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Real-World Inequality Problems

Use inequalities to solve budget, limit and goal problems and give answers that make sense.

Practice this skill 22 min Real-world inequalities Leer en español
Key idea

What you'll learn

Real life is full of limits: a budget you cannot go over, a weight an elevator cannot exceed, a score you need at least to pass. Those limits are inequalities. In this lesson you will learn to solve real-world problems with one-variable linear inequalities — from reading the story, to solving, to giving an answer that actually makes sense.

The process has four steps:

  1. Name the unknown with a letter.
  2. Write the inequality (the words tell you which symbol).
  3. Solve it with the rules you already know.
  4. Interpret the answer — round the right way, and answer the question that was asked.

Why it matters on the GED: budget and "how many can you buy?" problems are some of the most common items on the Mathematical Reasoning test. Many of them use a fixed fee plus a rate (a taxi fare, a phone plan, a gym membership). Step 4 is where most points are lost: the GED often lists the unrounded number, the wrongly rounded number, and the right one side by side.

Words to know

Key words

  • At least — this amount or more, symbol \ge (example: "at least $300" → \ge 300).
  • At most / no more than — this amount or less, symbol \le (example: "no more than 60 lb" → \le 60).
  • More than / exceeds — strictly bigger, symbol >.
  • Less than / fewer than — strictly smaller, symbol <.
  • Budget — the most money you can spend, usually a \le limit.
  • Fixed cost (flat fee) — a cost you pay once no matter how many items (example: a $3 taxi pickup fee).
  • Rate (per-unit cost) — the cost for each item, hour, or mile (example: $2 per mile).
  • Whole-number answer — when you count people, tickets or boxes, the answer must be a whole number.
  • Round down — go to the lower whole number (13.3 → 13). Use it for "the most you can afford".
  • Round up — go to the next whole number (23.3 → 24). Use it for "the least you need".
  • Break-even point — the value where two plans cost exactly the same.

All Math words in the glossary

From story to inequality to answer

Most GED inequality word problems follow the pattern

\text{fixed amount} + \text{rate} \times \text{quantity} \;\;(\text{symbol})\;\; \text{limit}

Example A — taxi. A taxi charges $3 to start plus $2 per mile. Ron has $25. How many miles can he ride?

  1. Let m = number of miles.
  2. Cost is 3 + 2m; he can spend no more than $25: 3 + 2m \le 25.
  3. Solve: subtract 3 → 2m \le 22; divide by 2 → m \le 11.
  4. Interpret: Ron can ride at most 11 miles.
Bar model of the taxi fare: a 3 dollars starting fee block followed by eleven 2 dollars blocks for 11 miles, totaling 22 dollars, which reaches Ron's 25 dollars limit exactly. So 3 + 2m <= 25 gives m <= 11. 3 + 2m ≤ 25 Ron has $25 11 miles × $2 = $22 $3 fee $0

If only whole miles count, the solutions are 0, 1, 2, …, 11:

Number line from -1 to 13 with separate closed dots at every whole number from 0 to 11: the whole-number solutions of 3 + 2m <= 25. −1 0 1 2 3 4 5 6 7 8 9 10 11 12 13

Example B — weight limit. An elevator holds at most 1,500 lb. The operator weighs 180 lb and each box weighs 45 lb.

180 + 45b \le 1500 \;\Rightarrow\; 45b \le 1320 \;\Rightarrow\; b \le 29.\overline{3}
Bar model of the elevator load: a 180 lb block for the operator plus 29 boxes of 45 lb each, 1,305 lb, for a total of 1,485 lb, just under the 1,500 lb limit line. A dashed red 30th box would bring the total to 1,530 lb, over the limit. 180 + 45b ≤ 1500 180 lb limit 1,500 lb operator 29 boxes × 45 lb = 1,305 lb total 1,485 lb ✓ 30 boxes: 1,530 lb ✗

Boxes come in whole numbers, so the answer is 29 boxes. (30 boxes would weigh 180 + 1350 = 1530 lb — too much.)

The fixed amount (the $3 or the 180-lb operator) must not be forgotten — it is the most common reason for a wrong answer.

Rounding the right way

When the answer must be a whole number, do not use ordinary rounding rules. Ask: which whole numbers actually satisfy the inequality?

Type of questionInequality looks likeExample resultCorrect whole number
"Most you can buy / carry / afford"x \le 13.313.313 (round down)
"Least you need to reach a goal"x \ge 23.323.324 (round up)
"Least number so it is more than…"x > 100100101 (100 itself fails)
"Most so it is less than…"x < 707069

Example C. Kai earns $15 per hour and wants to earn at least $350. 15h \ge 350 \Rightarrow h \ge 23.\overline{3}. 23 hours pays only $345 — not enough. He needs 24 hours ($360). Even though 23.3 "rounds" to 23, the correct answer is 24.

Number line from 20 to 27. Closed circle at 23.33 with shading and an arrow to the right: the graph of h >= 23.3 (repeating). The whole number 24 is marked as the least whole number of hours; 23 hours, paying 345 dollars, is marked as not enough. 15h ≥ 350 → h ≥ 23.3… → 24 hours 20 21 22 23 24 25 26 27 23.3 23 hours pay $345 ✗ (not enough) · 24 hours pay $360 ✓

Example D. Notebooks cost $3 and Jenna has $40. 3n \le 40 \Rightarrow n \le 13.\overline{3}. 14 notebooks cost $42 — too much. She can buy 13.

Number line from 9 to 16. Closed circle at 13.33 with shading and an arrow to the left: the graph of n <= 13.3 (repeating). The whole number 13 is marked as the greatest number of notebooks; 14 notebooks, costing 42 dollars, is marked as too many. 3n ≤ 40 → n ≤ 13.3… → 13 notebooks 9 10 11 12 13 14 15 16 13.3 13 notebooks cost $39 ✓ · 14 notebooks cost $42 ✗

Always test the whole number you choose and the next one past it.

Comparing plans, averages and two limits

Comparing two plans. "When is Plan A cheaper than Plan B?" means cost A < cost B.

Example E. Plan A costs $25 a month plus $0.15 per text; Plan B costs $40 with unlimited texts. Plan A is cheaper when

25 + 0.15t < 40 \;\Rightarrow\; 0.15t < 15 \;\Rightarrow\; t < 100.

So Plan A is cheaper for fewer than 100 texts. At exactly 100 texts the plans tie (the break-even point).

Graph of monthly cost against number of texts from 0 to 200. Plan A is a line starting at 25 dollars and rising 0.15 dollars per text; Plan B is a flat line at 40 dollars. The lines cross at 100 texts and 40 dollars, the break-even point. The region left of 100 texts is shaded and labelled: Plan A is cheaper when t < 100. Plan A: 25 + 0.15t | Plan B: 40 A cheaper (t < 100) 0 50 100 150 200 0 10 20 30 40 50 60 texts per month, t monthly cost ($) Plan A Plan B tie at 100 texts ($40)

When both plans have a variable part, put the variable on one side:

40 + 0.55m > 65 + 0.30m \;\Rightarrow\; 0.25m > 25 \;\Rightarrow\; m > 100

Averages. To have a mean of at least 85 on 4 tests, the total must be at least 4 \times 85 = 340. If the first three scores add to 253, the fourth score s must satisfy 253 + s \ge 340, so s \ge 87.

Percents and commissions. A salesperson earns $500 per week plus 6% of sales s. To earn at least $1,100: 500 + 0.06s \ge 1100 \Rightarrow 0.06s \ge 600 \Rightarrow s \ge 10{,}000.

Two limits at once. If a caterer's budget allows 69 guests but the hall holds 75, the real maximum is the smaller limit: 69. Always check every condition in the story.

Formula

Formulas you will use

\text{total cost} = (\text{number of units}) \times (\text{price per unit})

This one is on the GED formula sheet. Real-world inequalities usually add a fixed fee, which you must build yourself (not on the sheet):

\text{fixed fee} + (\text{rate}) \times x \;\le\; \text{budget} \quad\Rightarrow\quad x \le \frac{\text{budget} - \text{fixed fee}}{\text{rate}}
  • x = the quantity you are solving for (miles, hours, boxes, guests…).
  • rate = the cost (or weight, or pay) for one unit.
  • fixed fee = the one-time amount.

For goals use \ge instead: x \ge \dfrac{\text{goal} - \text{starting amount}}{\text{rate}}.

The mean formula \bar{x} = \dfrac{\text{sum of values}}{\text{number of values}} is also on the formula sheet.

Worked example

Worked example 1 (easy): how many can you buy?

Problem: Movie tickets cost $12 each. Amir has $70. What is the greatest number of tickets he can buy?

  1. Let t = number of tickets. Cost = 12t, which can be no more than 70: 12t \le 70.
  2. Divide by 12: t \le 5.8\overline{3}.
  3. Tickets are whole numbers, so round down: 5.

Answer: 5 tickets

Bar model: five 12 dollars ticket blocks totaling 60 dollars, which fits under Amir's 70 dollars limit line. A dashed red sixth ticket block brings the total to 72 dollars, over the limit. So the greatest number of tickets is 5. 12t ≤ 70 Amir has $70 5 tickets × $12 = $60 ✓ 6 tickets: $72 ✗ $0

Check: 5 \times 12 = 60 \le 70 ✓; 6 \times 12 = 72 > 70 ✗, so 6 is too many.

Worked example

Worked example 2: a goal with a starting amount

Problem: Sam has $25 saved and adds $15 each week. How many weeks until he has at least $200 for a bike?

  1. Let w = weeks. Savings = 25 + 15w; we need 25 + 15w \ge 200.
  2. Subtract 25: 15w \ge 175.
  3. Divide by 15: w \ge 11.\overline{6}.
  4. Weeks are whole numbers and this is a minimum, so round up: 12.

Answer: 12 weeks

Bar model of Sam's savings: a 25 dollars block already saved plus twelve 15 dollars weekly blocks, 180 dollars, for a total of 205 dollars, which passes the 200 dollars goal line. After 11 weeks he would have only 190 dollars. 25 + 15w ≥ 200 goal $200 $25 12 weeks × $15 = $180 total $205 ✓ $0

Check: After 12 weeks: 25 + 180 = 205 \ge 200 ✓. After 11 weeks: 25 + 165 = 190, not enough ✓.

Worked example

Worked example 3 (real world): profit goal

Problem: A club sells raffle tickets for $8 each. The prizes cost $150. The club wants a profit of at least $500. How many tickets must it sell?

  1. Let t = tickets. Profit = money in − costs = 8t - 150.
  2. Inequality: 8t - 150 \ge 500.
  3. Add 150: 8t \ge 650.
  4. Divide by 8: t \ge 81.25.
  5. Round up: 82 tickets.

Answer: 82 tickets

Number line from 78 to 86. Closed circle at 81.25 with shading and an arrow to the right: the graph of t >= 81.25. The whole number 82 is marked as the least number of tickets; 81 tickets gives only 498 dollars profit. 8t − 150 ≥ 500 → t ≥ 81.25 → 82 tickets 78 79 80 81 82 83 84 85 86 81.25 81 tickets: $498 profit ✗ · 82 tickets: $506 profit ✓

Check: 8(82) - 150 = 656 - 150 = 506 \ge 500 ✓. 8(81) - 150 = 498, short by $2 ✓.

Worked example

Worked example 4 (GED-hard): which company is cheaper?

Problem: Truck rental A charges $40 plus $0.55 per mile. Truck rental B charges $65 plus $0.30 per mile. What is the least whole number of miles for which B is cheaper?

  1. Let m = miles. "B is cheaper" means cost B < cost A: 65 + 0.30m < 40 + 0.55m
  2. Subtract 0.30m from both sides: 65 < 40 + 0.25m.
  3. Subtract 40: 25 < 0.25m.
  4. Divide by 0.25: 100 < m, i.e. m > 100.
  5. The inequality is strict, so 100 does not work. The least whole number is 101.

Answer: 101 miles

Graph of rental cost against miles from 0 to 200. Company A starts at 40 dollars and rises 0.55 dollars per mile; company B starts at 65 dollars and rises 0.30 dollars per mile. The lines cross at 100 miles and 95 dollars. The region to the right of 100 miles is shaded: B is cheaper when m > 100, so the least whole number is 101 miles. A: 40 + 0.55m | B: 65 + 0.30m B cheaper (m > 100) 0 50 100 150 200 0 20 40 60 80 100 120 140 160 miles driven, m rental cost ($) A B tie at 100 mi ($95)

Check: At 101 miles, A = 40 + 55.55 = 95.55 and B = 65 + 30.30 = 95.30; B is cheaper ✓. At 100 miles both cost $95 — a tie, not cheaper ✓.

Worked example

Worked example 5 (work backwards): the needed score

Problem: Maria's first three test scores are 78, 85 and 90. She wants her mean on four tests to be at least 85. What is the lowest score she can get on test 4?

  1. Mean \ge 85 over 4 tests means the total must be at least 4 \times 85 = 340.
  2. Her current total: 78 + 85 + 90 = 253.
  3. Let s = the fourth score: 253 + s \ge 340.
  4. Subtract 253: s \ge 87.

Answer: 87

Bar model of Maria's test total: blocks of 78, 85 and 90 add to 253; a fourth block s of at least 87 brings the total to the goal line of 340, which is 4 times 85. So s >= 87. Need total ≥ 4 × 85 = 340 78 85 90 s ≥ 87 goal 340 first three tests: 253

Check: \dfrac{78 + 85 + 90 + 87}{4} = \dfrac{340}{4} = 85 ✓ — exactly the goal. An 86 gives \dfrac{339}{4} = 84.75, too low ✓.

Calculator tip

Calculator tip (TI-30XS MultiView)

Word problems often have decimals ($0.15 per text, 6% commissions), so the calculator is allowed and useful here.

  • Use parentheses to do the "subtract the fixed fee, then divide" in one line: for 180 + 45b \le 1500 type ( 1500 − 180 ) ÷ 45 enter → 29.333\ldots
  • Convert percents to decimals before typing: 6% → 0.06.
  • If the answer shows as a fraction (like \tfrac{88}{3}), press toggle to see the decimal.
  • Check your whole-number answer by typing the cost for it (and for the next number): 180 + 45 × 29 enter → 1485 ✓; 180 + 45 × 30 enter → 1530 ✗.

The calculator gives 29.333 — it does not decide whether to round up or down. That is your job.

Common mistake

Common mistakes

  • Ignoring the fixed fee. Elevator: 45b \le 1500 \Rightarrow b \le 33 ✗. ✓ Include the 180-lb operator: b \le 29.
  • Rounding with "normal" rules. "At least $350 at $15/h" → 23.3 \to 23 ✗ (only $345). ✓ A minimum rounds up: 24.
  • Giving a decimal for a count. "13.33 notebooks" ✗. ✓ You can't buy part of a notebook: 13.
  • Mixing up "at least" and "at most". "At least" is \ge (a floor); "at most / no more than" is \le (a ceiling).
  • Including the boundary of a strict inequality. "B is cheaper" with m > 100 → answer 100 ✗. ✓ At 100 miles they tie, so 101.
  • Forgetting a second limit. Budget allows 69 guests but the room holds 75 → answer 75 ✗. ✓ Use the smaller limit, 69.
Test strategy

GED test strategy

  • Underline the limit words ("at least", "no more than", "exceeds") and the fixed fee before you start.
  • Plug the choices back in. In a "greatest number" question, start with the largest answer choice and test it; the first one that fits is the answer. In a "least number" question, start with the smallest.
  • Test the neighbor. After you choose a whole number, check that the next number past it fails. That catches rounding errors.
  • Estimate first. $25 at $2 per mile is about 12 miles; if your answer is 110, something went wrong.
  • Answer the question asked. If they ask for the cost, don't stop at the number of items.
  • Units: keep dollars with dollars and pounds with pounds; convert cents or percents to decimals first.

Check yourself

Real-world inequalities

Jenna has $40 to spend on notebooks. Each notebook costs $3. What is the greatest number of notebooks she can buy?

Answer choices
Hint

Write an inequality for the total cost. Can she buy part of a notebook?

Check yourself

Real-world inequalitiesFill in

A taxi charges a $3 pickup fee plus $2 per mile. Ron has $25. What is the greatest number of miles he can ride?

miles

Type your answer in the box. Fractions like 3/4 are fine.

Hint

The total cost is the pickup fee plus the cost of the miles. That total can be no more than $25.

Summary

  • Write the story as fixed amount + rate × quantity (symbol) limit, then solve.
  • "At least" → \ge; "at most / no more than" → \le; "more than" → >; "less than" → <.
  • For counts, round down for a maximum and round up for a minimum; with a strict inequality, the boundary itself does not count.
  • Comparing plans: set cost A < cost B and gather the variable on one side.
  • Always test your answer and the next whole number.
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