a whole number.
at least / no less than
≥
The stated boundary is included as a minimum.
at most / no more than
≤
The stated boundary is included as a maximum.
more than / greater than
>
The boundary itself is not enough.
less than / fewer than
<
The value must stay strictly below the boundary.

Story → constraint pipeline

Build the inequality from the real restriction, not from isolated keywords

Unknown Define x precisely.

State what x counts or measures, including units.

Expression Build the changing quantity.

For example, fixed fee + rate × quantity.

Boundary Identify the limit or requirement.

Budget, capacity, minimum score, maximum time, and so on.

Comparison Choose <, ≤, >, or ≥.

Ask whether equality at the boundary is permitted.

Interpret Return the solution to the story.

Apply units, whole-number rules, and realistic restrictions.

Budget case · maximum affordable quantity
Situation
A club has $80. There is a $14 setup fee and each ticket costs $9.
Variable
x = number of tickets
Constraint
14 + 9x ≤ 80
Solve
9x ≤ 66 → x ≤ 22/3 ≈ 7.33
Context answer
At most 7 whole tickets can be purchased.

Minimum / maximum dashboard

The wording tells you whether to search upward from a minimum or downward from a maximum

Minimum requirement

A student needs at least 70 points. She already has 46 and earns 4 points per assignment.

46 + 4x ≥ 70 → 4x ≥ 24 → x ≥ 6. She needs at least 6 more assignments.

minimum valid x = 6

Maximum allowed

A package may weigh no more than 32 kg. The container weighs 5 kg and each item weighs 3 kg.

5 + 3x ≤ 32 → 3x ≤ 27 → x ≤ 9. The package can contain at most 9 items.

maximum valid x = 9

Whole-number decision rule

Do not use ordinary rounding automatically when x counts discrete objects

Maximum problem
x ≤ 7.8
Largest whole-number solution is 7.
Minimum problem
x ≥ 7.2
Smallest whole-number solution is 8.
Continuous quantity
t ≤ 7.8 hours
7.8 can be meaningful if time is allowed to be continuous.

Capacity limit

Capacity problems combine a current amount with a repeated amount and a hard upper boundary

120-seat limit

Example: rows of seats

There are already 24 occupied seats. Each arriving group has 8 people. The room holds at most 120 people.

Let x be the number of additional groups:
24 + 8x ≤ 120
8x ≤ 96
x ≤ 12

The room can accept at most 12 complete additional groups.

Rate and time limits

When a repeated rate is involved, keep the units visible while building the inequality

Maximum time
A machine uses 6 kWh per hour and may use at most 48 kWh.
6t ≤ 48 → t ≤ 8.
The machine may run for at most 8 hours.
Minimum distance
A runner must cover at least 15 km at 5 km/h.
5t ≥ 15 → t ≥ 3.
At that constant rate, at least 3 hours are required.
Fixed + rate
$11 fixed charge + $4 per hour, total no more than $35.
11 + 4t ≤ 35 → t ≤ 6.
The service can be used for at most 6 hours.

Two-plan inequality

Some word problems ask when one linear model is cheaper, faster, or larger than another

Both sides may contain x because both plans change with the same quantity.

Plan A 18 + 3x

$18 fixed fee plus $3 per use.

<
Plan B 8 + 5x

$8 fixed fee plus $5 per use.

When is Plan A cheaper? 18 + 3x < 8 + 5x → 10 < 2x → x > 5. Plan A is cheaper for more than 5 uses. At x = 5, the plans cost the same, so 5 is excluded.

Boundary and test-value check

Test the boundary and a nearby allowed value in the original story

For the ticket problem, 14 + 9x ≤ 80 gives a maximum whole-number answer of 7.

x = 7 14 + 9(7) = 77
within budget
x = 8 14 + 9(8) = 86
too expensive
Conclusion 7 is the greatest feasible whole-number value.
context verified

Constraint-modeling error log

Most mistakes occur while translating the real-world restriction or interpreting the algebraic bound

Inclusive word made strict
“At most 40” → x < 40 x ≤ 40

The boundary value 40 is allowed.

Fixed cost multiplied
$14 fee + $9 each → 14x + 9x ≤ 80 14 + 9x ≤ 80

The setup fee occurs once; only the per-item charge repeats.

Wrong integer rounding
x ≤ 7.33 → 8 tickets At most 7 tickets.

Rounding upward would violate the upper bound.

Number without meaning
x ≤ 7 At most 7 tickets can be purchased.

Return the bound to the variable definition and units.

Constraint-planner diagnostics

Use missed questions to identify which modeling decision needs review

Language → sign Did “at least,” “at most,” “more than,” or “fewer than” produce the correct inequality symbol?
Linear model Were fixed quantities and repeated per-unit quantities represented correctly?
Boundary meaning Was equality included or excluded according to the wording of the problem?
Context answer Was the algebraic range converted into the correct whole-number, unit, minimum, or maximum statement?