Problems Involving Inequalities

How to solve word problems with inequalities: choosing the sign from the words, rounding down for the greatest number allowed and up for the least number needed, and checking the answer.

Before you start

From lesson 1.5: solve the inequality first, then read the whole numbers off its solution. The largest whole number in $x \le 7.5$ is 7, and the smallest in $x > 4.2$ is 5.

From lesson 1.1: "at least 10" is $x \ge 10$ , "at most 10" is $x \le 10$ , and "more than 10" is $x > 10$ .

Word Problems: Rounding Down or Up

In a word problem the unknown is often a number of things, such as rides, cards or buses. The solution of the inequality may not be a whole number, and the answer must be. Ordinary rounding can give the wrong answer: the words of the question decide which way to go.

Solving a word problem with an inequality

Step 1: Say what the letter stands for, with its unit.

Step 2: Write the inequality. Choose the sign from the words: "at most", "not more than", "within a budget of", "must not exceed" give $\le$ ; "at least", "no fewer than" give $\ge$ ; "more than", "cheaper than", "less than" give $>$ or $<$ .

Step 3: Solve it.

Step 4: Find the whole-number answer. For the greatest number allowed, take the largest whole number in the solution: round down. For the least number needed, take the smallest whole number in it: round up.

Step 5: Check the answer and the next whole number, and answer in a sentence with the unit.

Example 1: The greatest number: round down

A theme park charges $5 for entry and $3 for each ride. Aisha has $40 to spend. What is the greatest number of rides she can go on?

Solution

Let $r$ be the number of rides. She spends $5 + 3r$ dollars, and she cannot spend more than 40 dollars, so the cost is at most 40: $5 + 3r \le 40$ .

Subtract 5 from both sides: $3r \le 35$ . Divide by 3: $r \le 11\frac{2}{3}$ .

$r$ is a whole number, so the greatest value is 11, the first whole number to the left of $11\frac{2}{3}$ . Rounding $11\frac{2}{3}$ to the nearest whole number would give 12, which is not a solution.

Check: 11 rides cost $5 + 33 = 38$ dollars, which is within $40. 12 rides would cost $5 + 36 = 41$ dollars, which is more.

Aisha can go on at most 11 rides.

Example 2: The least number: round up

A class wants to raise at least $500 for charity by selling greeting cards at $3 each. What is the least number of cards the class must sell?

Solution

Let $c$ be the number of cards sold. They raise $3c$ dollars, and "at least 500" gives $3c \ge 500$ .

Divide by 3: $c \ge 166\frac{2}{3}$ .

$c$ is a whole number, so the least value is 167, the first whole number to the right of $166\frac{2}{3}$ . Rounding down to 166 would be wrong.

Check: 167 cards raise $3 \times 167 = 501$ dollars, which is at least $500. 166 cards raise only $498$ dollars.

The class must sell at least 167 cards.

Example 3: The mark needed

Ravi's marks in three tests were 68, 74 and 59. Each test is marked out of 100. What is the least mark he needs in a fourth test for the average of the four marks to be at least 70?

Solution

Let $m$ be his mark in the fourth test. The average of the four marks is $\frac{68 + 74 + 59 + m}{4} = \frac{201 + m}{4}\text{,}$ and it must be at least 70: $\frac{201 + m}{4} \ge 70$ .

Multiply both sides by 4: $201 + m \ge 280$ . Subtract 201: $m \ge 79$ .

The boundary value 79 is a whole number and is included. Check: $\frac{201 + 79}{4} = \frac{280}{4} = 70\text{,}$ which is at least 70. With 78 the average would be $\frac{279}{4} = 69.75$ .

Ravi needs at least 79 marks in the fourth test.

Comparing, Consecutive Numbers and Lengths

Example 4: Comparing two plans

A mobile phone plan A costs $20 a month plus $0.10 for each minute of calls. Plan B costs $35 a month with free calls. For what numbers of whole minutes of calls in a month is plan A cheaper than plan B?

Solution

Let $t$ be the number of minutes of calls in a month. Plan A costs $20 + 0.1t$ dollars and plan B costs 35 dollars.

"Cheaper" means "costs less": $20 + 0.1t < 35$ . Equal costs are not "cheaper", so the sign is $<$ , not $\le$ .

Subtract 20 from both sides: $0.1t < 15$ . Divide by 0.1: $t < 150$ .

Check: at 149 minutes plan A costs $20 + 14.90 = 34.90$ dollars, which is cheaper. At 150 minutes it costs exactly 35 dollars, the same as plan B.

Plan A is cheaper for fewer than 150 minutes of calls a month, that is, for at most 149 whole minutes.

Example 5: Consecutive integers

The sum of three consecutive integers is less than 50. Find the largest possible value of the largest of the three integers.

Solution

Let the smallest integer be $n$ . The three integers are $n$ , $n + 1$ and $n + 2$ , and their sum is $3n + 3$ .

$3n + 3 < 50$ . Subtract 3: $3n < 47$ . Divide by 3: $n < 15\frac{2}{3}$ .

The largest integer $n$ is 15, so the integers are 15, 16 and 17.

Check: $15 + 16 + 17 = 48$ , which is less than 50. The next three, 16, 17 and 18, add up to 51, which is not.

The largest possible value of the largest integer is 17 (the integers 15, 16 and 17).

Example 6: A rectangle

A rectangle is $(x + 3)$ cm long and $x$ cm wide, where $x$ is a whole number. Its perimeter is not more than 30 cm.
(a) Write an inequality in $x$ and solve it.
(b) Find the largest possible value of $x$ .
(c) List all the possible values of $x$ .

Solution

(a) The perimeter is $2(x + 3 + x) = 2(2x + 3) = 4x + 6$ centimetres. "Not more than 30" gives $4x + 6 \le 30$ . Subtract 6: $4x \le 24$ . Divide by 4: $x \le 6$ .

(b) 6 is included. The rectangle is then 9 cm by 6 cm, with perimeter exactly 30 cm.

(a) $4x + 6 \le 30$ , so $x \le 6$ (b) 6 (c) 1, 2, 3, 4, 5, 6

Common Mistakes

Rounding to the nearest whole number. $r \le 11\frac{2}{3}$ rides gives at most 11 rides, not 12: 12 rides cost $41. For the greatest number allowed, round down.
Rounding down when the question asks for the least number. $c \ge 166\frac{2}{3}$ cards gives at least 167 cards: 166 cards raise only $498. For the least number needed, round up.
Using $\le$ or $\ge$ when the words leave the boundary out. "Cheaper than" and "more than" do not allow equality: at 150 minutes plan A costs the same as plan B, so it is not cheaper.
Stopping at the solution of the inequality. $r \le 11\frac{2}{3}$ does not answer "how many rides?". Give a whole number, in a sentence with its unit.
Forgetting a condition in the question, such as a length being positive: the rectangle of Example 6 can be 1 to 6 cm wide, but not 0 cm.

Practice

Work each question on paper before you open its solution. Answer each word problem in a sentence with its unit.

Question 1

A school can spend at most $10 000 on laptops. Each laptop costs $1650. What is the greatest number of laptops the school can buy?

ShowHide solution

Let $n$ be the number of laptops. $1650n \le 10\,000$ . Divide by 1650: $n \le 6.06\ldots$

The greatest whole number is 6. Check: 6 laptops cost $9900; 7 would cost $11 550.

The school can buy at most 6 laptops.

Question 2

A lift can carry a load of at most 1000 kg. A worker of mass 75 kg takes boxes up in the lift. Each box has a mass of 24 kg. What is the greatest number of boxes he can take up with him in one trip?

ShowHide solution

Let $b$ be the number of boxes. The total mass is $75 + 24b$ kilograms: $75 + 24b \le 1000$ .

Subtract 75: $24b \le 925$ . Divide by 24: $b \le 38.5\ldots$ The greatest whole number is 38.

Check: $75 + 24 \times 38 = 987$ kg, within the limit. With 39 boxes the mass is 1011 kg.

He can take at most 38 boxes in one trip.

Question 3

Each school bus can carry 45 students. What is the least number of buses needed to take 320 students on a learning journey?

ShowHide solution

Let $b$ be the number of buses. They must carry at least 320 students: $45b \ge 320$ .

Divide by 45: $b \ge 7.1\ldots$ The least whole number is 8.

Check: 7 buses carry 315 students, which is not enough. 8 buses carry 360.

At least 8 buses are needed.

Question 4

A class needs to collect at least 150 recycled bottles. It collected 57 bottles on the first day and 28 on the second. What is the least number of bottles it must collect on the third day?

ShowHide solution

Let $b$ be the number of bottles collected on the third day. $57 + 28 + b \ge 150$ , so $85 + b \ge 150$ .

Subtract 85 from both sides: $b \ge 65$ . The boundary value 65 is a whole number and is included.

Check: $85 + 65 = 150$ , which is at least 150; $85 + 64 = 149$ , which is not.

The class must collect at least 65 bottles on the third day.

Question 5

Wei Ming's marks in four quizzes were 14, 17, 12 and 15. Each quiz is marked out of 20. What is the least mark he needs in a fifth quiz for the average of the five marks to be at least 15?

ShowHide solution

Let $m$ be his mark in the fifth quiz. The five marks add up to $14 + 17 + 12 + 15 + m = 58 + m\text{,}$ and their average must be at least 15: $\frac{58 + m}{5} \ge 15$ .

Multiply both sides by 5: $58 + m \ge 75$ . Subtract 58: $m \ge 17$ .

Check: $\frac{58 + 17}{5} = \frac{75}{5} = 15\text{,}$ which is at least 15. With 16 the average would be $\frac{74}{5} = 14.8$ . A mark of 17 is possible, because the quiz is out of 20.

Wei Ming needs at least 17 marks in the fifth quiz.

Question 6

A school team raises $12 for every kilometre it runs in a charity run. Find the least whole number of kilometres the team must run to raise
(a) more than $480,
(b) at least $480.

ShowHide solution

Let $k$ be the number of kilometres. The team raises $12k$ dollars.

(a) "More than 480" does not allow 480 itself: $12k > 480$ . Divide both sides by 12: $k > 40$ . 40 is not included, so the least whole number is 41. Check: for 40 km the team raises $480, which is not more than $480; for 41 km it raises $492.

(b) "At least 480" allows 480: $12k \ge 480$ , so $k \ge 40$ , and 40 is included. Check: for 40 km it raises exactly $480; for 39 km only $468.

(a) The team must run at least 41 km. (b) The team must run at least 40 km.

Question 7

A gym offers two ways to pay. Plan A: $10 for each visit. Plan B: $60 a month, plus $4 for each visit. What is the least number of visits in a month for which plan B is cheaper?

ShowHide solution

Let $v$ be the number of visits in a month. Plan B is cheaper when $60 + 4v < 10v$ .

Subtract $4v$ from both sides: $60 < 6v$ . Divide by 6: $10 < v$ , which is $v > 10$ .

10 is not included, so the least whole number is 11. Check: at 10 visits both plans cost $100; at 11 visits plan A costs $110 and plan B costs $104.

Plan B is cheaper for 11 or more visits a month.

Question 8

The sum of two consecutive even numbers is at least 75. Find the smallest possible pair.

ShowHide solution

Let the smaller number be $n$ . The next even number is $n + 2$ , and $n + (n + 2) \ge 75$ , so $2n + 2 \ge 75$ .

Subtract 2: $2n \ge 73$ . Divide by 2: $n \ge 36.5$ . $n$ must be even, and the smallest even number from 36.5 upwards is 38.

Check: $38 + 40 = 78$ , which is at least 75. The pair before, 36 and 38, adds up to 74.

The smallest possible pair is 38 and 40.

Question 9

A rectangular garden has sides of $(2x - 3)$ m and $x$ m, where $x$ is a whole number and $x \ge 2$ . Its perimeter is at most 42 m.
(a) Write an inequality in $x$ and solve it.
(b) Find the largest possible area of the garden.

ShowHide solution

(a) The perimeter is $2(2x - 3 + x) = 6x - 6$ metres, and $6x - 6 \le 42$ . Add 6: $6x \le 48$ . Divide by 6: $x \le 8$ .

(b) $x$ is a whole number from 2 to 8. As $x$ grows, both sides grow, so the area is largest when $x = 8$ : the sides are 13 m and 8 m, and the area is $13 \times 8 = 104$ m².

(a) $6x - 6 \le 42$ , so $x \le 8$ (b) The largest possible area is 104 m².

Key Points

Say what the letter stands for, with its unit, and choose the sign from the words: "at most", "not more than" and "within" give $\le$ ; "at least" and "no fewer than" give $\ge$ ; "more than", "cheaper than" and "less than" give $>$ or $<$ .
Solve the inequality, then find the whole-number answer: round down for the greatest number allowed and up for the least number needed, never to the nearest.
When the boundary value is a whole number, the words decide whether it is allowed: "more than 480" leaves 480 out, and "at least 480" keeps it.
Check the answer and the next whole number.
Answer in a sentence with the unit, and remember any other condition in the question, such as a length being positive.