Skip to content

Linear Inequality Word Problems Assignment Day One

Solving Inequality Word Questions

(You might like to read Introduction to Inequalities and Solving Inequalities first.)


In Algebra we have "inequality" questions like:

Sam and Alex play in the same soccer team.
Last Saturday Alex scored 3 more goals than Sam, but together they scored less than 9 goals.

What are the possible number of goals Alex scored?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

Turning English into Algebra

To turn the English into Algebra it helps to:

  • Read the whole thing first
  • Do a sketch if needed
  • Assign letters for the values
  • Find or work out formulas

We should also write down what is actually being asked for, so we know where we are going and when we have arrived!

 

The best way to learn this is by example, so let's try our first example:

Sam and Alex play in the same soccer team.
Last Saturday Alex scored 3 more goals than Sam, but together they scored less than 9 goals.

What are the possible number of goals Alex scored?

 

Assign Letters:

  • the number of goals Alex scored: A
  • the number of goals Sam scored: S

We know that Alex scored 3 more goals than Sam did, so: A = S + 3

And we know that together they scored less than 9 goals: S + A < 9

We are being asked for how many goals Alex might have scored: A

 

Solve:

Start with:S + A < 9

A = S + 3, so:S + (S + 3) < 9

Simplify:2S + 3 < 9

Subtract 3 from both sides:2S < 9 − 3

Simplify:2S < 6

Divide both sides by 2:S < 3

Sam scored less than 3 goals, which means that Sam could have scored 0, 1 or 2 goals.

Alex scored 3 more goals than Sam did, so Alex could have scored 3, 4, or 5 goals.

 

Check:

  • When S = 0, then A = 3 and S + A = 3, and 3 < 9 is correct
  • When S = 1, then A = 4 and S + A = 5, and 5 < 9 is correct
  • When S = 2, then A = 5 and S + A = 7, and 7 < 9 is correct
  • (But when S = 3, then A = 6 and S + A = 9, and 9 < 9 is incorrect)

Lots More Examples!

Example: Of 8 pups, there are more girls than boys.

How many girl pups could there be?

Assign Letters:

  • the number of girls: g
  • the number of boys: b

We know that there are 8 pups, so: g + b = 8, which can be rearranged to

b = 8 − g

We also know there are more girls than boys, so:

g > b

We are being asked for the number of girl pups: g

Solve:

Start with:g > b

b = 8 − g, so:g > 8 − g

Add g to both sides:g + g > 8

Simplify:2g > 8

Divide both sides by 2:g > 4

So there could be 5, 6, 7 or 8 girl pups.

Could there be 8 girl pups? Then there would be no boys at all, and the question isn't clear on that point (sometimes questions are like that).

Check

  • When g = 8, then b = 0 and g > b is correct (but is b = 0 allowed?)
  • When g = 7, then b = 1 and g > b is correct
  • When g = 6, then b = 2 and g > b is correct
  • When g = 5, then b = 3 and g > b is correct
  • (But if g = 4, then b = 4 and g > b is incorrect)

A speedy example:

Example: Joe enters a race where he has to cycle and run.

He cycles a distance of 25 km, and then runs for 20 km. His average running speed is half of his average cycling speed.

Joe completes the race in less than 2½ hours, what can we say about his average speeds?

Assign Letters:

  • Average running speed: s
  • So average cycling speed: 2s

Formulas:

  • Speed = Distance / Time
  • Which can be rearranged to: Time = Distance / Speed

We are being asked for his average speeds: s and 2s

 

The race is divided into two parts:

1. Cycling

  • Distance = 25 km
  • Average speed = 2s km/h
  • So Time = Distance/Average speed = 25/2s hours

2. Running

  • Distance = 20 km
  • Average speed = s km/h
  • So Time = Distance/Average speed = 20/s hours

Joe completes the race in less than 2½ hours

  • The total time < 2½
  • 25/2s + 20/s < 2½

Solve:

Start with:25/2s + 20/s < 2½

Multiply all terms by 2s:25 + 40 < 5s

Simplify:65 < 5s

Divide both sides by 5:13 < s

Swap sides:s > 13

So his average speed running is greater than 13 km/h and his average speed cycling is greater than 26 km/h

In this example we get to use two inequalities at once:

Example: The velocity v m/s of a ball thrown directly up in the air is given by v = 20 − 10t, where t is the time in seconds.

At what times will the velocity be between 10 m/s and 15 m/s?

Letters:

  • velocity in m/s: v
  • the time in seconds: t

Formula:

We are being asked for the time t when v is between 5 and 15 m/s:

10  <  v  <  15

10  <  20 − 10t  <  15

Solve:

Start with:10  <  20 − 10t  <  15

Subtract 20 from each:10 − 20  <  20 − 10t − 20  <  15 − 20

Simplify:−10  < −10t  <  −5

Divide each by 10:−1  < −t  <  −0.5

Change signs and reverse inequalities:1  >  t  >  0.5

It is neater to show the smaller
number first, so swap over:0.5  <  t  <  1

So the velocity is between 10 m/s and 15 m/s between 0.5 and 1 second after.

And a reasonably hard example to finish with:

Example: A rectangular room fits at least 7 tables that each have 1 square meter of surface area. The perimeter of the room is 16 m.
What could the width and length of the room be?

Make a sketch: we don't know the size of the tables, only their area, they may fit perfectly or not!

Assign Letters:

  • the length of the room: L
  • the width of the room: W

The formula for the perimeter is 2(W + L), and we know it is 16 m

  • 2(W + L) = 16
  • W + L = 8
  • L = 8 − W

We also know the area of a rectangle is the width times the length: Area = W × L

And the area must be greater than or equal to 7:

We are being asked for the possible values of W and L

 

Let's solve:

Start with:W × L ≥ 7

Substitute L = 8 − W:W × (8 − W) ≥ 7

Expand:8W − W2 ≥ 7

Bring all terms to left hand side:W2 − 8W + 7 ≤ 0

This is a quadratic inequality. It can be solved many way, here we will solve it by completing the square:

Move the number term 7 to the right side of the inequality:W2 − 8W ≤ −7

Complete the square on the left side of the inequality and balance this by adding the same value to the right side of the inequality:W2 − 8W + 16 ≤ −7 + 16

Simplify:(W − 4)2 ≤ 9

Take the square root on both sides of the inequality:−3 ≤ W − 4 ≤ 3

Yes we have two inequalities, because 32 = 9 AND (−3)2 = 9

Add 4 to both sides of each inequality:1 ≤ W ≤ 7

So the width must be between 1 m and 7 m (inclusive) and the length is 8−width.

 

Check:

  • Say W = 1, then L = 8−1 = 7, and A = 1 x 7 = 7 m2 (fits exactly 7 tables)
  • Say W = 0.9 (less than 1), then L = 7.1, and A = 0.9 x 7.1 = 6.39 m2 (7 won't fit)
  • Say W = 1.1 (just above 1), then L = 6.9, and A = 1.1 x 6.9 = 7.59 m2 (7 fit easily)
  • Likewise for W around 7 m

 

 

InequalitiesSolving InequalitiesGraphing Linear InequalitiesInequality GrapherAlgebra Index

Copyright © 2016 MathsIsFun.com

Solving Word Problems in Algebra
Inequality Word Problems

How are you with solving word problems in Algebra? Are you ready to dive into the "real world" of inequalities? I know that solving word problems in Algebra is probably not your favorite, but there's no point in learning the skill if you don't apply it.

I promise to make this as easy as possible. Pay close attention to the key words given below, as this will help you to write the inequality. Once the inequality is written, you can solve the inequality using the skills you learned in our past lessons.

I've tried to provide you with examples that could pertain to your life and come in handy one day. Think about others ways you might use inequalities in real world problems. I'd love to hear about them if you do!

Before we look at the examples let's go over some of the rules and key words for solving word problems in Algebra (or any math class).

Word Problem Solving Strategies

  • Read through the entire problem.
  • Highlight the important information and key words that you need to solve the problem.
  • Identify your variables.
  • Write the equation or inequality.
  • Solve.
  • Write your answer in a complete sentence.
  • Check or justify your answer.

I know it always helps too, if you have key words that help you to write the equation or inequality. Here are a few key words that we associate with inequalities! Keep these handy as a reference.


Inequality Key Words


  • at least - means greater than or equal to
  • no more than - means less than or equal to
  • more than - means greater than
  • less than - means less than

Ok... let's put it into action and look at our examples.


Example 1: Inequality Word Problems


Keith has $500 in a savings account at the beginning of the summer. He wants to have at least $200 in the account by the end of the summer. He withdraws $25 each week for food, clothes, and movie tickets.

  • Write an inequality that represents Keith's situation.
  • How many weeks can Keith withdraw money from his account? Justify your answer.

Solution

Step 1: Highlight the important information in this problem.

Note:  At least is a key word that notes that this problem must be written as an inequality.


Step 2: Identify your variable. What don't you know? The question verifies that you don't know how many weeks.

Let w = the number of weeks

Step 3: Write your inequality.

500 - 25w > 200


I know you are saying, "How did you get that inequality?"


I know the "at least" part is tricky. You would probably think that at least means less than.

But... he wants the amount in his account to be at least $200 which means $200 or greater. So, we must use the greater than or equal to symbol.

Step 4: Solve the inequality.

The number of weeks that Keith can withdraw money from his account is 12 weeks or less.

Step 5: Justify (prove your answer mathematically).

I'm going to prove that the largest number of weeks is 12 by substituting 12 into the inequality for w. You could also substitute any number less than 12.

Since 200 is equal to 200, my answer is correct. Any more than 12 weeks and his account balance would be less than $200.  Any number of weeks less than 12 and his account would stay above $200.


That wasn't too bad, was it? Let's take a look at another example.

Example 2: More Inequality Word Problems


Yellow Cab Taxi charges a $1.75 flat rate in addition to $0.65 per mile. Katie has no more than $10 to spend on a ride.

  • Write an inequality that represents Katie's situation.
  • How many miles can Katie travel without exceeding her budget? Justify your answer.

Solution

Step 1: Highlight the important information in this problem.

Note:  No more than are key words that note that this problem must be written as an inequality.

Step 2: Identify your variable. What don't you know? The question verifies that you don't know the number of miles Katie can travel.

Let m = the number of miles


Step 3:  Write the inequality.

0.65m + 1.75 < 10


Are you thinking, "How did you write that inequality?"


The "no more than" can also be tricky. "No more than" means that you can't have more than something, so that means you must have less than!

Step 4: Solve the inequality.

Since this is a real world problem and taxi's usually charge by the mile, we can say that Katie can travel 12 miles or less before reaching her limit of $10.

Step 5: Justify (prove your answer mathematically).

Are you ready to try some on your own now? Yes... of course you are! Click here to move onto the word problem practice problems.


Take a look at the questions that other students have submitted:


  1. Home
  2. >
  3. Inequalities
  4. >
  5. Inequality Word Problems